Fundamentals of Electromagnetics with Engineering Applications Solutions Manual

Fundamentals of Electromagnetics with Engineering Applications BY Wentworth

2-1 Solutions for Chapter 2 Problems 1. Vectors in the Cartesian Coordinate System P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q. APQ = 2 ax + 4 ay + 6 az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az Qx-Px=Qx-4=2; Qx=6 Qy-Py=Qy-2=4; Qy=6 Qz-Pz=Qz-1=6; Qz=7 Ans: Q(6,6,7)

P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, fill in the table and make a sketch of the vectors found in (a) through (f).

Vector Mag Unit Vector a. Find the vector A AOP = 4 ax + 1 ay 4.12 AOP = 0.97 ax + 0.24 ay from the origin to P b. Find the vector B BOQ = 1 ax + 3 ay 3.16 aOQ = 0.32 ax + 0.95 ay from the origin to Q c. Find the vector C CPQ = -3 ax + 2 ay 3.61 aPQ = -0.83 ax + 0.55 ay from P to Q d. Find A + B A + B = 5 ax + 4 ay 6.4 a = 0.78 ax + 0.62 ay e. Find C – A C - A = -7 ax + 1 ay 7.07 a = -0.99 ax + 0.14 ay f. Find B - A B - A = -3 ax + 2 ay 3.6 a = -0.83 ax + 0.55 ay a. AOP = (4-0)ax + (1-0)ay + (0-0)az = 4 ax + 1 ay.

AOP = 42 + 12 = 17 = 4.12 4 1 ax + a y = 0.97a x + 0.24a y 17 17 (see Figure P2.2ab) aOP =

b. BOQ =(1-0)ax + (3-0)ay + (0-0)az = 1 ax + 3 ay.

BOQ = 12 + 32 = 10 = 3.16

Fig. P2.2ab

1 3 ax + a y = 0.32ax + 0.95a y 10 10 (see Figure P2.2ab) aOQ =

c. CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.

CPQ = 32 + 22 = 13 = 3.61 −3 2 ax + a y = −0.83ax + 0.55a y 13 13 (see Figure P2.2cd) a PQ =

Fig. P2.2cd

2-2 d. A + B = (4+1)ax + (1+3)ay + (0-0)az = 5 ax + 4 ay.

A + B = 52 + 42 = 41 = 6.4 5 4 ax + a y = 0.78ax + 0.62a y 41 41 (see Figure P2.2cd) a=

e. C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + 1 ay.

C − A = 72 + 12 = 50 = 7.07 −7 1 ax + a y = −0.99ax + 0.14a y 50 50 (see Figure P2.2ef) a=

FigP2.2ef

f. B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.

B − A = 32 + 22 = 13 = 3.6 −3 2 ax + a y = −0.83ax + 0.55a y 13 13 (see Figure P2.2ef) a=

P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary points in the Cartesian Coordinate System.

A program or function for this task is really overkill, as it is so easy to perform the task.

Enter points P and Q (for example, P=[1 2 3]; Q=[6 5 4]). Then, the vector from P toQ is simply given by Q-P. As a function we could have: function PQ=vector(P,Q) % Given a pair of Cartesian points % P and Q, the program determines the % vector from P to Q. PQ=Q-P; Running this function we have: >> P=[1 2 3]; >> Q=[6 5 4]; >> PQ=vector(P,Q) PQ = 5

3

1

Alternatively, we could simply perform the math in the command line window:

2-3 >> PQ=Q-P PQ = 5 3 1 >> 2. Coulomb’s Law, Electric Field Intensity, and Field Lines P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m, 0.0, 0.0) = 1.0nC. (a) Find the total force acting on the charge Q3. (b) Repeat the problem after changing the charge of Q2 to –4.0nC. (c) Find the electric field intensity for parts (a) and (b). (a) F13 =

Q1Q2 a13 , where R13 = 4 ax + 3 ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay.

4 o R132

so

( 4a + 3a ) 4 x10 C )(1x10 C ) ( 5 FV NM F = C VC 4 (10 F 36 m ) ( 5m ) −9

13

−9

x

−9

y

2

= 1.15 x10−9 ax + 0.86 x10−9 a y N .

Similarly, F23 = 1.15x10−9 ax − 0.86 x10−9 a y N , so FTOT = 2.3a x nN (b) with Q2 = -4 nC, F13 is unchanged but F23 = −1.15x10−9 ax + 0.86 x10−9 a y N , so

FTOT = 1.7a y nN . −9 FTOT ( 2.3x10 ax N ) VC V (c) Ea = = = 2.3ax .

-9 Q3 m (1x10 C ) Nm

Likewise, Eb = 1.7a y

V . m

Fig. P2.4 P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 C on Q2(6.0m, 9.0m, 3.0m) = 10. nC. Q1Q2 a12 , where 4 o R122 R12 = (6-3)ax + (9-3)ay + (3-3)az = 3 ax + 6 ay m 3a + 6ay R12 = 32 + 62 = 45m, a12 = x ,and 45 F12 =

2-4

(1x10 C )(10 x10 C ) 3a + 6a FV NM , so F = 0.89a + 1.8a  N .

F = C VC 45 4 (10 F 45m ) 36 m ) ( −6

−9

x

−9

12

y

12

2

x

y

Fig. P2.5

P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0 cm. Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P. We arbitrarily label the charges as shown in Figure P2.6. Then ROP = 0.1 ax + 0.1 ay ROP = 0.141 m aOP = 0.707 ax + 0.707 ay.

FOP =

(10nC )(10nC )( 0.707 ) ( ax + ay ) −9 2 ( 4 ) 10 F 36 m ( 0.141m )

(

)

= 32 ( ax + a y )  N

FTP =

FSP =

(10nC )(10nC ) ay = 90a y  N −9 2 10 F 0.1m ) ( 4 ) 36 m (

(

)

(

)

Fig. P2.6

(10nC )(10nC ) ax = 90ax  N −9 2 ( 4 ) 10 F 36 m ( 0.1m )

and then the total (adjusting to 2 significant digits) is: FTOT = 120 ( ax + ay )  N .

2-5 P2.7: 1.00 nC point charges are located at (0.00, -2.00, 0.00)m, (0.00, 2.00, 0.00)m, (0.00, 0.00, -2.00)m and (0.00, 0.00, +2.00)m. Find the total force acting on a 1.00 nC charge located at (2.00, 0.00, 0.00)m.

Figure P2.7a shows the situation, but we need only find the x-directed force from one of the charges on Qt (Figure P2.7b) and multiply this result by 4. Because of the problem’s symmetry, the rest of the components cancel.

2a + 2a y QQt F1t = aR , R = 2a x + 2a y , R = 8 m, aR = x , 4 o R 8 (1x10−9 C )(1x10−9 C )  2a x + 2a y 8  so F1t = = 796 x10−12 ( a x + a y ) N −9 4 10 F 8m 2 ) 36 m ( The force from all charges is then FTOT = ( 4 ) 796 x10−12 ax nN = 3.2ax nN .

)

(

(

)

Fig. P2.7a

Fig. P2.7b

P2.8: A 20.0 nC point charge exists at P(0.00,0.00,-3.00m). Where must a 10.0 nC charge be located such that the total field is zero at the origin?

For zero field at the origin, we must cancel the +az directed field from QP by placing Q at the point Q(0,0,z) (see Figure P2.8). Then we have Etot = EP + EQ = 0.

20 x10−9 C ) a z ( QP FV V a = = 20 a So, E P = z 2 R −9 2 4 o R m 4 10 F 3m C 36 m ( )

(

)

2-6 and

EQ =

Q 4 o R 2

aR

(10 x10 C ) (−a ) −9

=

(

z

−9

4 10 F

) z ( m) )

36 m (

2

=

−90 az z2

So then 90 20a z − 2 a z = 0, z 90 z 2 = , z = 2.12.

20 Thus, Q(0,0,2.12m).

Fig. P2.8

3. The Spherical Coordinate System P2.9: Convert the following points from Cartesian to Spherical coordinates: a. P(6.0, 2.0, 6.0) b. P(0.0, -4.0, 3.0) c. P(-5.0,-1.0, -4.0)

 6  o −1  2  o (a) r = 62 + 22 + 62 = 8.7, = cos −1   = 47 ,  = tan   = 18  8.7  6  3  −4  (b) r = 02 + 42 + 32 = 5, = cos −1   = 53o ,  = tan −1   = −90o 5  0   −1  o −1  −1  o (c) r = 52 + 12 + 42 = 6.5, = cos −1   = 130 ,  = tan   = 190  6.5   −5  P2.10: Convert the following points from Spherical to Cartesian coordinates: a. P(3.0, 30., 45.) b. P(5.0, /4, 3/2) c. P(10., 135, 180) (a) x = r sin  cos  = 3sin 30o cos 45o = 1.06

y = r sin  sin  = 3sin 30o sin 45o = 1.06 z = r cos  = 3cos 30o = 2.6 so P(1.1,1.1, 2.6). (b)

2-7

x = r sin  cos  = 5sin 45o cos 270o = 0 y = r sin  sin  = 5sin 45o sin 270o = −3.5 z = r cos  = 5cos 45o = 3.5 so P(0, −3.5,3.5). (c) x = r sin  cos  = 10sin135o cos180o = −7.1 y = r sin  sin  = 10sin135o sin180o = 0 z = r cos  = 10 cos135o = −7.1 so P(−7.1, 0, −7.1).

P2.11: Given a volume defined by 1.0m ≤ r ≤ 3.0m, 0 ≤  ≤ 0, 90 ≤  ≤ 90, (a) sketch the volume, (b) perform the integration to find the volume, and (c) perform the necessary integrations to find the total surface area. (a)

Fig. P2.11 (b) 3

90o

 2

0

0

V =  r sin  drd d =  r dr  sin  d  d = 2

2

1

13 = 13.6m3 .

3

So volume V = 14 m3. (c) There are 5 surfaces: an inner, an outer, and 3 identical sides.

3

 2

1

0

S side =  rdrd =  rdr  d = 2 m2 ; S sides = 6 m 2 90o

 2

0

0

Souter =  r sin  d d = 3  sin  d  d = 2



2

m2 ; STOT = 11 m2 = 34.6m2 2 So Stotal = 35 m2.

Sinner =

9 2 m 2

2-8 4. Line Charges and the Cylindrical Coordinate System P2.12: Convert the following points from Cartesian to cylindrical coordinates: a. P(0.0, 4.0, 3.0) b. P(-2.0, 3.0, 2.0) c. P(4.0, -3.0, -4.0) 4 (a)  = 02 + 42 = 4,  = tan −1   = 90o , z = 3, so P(4.0,90o ,3.0) 0  3  (b)  = 22 + 32 = 3.6,  = tan −1   = 124o , z = 2, so P(3.6,120o , 2.0)  −2   −3  (c)  = 42 + 32 = 5,  = tan −1   = −37o , z = −4, so P(5.0, −37o , −4.0)  4  P2.13: Convert the following points from cylindrical to Cartesian coordinates: a. P(2.83, 45.0, 2.00) b. P(6.00, 120., -3.00) c. P(10.0, -90.0, 6.00) (a) x =  cos  = 2.83cos 45o = 2.00

y =  sin  = 2.83sin 45o = 2.00 z = z = 2.00 so P(2.00, 2.00, 2.00). (b) x =  cos  = 6.00 cos120o = −3.00 y =  sin  = 6.00sin120o = 5.20 z = z = −3.00 so P(−3.00,5.20, −3.00). (c) x =  cos  = 10.0 cos(−90.0o ) = 0

y =  sin  = 10.0sin(−90.0o ) = −10.0 z = z = 6.00 so P(0, −10.0, 6.00).

P2.14: A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diameter of 6.00 cm. a. Sketch the pipe conveniently overlaying the cylindrical coordinate system, lining up the length direction with the z-axis b. Determine the total surface area (this could actually be useful if, say, you needed to do an electroplating step on this piece of pipe) c. Determine the weight of the pipe given the density of copper is 8.96 g/cm3

2-9 (a) See Figure P2.14 (b) The top area, Stop, is equal to the bottom area. We must also find the inner area, Sinner, and the outer area, Souter.

3

2

2

0

Stop =   d  d =   d   d = 5 cm 2 .

Sbottom = Stop .

2

20

0

0

Souter =   d dz = 3  d  dz = 120 cm 2 2

20

0

0

Sinner ==   d dz = 2  d  dz = 80 cm 2 The total area, then, is 210 cm2, or Stot = 660 cm2. (c) Determining the weight of the pipe requires the volume: V =   d  d dz 3

2

20

2

0

0

=   d   d  dz = 100 cm3 . g   M pipe =  8.96 3  (100 cm3 ) cm   = 2815 g.

So Mpipe = 2820g.

Fig. P2.14 P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00. (a) A charge Q = 8.00 nC exists somewhere along the y-axis. Where must you locate Q so that the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of part (a) that you locate a charge Q at (0.00, 6.00m, 0.00). What value of Q will result in a total electric field intensity of zero at the origin? (a) The contributions to E from the line and point charge must cancel, or E = EL + EQ .

For the line: E L =

( 2nC / m ) L V a = a y = 18 a y −9 2 o  m 2 10 F 2m 36 m ( )

(

)

2-10 and for the point charge, where the point is located a distance y along the y-axis, we (8nC ) ( −a y ) Q 72 have: EQ = − a = = 2 ( −a y ) ( ) y 2 −9 4 o y 4 10 F y2 y 36 m Therefore: 72 72 = 18, or y = = 2m.

2 y 18

)

(

So Q ( 0,2.0m,0 ) (b)

Q 4 o ( 6 ) Q=

2

= 18,

(18)( 36 ) = 72nC. 9

Fig. P2.15

P2.16: You are given two z-directed line charges of charge density +1 nC/m at x = 0, y = -1.0 m, and charge density –1.0 nC/m at x = 0, y = 1.0 m. Find E at P(1.0m,0,0).

The situation is represented by Figure P2.16a. A better 2-dimensional view in Figure P2.16b is useful for solving the problem.  a + ay   L1 E1 = a  , and a  = 2  x .

2 o  2  

(1x10 C ) −9

E1 =

(

−9

2 10 F

2m ) 36 m ) (

( a + a ) FV = 9 a + a V , and E = 9 -a + a V . ( )m ( )m C 2 x

y

x

y

So ETOT = 18 ay V/m.

Fig. P2.16a

Fig. P2.16b

2

x

y

2-11 P2.17: MATLAB: Suppose you have a segment of line charge of length 2L centered on the z-axis and having a charge distribution L. Compare the electric field intensity at a point on the y-axis a distance d from the origin with the electric field at that point assuming the line charge is of infinite length. The ratio of E for the segment to E for the infinite line is to be plotted versus the ratio L/d using MATLAB. This is similar to MATLAB 2.3. We have for the ideal case

Eideal =

L L a = a .

2 o  2 o d

For the actual 2L case, we have an integration to perform (Equation (2.35) with different limits): +L

 L da y  L a  + L  dz z Eactual = =   3 2  2 2 2 4 o − L (  2 + z 2 ) 4 o  d z + d  − L Eactual =

 La y   L  2 .

2 2 o d  L + d 

Now we manipulate these expressions to get the following ratio: L Eactual d = .

2 Eideal L 1+ d In the program, the actual to ideal field ratio is termed “Eratio” and the charged line halflength L ratioed to the distance d is termed “Lod”.

( )

% % % % % % % % % % % % % % %

M-File: MLP0217 This program is similar to ML0203.

It compares the E-field from a finite length segment of charge (from -L to +L on the z-axis) to the E-field from an infinite length line of charge. The ratio (E from segment to E from infinite length line) is plotted versus the ratio Lod=L/d, where d is the distance along the y axis.

Wentworth, 12/19/02 Variables: Lod Eratio

clc clear

the ratio L/d ratio of E from segment to E from line

%clears the command window %clears variables

% Initialize Lod array and calculate Eratio Lod=0.1:0.01:100;

2-12 Eratio=Lod./(sqrt(1+Lod.^2)); % Plot Eratio versus Lod semilogx(Lod,Eratio) grid on xlabel('Lod=L/d') ylabel('E ratio: segment to line') Executing the program gives Figure P2.17.

Fig. P2.17 So we see that the field from a line segment of charge appears equivalent to the field from an infinite length line if the test point is close to the line.

P2.18: A segment of line charge L =10 nC/m exists on the y-axis from the origin to y = +3.0 m. Determine E at the point (3.0, 0, 0)m.

It is clear from a sketch of the problem in Figure P2.18a that the resultant field will be directed in the x-y plane. The situation is redrawn in a temporary coordinate system in Figure P2.18b.

We have from Eqn (2.34) E = 

 L dz  a  − za z = E a  + E z a z .

4 o  2 + z 2 3 2

(

)

For E we have: 3

   L   dz z E = L  =   4 o  2 + z 2 3 2 4 o   2 z 2 +  2  ( ) 0 With  = 3, we then have E = 21.2 V/m.

For Ez:

2-13

Fig. P2.18b

Fig. P2.18a 3

 −L L  zdz 1 V   = −8.79 Ez = = 3  2 2   4 o  2 + z 2 2 4 o   + z  m ( ) z =0 Thus we have ETOT = 21 a – 8.8 az V/m.

Converting back to the original coordinates, we have ETOT = 21 ax – 8.8 ay V/m.

5. Surface and Volume Charge P2.19: In free space, there is a point charge Q = 8.0 nC at (-2.0,0,0)m, a line charge L = 10 nC/m at y = -9.0m, x = 0m, and a sheet charge s = 12. nC/m2 at z = -2.0m.

Determine E at the origin.

The situation is represented by Figure P2.19, and the total field is ETOT = EQ + EL + ES. 8 x10−9 C ) a x ( Q EQ = aR = −9 2 4 o R 2 4 10 F 2m 36 m ( ) V = 18a x m 10 x10−9 C m ) a y ( L EL = a = −9 2 o  2 10 F 9m 36 m ( ) V = 20a y m 12 x10−9 C m 2 ) ( s Es = aN = az −9 2 o 2 10 F 36 m Fig. P2.19 V = 679a z m So: Etot = 18 ax + 20 ay + 680 az V/m.

)

(

)

(

(

)

2-14 P2.20: An infinitely long line charge (L = 21 nC/m) lies along the z-axis. An infinite area sheet charge (s = 3 nC/m2) lies in the x-z plane at y = 10 m. Find a point on the yaxis where the electric field intensity is zero.

We have ETOT = EL + ES. a a EL = L  = L y 2 o  2 o y

( 21 x10 C / m ) a = 378 a = −9

(

−9 2 10 F

36 m

)

y

y

y

y

( 3x10 C m ) ( −a y ) a Es = s N = −9 2 o 2 10 F 36 m = −54 a y −9

2

)

(

so 378 − 54 = 0, or y = 7. y Therefore, P(0, 7m, 0).

Fig. P2.20

P2.21: Sketch the following surfaces and find the total charge on each surface given a surface charge density of s = 1nC/m2. Units (other than degrees) are meters. (a) –3 ≤ x ≤ 3, 0 ≤ y ≤ 4, z = 0 (b) 1 ≤ r ≤ 4, 180 ≤  ≤ 360,  = /2 (c) 1 ≤  ≤ 4, 180 ≤  ≤ 360, z = 0

Fig. P2.21b&c

Fig. P2.21a

3

4

−3

0

(a) Q =   s dS =  s  dx  dy = 24 s = 24nC

2-15 4

2

1



(b) Q =  s  rdr  d = 4

2

1



15  s = 24nC 2

(c) Q =  s   d   d = 24nC P2.22: Consider a circular disk in the x-y plane of radius 5.0 cm. Suppose the charge density is a function of radius such that s = 12 nC/cm2 (when  is in cm). Find the electric field intensity a point 20.0 cm above the origin on the z-axis.

From section 4 for a ring of charge of radius a, E =

 L aha z

2 o ( a + h ) 2

L=sd and dE =

A d   ha z

2 o (  2 + h 2 ) 2 3

2

3

. Now we have 2

, where s = A nC/cm2. Now the total field is given

by the integral: Aha z  2d  E= .

2 o   2 + h 2 3 2

(

)

This can be solved using integration by parts, where u = , du = d, −1 d  v= , and dv = . This leads to 2 2  +h  2 + h2

 a + a 2 + h2  Ah  −a  + ln   a z .   2 o  a 2 + h2 h    Plugging in the appropriate values we arrive at E = 6.7 kV/cm az.

E=

P2.23: Suppose a ribbon of charge with density s exists in the y-z plane of infinite length in the z direction and extending from –a to +a in the y direction. Find a general expression for the electric field intensity at a point d along the x-axis.

The problem is represented by Figure P2.23a. A better representation for solving the problem is shown in Figure P2.23b. L a  , where L = sdy. Then, since We have dE = 2 o  da − ya y a  = x , d 2 + y2 the integral becomes da x − ya y  s dy E= .

2 o d 2 + y 2 d 2 + y 2

2-16 It may be noted that the ay component will cancel by symmetry. The ax integral is found from the appendix and we have  a E = s tan −1   a x .  o d 

FigP2.23b

FigP2.23a

P2.24: Sketch the following volumes and find the total charge for each given a volume charge density of v = 1nC/m3. Units (other than degrees) are meters. (a) 0 ≤ x ≤ 4, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6 (b) 1 ≤ r ≤ 5, 0 ≤  ≤ 60 (c) 1 ≤  ≤ 5, 0 ≤  ≤ 90, 0 ≤ z ≤ 5 4

5

6

0

0

0

(a) Q =  v dv = v  dx  dy  dz = 120nC (b) Q =  v dv 5

60

2

0

0

= v  r dr  sin  d  d = 130nC 2

1

(c) Q =  v dv 5

 2

5

1

0

0

Fig. P2.24a

= v   d   d  dz = 94nC

Fig. P2.24b

2-17

Fig. P2.24c P2.25: You have a cylinder of 4.00 inch diameter and 5.00 inch length (imagine a can of tomatoes) that has a charge distribution that varies with radius as v = (6 ) nC/in3 where  is in inches. (It may help you with the units to think of this as v (nC/in3)= 6 (nC/in4) (in)) Find the total charge contained in this cylinder.

2

2

5

0

0

Q =  v dv =  ( 6  )  d  d dz = 6  d   d  dz = 160 nC = 503nC 2

0

P2.26: MATLAB: Consider a rectangular volume with 0.00 ≤ x ≤ 4.00 m, 0.00 ≤ y ≤ 5.00 m and –6.00 m ≤ z ≤ 0.00 with charge density v = 40.0 nC/m3. Find the electric field intensity at the point P(0.00,0.00,20.0m). % %

MLP0226 calculate E from a rectangular volume of charge

% % % % % % % % % % % % % % %

variables xstart,xstop ystart,ystop zstart,zstop xt,yt,zt rhov Nx,Ny,Nz dx,dy,dz dQ eo dEi dEix,dEiy,dEiz dEjx,dEjy,dEjz dEkx,dEky,dEkz Etot

limits on x for vol charge (m) test point (m) vol charge density, nC/m^3 discretization points differential lengths differential charge, nC free space permittivity (F/m) differential field vector x,y and z components of dEi of dEj of dEk total field vector, V/m

2-18 clc clear % initialize variables xstart=0;xstop=4; ystart=0;ystop=5; zstart=-6;zstop=0; xt=0;yt=0;zt=20; rhov=40e-9; Nx=10;Ny=10;Nz=10; eo=8.854e-12; dx=(xstop-xstart)/Nx; dy=(ystop-ystart)/Ny; dz=(zstop-zstart)/Nz; dQ=rhov*dx*dy*dz; for k=1:Nz for j=1:Ny for i=1:Nx xv=xstart+(i-0.5)*dx; yv=ystart+(j-0.5)*dy; zv=zstart+(k-0.5)*dz; R=[xt-xv yt-yv zt-zv]; magR=magvector(R); uvR=unitvector(R); dEi=(dQ/(4*pi*eo*magR^2))*uvR; dEix(i)=dEi(1); dEiy(i)=dEi(2); dEiz(i)=dEi(3); end dEjx(j)=sum(dEix); dEjy(j)=sum(dEiy); dEjz(j)=sum(dEiz); end dEkx(k)=sum(dEjx); dEky(k)=sum(dEjy); dEkz(k)=sum(dEjz); end Etotx=sum(dEkx); Etoty=sum(dEky); Etotz=sum(dEkz); Etot=[Etotx Etoty Etotz] Now to run the program: Etot =

2-19

-6.9983 -8.7104 79.7668 >> So E = -7.0 ax -8.7 ay + 80. az V/m P2.27: MATLAB: Consider a sphere with charge density v = 120 nC/m3 centered at the origin with a radius of 2.00 m. Now, remove the top half of the sphere, leaving a hemisphere below the x-y plane. Find the electric field intensity at the point P(8.00m,0.00,0.00). (Hint: see MATLAB 2.4, and consider that your answer will now have two field components.) % % % % % % % % % % % % % % % % % % %

M-File: MLP0227 This program modifies ML0204 to find the field at point P(8m,0,0) from a hemispherical distribution of charge given by rhov=120 nC/m^3 from 0 < r < 2m and pi/2 < theta < pi.

Wentworth, 12/23/02 Variables: d a dV

% % % % % %

y axis distance to test point (m) sphere radius (m) differential charge volume where dV=delta_r*delta_theta*delta_phi eo free space permittivity (F/m) r,theta,phi spherical coordinate location of center of a differential charge element x,y,z cartesian coord location of charge % element R vector from charge element to P Rmag magnitude of R aR unit vector of R dr,dtheta,dphi differential spherical elements dEi,dEj,dEk partial field values Etot total field at P resulting from charge

clc clear

%clears the command window %clears variables

% Initialize variables eo=8.854e-12; d=8;a=2;

2-20 delta_r=40;delta_theta=72;delta_phi=144; % Perform calculation for k=(1:delta_phi) for j=(1:delta_theta) for i=(1:delta_r) r=i*a/delta_r; theta=(pi/2)+j*pi/(2*delta_theta); phi=k*2*pi/delta_phi; x=r*sin(theta)*cos(phi); y=r*sin(theta)*sin(phi); z=r*cos(theta); R=[d-x,-y,-z]; Rmag=magvector(R); aR=R/Rmag; dr=a/delta_r; dtheta=pi/delta_theta; dphi=2*pi/delta_phi; dV=r^2*sin(theta)*dr*dtheta*dphi; dQ=120e-9*dV; dEi=dQ*aR/(4*pi*eo*Rmag^2); dEix(i)=dEi(1); dEiy(i)=dEi(2); dEiz(i)=dEi(3); end dEjx(j)=sum(dEix); dEjy(j)=sum(dEiy); dEjz(j)=sum(dEiz); end dEkx(k)=sum(dEjx); dEky(k)=sum(dEjy); dEkz(k)=sum(dEjz); end Etotx=sum(dEkx); Etoty=sum(dEky); Etotz=sum(dEkz); Etot=[Etotx Etoty Etotz] Now to run the program: Etot = 579.4623 0.0000 56.5317 So E = 580 ax + 57 az V/m.

6. Electric Flux Density

2-21 P2.28: Use the definition of dot product to find the three interior angles for the triangle bounded by the points P(-3.00, -4.00, 5.00), Q(2.00, 0.00, -4.00), and R(5.00, -1.00, 0.00).

Here we use A B = A B cos  AB .

PR = ( 5 − −3) a x + ( −1 − −4 ) a y + ( 0 − 5 ) a z PR = 8a x + 3a y − 5a z , PR = 9.9

PQ = 5a x + 4a y − 9a z , PQ = 11.0 PR PQ = ( 8 )( 5 ) + ( 3)( 4 ) + ( −5 )( −9 ) = 97 = 9.9 11.0 cos  P 

 97  = 27 9.9 11 ( )( )  

 p = cos −1 

Fig. P2.28

RQ = −3a x + 1a y − 4a z , RQ = 5.1 RP = −8a x − 3a y + 5a z , RP = 9.9 RQ RP = 1 = ( 5.1)( 9.9 ) cos  R , R = 89

Q = 180 − 27 − 89 = 64 P2.29: Given D = 2 a + sin  az C/m2, find the electric flux passing through the surface defined by 2.0 ≤  ≤  m, 90. ≤  ≤ 180, and z = 4.0 m.

 =  E dS, dS =  d  da z 4



2

 2

 =  ( 2  a  + sin a z )  d  da z =   d   sin  d = 6C

P2.30: Suppose the electric flux density is given by D = 3r ar –cos  a + sin2 a C/m2.

Find the electric flux through both surfaces of a hemisphere of radius 2.00 m and 0.00 ≤  ≤ 90.0˚.

1 =  D dS, dS1 = r 2 sin  d dar

1 =  ( 3ra r − cos  a + sin 2  a ) ( r 2 sin  d d a r )  2

2

0

0

= 3r  sin  d  d = 48 C 3

dS2 = rdrda

2-22

 2 =  − cos  a rdrd a 2

 r2 2   sin  = 0 = −  2 0   0  = 48 C

Fig. P2.30 7. Gauss’s Law and Applications P2.31: Given a 3.00 mm radius solid wire centered on the z-axis with an evenly distributed 2.00 coulombs of charge per meter length of wire, plot the electric flux density D versus radial distance from the z-axis over the range 0 ≤  ≤ 9 mm.

2 (C m) C = 70.7 x103 3 , ( a = 3mm = .003m ) 2 a m Qenc =  D dS =  D a   d dza  = 2 LD , where L is the length of the Gaussian

For a 1 m length, v =

surface. Note that this expression for Qenc is valid for both Gaussian surfaces.

GS1 ( < a): 

2

L

0

0

0

Qenc =  v dv = v  d   d  dz = v  2 L so D =

v   L v =  for   a.

2 L 2 2

GS2 ( > a):

Qenc = v a 2 L, D =

v a 2 1 for   a.

2 

This is plotted with the following Matlab routine: % M-File: MLP0231 % % Gauss's Law Problem % solid cylinder with even charge % % Variables % rhov charge density (C/m^3) % a radius of cylinder (m) % rho radial distance from z-axis % rhomm rho in mm % D electric flux density (C/m^3)

2-23 % %

N maxrad

number of data points max radius for plot (m)

clc;clear; % initialize variables rhov=70.7e3; a=0.003; maxrad=.009; N=100; bndy=round(N*a/maxrad); for i=1:bndy rho(i)=i*maxrad/N; rhomm(i)=rho(i)*1000; D(i)=rhov*rho(i)/2; end for i=bndy+1:N rho(i)=i*maxrad/N; rhomm(i)=rho(i)*1000; D(i)=(rhov*a^2)/(2*rho(i)); end plot(rhomm,D) xlabel('radial distance (mm)') ylabel('elect. flux density (C/m^2)') grid on P2.32: Given a 2.00 cm radius solid wire centered on the z-axis with a charge density v = 6 C/cm3 (when  is in cm), plot the electric flux density D versus radial distance from the z-axis over the range 0 ≤  ≤ 8 cm.

Fig. P2.31

2-24

Choose Gaussian surface length L, and as usual we have Qenc =  D dS =  D a   d dza  = 2 L D , valid for both Gaussian surfaces.

In GS1 ( < a): Qenc =  v dv =  6 2 d  d dz = 4 L 3 ,

4 L 3 so D = = 2  2 for   a.

2 L For GS2 ( > a): Qenc = 4 La3 , D =

2a 3



for   a.

This is plotted for the problem values in the following Matlab routine. % % % % % % % % % % %

M-File: MLP0232 Gauss's Law Problem solid cylinder with radially-dependent charge Variables a radius of cylinder (cm) rho radial distance from z-axis D electric flux density (C/cm^3) N number of data points maxrad max radius for plot (cm)

clc;clear; % initialize variables a=2; maxrad=8; N=100; bndy=round(N*a/maxrad); for i=1:bndy rho(i)=i*maxrad/N; D(i)=2*rho(i)^2; end for i=bndy+1:N rho(i)=i*maxrad/N; D(i)=(2*a^3)/rho(i); end plot(rho,D) xlabel('radial distance (cm)') ylabel('elect. flux density (C/cm^2)') grid on

2-25

Fig. P2.32

P2.33: A cylindrical pipe with a 1.00 cm wall thickness and an inner radius of 4.00 cm is centered on the z-axis and has an evenly distributed 3.00 C of charge per meter length of pipe. Plot D as a function of radial distance from the z-axis over the range 0 ≤  ≤ 10 cm.

Qenc =  D dS =  D a   d dza  = 2 h D ; this is true for all the Gaussian surfaces.

GS1 ( < a): since Qenc = 0, D = 0.

GS2(a <  < b): 3h 3 v = = 2 2   d  d dz  (b − a )

Qenc =  v dv 

=

2

h

3  d   d  dz 2  ( b − a 2 ) a 0 0

( − a ) = 3h (b − a ) 2

2

2

2

So, Fig. P2.33a

3h (  2 − a 2 )

(  − a ) for a    b.

D = = 2 h ( b − a ) 2 ( b − a ) 

GS3( > b):

2

2

3

2

2

2

2

2-26

3

for   b.

2 A plot with the appropriate values is generated by the following Matlab routine: Qenc = 3h, D =

% M-File: MLP0233 % Gauss's Law Problem % cylindrical pipe with even charge distribution % % Variables % a inner radius of pipe (m) % b outer radius of pipe (m) % rho radial distance from z-axis (m) % rhocm radial distance in cm % D electric flux density (C/cm^3) % N number of data points % maxrad max radius for plot (m) clc;clear; % initialize variables a=.04;b=.05;maxrad=0.10;N=100; bndya=round(N*a/maxrad); bndyb=round(N*b/maxrad); for i=1:bndya rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=0; end for i=bndya+1:bndyb rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=(3/(2*pi*rho(i)))*((rho(i)^2-a^2)/(b^2-a^2)); end for i=bndyb+1:N rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=3/(2*pi*rho(i)); end plot(rhocm,D) xlabel('radial distance (cm)') ylabel('elect. flux density (C/m^2)') grid on

2-27

Fig. P2.33b

P2.34: An infinitesimally thin metallic cylindrical shell of radius 4.00 cm is centered on the z-axis and has an evenly distributed charge of 100. nC per meter length of shell. (a) Determine the value of the surface charge density on the conductive shell and (b) plot D as a function of radial distance from the z-axis over the range 0 ≤  ≤ 12 cm.

s =

Q 100nC 100nC nC = = = 398 2 .

S   d dz (.04m )( 2 )(1m ) m

For all Gaussian surfaces, of height h and radius , we have: Qenc =  D dS, where dS =  d dza  ,

 Qenc = 2 hD .

GS1 ( < a): Qenc = 0 so D = 0 GS2 ( > a): Qenc =   s dS =  s   d dz = 2 ah s ,

D =  s % % % % % % % % %

a



for   a.

M-File: MLP0234 Fig. P2.34a Gauss's Law Problem cylindrical shell of charge Variables a radius of cylinder (m) Qs surface charge density (nC/m^2) rho radial distance from z-axis (m)

2-28 % % % %

rhocm D N maxrad

radial distance in cm electric flux density (nC/cm^3) number of data points max radius for plot (cm)

clc;clear; % initialize variables a=.04;Qs=398;maxrad=0.12;N=100; bndy=round(N*a/maxrad); for i=1:bndy rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=0; end for i=bndy+1:N rho(i)=i*maxrad/N; rhocm(i)=rho(i)*100; D(i)=Qs*a/rho(i); end plot(rhocm,D) xlabel('radial distance (cm)') ylabel('elect. flux density (nC/m^2)') grid on

Fig. P2.34b

2-29 P2.35: A spherical charge density is given by v = o r/a for 0 ≤ r ≤ a, and v = 0 for r > a. Derive equations for the electric flux density for all r.

Qenc =  D dS =  Dr ar r 2 sin  d dar = 4 r 2 Dr . surface.

GS1 (r < a): Qenc =  v dv =

o r 3



2

0

0

This is valid for each Gaussian

 r dr  sin  d  d =

a 0

o r 4 o 2 = r for r  a.

So Dr = a 4 r 2 4a  a3 GS2 (r > a): Qenc = o a 3 , Dr = o 2 4r

o r 4 a

.

for r  a.

P2.36: A thick-walled spherical shell, with inner radius 2.00 cm and outer radius 4.00 cm, has an evenly distributed 12.0 nC charge. Plot Dr as a function of radial distance from the origin over the range 0 ≤ r ≤ 10 cm.

Here we’ll let a = inner radius and b = outer radius. Then Qenc =  D dS =  Dr ar r 2 sin  d dar = 4 r 2 Dr ; This is true for each Gaussian surface.

The volume containing charge is b  2 4 v =  r 2 dr  sin  d  d =  ( b3 − a 3 ) .

3 a 0 0 So v =

Q 3Q = . v 4 ( b3 − a 3 )

Now we can evaluate Qenc for each Gaussian surface.

GS1 (r < a): Qenc = 0 so Dr = 0. r



2

GS2 (a < r < b): Qenc =  v dv = v  r 2 dr  sin  d  d = a

0

0

v 4 3

(r − a ).

3

3

Inserting our value for v, we find 3 3 Q (r − a ) Dr = for a  r  b.

4 r 2 ( b3 − a3 )

Q , for r  b.

4 r 2 This is plotted for appropriate values using the following Matlab routine: GS3 (r >b): Qenc = Q, Dr =

% % % %

M-File: MLP0236 Gauss's Law Problem thick spherical shell with even charge

2-30 % % % % % % % % %

Variables a inner radius of sphere (m) b outer radius of sphere (m) r radial distance from origin (m) rcm radial distance in cm D electric flux density (nC/cm^3) N number of data points maxr max radius for plot (m) Q charge (nC)

clc;clear; % initialize variables a=.02;b=.04; Q=12; maxrad=0.10; N=100; bndya=round(N*a/maxrad); bndyb=round(N*b/maxrad); for i=1:bndya r(i)=i*maxrad/N; rcm(i)=r(i)*100; D(i)=0; end Fig. P2.36 for i=bndya+1:bndyb r(i)=i*maxrad/N; rcm(i)=r(i)*100; D(i)=(Q/(4*pi*r(i)^2))*(r(i)^3-a^3)/(b^3-a^3); end for i=bndyb+1:N r(i)=i*maxrad/N; rcm(i)=r(i)*100; D(i)=Q/(4*pi*r(i)^2); end plot(rcm,D) xlabel('radial distance (cm)') ylabel('elect. flux density (nC/m^2)') grid on

2-31 P2.37: Given a coaxial cable with solid inner conductor of radius a, an outer conductor that goes from radius b to c, (so c > b > a), a charge +Q that is evenly distributed throughout a meter length of the inner conductor and a charge –Q that is evenly distributed throughout a meter length of the outer conductor, derive equations for the electric flux density for all . You may orient the cable in any way you wish.

We conveniently center the cable on the z-axis. Then, for a Gaussian surface of length L, Qenc =  D dS = 2 LD ; valid for all Gaussian surfaces.

GS1: ( < a): v =

Q

(1m ) ( a 2 )



2

L

0

0

0

;

Q1 =  v dv = v   d   d  dz = D =

QL 2  ; a2

QL Q =  for   a 2 a 2 L 2 a 2 2

GS2 (a <  < b): Q2 = QL; D =

QL Q = 2 L 2

GS3 (b <  < c): Q3 = Q +  vo dv, where vo =

for a    b.

−Q (1m )  ( c 2 − b2 )

 2 L c2 −  2 ) ( −Q Q3 = Q +  d   d  dz = Q 2 2  ( c 2 − b2 ) b (c − b ) 0 0

2 2 Q (c −  ) so D = for b    c.

2 ( c 2 − b2 )

GS4 ( > c): Qenc = 0, D = 0.

8. Divergence and the Point Form of Gauss’s Law P2.38: Determine the charge density at the point P(3.0m,4.0m,0.0) if the electric flux density is given as D = xyz az C/m2.

Dz  ( xyz ) = = xy = v . z z v(3,4,0)=(3)(4)=12 C/m3.  D=

P2.39: Given D = 3ax +2xyay +8x2y3az C/m2, (a) determine the charge density at the point P(1,1,1). Find the total flux through the surface of a cube with 0.0 ≤ x ≤ 2.0m, 0.0

2-32 ≤ y ≤ 2.0m and 0.0 ≤ z ≤ 2.0m by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.

 C ( 2 xy ) = 2 x, v (1,1,1) = 2 3 . y m

(a)  D =

(b)  D dS =   Ddv =  + top

 + +  +  + 

bottom

2

left

right

front

back

2

 =  8x y a dxdya =8 x dx y dy = 85.3C 2

3

2

z

top



bottom

3

z

0

0

=  8 x 2 y 3a z ( −dxdya z ) = − 85.3C

 =  2x y

y=0

a y ( −dxdza y ) =0

left



=  2 x y y=2 a y dxdza y =16C



=  3a x dydza x =12C



=  3a x ( −dydza x ) = − 12C

right

front

back

Qenc =  D dS = 16C. (c)  D =

 ( 2 xy ) = 2 x; y

2

2

2

0

0

0

  Ddv = 2 xdx  dy  dz = 16C.

P2.40: Suppose D = 6cos a C/m2. (a) Determine the charge density at the point (3m, 90, -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤  ≤ 4m, 0 ≤  ≤ 90, and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.

1 D 1  ( 6  cos  ) = == −6sin .     C v ( 3,90 , −2 ) = −6 3 . m

(a) (  D )cylinder =

(b)  D dS =  +  +  +  +   =0

 =90

top

bottom

outside

,

2-33 note that the top, bottom and outside integrals yield zero since there is no component of D in the these dS directions.



=  6  cos 



=  6  cos 

 =0

 =90

 =0

a ( −d  dza ) = −192C

 =90

a ( d  dza ) = 0

So,  D dS = −192C. (c)  D = −6sin  , dv =  d  d dz 90

4

0

0

0

−4

  Ddv = −6  sin  d   d   dz = −192C. P2.41: Suppose D = r2sin ar + sincos a C/m2. (a) Determine the charge density at the point (1.0m, 45, 90). Find the total flux through the surface of a volume defined by 0.0 ≤ r ≤ 2.0 m, 0.0 ≤  ≤ 90., and 0.0 ≤  ≤ 180 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.

The volume is that of a quartered-sphere, as indicated in Figure P2.41. (a)

1  2 1 D sin  r D + = 4r sin  − =v , ( ) r 2 r r r sin   r C v (1, 45 ,90 ) = 1.83 3 m

 D=

(b)  D dS =  +  +  +  ; note that  =0

 =180

 =90

r =2



=  sin  cos   =0 a ( −rdrd a ) = −2C



=  sin  cos   =180 a rdrd a = −2C

 =0

 =180



r =2

=  r sin  ar r sin  d d ar = r 2

2

4

 = 0 since D = 0.

 =90

 2



90

0

0

0

2 2  sin  d  d = 8  (1 − cos 2 ) d =4 C

Summing these terms we have Q = 4( – 1)C = 35.5C. 2

(c)

2-34

sin   2



  Ddv =   4r sin  − r  r sin  drd d  2

2



2

 2



0

0

0

0

= 4 r dr  sin  d  d −  rdr  sin  d  sin  d = 4 2 − 4 = 35.5C. 3

0

2

0

Fig. P2.41

9. Electric Potential P2.42: A sheet of charge density s = 100 nC/m2 occupies the x-z plane at y = 0. (a) Find the work required to move a 2.0 nC charge from P(-5.0m, 10.m, 2.0m) to M(2.0m, 3.0m, 0.0). (b)Find VMP. M

(a) W = −Q  E dL; so we need E for the sheet charge.

P

(100 x10 C ) FV a = 5.65x103 V a  E = s aN = y y 2 o m 2 (8.854 x10−12 F m ) C −9

Notice that we are only concerned with movement in the y-direction. We then have: y =3 V    J  −9 W = −2 x10 C   5.65 x103 a y  dya y   = 79 J m   CV  y =10  (b) VMP =

( 79 J ) CV = 39.5kV ; so V = 40kV .

W = MP Q ( 2 x10−9 C ) J

P2.43: A surface is defined by the function 2x + 4y2 –ln z = 12. Use the gradient equation to find a unit vector normal to the plane at the point (3.00m,2.00m,1.00m).

2-35 Let F = 2 x + 4 y 2 − ln z = 12, then F 1 ax = ; F = 2a x + 8 ya y − a z , F z At (3,2,1),

F = 2a x + 16a y − a z , F = 22 + 162 + 12 = 16.16, a N = 0.124a x + 0.990a y − 0.062a z

P2.44: For the following potential distributions, use the gradient equation to find E. (a) V = x+y2z (V) (b) V = 2sin(V) (c) V = r sin cos (V). (a) E = −V = −a x − 2 yza y − y 2a z

 V 1 V V  a + a + a z  = −2  sin  a  −  cos  a (b) E = −V = −    z    (c)  V 1 V 1 V  E = −V = −  a + a + a  = − sin  cos  a  − cos  cos  a + sin  a r  r sin     r

P2.45: A 100 nC point charge is located at the origin. (a) Determine the potential difference VBA between the point A(0.0,0.0,-6.0)m and point B(0.0,2.0,0.0)m. (b) How much work would be done to move a 1.0 nC charge from point A to point B against the electric field generated by the 100 nC point charge?

A

(a) VBA = −  E dL. A

The potential difference is only a function of radial distance from the origin. Letting ra = 6m and rb = 2m, we then have rb Q Q 1 1 VBA = −  a dra r =  −  = 300V .

2 r 4  r 4  o o  rb ra  ra (b) W = Q2VBA = (10−9 C ) ( 300V )

J = 300nJ CV

P2.46: MATLAB: Suppose you have a pair of charges Q1(0.0, -5.0m, 0.0) = 1.0 nC and Q2(0.0, 5.0m, 0.0) = 2.0 nC. Write a MATLAB routine to calculate the potential VRO moving from the origin to the point R(5.0m, 0.0, 0.0). Your numerical integration will involve choosing a step size L and finding the field at the center of the step. You should try several different step sizes to see how much this affects the solution.

2-36

% % % % % % % % % % % % % % % % % % % % % % % % % %

M-File: MLP0246 Modify ML0207 to calculate the potential difference going from the origin (O) to the point R(5,0,0) given a pair of point charges Q1(0,-5,0)=1nC and Q2(0,5,0)=2nC. The approach will be to break up the distance from O to R into k sections. The total field E will be found at the center of each section (located at point P) and then dot(Ep,dLv) will give the potential drop across the kth section. Total potential is found by summing the potential drops.

Wentworth, 1/7/03 Variables: Q1,Q2 k dL dLv x(n) R1,R2 E1,E2 Etot V(n)

clc clear

the point charges, in nC number of numerical integration steps magnitude of one step vector for a step x location at center of section at P vector from Q1,Q2 to P electric fields from Q1 & Q2 at P total electric field at P portion of dot(Etot,dL) at P

%clears the command window %clears variables

% Initialize variables k=64; Q1=1; Q2=2; dL=5/k; dLv=dL*[1 0 0]; % Perform calculation for n=1:k x(n)=(n-1)*dL+dL/2; R1=[x(n) 5 0]; R2=[x(n) -5 0]; Rmag1=magvector(R1); Rmag2=magvector(R2); E1=9*Q1*R1/Rmag1^3;

2-37 E2=9*Q2*R2/Rmag2^3; Etot=E1+E2; V(n)=dot(Etot,dLv); end Vtot=sum(-V) Now running the program: Vtot = -1.5817 So VRO = -1.6 V. P2.47: For an infinite length line of charge density L = 20 nC/m on the z-axis, find the potential difference VBA between point B(0, 2m, 0) and point A(0, 1m, 0).

B

VBA = −  E dL; E = A

L a  , dL = d  a  , 2 o 

L −L a d a = ln ( 2 ) = −250V 2 o  2 o A

B

so VBA = − 

P2.48: Find the electric field at point P(0.0,0.0,8.0m) resulting from a surface charge density s = 5.0 nC/m2 existing on the z = 0 plane from  = 2.0 m to  = 6.0 m. Assume V = 0 at a point an infinite distance from the origin.

(Method 1) For a ring of charge it was previously found that  L aha z E= .

3 2 o ( a 2 + h 2 ) 2 We can then break up our disk into differential rings (see Figure P2.48), each contributing dE as:  h d  dE = S a z , where we've used  L =  S d  .

2 o  2 + h 2 3 2

(

)

So we then have  ha d  E= S z  .

3 2 o  2 + h2 2

(

)

This is easy to integrate if we let u = 2 + h2, then du = 2  d, and we have −3  ha  ha −2 −  S ha z E = S z  u 2 du = S z = 4 o 4 o 2 o u

b

1

 2 + h2 a

2-38 Solving, we arrive at  − h  1 1 E= S  − .

2 o  b2 + h2 a 2 + h2  Upon inserting the appropriate values we find E = 48 V/m az.

Fig. P2.48 (method 2) Find an expression for potential and then evaluate the gradient at the point.  S  d  d dQ V = , R =  2 + h 2 , dQ =  S  d  d , so V=  4 o R 4 o  2 + h 2

V=

b S b  d  S  2 2 =  + h = S  b2 + h2 − a 2 + h2  .   a 2 o a  2 + h2 2 o 2 o 

Now we let h = z and E = −V ; 1 1     E = − S  ( b2 + z 2 ) 2 − ( a 2 + z 2 ) 2  a z 2 o  z z 

=−

  S  1 2 2 − 12   1 2 2 − 12  z z b + z 2 z − a + z ) 2z  az = − S  − ( ) (  az  2 o  2 2 2 o  b 2 + z 2  a2 + z 2 

Plugging in the values we find E = 48 V/m az.

P2.49: Suppose a 6.0 m diameter ring with charge density 5.0 nC/m lies in the x-y plane with the origin at its center. Determine the potential difference Vho between the point h(0.0,0.0,4.0)m and the origin. (Hint: first find an expression for E on the z-axis as a general function of z.)

For the ring of charge, replacing h with z, we have  aa z E= L z 2 o a 2 + z 2 3 2

(

)

2-39 h

Vho = −  E dL = − 0

2

L a zdz .  2 o a 2 + z 2 3 2

(

)

2

Letting u = a + z , du = 2z dz, we have  a −3 a 1 Vho = − L  u 2 du = L .

4 o 2 o u Replacing u and evaluating from 0 to h,  a 1 1 Vho = L  −  2 o  a 2 + h 2 a  = −36 V = −113V .

Fig. P2.49

10. Conductivity and Current P2.50: A columnular beam of electrons from 0 ≤  ≤ 1 mm has a charge density v =-0.1 cos(/2) nC/mm3 (where  is in mm) and a velocity of 6 x 106 m/sec in the +az direction. Find the current.

   Let’s let v = o cos   , where o = -0.1 nC/mm3. Then we’ll let u = uoaz, where uo =  2  9 6x10 mm/s. Notice we convert the units to mm. Now,    J = vu = ouo cos   a z ,  2  and with dS =  d d az we then have a 2    I =  J dS = ouo   cos   d   d .  2  0 0 This becomes a    I = A  cos   d  , where A = 2ouo.  2  0 Now we can integrate by parts, or

 udv = uv −  vdu, where u = A, du = Ad,

      sin   , and dv = cos   d  .   2   2  We then have 2 Aa   a  4A    a   I= sin   + 2 cos   − 1 .   2     2   To evaluate, we first find A = 2(-0.1x10-9)(6x109)=3.77, and then I = 2.40-1.53=0.87A. v=

2

I = 0.87A.

2-40 P2.51: Two spherical conductive shells of radius a and b (b > a) are separated by a material with conductivity . Find an expression for the resistance between the two spheres.

First find E for a < r < b, assuming +Q at r = a and –Q at r = b. From Gauss’s law: Q E= ar 4 o r 2 Now find Vab: a a Q Vab = −  E dL = −  a dra r 2 r 4  r o b b

−Q dr Q 1 Q 1 1 = = =  − .

2  4 o b r 4 o r b 4 o  a b  Now can find I: Q 1 I =  J dS =   E dS =   a r 2 sin  d da r 2 r 4 o r a

a

2 Q  Q = sin  d  d = .  4 o 0 o 0 V 1 1 1 Finally, R = ab =  −  I 4  a b 

P2.52: The typical length of each piece of jumper wire on a student’s protoboard is 5.0 cm. Assuming AWG-20 (wire diameter 0.812 mm) copper wire, (a) determine the resistance for this length of wire. (b) Determine the power dissipated in the wire for 10. mA of current. (a) R =

1 L 1 0.05m = = 1.67m 2 7   a 5.8 x10 ( S m )  ( 0.406 x10−3 m )2

so R = 1.7 m (b) P = I 2 R = (10 x10−3 A) (1.7 x10−3  ) = 170nW 2

P2.53: A densely wrapped coil of AWG-22 (0.644 mm diameter) copper magnet wire is 150 m long. The wire has a very thin insulative sheath. Determine the resistance for this length of wire.

R=

1 L 1 150m = = 7.94 2 7   a 5.8 x10 S m  ( 0.322 x10−3 m )2

so R = 7.9

2-41

P2.54: Determine an expression for the power dissipated per unit length in coaxial cable of inner radius a, outer radius b, and conductivity between the conductors  if a potential difference Vab is applied. b ln   2 L  a  Now for a given potential difference Vab we have V 2 2 LVab2 P 2 Vab2 P = ab = , so = .

R L ln b ln b a a

From Eqn(2.84) we have R =

( )

1

( )

P2.55: Find the resistance per unit length of a stainless steel pipe of inner radius 2.5 cm and outer radius 3.0 cm.

R=

1

L ,   (b − a2 ) 2

   R 1 1 1 1 m   = 1.05 = =   6 2 2 2 2 2   L   ( b − a )  1.1x10 S m   (.030 − .025 ) m m   so R/L = 1.0 m/m so we have

P2.56: A nickel wire of diameter 5.0 mm is surrounded by a 0.50 mm thick layer of silver. What is the resistance per unit length for this wire? Assuming 1.0 m of this wire carries 1.0 A of current, determine the power dissipated in the nickel portion and in the silver portion of the wire.

We can treat this wire as two resistors in parallel. We have RNi 1 1  = = 3.4 x10−3 2 7 L 1.5 x10  ( 2.5 x10−3 ) m

1 1  = 1.87 x10−3 7 2 2 L 6.2 x10  ( 3x10−3 ) − ( 2.5 x10−3 )  m   Rtotal RNi RAg m = = 1.2 L L L m To find the power dissipated, we first find the potential difference: V = IRtotal = 1.2mV then V2 V2 PNi = = 0.42mW , PAg = = 0.77mW RNi RAg RAg

=

2-42

11. Dielectrics P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m.

Determine the electric flux density.

(

D =  E = 194.5 x10−12 F

m

a )(12V m) FVC = 2.3 nC m 2

x

P2.58: MATLAB: A 20 nC point charge at the origin is embedded in Teflon (r = 2.1).

Find and plot the magnitudes of the polarization vector, the electric field intensity and the electric flux density at a radial distance from 0.1 cm out to 10 cm.

We use the following equations: Q E= , P =  e o E, D =  r  o E 4 r  o r 2 % % % % % % % % % % % % %

M-File: MLP0258 Plot E, P and D vs distance r from a point charge Q at the origin with a dielectric.

Variables Q charge (C) eo free space permittivity (F/m) r radial distance (m) Chi electric susceptibility E electric field intensity(V/m) D electric flux density (C/m^2) P polarization vector (C/m^2)

% initialize variables Q=20e-9; er=2.1; eo=8.854e-12; Chi=er-1; % perform calculations r=0.001:.001:0.100; rcm=r.*100; E=Q./(4*pi*r.^2); P=Chi*eo*E; D=er*eo*E; %

plot data

2-43 subplot(2,1,1) loglog(rcm,P,'--k',rcm,D,'-k') legend('P','D') ylabel('C/m^2') grid on subplot(2,1,2) loglog(rcm,E) ylabel('V/m') xlabel('radial distance (cm)') grid on

Fig. P2.58

P2.59: Suppose the force is very carefully measured between a pair of point charges separated by a dielectric material and is found to be 20 nN. The dielectric material is removed without changing the position of the point charges, and the force has increased to 100 nN. What is the relative permittivity of the dielectric?

F1 =

Q1Q2 QQ , F2 = 1 2 2 , 2 4 r  o R 4 o R

F2 100 = r = =5 F1 20

P2.60: The potential field in a material with r = 10.2 is V = 12 xy2 (V). Find E, P and D.

E = −V = −

 (12 xy 2 ) x

ax −

 (12 xy 2 )

y nC D =  r  o E = -1.1y 2a x − 2.2 xya y 2 m  e =  r − 1 = 9.2

a y = −12 y 2a x − 24 xya y

P =  e o E = ( 9.2 ) (8.854 x10−12 ) E = -9.8 y 2a x − 2.00 xya y

V m

nC m2

P2.61: In a mineral oil dielectric, with breakdown voltage of 15 MV/m, the potential function is V = x3 – 6x2 –3.1x (MV). Is the dielectric likely to breakdown, and if so, where?

2-44

E = −V = ( −3x 2 + 12 x + 3.1) a x

MV m

dE d 2E = −6 x + 12, = −6, so from 6x – 12 = 0 we find the maximum electric field dx dx 2 occurs at x = 2m.

At x = 2m, we have E = -12+24+3.1 = 15.1 MV/m, exceeding the breakdown voltage.

12. Boundary Conditions P2.62: For y < 0, r1 = 4.0 and E1 = 3ax + 6ay + 4az V/m. At y = 0, s = 0.25 nC/m2. If r2 = 5.0 for y > 0, find E2. (g) E2 = 3ax + 20.7ay + 4az V/m E1 = 3ax + 6ay + 4az V/m (a) EN1 = 6ay (f) EN2 = DN2/5o = 20.7ay (b) ET1 = 3ax + 4az (c) ET2 = ET1 = 3ax + 4az (e) DN2 = 0.92 ay (d) DN1 = r1oEN1 = 24o ay (e) a21 ( D1 − D2 ) =  s , -a y ( DN 1 − DN 2 ) a y = s , DN 2 − DN 1 = s

DN 2 =  s + DN 1 = 0.25

 10−9 F  nC nC nC + 24    2 = 0.92 2 2 m m  36 m  m

P2.63: For z ≤ 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes a 30 angle with a normal to the surface, what angle does E2 make with a normal to the surface?

Refer to Figure P2.63.

ET 1 = E1 sin 1 , ET 2 = E2 sin 2 , and ET 1 = ET 2 also DN 1 =  r1 o E1 cos 1 , DN 2 =  r 2 o E2 cos 2 , and DN 1 = DN 2 ( since  s = 0 ) Therefore   ET 1 ET 2 = , and after routine math we find  2 = tan −1  r 2 tan 1  DN 1 DN 2   r1  Using this formula we obtain for this problem 2 = 14°.

Fig. P2.63

2-45

P2.64: A plane defined by 3x + 2y + z = 6 separates two dielectrics. The first dielectric, on the side of the plane containing the origin, has r1 = 3.0 and E1 = 4.0az V/m. The other dielectric has r2 = 6.0. Find E2.

We first use gradient to find a normal to the planar surface.

Let F = 3x + 2y + z – 6 = 0. F = 3a x + 2a y + a z , and F = 14,

F = 0.802a x + 0.534a y + 0.267a z . F Now we can work the boundary condition problem.

E1 = 4a z , EN1 = ( E1 a N ) a N = 0.857a x + 0.570a y + 0.285a z . so a N =

ET 1 = E1 − EN 1 = −0.857a x − 0.570a y + 3.715a z , ET 2 = ET 1

DN 1 =  r1 o EN 1 =  o 2.571a x + 1.710a y + 0.855a z  , and DN 2 = DN 1 EN 2 =

DN 2

 r 2 o

=

DN 2 = 0.429a x + 0.285a y + 0.143a z 6 o

Finally we have E2 = ET 2 + E N 2 = −0.43a x − 0.29a y + 3.8a z

V . m

P2.65: MATLAB: Consider a dielectric-dielectric charge free boundary at the plane z = 0. Construct a program that will allow the user to enter r1 (for z < 0), r2, and E1, and will then calculate E2. (Just for fun, you may want to have the program calculate the angles that E1 and E2 make with a normal to the surface). % M-File: MLP0265 % % Given E1 at boundary between a pair of % dielectrics with no charge at boundary, % calculate E2. Also calculates angles. % clc clear % enter variables disp('enter vector quantities in brackets,') disp('for example: [1 2 3]') er1=input('relative permittivity in material 1: '); er2=input('relative permittivity in material 2: '); a12=input('unit vector from mtrl 1 to mtrl 2: '); E1=input('electric field intensity vector in mtrl 1: '); %

perform calculations

2-46 En1=dot(E1,a12)*a12; Et1=E1-En1; Et2=Et1; Dn1=er1*En1; %ignores eo since it will factor out Dn2=Dn1; En2=Dn2/er2; E2=Et2+En2 % calculate the angles th1=atan(magvector(Et1)/magvector(En1)); th2=atan(magvector(Et2)/magvector(En2)); th1r=th1*180/pi th2r=th2*180/pi Now run the program: enter vector quantities in brackets, for example: [1 2 3] relative permittivity in material 1: 2 relative permittivity in material 2: 5 unit vector from mtrl 1 to mtrl 2: [0 0 1] electric field intensity vector in mtrl 1: [3 4 5] E2 = 3

4

2

th1r = 45

th2r = 68.1986

P2.66: A 1.0 cm diameter conductor is sheathed with a 0.50 cm thickness of Teflon and then a 2.0 cm (inner) diameter outer conductor. (a) Use Laplace’s equations to find an expression for the potential as a function of  in the dielectric. (b) Find E as a function of . (c) What is the maximum potential difference that can be applied across this coaxial cable without breaking down the dielectric? (a) Since V is only a function of ,

2-47

 2Vcyl =

1   V   =0     

V =A  or V = A ln  + B where A and B are constants.

Now we apply boundary conditions.

BC1: 0 = A ln b + B, B = − A ln b, so 

 V = A ln   b

Va a BC2: Va = A ln   , A = b ln a

( b)

, V = Va

Fig. P2.66

( b)

ln 

( b)

ln a

or V = −1.443Va ln (100  ) . (b) E = −V = −

1.443Va V a = a  

(c)

1.443Va = 288.5Va = Ebr = 60 x106 , .005 60 x106 so Va = = 208kV ,  (Vab )max = 210kV 288.5 Emax =

P2.67: A 1.0 m long carbon pipe of inner diameter 3.0 cm and outer diameter 5.0 cm is cut in half lengthwise. Determine the resistance between the inner surface and the outer surface of one of the half sections of pipe.

One approach is to consider the resistance for the half-section of pipe is twice the resistance for a complete cylindrical section, given by Eqn. (2.84). But we’ll used the LaPlace equation approach instead.

Laplace:  2Vcyl =

1   V   = 0 ; here we see V only depends on      

V = A; V = A ln  + B ;  where A and B are constants.

Now apply boundary conditions.

BC1: So: 

2-48

Vb = 0 = A ln b + B; B = − A ln b;  V = A ln   b BC2: Va a Va = A ln   ; A = b ln a V = Va

( b)

( b)

;

ln 

( b)

ln a

E = −V = −

Va V a = −  ln a

1

( b) 

a Fig. P2.67

J=E

I =  J dS = −

 Va



1

a

I

 L Va

   d  dz = ln b

( b) ln ( b ) V a = 5.4.

R= = ln a

L

0

0

( a)

 L

P2.68: For a coaxial cable of inner conductor radius a and outer conductor radius b and a dielectric r in-between, assume a charge density v = o  is added in the dielectric region. Use Poisson’s equation to derive an expression for V and E. Calculate s on each plate.

 2V = −

v 1   V  − o =−  =       

so  V  o   V  o V o =  + A , where A is a constant.   = ; d  =  d ;               V o A A = + ; dV =  o d  +  d  ; V = o  + A ln  + B , where B is a constant.       Now apply boundary conditions: V = Va at  = a and V = 0 at  = b Applying the second one gives us:

V=

( )

o (  − b ) + A ln  b . 

Applying the first one:

2-49

Va =

( )

o ( a − b ) + A ln a b ; A = 

Va +

o (b − a ) 

( b)

ln a

Therefore,

V=

o (b − a )      ln   + o (  − b ) . a b  ln

Va +

( b)

E = −V = −

  V     a  = −  K ln   + o  − o b  a       b 

 K   E =  − − o  a     where K=

Va +

o (b − a )  ,

( b)

ln a

  o    −  Va +  ( b − a )     − o a . so E =   a        ln     b   o    −  Va +  ( b − a )    − o= DN =  s ; DNa =  E  = a =   sa a      a ln     b   o    −  Va +  ( b − a )    − o= DNb =  E  =b =   sb a      b ln     b

P2.69: For the parallel plate capacitor given in Figure 2.51, suppose a charge density z  v = o sin    2d  is added between the plates. Use Poisson’s equation to derive a new expression for V and E. Calculate s on each plate.

2-50

(

)

z  2V ( z ) − v − o sin 2d = = 2 z   2 d V ( z ) − o = sin  z dz = o cos  z +A  2d 2d z   2 o d 2 o d 2  z V ( z) = cos dz + A dz = 2 sin  z + Az + B 2d 2d     Now apply the boundary conditions: 2 d 2 Vd − o2 2 2 d   Va = 0 = B; Vd = o2 sin  d + Ad ; A = 2d   d 2 d 2  V 2 d  V ( z ) = o2 sin  z +  d − 2o  z 2 d      d

( (

) )

(

(

(

(

E = −V = −

)

)

)

)

   2 d 2  V    V 2 d   a z =  −  o2 sin  z + −   d − 2o  z   a z  2d z z   d       z   

(

(

)

)

V 2 d   d E =  − o cos  z − d + 2o  a z 2 d d     

o d Vd 2 o d − + 2 = s .  d  V 2  d at z = d, DN =  E , so  s z =d = − d + o2 =  s . d  at z = 0, DN =  E, so  s z =0 = −

13. Capacitors P2.70: A parallel plate capacitor is constructed such that the dielectric can be easily removed. With the dielectric in place, the capacitance is 48 nF. With the dielectric removed, the capacitance drops to 12 nF. Determine the relative permittivity of the dielectric.

C1 =

 r o A d

; C2 =

o A d

;

C1 48 = r = = 4.0 C2 12

P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate separation, and a dielectric with relative permittivity of 1200 has a 12. V potential difference across the plates. (a) What is the minimum allowed dielectric strength for this capacitor? Calculate (b) the capacitance, and (c) the magnitude of the charge density on one of the plates. (a) E =

12V kV kV =6 ; (a) Ebr min = 6 0.002m m m

2-51

(b) C = (c) C =

 r o A d

=

(1200 ) (8.854 x10−12 F / m )(1m2 ) 0.002m

= 5.3 F

Q C ; Q = CV = ( 5.3x10−6 F ) (12V ) = 64C V FV

P2.72: A conical section of material extends from 2.0 cm ≤ r ≤ 9.0 cm for 0 ≤  ≤ 30 with r = 9.0 and  = 0.020 S/m. Conductive plates are placed at each radial end of the section. Determine the resistance and capacitance of the section.

1   2 V  A 2 V = A; V = − + B , where A and B are constants. r  = 0; r 2 r r  r  r r Boundary conditions: r = a, V = 0 and r = b, V = Vb 1 1  −  a r V = Vb  1 1  −  a b V E = −V = − ar r −Vba r −  V a E= ,D = r o b r 1 1 2 1 1 2  − r  − r a b a b 2V =

Fig. P2.72

 sb =

− r  oVb ; 1 1 2  − b a b

Q =   s dS ; dS = r 2 sin  d d

2-52

Qb =

 r  oVb

2

30o

b  sin  d  d 2

1 1 2 0 0  − b a b   −12 = 1.73x10 Vb .

Q  1  C = b = 1.7 pF ; RC = ; R = = 2.3k  Vb  C

P2.73: An inhomogeneous dielectric fills a parallel plate capacitor of surface area 50. cm2 and thickness 1.0 cm. You are given r = 3(1 + z), where z is measured from the bottom plate in cm. Determine the capacitance.

Place +Q at z = d and –Q at z = 0.

Q Q Q s = , D = − a z , E = − a S S  r o S z d

d

Vdo = −  E dL = −  0

0

-Q

 r o S

d

a z dza z =

Q dz  o S 0  r

evaluating the integral: d d 1 dz dz 1 1 0  r =0 3 (1 + z ) = 3 ln (1 + z ) 0 = 3 ln 2 cm

−12 2 2 Q 3 o S 3 ( 8.854 x10 F m )( 50cm )  m  C= = =   = 19 pF Vdo ln 2 ( ln ( 2 ) cm )  100cm 

P2.74: Given E = 5xyax + 3zaz V/m, find the electrostatic potential energy stored in a volume defined by 0 ≤ x ≤ 2 m, 0 ≤ y ≤ 1 m, and 0 ≤ z ≤ 1 m. Assume  = o.

1 1 WE =  o  E Edv =  o   25 x 2 y 2 dxdydz +  9 z 2 dxdydz  2 2 2 1 1 2 1 1  1  WE =  o  25 x 2 dx  y 2 dy  dz + 9 dx  dy  z 2 dz  = 125 pJ 2  0 0 0 0 0 0 

P2.75: Suppose a coaxial capacitor with inner radius 1.0 cm, outer radius 2.0 cm and length 1.0 m is constructed with 2 different dielectrics. When oriented along the z-axis, r for 0 ≤  ≤ 180 is 9.0, and for 180 ≤  ≤ 360 is 4.0. (a) Calculate the capacitance. (b) If 9.0 V is applied across the conductors, determine the electrostatic potential energy stored in each dielectric for this capacitor. (a) a coaxial line,

2-53

2 L r  o ln b a But for only half the line,  L r  o C= ln b a So  L r1 o C1 = = 361 pF ln b a and  L r 2 o C2 = = 161 pF ln b a So Fig. P2.75 CTOT = C1 + C2 = 522 pF 1 1 (b) WE1 = C1V 2 = 14.6nJ ; WE 2 = C2V 2 = 6.5nJ 2 2 C=

( )

( )

( )

( )

3-1 Solutions for Chapter 3 Problems 1. Magnetic Fields and Cross Products P3.1: Find AxB for the following: a. A = 2ax – 3ay + 4az, B = 5ay - 1az b. A = a + 2a + 4az, B = 2a + 6az c. A = 2ar + 5a + 1a, B = ar + 3a (a)

ax AB = 2 0

a y az −3 4 = ( 3 − 20 ) a x − ( −2 ) a y + (10 ) a z = −17a x + 2a y + 10a z 5 −1

(b)

a AB = 1 2

a 2 0

az 4 = 12a  − ( 6 − 8 ) a + ( 0 − 4 ) a z = 12a  + 2a − 4a z 6

a 5 0

a 1 = 15a r − ( 6 − 1) a + ( 0 − 5 ) a = 15a r − 5a − 5a 3

(c)

ar AB = 2 1

P3.2: If a parallelogram has a short side a, a long side b, and an interior angle  (the smaller of the two interior angles), the area of the parallelogram is given by area = ab sin  .

Determine how you would use the cross product of a pair of vectors to find the area of a parallelogram defined by the points O(0,0,0), P(6,0,0), Q(8,12,0) and R(2,12,0).

(Assume dimensions in meters)

area = ab sin  = A  B = A B sin  A = 6ax, B = 2ax + 12ay A x B = 72az, Area = 72 m2

Fig. P3.2

3-2 P3.3: Given the vertices of a triangle P(1,2,0), Q(2,5,0) and R(0,4,7), find (a) the interior angles, (b) a unit vector normal to the surface containing the triangle and (c) the area of the triangle. (a) PQ = 1a x + 3a y ; PQ = 3.16

PR = −1a x + 2a y + 7a z ; PR = 7.348

PQ  PR = 21a x − 7a y + 5a z PQ  PR = 22.69 A  B = A B sin  ab sin  P =

22.69 ; ( 3.16 )( 7.348)

 P = 78 QP = −1a x − 3a y ; QP = 3.16

Fig. P3.3

QR = −2a x − 1a y + 7a z ; QR = 7.348

QP  QR = 22.69; Q = 78 ;  R = 180 −  P − Q = 24 (b) an =

PQ  PR = 0.93a x − 0.31a y + 0.22a z PQ  PR

(c) area =

1 PQ  PR = 11.4m2 2

2. Biot-Savart’s Law P3.4: A segment of conductor on the z-axis extends from z = 0 to z = h. If this segment conducts current I in the +az direction, find H(0,y,0). Compare your answer to that of Example 3.2.

We use Eqn. (3.7) and change the limits: h

 I  a  z H=  2 2  4   z +  2  0 =

 Iy ( −a x )  h   4  y 2 h 2 + y 2 

H=

− Iha x 4 y h 2 + y 2

Note that if the line of current is semiinfinite (goes from z = 0 to z = ∞), we’d have:

Fig. P3.4

3-3

H=

− Ia x 4 y

P3.5: An infinite length line with 2.0 A current in the +ax direction exists at y = -3.0 m, z = 4.0 m. A second infinite length line with 3.0 A current in the +az direction exists at x = 0, y = 3.0 m. Find H(0, 0, 0).

This situation is shown in Figure P3.5a.

Ho = H1 + H2 Referring to the figure, R = 3ay – 4az, R = 5, aR = 0.6ay – 0.8az a = ax x aR = 0.80ay + 0.60az I H1 = 1 a 2

2 A ( 0.80a y + 0.60a z ) 2 5m = 51a y + 38a z mA m I H 2 = 2 a 2 3A ax = = 159a x mA m 2 3m Ho = 159ax + 51ay +38az mA/m =

Fig. P3.5a

Or with 2 significant digits Ho = 160ax + 51ay +38az mA/m Fig. P3.5b

P3.6: A conductive loop in the shape of an equilateral triangle of side 8.0 cm is centered in the x-y plane. It carries 20.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 16cm).

The situation is illustrated in Figure P3.6a. The sketch in Figure P3.6b is used to find the +x-axis intercept for the triangle. By simple trigonometry we have: b = tan 30 ;and since a = 4 cm we find b = 2.31 cm. a

3-4 Now for one segment we adapt Eqn. (3.7): +a

 Ia  z H1 =    4  z 2 +  2  −a with a = RaR, R = −2.31a x + 16a z , R = 16.17cm, a = −a y  a R = −0.99a x − 0.143a z

( 20 x10 A) ( −0.99a − 0.143a )  2 ( 4)  H =   4 (16.17 x10 m )  4 + 16.17  −3

x

z

−2

1

2

2

H1 = -4.7ax – 0.68az mA/m Now by symmetry the total H contains only the az component: Htot = -2.0 mA/m az.

Fig. P3.6a

Fig. P3.6b

P3.7: A square conductive loop of side 10.0 cm is centered in the x-y plane. It carries 10.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 10cm).

We find H for one section of the square by adopting Eqn. (3.7): +a

 Ia  z H=    4  z 2 +  2  −a Iaa H= 2 a 2 +  2

With a = RaR, we have R = -5ax + 10az, |R| = 11.18x10-2 m aR = -0.447ax + 0.894az, a = -ay x aR = -0.894 ax – 0.447 az

Fig. P3.7

3-5

(10 x10 ) (5)( −0.894a − 0.447a ) = −5.2a − 2.6a mA H = m 2 (11.18 x10 ) 5 + 11.18 −3

x

1

−2

2

z

2

x

z

Now by symmetry the total H contains only the az component: HTOT = -10.4az mA/m

P3.8: A conductive loop on the x-y plane is bounded by  = 2.0 cm,  = 6.0 cm,  = 0 and  = 90. 1.0 A of current flows in the loop, going in the a direction on the  = 2.0 cm arm. Determine H at the origin.

By inspection of the figure, we see that only the arc portions of the loop contribute to H. From a ring example we have: 2 Ia 2a z H= d 3  2 2 2 0 4 ( h + a ) For the  = a segment of the loop:  /2 Ia 2a z I Ha = d = a z 3  4 a 0 8a Ib 2a z −I d = az ; At  = b: H b = 3  4 b  / 2 8b 0

Fig. P3.8

I 1 1 1 1 1  A So HTOT =  −  a z =  −  a z = 4.2a z 8a b 8  0.02 0.06  m

P3.9: MATLAB: How close do you have to be to the middle of a finite length of currentcarrying line before it appears infinite in length? Consider Hf(0, a, 0) is the field for the finite line of length 2h centered on the z-axis, and that Hi(0, a, 0) is the field for an infinite length line of current on the z-axis. In both cases consider current I in the +az direction. Plot Hf/Hi vs h/a.

Adapting Eqn. (3.7), for the finite length line we have:

Hf =

 Ia  h  2  2  h +  2   =a

For the infinite length of line: Ia Hi =  2

3-6 The ratio we wish to plot is: h Hf h a = = 2 2 2 Hi h +a h +1 a

( ) ( )

The MATLAB routine follows. % M-File: MLP0309 %Consider the field for a finite line of length 2h %oriented on z-axis with current I in +z direction. %The field is to be found a distance a away from %the current on the y axis (point (0,a,0)). %We want to compare this field with that of an infinite %length line of current. %Plot Hf/Hi versus h/a. We expect that as h/a grows large, %the line will appear more 'infinite' to an observation %point at (0,a,0). hova=0.01:.01:100; HfovHi=hova./sqrt(1+(hova).^2); semilogx(hova,HfovHi) xlabel('h/a') ylabel('Hf/Hi') grid on

Fig. P3.9a

Fig. P3.9b

3-7 P3.10: MATLAB: For the ring of current described in MATLAB 3.2, find H at the following points (a) (0, 0, 1m), (b) (0, 2m, 0), and (c) (1m, 1m, 0). %M-File: MLP0310 %Find the magnetic field intensity at any observation point %resulting from a ring of radius a and current I, %in the aphi direction centered in the x-y plane. df=1; %increment in degrees a=1; %ring radius in m I=1; %current in A Ro=input('vector location of observation point: '); for j=1:df:360; Fr=j*pi/180; Rs=[a*cos(Fr) a*sin(Fr) 0]; as=unitvector(Rs); dL=a*df*(pi/180)*cross([0 0 1],as); Rso=Ro-Rs; aso=unitvector(Rso); dH=I*cross(dL,aso)/(4*pi*(magvector(Rso))^2); dHx(j)=dH(1); dHy(j)=dH(2); dHz(j)=dH(3); end H=[sum(dHx) sum(dHy) sum(dHz)] Now to run the program: >> MLP0310 vector location of observation point: [0 0 1] H = -0.0000 -0.0000

0.1768

>> MLP0310 vector location of observation point: [0 2 0] H=

0

0 -0.0431

>> MLP0310 vector location of observation point: [1 1 0] H=

0

0 -0.1907

3-8 >> MLP0310 vector location of observation point: [0 1 1] H=

0.0000

0.0910

0.0768

>> So we see: (a) H = 0.18 az A/m (b) H = -0.043 az A/m (c) H = -0.19 az A/m (extra) H = 9.1ay + 7.7 az mA/m

P3.11: A solenoid has 200 turns, is 10.0 cm long, and has a radius of 1.0 cm. Assuming 1.0 A of current, determine the magnetic field intensity at the very center of the solenoid.

How does this compare with your solution if you make the assumption that 10 cm >> 1 cm?

Eqn. (3.10):

 NI  H= 2h  

 a + 2 2 2  z 2 z + a (h − z) + a   200 (1A)  0.1 − 0.05 0.05  a = 1961 A a H= + z z 2 ( 0.1m )  ( 0.1 − 0.05 )2 + 0.012 m 0.052 + 0.012    h−z

z

Or H = 1960 A/m az The approximate solution, assuming 10cm >> 1cm, is 200 (1A ) NI A H= az = a z = 2000 a z h 0.1m m

P3.12: MATLAB: For the solenoid of the previous problem, plot the magnitude of the field versus position along the axis of the solenoid. Include the axis 2 cm beyond each end of the solenoid. % % % %

M-File: MLP0312 Plot H vs length thru center of a solenoid

3-9 clc clear % initialize variables N=200; %number of turns h=0.10; %height of solenoid a=0.01; %radius of solenoid I=1; %current dz=0.001; %step change in z z=-.02:dz:h+.02; zcm=z.*100; A1=(h-z)./sqrt((h-z).^2+a^2); A2=z./sqrt(z.^2+a^2); Atot=A1+A2; H=N*I.*Atot/(2*h); % generate plot plot(zcm,H) xlabel('z(cm)') ylabel('H (A/m)') grid on

Fig. P3.12

P3.13: A 4.0 cm wide ribbon of current is centered about the y-axis on the x-y plane and has a surface current density K = 2 ay A/m. Determine the magnetic field intensity at the point (a) P(0, 0, 2cm), (b) Q(2cm, 2cm, 2cm). (a) Because of the symmetry (Figure P3.13a), we can use a modified Eqn. (3.14): K d  H = y tan −1   a x  a

2

A 2 tan −1   a x = 1.57a x  m 2 (b) Referring to Figure P3.13b; I dH = a ; 2 =

Ra r =  a  ; where R = ( d − x ) a x + aa z

Fig. P3.13a

3-10

( d − x ) + a 2 , and ( d − x ) a x + aa z . So aR = 2 ( d − x ) + a2 aa + ( x − d ) a z a = a y  a R = x 2 ( d − x ) + a2 K ( aa x + xa z − da z ) dx H= y  2 2 ( d − x ) + a2 so  =

2

Fig. P3.13b

This is separated into 3 integrals, each one solved via numerical integration, resulting in: H=1.1083ax +0.3032az – 1.1083az; or H = 1.1ax – 0.80az A/m

3. Ampere’s Circuit Law P3.14: A pair of infinite extent current sheets exists at z = -2.0 m and at z = +2.0 m. The top sheet has a uniform current density K = 3.0 ay A/m and the bottom one has K = -3.0 ay A/m. Find H at (a) (0,0,4m), (b) (0,0,0) and (c) (0,0,-4m).

1 We apply H = K  a N , 2 1 1 (a) H = ( 3a y )  a z + ( -3a y )  a z = 0 2 2 (b) 1 1 H = ( 3a y )  −a z + ( -3a y )  a z 2 2 A = −3a x m (c) H = 0

Fig. P3.14

P3.15: An infinite extent current sheet with K = 6.0 ay A/m exists at z = 0. A conductive loop of radius 1.0 m, in the y-z plane centered at z = 2.0 m, has zero magnetic field intensity measured at its center. Determine the magnitude of the current in the loop and show its direction with a sketch.

Htot = HS + HL

3-11

1 1 A H S = K  a N = ( 6a y )  a z = 3a x 2 2 m For the loop, we use Eqn. (3.10): I H= az 2a where here I −I HL = ( −a x ) = a x 2a 2 (sign is chosen opposite HS).

So, I/2 = 3 and I = 6A. Fig. P3.15

P3.16: Given the field H = 3y2 ax, find the current passing through a square in the x-y plane that has one corner at the origin and the opposite corner at (2, 2, 0).

Referring to Figure P3.6, we evaluate the circulation of H around the square path.

 H dL = I

enc

b

c

d

a

a

b

c

d

=  + + +

b

 3 ( 0 ) a dxa = 0 2

x

x

a c

 3 y a dya = 0 2

x

y

b

d

0

 3 ( 2 ) a x dxa x = 12 dx = −24 2

c

2

a

 =0 d

So we have Ienc = 24 A. The negative Sign indicates current is going in the -az direction.

Fig. P3.16

P3.17: Given a 3.0 mm radius solid wire centered on the z-axis with an evenly distributed 2.0 amps of current in the +az direction, plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 ≤  ≤ 9 mm.

Figure P3.17 shows the situation along with the Amperian Paths. We have:

 H dL = I , where H = H a and dL = da ; enc

2 H = I enc

3-12 This will be true for each Amperian path. 

2

I I I 2 AP1: I enc =  J dS, J = 2 a z , I enc = 2   d   d = 2 a a 0 a 0 I  a for   a 2 a 2 I a for   a AP2: Ienc = I, H = 2 So: H =

Fig. P3.17a % %

Fig. P3.17b

MLP0317 generate plot for ACL problem

a=3e-3; I=2; N=30; rmax=9e-3; dr=rmax/N;

%radius of solid wire (m) %current (A) %number of data points to plot %max radius for plot (m)

for i=1:round(a/dr) r(i)=i*dr; H(i)=(I/(2*pi*a^2))*r(i); end for i=round(a/dr)+1:N r(i)=i*dr; H(i)=I/(2*pi*r(i)); end plot(r,H) xlabel('rho(m)') ylabel('H (A/m)') grid on

3-13

P3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J = 3 A/cm2 az (for  in cm) plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 ≤  ≤ 8 cm.

We’ll let a = 2 cm.

 H dL = I , where H = H a and dL = da ; 2 H = I AP1 ( < a): I =  J dS =  3a   d  da = 2 enc

enc

3

enc

z

and H =  2a for   a AP2 ( > a): Ienc = 2a3, so H =

a3



a for   a

The MATLAB plotting routine is as follows: % MLP0318 % generate plot for ACL problem a=2; %radius of solid wire (cm) N=40; %number of data points to plot rmax=8; %max radius for plot (cm) dr=rmax/N; for i=1:round(a/dr) r(i)=i*dr; H(i)=r(i)^2; end for i=round(a/dr)+1:N r(i)=i*dr; H(i)=a^3/r(i); end plot(r,H) xlabel('rho(cm)') ylabel('H (A/cm)') grid on

Fig. P3.18

P3.19: An infinitesimally thin metallic cylindrical shell of radius 4.0 cm is centered on the z-axis and carries an evenly distributed current of 10.0 mA in the +az direction. (a) Determine the value of the surface current density on the conductive shell and (b) plot H as a function of radial distance from the z-axis over the range 0 ≤  ≤ 12 cm.

3-14 (a) K s =

I 2 a

=

10mA mA mA = 39.8 ; so K = 40a z 2 ( 0.04m ) m m

(b) for  < a, H = 0. For  > a we have: I H= a 2 The MATLAB routine to generate the plot is as follows: % MLP0319 % generate plot for ACL problem a=4; %radius of solid wire (cm) N=120; %number of data points to plot I=10e-3; %current (A) rmax=12; %max plot radius(cm) dr=rmax/N; for i=1:round(a/dr) r(i)=i*dr; H(i)=0; end for i=round(a/dr)+1:N r(i)=i*dr; H(i)=100*I/(2*pi*r(i)); end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on

Fig. P3.19a

Fig. P3.19b

3-15 P3.20: A cylindrical pipe with a 1.0 cm wall thickness and an inner radius of 4.0 cm is centered on the z-axis and has an evenly distributed 3.0 amps of current in the +az direction. Plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 ≤  ≤ 10 cm.

For each Amperian Path:

 H dL = I , where H = H a and dL = da ; enc

2 H = I enc

Now, for  < a, Ienc = 0 so H = 0.

For a <  < b, I enc =  J dS, where J =

I a z and dS = d  d a z  (b − a2 ) 2

( − a ) a I  2 − a2 I I enc =  d  d  = I ,H =  2 2  2 2  2 2 b −a   (b − a ) a 2 ( b − a ) 0 

2

Fig. P3.20a % %

2

2

Fig. P3.20b

MLP0320 generate plot for ACL problem

a=4; %inner radius of pipe (cm) b=5; %outer radius of pipe(cm) N=120; %number of data points to plot I=3; %current (A) rmax=10; %max radius for plot (cm) dr=rmax/N; aoverdr=a/dr boverdr=b/dr for i=1:round(a/dr) r(i)=i*dr; H(i)=0;

3-16 end for i=round(a/dr)+1:round(b/dr) r(i)=i*dr; num(i)=I*(r(i)^2-a^2); den(i)=2*pi*(b^2-a^2)*r(i); H(i)=100*num(i)/den(i); end for i=round(b/dr)+1:N r(i)=i*dr; H(i)=100*I/(2*pi*r(i)); end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on

P3.21: An infinite length line carries current I in the +az direction on the z-axis, and this is surrounded by an infinite length cylindrical shell (centered about the z-axis) of radius a carrying the return current I in the –az direction as a surface current. Find expressions for the magnetic field intensity everywhere. If the current is 1.0 A and the radius a is 2.0 cm, plot the magnitude of H versus radial distance from the z-axis from 0.1 cm to 4 cm.

I

 H dL = I ; for 0<  a, H = 2 a and for   a, H = 0. enc

Fig. P3.21a

Fig. P3.21b

The MATLAB routine used to generate Figure P3.21b is as follows: % MLP0321 % generate plot for ACL problem

3-17 clc clear a=2; %inner radius of cylinder(cm) N=80; %number of data points to plot I=1; %current (A) rmax=4; %max radius for plot (cm) dr=rmax/N; for i=1:40 r(i)=.1+(i-1)*dr; H(i)=100*I/(2*pi*r(i)); end for i=40:N r(i)=i*dr; H(i)=0; end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on

P3.22: Consider a pair of collinear cylindrical shells centered on the z-axis. The inner shell has radius a and carries a sheet current totaling I amps in the +az direction while the outer shell of radius b carries the return current I in the –az direction. Find expressions for the magnetic field intensity everywhere. If a = 2cm, b = 4cm and I = 4A, plot the magnitude of H versus radial distance from the z-axis from 0 to 8 cm.

 H dL = I ; for 0<  a, H = 0; enc

for a<  b, H =

I

a=2; b=4; N=160;

%inner radius of coax (cm) %outer radius of coax(cm) %number of data points to plot

a ; 2 and for   b, H = 0.

The MATLAB routine used to generate Figure P3.22b is as follows: % MLP0322 % generate plot for ACL problem

3-18

Fig. P3.22a

Fig. P3.22b

I=4; %current (A) rmax=8; %max radius for plot (cm) dr=rmax/N; aoverdr=a/dr boverdr=b/dr for i=1:round(a/dr) r(i)=i*dr; H(i)=0; end for i=round(a/dr)+1:round(b/dr) r(i)=i*dr; H(i)=100*I/(2*pi*r(i)); end for i=round(b/dr)+1:N r(i)=i*dr; H(i)=0; end plot(r,H) xlabel('rho(cm)') ylabel('H (A/m)') grid on

P3.23: Consider the toroid in Figure 3.55 that is tightly wrapped with N turns of conductive wire. For an Amperian path with radius less than a, no current is enclosed and therefore the field is zero. Likewise, for radius greater than c, the net current

3-19 enclosed is zero and again the field is zero. Use Ampere’s Circuital Law to find an expression for the magnetic field at radius b, the center of the toroid.

 H dL = I ; enc

Within the toroid, H = H a, so

 H dL =  H a    da = 2 bH . =b

Then, Ienc by the Amperian path is: Ienc = NI. NI H = a .

2 b

4. Curl and the Point Form of Ampere’s Circuital Law P3.24: Find   A for the following fields: a. A = 3xy2/z ax b. A = sin2 a – 2 z cos a c. A = r2sin ar + r/cos  a (a)  ( 3xy 2 z ) a x =

Ax A −3xy 2 6 xy a y − x az = ay − az 2 z y z z

(b)  (  sin 2  a  −  2 z cos  a ) = −

=  2 cos  a  + 0a −

3 2 z cos 

A z

a +

A z

a +

A  1  A ) −   a z (      

1

 2sin  cos  a z  =  2 cos a  − 3 z cos  + 2sin  cos   a z 

az −

(c)

  r    r 2 sin  a r + a  cos    =

−1  A  1  1 Ar  1   ( rA ) Ar  ar +  a +  −  a    r sin     r  sin    r  r  

=

−r  ( cos  ) 1   r2  1  2 ar + ( r sin  ) a   a − r sin   r r  cos   r 

=

 2  − sin  ar +  − r cos   a 2 sin  cos   cos  

−1

3-20 P3.25: Find J at (3m, 60, 4m) for H = (z/sin) a – (2/cos) az A/m.

 A A  1 Az −1 A a  +   − z  a + a     z  z  

 H = =

 1 −  sin  2  z cos  a +  + a  a + 2 cos   sin 2  z  sin  cos  

Now find J by evaluating   H at the given point: A J = −10a  + 13a + 0.89a z 2 m P3.26: Suppose H = y2ax + x2ay A/m. a. Calculate

 H dL around the path A → B → C → D → A , where A(2m,0,0),

B(2m,4m,0), C(0,4m,0) and D(0,0,0). b. Divide this  H dL by the area S (2m*4m = 8m2). c. Evaluate   H at the center point. d. Comment on your results for (b) & (c).

B

C

D

A

A

B

C

D

(a) Referring to the figure, we evaluate  H dL = I enc =  +  +  +  B

4

 =x A

0

C

0

 =y B

2

D

0

 =x

C

A



D

dy = 4 ( 4 ) = 16C

2 x=2

dx = −32C

2 y =4

dy = 0

2

4

x =0

2

=  y2 0

dy = 0 y =0

So we have  H dL = −16C

Fig. P3.26

(b) dividing by S = 8m2, we have -2 C/m2 (c) Evaluating the curl of H:  A A   H =  y − x  a z = ( 2 x − 2 y ) a z , and at the center point (x = 1 and y = 2) we have  x y 

C az m2 (d) In this particular case,  H =  H dL S , even though S is of appreciable size.  H center = −2

3-21 P3.27: For the coaxial cable example 3.8, we found: I for   a, H= a , 2 a 2 I for a<  b, H = a , 2

for b<  c,

H=

 c2 −  2    a , 2 c 2 − b 2   

(

I

)

and for c< , H =0. a. Evaluate the curl in all 4 regions. b. Calculate the current density in the conductive regions by dividing the current by the area. Are these results the same as what you found in (a)? (a)  H =

 H =

1   I2  I a = 2 a z for   a  2  z    2 a  a

1   I  a = 0 for a    b    2  z

2 2 1   I (c −  )  −I   az = H = a z for b<  c 2 2 2 2    2 ( c − b )   c − b ( )    H = 0 for   c I (b) for   a, S =  a 2 , J = I a z = 2 a z S a −I for b<  c, S =  ( c 2 − b 2 ) , J = az  ( c2 − b2 )

( )

Comment:   H = J is confirmed.

P3.28: Suppose you have the field H = r cos  a A/m. Now consider the cone specified by  = /4, with a height a as shown in Figure 3.56. The circular top of the cone has a radius a. a. Evaluate the right side of Stoke’s theorem through the dS = dSa surface. b. Evaluate the left side of Stoke’s theorem by integrating around the loop.

1  1  ( rH )  sin  H a − a ( )   r r sin    r r  ar derivative: (a)  H =

 ( r sin  cos ) = r ( cos2  − sin 2  ) ; 

( cos  − sin  ) a 1   sin  H a = ( )  r r  r sin    sin  2

2

3-22

Fig. P3.28 a derivative: 2 −1  ( r cos  )

r

r

a = −2 cos  a

( cos  − sin  ) a − 2 cos a So,   H = 2

2



sin  Now we must integrate this over the a surface:  ( cos 2  − sin 2  )     H d S = a − 2 cos  a r   r sin  drd ( −a )   sin    r

2a

2

0

0

=  2r sin  cos  drd =2 ( sin  cos  )  =   rdr  d = 2 a 2 4

2

(b)  H dL =  r r = a 2 cos  a ad a = a 2 cos   =   d = 2 a 2 4 0

Clearly in this case the circulation of H is the easiest approach.

5. Magnetic Flux Density P3.29: An infinite length line of 3.0 A current in the +ay direction lies on the y-axis. Find the magnetic flux density at P(7.0m,0,0) in (a) Teslas, (b) Wb/m2, and (c) Gauss.

H=

I 2

a =

3A mA ( −a z ) = −68 a z 2 ( 7m ) m

H  A   Wb   −9 Wb B = o H =  4 x10−7  −68  a z  a z = −86 x10−9 Ta z  = −86 x10 2 m  m   HA  m  10, 000G B = ( −86 x10−9 Ta z ) = −860 x10−6 Ga z T

3-23 P3.30: Suppose an infinite extent sheet of current with K = 12ax A/m lies on the x-y plane at z = 0. Find B for any point above the sheet. Find the magnetic flux passing through a 2m2 area in the x-z plane for z > 0.

1 H = K  aN ; 2 B=

o

( 4 x10 H m ) 12a A  a  = −7.54 x10 Wb a K a = −7

x  2 2 m  This is valid at any point above the sheet.

N

z

−6

 

(

m2

y

)

Wb   Now,  =  B dS = B S =  -7.54x10-6 2 a y  2m2 ( −a y ) = 15Wb m  

P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of radius a carrying current I in the +az direction and outer shell of radius b carrying the return current. Find the magnetic flux passing through an area of length h along the zaxis bounded by radius between a and b.

H=

I 2

a , B =

For a <  < b,  =  B dS =

=

o I a , 2

b o I a I d  dza = o ln (  ) a h  2  2

o Ih  b  ln   Wb 2 a

6. Magnetic Forces P3.32: A 1.0 nC charge with velocity 100. m/sec in the y direction enters a region where the electric field intensity is 100. V/m az and the magnetic flux density is 5.0 Wb/m2 ax.

Determine the force vector acting on the charge. m Wb Wb F = q ( E + u  B ) ; u  B = 100 a y  5 2 = −500 az s m ax sm

V Wb Vs  mN  F = 10−9 C 100 a z + −500 az  = −400nNa z m sm Wb  VC 

P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m2 ay.

Determine the force vector acting on the charge.

3-24

 V m Wb  F = q ( E + u  B ) = 10 x10−9 C  800 a x + 100 a z 12 2  = −4  Na x  m s m a y  

P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it enters a magnetic field B = 1000. T ay (recall that a tesla T = Wb/m2). Calculate the force vector on the charge.

m Wb   F = q ( u  B ) = 10 x10−9 C ( 3a x + 4a y + 5a z ) 1000 2 a y  s m   The cross-product: a x a y a z  3 4 5  = −5000a x + 3000a z   0 1000 0  Evaluating we find: F = -50ax + 30az N

P3.35: What electric field is required so that the velocity of the charged particle in the previous problem remains constant?

dv = 0 (constant velocity) dt F = q ( E + u  B ) = 0;

F = ma = m

E = −u  B = - ( 3a x + 4a y + 5a z ) E = 5a x − 3a z

m Wb V V 1000 2 a y = −3000 a z + 5000 a x s m m m

kV m

P3.36: An electron (with rest mass Me= 9.11x10-31kg and charge q = -1.6 x 10-19 C) has a velocity of 1.0 km/sec as it enters a 1.0 nT magnetic field. The field is oriented normal to the velocity of the electron. Determine the magnitude of the acceleration on the electron caused by its encounter with the magnetic field.

F = ma = q ( u  B ) ; a=

q (u  B) ; m

3-25

quB ( −1.6 x10 a= = m

−19

(

C ) 1000 m

)

)(

10−9 Wb 2 s m = 175 x103 m −31 s2 ( 9.11x10 kg )

P3.37: Suppose you have a surface current K = 20. ax A/m along the z = 0 plane. About a meter or so above this plane, a 5.0 nC charged particle is moving along with velocity v = -10.ax m/sec. Determine the force vector on this particle.

1 1 A H = K  a N = 20a x  a z = −10a y 2 2 m −7 B = o H = −10 ( 4 x10 ) a y Wb 2 = −40 x10−7 a y Wb 2 m m m Wb   F = qu  b = ( 5 x10−9 C )  −10a x  −40 x10−7 a y 2  = 0.63 pNa z s m  

P3.38: A meter or so above the surface current of the previous problem there is an infinite length line conducting 1.0 A of current in the –ax direction. Determine the force per unit length acting on this line of current.

0

F12 =  I 2 dL2  B1 = I 2  dxa x  −40 x10−7 a y L

Wb m2

F12 = I 2 ( − L ) ( −40 x10−7 a z ) ; F12 N = 40 x10−7 a z = 12.6 az L m

P3.39: Recall that the gravitational force on a mass m is F = mg, where, at the earth’s surface, g = 9.8 m/s2 (-az). A line of 2.0 A current with 100. g mass per meter length is horizontal with the earth’s surface and is directed from west to east. What magnitude and direction of uniform magnetic flux density would be required to levitate this line?

 Ns 2 kg  m  Fg = mg; for 1 m Fg = (100 g )  9.8 2  ( −a z )   = −0.98 Na z s    kg m 1000 g  1m

F =  IdL  B = − Fg = +0.98 Na z 0

By inspection, B = Bo(-ax)

3-26 L

 Idya  B ( −a ) = ILB a y

o

x

o

z

o

 Wb  = ( 2 A )(1m )  Bo 2  = 2 Bo ( N )  m  The unit conversion to arrive at Newtons is as follows: Am Wb Vs W J Nm =N m2 Wb VA Ws J So we have Bo = 0.490 Wb/m2, and B = 0.490 Wb/m2 (-ax) (directed north)

Fig. P3.39

P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of 0.10 kg/m. One line sits on the ground and conducts 200. A in the +ax direction, and the other one, 1.0 cm above the first (and parallel), has sufficient current to levitate. Determine the current and its direction for line 2.

Here we will use

F12 o I1 I 2 4 x10−7 H m 200 A = ay = I 2a z = 4 x10−3 I 2a z L 2 y 2 0.01m

F mg Ns 2 N = = ( 0.10 kg m ) ( 9.8 m s 2 ) = 0.98 L L kgm m So solving for I2: 0.98 I2 = = 245 A in the -a x direction.

4 x10−3

P3.41: In Figure 3.57, a 2.0 A line of current is shown on the z-axis with the current in the +az direction. A current loop exists on the x-y plane (z = 0) that has 4 wires (labeled 1 through 4) and carries 1.0 mA as shown. Find the force on each arm and the total force acting on the loop from the field of the 2.0 A line.

o I1 a 2 A → B : dL2 = 5d a , a  a = 0

F12 =  I 2 dL2  B1 ; B1 =

C → D : a  a = 0 3

B → C : F12 = I 2  d  a   5

o I1  I I 3 a = o 1 2 ln   a z 2 2 5

3-27

Fig. P3.41

4 x10−7 H m 3 F12 = ( 2 A) (10−3 A) ln   a z = −204 pNa z 2 5 So for B to C: F12 = -0.20 nN az Likewise, from D to A: F12 = +0.20 nN az

P3.42: MATLAB: Modify MATLAB 3.4 to find the differential force acting from each individual differential segment on the loop. Plot this force against the phi location of the segment. %MLP0342 %modify ML0304 to find dF acting from the field % of each segment of current; plot vs phi clear clc I=1; %current in A a=1; %loop radius, in m mu=pi*(4e-7); %free space permeability az=[0 0 1]; %unit vector in z direction DL1=a*2*(pi/180)*[0 1 0]; %Assume 2 degree increments %DL1 is the test element vector %F is the angle phi in radians %xi,yi is location of ith element on the loop %Ai & ai = vector and unit vector from origin % to xi,yi %DLi is the ith element vector %Ri1 & ri1 = vector and unit vector from ith % point to test point for i=1:179

3-28 phi(i)=i*2; F=2*i*pi/180; xi=a*cos(F); yi=a*sin(F); Ai=[xi yi 0]; ai=unitvector(Ai); DLi=(pi*a/90)*cross(az,ai); Ri1=[a-xi -yi 0]; ri1=unitvector(Ri1); num=mu*I*cross(DLi,ri1); den=4*pi*(magvector(Ri1)^2); B=num/den; dFvect=I*cross(DL1,B); dF(i)=dFvect(1); end plot(phi,dF) xlabel('angle in degrees') ylabel('the differential force, N')

Fig. P3.42

P3.43: MATLAB: Consider a circular conducting loop of radius 4.0 cm in the y-z plane centered at (0,6cm,0). The loop conducts 1.0 mA current clockwise as viewed from the +x-axis. An infinite length line on the z-axis conducts 10. A current in the +az direction.

Find the net force on the loop.

The following MATLAB routine shows the force as a function of radial position around the loop. Notice that while there is a net force in the -y direction, the forces in the zdirection cancel.

3-29

% % %

MLP0343 find total force and torque on a loop of current next to a line of current

% % % % % % % % % % % % % %

variables I1,I2 current in the line and loop (A) yo center of loop on y axis (m) uo free space permeability (H/m) N number of segements on loop a loop radius (m) dalpha differential loop element dL length of differential section DL diff section vector B1 I1's mag flux vector (Wb/m^2) Rv vector from center of loop to the diff segment ar unit vector for Rv y,z the location of the diff segment

clc clear % initialize variables I1=10; I2=1e-3; yo=.06; uo=pi*4e-7; N=180; a=0.04; dalpha=360/N; dL=a*dalpha*pi/180; ax=[1 0 0];

% perform calculations for i=1:N dalpha=360/N; Fig. P3.43 alpha=(i-1)*dalpha; phi(i)=alpha; z=a*sin(alpha*pi/180); y=yo+a*cos(alpha*pi/180); B1=-(uo*I1/(2*pi*y))*[1 0 0];

3-30 Rv=[0 y-.06 z]; ar=unitvector(Rv); aL=cross(ar,ax); DL=dL*aL; dF=cross(I2*DL,B1); dFx(i)=dF(1); dFy(i)=dF(2); dFz(i)=dF(3); end plot(phi,dFy,phi,dFz,'--k') legend('dFy','dFz') Fnet=sum(dFy) Running the program we get: Fnet = -4.2932e-009 >> So Fnet = -4.3 nN ay

P3.44: MATLAB: A square loop of 1.0 A current of side 4.0 cm is centered on the x-y plane. Assume 1 mm diameter wire, and estimate the force vector on one arm resulting from the field of the other 3 arms. % MLP0344 V2 % % Square loop of current is centered on x-y plane.

Viewed % from the +z axis, let current go clockwise. We want to % find the force on the arem at x = +2 cm resulting from the % current in arms at y = -2 cm, x = -2 cm and y = +2 cm. %

Wentworth, 12/3/03

% % % % % % % %

Variables a side length (m) b wire radius (m) I current in loop (A) uo free space permeability (H/m) N number of segments for each arm xi,yi location of test arm segment (at x = +2 cm) xj,yj location of source arm segment (at y = -2 cm)

3-31 % % % % % %

xk,yk location of source arm segment (at x = -2 cm) xL,yL location of source arm segment (at y = +2 cm) dLi differential test segment vector dLj, dLk, dLL diff vectors on sources Rji vector from source point j to test point i aji unit vector of Rji

clc;clear; a=0.04; b=.0005; I=1; uo=pi*4e-7; N=80; for i=1:N xi=(a/2)+b; yi=-(a/2)+(i-0.5)*a/N; ypos(i)=yi; dLi=(a/N)*[0 -1 0]; for j=1:N xj=-(a/2)+(j-0.5)*a/N; yj=-a/2-b; dLj=(a/N)*[-1 0 0]; Rji=[xi-xj yi-yj 0]; aji=unitvector(Rji); num=I*CROSS(dLj,aji); den=4*pi*(magvector(Rji))^2; H=num/den; dHj(j)=H(3); end for k=1:N yk=(-a/2)+(k-0.5)*a/N; xk=-(a/2)-b; dLk=(a/N)*[0 1 0]; Rki=[xi-xk yi-yk 0]; aki=unitvector(Rki); num=I*CROSS(dLk,aki); den=4*pi*(magvector(Rki))^2; H=num/den; dHk(k)=H(3); end for L=1:N

3-32 xL=(-a/2)+(L-0.5)*a/N; yL=(a/2)+b; dLL=(a/N)*[1 0 0]; RLi=[xi-xL yi-yL 0]; aLi=unitvector(RLi); num=I*CROSS(dLL,aLi); den=4*pi*(magvector(RLi))^2; H=num/den; dHL(L)=H(3); end H=sum(dHj)+sum(dHk)+sum(dHL); B=uo*H*[0 0 1]; F=I*CROSS(dLi,B); dF(i)=F(1); end Ftot=sum(dF) plot(ypos,dF) Running the program: Ftot = 7.4448e-007 Fig. P3.44

So Ftot = 740 nN

P3.45: A current sheet K = 100ax A/m exists at z = 2.0 cm. A 2.0 cm diameter loop centered in the x-y plane at z = 0 conducts 1.0 mA current in the +a direction. Find the torque on this loop.

 = m  B;

(

)

m = ISa N = (10−3 A )  ( 0.01m ) a z = 314 x10−9 Am 2a z B =  o H; H =

1 1 A A K  a N = 100a x   ( −a z ) = 50a y ; 2 2 m m

Wb ; m2  = m  B = -20 pNma x B = 50oa y

2

3-33 P3.46: 10 turns of insulated wire in a 4.0 cm diameter coil are centered in the x-y plane.

Each strand of the coil conducts 2.0 A of current in the a direction. (a) What is the magnetic dipole moment of this coil? Now suppose this coil is in a uniform magnetic field B = 6.0ax + 3.0ay + 6.0az Wb/m2, (b) what is the torque on the coil?

(

(a) m = NISa x = (10 )( 2 A)  ( 0.02m )

2

) = 25.1mAm a

(b)  = m  B = 25.1mAa z  ( 6a x + 3a y + 6a z )

2

z

Wb = 0.151a y − 0.075a x Nm m2

 = −75a x + 151a y mNm

P3.47: A square conducting loop of side 2.0 cm is free to rotate about one side that is fixed on the z-axis. There is 1.0 A current in the loop, flowing in the –az direction on the fixed side. A uniform B-field exists such that when the loop is positioned at  = 90, no torque acts on the loop, and when the loop is positioned at  = 180 a maximum torque of 8.0 N-m az occurs. Determine the magnetic flux density.

At  = 90°, m = ISa N = 0.0004 Am2a x . Also, since  = m  B = 0, B is in direction of m, and therefore B = ±Boax.

At  = 180°, m = ISa N = 0.0004 Am2a y , and  = m  B = 8 x10−6 a z Nm. Therefore, B = -Boax and mBo = 8x10-6, so 8 x10−6 mWb mWb Bo = = 20 2 , and B = −20 2 a x .

0.0004 m m

7. Magnetic Materials P3.48: A solid nickel wire of diameter 2.0 mm evenly conducts 1.0 amp of current.

Determine the magnitude of the magnetic flux density B as a function of radial distance from the center of the wire. Plot to a radius of 2 mm.

J=

I 1A kA a = a z = 31.8 2 a z 2 2 z a m  (1x10−3 m )

 H dL = I

enc

=  J dS

I I  d  d = 2  2 2  a a   I I for   a H = a ; B = r o 2 a 2  2 a 2 a 2 H =

3-34

for   a H = % %

I 2

a ; B =

o I a 2

MLP0348 generate plot for ACL problem

a=2e-3; %radius of solid wire (m) I=1; %current (A) N=30; %number of data points to plot rmax=4e-3; %max radius for plot (m) dr=rmax/N; uo=pi*4e-7; ur=600; for i=1:round(a/dr) r(i)=i*dr; B(i)=(ur*uo*I/(2*pi*a^2))*r(i); end for i=round(a/dr)+1:N r(i)=i*dr; B(i)=uo*I/(2*pi*r(i)); End rmm=r*1000; plot(rmm,B) xlabel('rho(cm)') ylabel('B (Wb/m^2)') grid on Fig. P3.48 8. Boundary Conditions P3.49: A planar interface separates two magnetic media. The magnetic field in media 1 (with r1) makes an angle 1 with a normal to the interface. (a) Find an equation for 2, the angle the field in media 2 (that has r2) makes with a normal to the interface, in terms of 1 and the relative permeabilities in the two media. (b) Suppose media 1 is nickel and media 2 is air, and that the magnetic field in the nickel makes an 89 angle with a normal to the surface. Find 2.

H1 = H1N a N + H1T aT ; H 2T aT = H1T aT ; B1N a N = r1o H1N a N B2 N a N = B1N a N = r1o H1N a N = r 2 o H 2 N a N ;  H 2 N =

 r1 H r 2 1N

3-35

tan  2 =

H 2T H1T = ; H2N  r1 H1N

r 2

tan  2 =

r 2 H1T r 2 = tan 1 r1 H1N r1

   2 = tan −1  r 2 tan 1    r1   1  tan 89  = 5.5  600 

 2 = tan −1 

Fig. P3.49

P3.50: MATLAB: Suppose the z = 0 plane separates two magnetic media, and that no surface current exists at the interface. Construct a program that prompts the user for r1 (for z < 0), r2 (for z > 0), and one of the fields, either H1 or H2. The program is to calculate the unknown H. Verify the program using Example 3.11. % M-File: MLP0350 % % Given H1 at boundary between a pair of % materials with no surface current at boundary, % calculate H2. % clc clear % enter variables disp('enter vectors quantities in brackets,') disp('for example: [1 2 3]') ur1=input('relative permeability in material 1: '); ur2=input('relative permeability in material 2: '); a12=input('unit vector from mtrl 1 to mtrl 2: '); F=input('material where field is known (1 or 2): '); Ha=input('known magnetic field intensity vector: '); if F==1 ura=ur1; urb=ur2; a=a12; else ura=ur2; urb=ur1;

3-36 a=-a12; end % perform calculations Hna=dot(Ha,a)*a; Hta=Ha-Hna; Htb=Hta; Bna=ura*Hna; %ignores uo since it will factor out Bnb=Bna; Hnb=Bnb/urb; display('The magnetic field in the other medium is: ') Hb=Htb+Hnb Now run the program (for Example 3.11): enter vectors quantities in brackets, for example: [1 2 3] relative permeability in material 1: 6000 relative permeability in material 2: 3000 unit vector from mtrl 1 to mtrl 2: [0 0 1] material where field is known (1 or 2): 1 known magnetic field intensity vector: [6 2 3] ans = The magnetic field in the other medium is: Hb = 6 2 6 For a second test, run the program for problem P3.52(a). enter vectors quantities in brackets, for example: [1 2 3] relative permeability in material 1: 4 relative permeability in material 2: 1 unit vector from mtrl 1 to mtrl 2: [0 0 -1] material where field is known (1 or 2): 1 known magnetic field intensity vector: [3 0 4] ans = The magnetic field in the other medium is: Hb = 3 0 16

3-37 P3.51: The plane y = 0 separates two magnetic media. Media 1 (y < 0) has r1 = 3.0 and media 2 (y > 0) has r2 = 9.0. A sheet current K = (1/o) ax A/m exists at the interface, and B1 = 4.0ay + 6.0az Wb/m2. (a) Find B2. (b) What angles do B1 and B2 make with a normal to the surface? (a) B N 1 = 4a y (b) B N 2 = B N 1 = 4a y (c) BT 1 = 6a z (d) HT 1 = (e) HT 2 =

BT 1

 r 1 o 3

=

a

2

o

az

(see below)

o z (f) BT 2 = r o HT 2 = 27a z (g) B 2 = 4a y + 27a z

Wb m2

Fig. P3.51

Now for step (e):

a21  ( H1 − H 2 ) = K; − ( HT 1 − HT 2 ) a x =

1

o

− a y  ( HT 1a z − HT 2a z ) = a x ; HT 2 − HT 1 =

1

o

1

o

ax

; HT 2 =

1

o

+

2

o

=

3

o

 B   B  Angles: 1 = tan −1  T 1  = 56 ;  2 = tan −1  T 2  = 82  BN 1   BN 2 

P3.52: Above the x-y plane (z > 0), there exists a magnetic material with r1 = 4.0 and a field H1 = 3.0ax + 4.0az A/m. Below the plane (z < 0) is free space. (a) Find H2, assuming the boundary is free of surface current. What angle does H2 make with a normal to the surface? (b) Find H2, assuming the boundary has a surface current K = 5.0 ax A/m. (a) (1) H N1 = 4a z , (2) B N1 = 16oa z , (3) B N 2 = B N 1 = 16 oa z ,(4) H N 2 = (5) HT 1 = 3a x , (6) HT 2 = HT 1 = 3a x , (7) H 2 = 3a x + 16a z

A m

BN 2

o

= 16a z

3-38

 HT 2   = 10.6 H N 2  

 2 = tan −1 

(b) Now step (6) becomes a21  ( H1 − H 2 ) = K , shere a21 = az.

Let’s let H 2 = Aa x + Ba y + 16a z , then a21  ( 3a x + 4a z − Aa x − Ba y − 16a z ) = 5a x Solve for A and B: ay az   ax  0 0 1  = Ba x + ( A − 3) a y = 5a x ; so A = 3 and B = 5  3 − A − B −12  Finally, H 2 = 3a x + 5a y + 16a z

A . m

P3.53: The x-z plane separates magnetic material with r1 = 2.0 (for y < 0) from magnetic material with r2 = 4.0 (for y > 0). In medium 1, there is a field H1 = 2.0ax + 4.0ay + 6.0az A/m. Find H2 assuming the boundary has a surface current K = 2.0ax – 2.0az A/m.

(1)H1N = 4a y ,(2)B1N = 8oa y ,(3)B2 N = B1N = 8oa y ,(4)H2 N =

8o a y = 2a y 4o

(5)H1T = 2a x + 6a z ,(6)a21  ( H1 − H2 ) = K , let HT 2 = H xa x + H z a z ,

so − a y  ( ( 2 − H x ) a x + ( 6 − H z ) a z ) = 2a x − 2a z , ax 0 2 − Hx

ay az −1 0 = ( H z − 6 ) a x + ( 2 − H x ) a z = 2a x − 2a z , so H z = 8, H x = 4 0 6 − Hz

(7)H2T = 4a x + 8a z ,

 H2 = 4a x + 2a y + 8a z

A m

P3.54: An infinite length line of 2 A current in the +az direction exists on the z-axis.

This is surrounded by air for  ≤ 50 cm, at which point the magnetic media has r2 = 9.0 for  > 50 cm. If the field in media 2 at  = 1.0 m is H = 5.0a A/m, find the sheet current density vector at  = 50. cm, if any.

Method 1: From just the line of current we would have H1 =

I1

2 (1)

a = 1a .

3-39 Now, since HTOT = 5a = H1 + H 2 , then H 2 = 4a is the contribution from the sheet current.

H2 =

I2

2 (1)

K = 8

a = 4a , so I 2 = 8 A, then K =

I2 8 A az = = 8 az 2 a 2 ( 0.5) m

A az m

Method 2: From I1 at boundary we have 2 a H1 = = 2a , but 5a at  = 1.0m corresponds to 10a at  =0.5m since H varies 2 ( 0.5 ) as 1/. So

−a   ( 2a − 10a ) = K,

K = 8

− a   ( −8a ) = 8a z = K

A az m

9. Inductance and Magnetic Energy P3.55: Consider a long pair of straight parallel wires, each of radius a, with their centers separated by a distance d. Assuming d >> a, find the inductance per unit length for this pair of wires.

Each wire can be solved using the energy approach: L   =2 = h wires 8 4 The fields are not confined to a volume, so we must use the flux linkage approach to find inductance outside the wires.

I I H1 = a , H 2 = a , 2 2 ( d −  )

1 =  B dS, TOT = 2 B dS,

Fig. P3.55

d −a h I 2 I d I  d − a  TOT = 2 a d  dza = dz = ln h   2 2 a  0   a 

3-40

Finally, with L =

TOT I

, we arrive at

L  1  d − a  =  + ln   . h TOT   4  a 

P3.56: In problem P3.23 the task was to find the field at the center (radius b) of an N-turn toroid. If the radius of the toroid is large compared to the diameter of the coil (that is, if b >> c-a), then the field is approximately constant from radius a to radius c. (a) Obtain an expression for the toroid’s inductance. (b) Find L if there are 600 turns around a 99.8% iron core with a = 8.0 cm and c = 9.0 cm.

From P3.23 we found: H =

 =  B dS = 

 NI a dSa 2 b

NI a at radius b for the toroid.

2 b

and with area  ( b − a ) , we have  = 2

 = N =

N I 2

(b − a )

 NI 2b

(b − a ) .

2

2

2b  N 2 2 L= = (b − a ) I 2b

( 5000) ( 4 x10−7 ) ( 600 ) (.085 − .080) Plugging in our numbers we have: L = = 0.33H 2 ( 0.085) 2

2

P3.57: MATLAB: Consider that a solid wire of radius a = 1.0 mm is bent into a circular loop of radius 10. cm. Neglecting internal inductance of the wire, write a program to find the inductance for this loop. % % % % % % % % % % % %

M-File: MLP0357 Inductance inside a conductive loop This modifies ML0302 to calculate inductance of a conductive loop. It does this by calculating the mag field at discrete points along a pie wedge, then calculates flux through each portion of the wedge. Then it multiplies by the number of wedges in the 'pie'.

Wentworth, 1/19/03

3-41 % % % % % % % % % % % % % % % % % % % % % % %

Variables: I current(A) in +phi direction on ring a ring radius (m) b wire radius (m) Ndeg number of increments for phi f angle of phi in radians df differential change in phi dL differential length vector on the ring dLmag magnitude of dL dLuv unit vector in direction of dL [xL,yL,0] location of source point Ntest number of test points Rsuv unit vector from origin to source point R vector from the source to test point Ruv unit vector for R Rmag magnitude of R dH differential contribution to H dHmag magnitude of dH radius radial distance from origin Hz total magnetic field at test point Bz total mag flux density at test point flux flux through each differential segment

clc clear

%clears the command window %clears variables

% Initialize Variables a=0.1; b=1e-3; I=1; Ndeg=180; Ntest=60; uo=pi*4e-7; df=360/Ndeg; dLmag=(df*pi/180)*a; dr=(a-b)/Ntest; % Calculate flux thru each segment of pie wedge for j=1:Ntest x=(j-0.5)*dr; for i=(df/2):df:360 f=i*pi/180;

3-42 xL=a*cos(f); yL=a*sin(f); Rsuv=[xL yL 0]/a; dLuv=cross([0 0 1],Rsuv); dL=dLmag*dLuv; R=[x-xL -yL 0]; Rmag=magvector(R); Ruv=R/Rmag; dH=I*cross(dL,Ruv)/(4*pi*Rmag^2); dHmag(i)=magvector(dH); end Hz(j)=sum(dHmag); Bz(j)=uo*Hz(j); dSz(j)=x*df*(pi/180)*dr; flux(j)=Bz(j)*dSz(j); end fluxwedge=sum(flux); Inductance=Ndeg*fluxwedge

Now run the program: Inductance = 5.5410e-007 or L = 550 nH

P3.58: Find the mutual inductance between an infinitely long wire and a rectangular wire with dimensions shown in Figure 3.58.

H1 =

I1

2

a , B1 =

12 =  B1 dS2 =

o I1 a 2 o + a

b o I 1 d  I b   +a dz = o 1 ln  o    2   0 2  o  o

M 12 =

12 I1

=

ob  o + a  ln   2  o 

P3.59: Consider a pair of concentric conductive loops, centered in the same plane, with radii a and b. Determine the mutual inductance between these loops if b >> a.

3-43 In this case will drive the b-radius loop with current I. Here, at the center of the b-radius loop we have from Eqn. (3.10): I B1 = o a z 2b a

o I1

2o I1  2 oa 2 I1 Then, 12 =  B1 dS2 =  d  d  = = 2b  2b 2  =0 2b So M 12 =

12 I1

=

o a 2 2b

P3.60: A 4.0 cm diameter solid nickel wire, centered on the z-axis, conducts current with a density J = 4 A/cm2 az (where  is in cm). Find the internal inductance per unit length for the wire with this current distribution. 

2

8

 H dL = I , 2 H = 4  d   d = 3  2

3

enc

0

H =

0

8 4 4 2 =  2 , B =  6 3 3 3

2

1 4  2 4  2 1 16 8 Wm =  a a  d  d dz =    5d   d  dz =  a 6 h 2 3 3 2 9 0 27 0 0 a

Wm =

h

1 2 LI ; 2

8 a 3 , and then solve for L: Now we solve for I: I =  J dS = 3 1 h L= 12  L 1  1 4 x10−7 H =L= = 600 = 20 . h 12  12  m

10. Magnetic Circuits P3.61: Referring to Figure 3.48(a), suppose 2.0 Amps flows through 80 turns of a toroid that has a core cross sectional area of 2.0 cm2 and a mean radius of 80. cm. The core is 99.8% pure iron. (a) How much magnetic flux exists in the toroid? (b) How much energy is stored in the magnetic field contained by the toroid?

3-44

( 5000 ) ( 4 x10 ) ( 80 )( 2 )  NI (a)  = BA = A; B = = 0.200Wb 2 ,  = 40Wb m 2o 2 ( 0.8 ) −7

(b) Wm =

1 1 B2 BH  dv = 2o A = 3.2mJ 2 2 

P3.62: In Figure 3.59, a 2.0 cm diameter toroidal core with r1 = 10,000 is wrapped with a 1.0 cm thick layer of r2 = 3000. The toroid has a 1.0 m mean radius. For 20. A of current driven through 50 loops of wire, find the magnetic field intensity in each material of the toroid.

This toroid can be modeled with the circuit of Figure P3.62.

Fig. P3.62 Vm = NI = (50)(20A)=1000 A-turns 2o 2o l R= ; R1 = = 1.592 x106 ; R2 = = 1.768 x106 ; 2 2 2 A r1o a r 2 o ( b − a )

1 =

Vm

2 =

Vm

R1

R2

H2 =

= 628 x10−6 = BA; B =

= 566 x10−6 = BA; B =

B

 r 2 o

= 159

A m

628 x10−6

 ( 0.01)

(

2

= 2; H1 =

566 x10−6

 ( 0.02 ) − ( 0.01) 2

2

)

B

 r 1 o

= 159

= 0.6001

A m

3-45 P3.63: Referring to Figure 3.49(a), the 2.0 cm diameter core is characterized by the magnetization curve of Figure 3.60. The toroid has a mean radius of 60. cm. For 10. A of current driven through 100 loops of wire, find the magnetic field intensity in the 1.0 mm gap.

Referring to the model of Figure 3.49(b), Vm = NI = (100)(10A)=1000 A-turns; A = a2 = 314x10-6

RG =

lG = 2.53x106 ; o A

Approach: (1) Assume B: B = 0.4 (2) Read H from chart: H = 200 2o 12 x103 H (3) Evaluate RC = : Rc =6x106 = A B (4) RTOT = Rc + 2.53x106: RTOT =8.53x106 (5)  = Vm/ RTOT:  =117x10-6 (6) B = /A: B = 0.373 Insert this value of B into step (1) and repeat. After 4 or 5 iterations the routine converges to a solution B = 0.382 Wb/m2 and H = 190 kA/m. Then, in the gap, we have B kA H= = 300 . o m

P3.64: Referring to Figure 3.52, suppose the cross sectional area of the bar is 3.0 cm 2, while that of the electromagnet core is 2.0 cm2. Also, the bar has a relative permittivity of 3000, while that of the magnetic core is 10,000. The dimensions for h and w are 12. cm and 16. cm, respectively. If the mass of the bar is 20. kg, how much current must be driven through 24 loops to hold up the bar against gravity?

We will follow an approach similar to Example 3.18, but realize that Ac ≠ Ab.

l 2(.12) + 0.16  100cm  3 Rc = =  = 159 x10 2   Ac 10, 000o (2cm )  m  2

l 0.16  100cm  3 Rb = =  = 141x10 2   Ab 3, 000o (3cm )  m  2

Vm = NI ,  =

Vm

RTOT

=

NI

RTOT

= o HAc ; H=

NI

RTOT o Ac

3-46 2

2

   NI  1 NI F = o H Ac = o  = mg  Ac =    RTOT o Ac   RTOT  o Ac 2

Solve for I: I = mg o Ac

RTOT N

= 2.8 A

P3.65: Consider a 1.0 mm air gap in Figure 3.49(a). The toroid mean radius and cross sectional area are 50. cm and 2.0 cm2, respectively. If the magnetic core has a r = 6000, and 4.0 A is being driven through 30 loops, determine the force pulling the gap closed.

F = o H 2 A, RTOT =

=

Vm

RTOT

=

( A)( turns ) 2o l + = 6.06 x106  r o A o A Wb

( 30 )( 4 ) = 19.79 x10−6 = BA

6.06 x106

In the gap, B = 19.79x10-6/A = 98.97x10-3 = oH, H = 78.76x103A/m 2

2  1  F = o H 2 A = ( 4 x10−7 )( 78.76 x103 ) ( 2 )   = 1.6  100  F = 1.6N

4-1 Solutions for Chapter 4 Problems 1. Current Continuity and Relaxation Time P4.1: How long does it take for charge density to drop to 1% of its initial value in polystyrene?

Polystyrene has r = 2.56 and  = 10-17 S/m. v = oe−( /  )t

  −9 −10−17 0.01o = o exp  t  = o e −441 x10 t − 12  ( 2.56 ) (8.854 x10 )    Solving for t we get t = 10.4x106 sec = 120 days P4.2: At a particular point in a slab of silver, a charge density of 109 C/m3 is introduced.

Plot v versus time for a duration of 10 relaxation times.

For silver we have r = 1 and  = 6.2x107 S/m 18 C v = oe −( /  )t = 109 e−7 x10 t 3 m The relaxation time is:  =

 = 142 x10−21 sec 

The plot is obtained with the following MATLAB routine. % % % %

MLP0402 Plot rhov vs time for a charge placed in silver

rhovo=1e9; %initial charge density, C/m^3 er=1; %relative permittivity of silver sig=6.2e7; %conductivity, S/m eo=8.854e-12; tau=er*eo/sig; t=tau/10:tau/10:10*tau; rho=rhovo*exp(-t./tau); subplot(2,1,1) plot(t,rho) xlabel('t (s)') ylabel('rhov (C/m^3)') grid on

4-2

subplot(2,1,2) loglog(t,rho) xlabel('t (s)') ylabel('rhov (C/m^3)') grid on

Fig. P4.02

P4.3: A current density is given by J = e-.01ta A/m2. Find the charge density after 10 seconds if it has an initial value of zero.

v 1  2 −0.01t =  e = 2e−0.01t ( ) t  t −0.01t d v = −2e dt; v = 200e−0.01t + C At t = 0, v = 0 = 200 + C, C = -200.  J=−

So we have v = 200 ( e −0.01(10) − 1) = −19

C m3

P4.4: At t = 0 seconds, 60.0 C is evenly distributed throughout a 2.00 cm diameter pure silicon sphere. (a) Find the initial charge density. (b) How long does it take to drop to 10% of its initial value? (c) What will be the final surface charge density?

4 Q C 3 (a) Volume v =  ( 0.01m ) = 4.19 x10−6 m3 ; vo = = 14.3 3 3 v m −t / (b) v = 0.10 vo = voe

 (11.8) (8.854 x10 For pure silicon,  = =  ( 4.4 x10−4 )

−12

) = 237.4 x10 sec −9

4-3

ln 0.10 = (c)  s =

−t



= −2.303; t = ( 2.303) = 547ns

Q 60C ; area=4 r 2 = 1.257 x10−3 m2 ;  s = = 47.8 mC 2 m area 1.257 x10−3 m2

2. Wave Fundamentals P4.5: A propagating electric field is given by

 V  E ( z, t ) = 100.e−.01z cos    107 t +  z −  .

4m  (a) Determine the attenuation constant, the wave frequency, the wavelength, the propagation velocity and the phase shift. (b) How far must the wave travel before its amplitude is reduced to 1.0 V/m? (a)  = 0.01 Np m ; f = 5MHz;  = 2m; u p = 1x107

m − ;= = −45 s 4

 1  (b) E ( z ) = 100e−0.01z = 1; − 0.01z = ln   ; z = 460m  100 

P4.6: A 10.0 MHz magnetic field travels in a fluid for which the propagation velocity is 1.0x108 m/sec. Initially, we have H(0,0)=2.0 ax A/m. The amplitude drops to 1.0 A/m after the wave travels 5.0 meters in the y direction. Find the general expression for this wave.

The general expression for the wave is: H( y, t ) = H oe − y cos ( t −  y +  ) a x

A m

 = 2 f = 2 (10 x106 ) = 20 x106

rad s u 2 rad u p = 1x108 m s ;  = p = 10m;  = = 0.2 f  m A H(0,0) = 2.0a x ;  H o = 2.0,  = 0 m H ( y = 5) = 1 = H oe− y = 2.0e− (5) ; solving we get  = 0.14 Finally, A H( y, t ) = 2.0e−0.14 y cos ( 20 x106 t − 0.2 y ) a x m

P4.7: MATLAB: Modify the simple wave program in MATLAB 4.1 to include attenuation. Generate a plot for the case where the amplitude is 4 V/m, the attenuation

4-4 constant is 0.001 Np/m, and the frequency is 1 MHz. Take your snapshot in time at 0 seconds, and let your phase shift be 0. % % % % % % % % % % % % % % % % % % % % %

M-File: MLP0407 This is a modification of ML0401 that includes attenuation. It plots a wave (in vac) versus position (in z direction) for a fixed time.

Wentworth, 1/19/03 Variables: atten Eo f omega t phi phir c lambda B E z

clc clear

attenuation constant wave amplitude (V/m) frequency (Hz) angular frequency (rad/sec) time snapshot (sec) phase constant (degrees) phase constant (radians) speed of light in vacuum (m/s) wavelength (m) phase constant (1/m) electric field intensity position

%clears the command window %clears variables

% Initialize Variables atten=0.001; Eo=4; f=1e6; t=0; phi=0; phir=phi*pi/180; c=2.998e8; lambda=c/f; B=2*pi/lambda; omega=2*pi*f; % Perform Calculation z=0:4*lambda/100:4*lambda; E=Eo*exp(-atten.*z).*cos(omega*t-B*z+phir);

4-5

% Generate the Plot plot(z,E) axis('tight') %sets axes min & max data values grid xlabel('z(m)') ylabel('E(V/m)')

Fig. P4.7

P4.8: MATLAB: Modify the traveling wave program in MATLAB 4.2 to include attenuation. Use the parameters from problem P4.7, except for the fixed time of course. % % % % % % % % % % % % % % % %

M-File: MLP0408 This program illustrates a traveling wave including attenuation. It modifies ML0402.

Wentworth, 1/19/03 Variables: atten Eo f omega t phi phir c

attenuation (Np/m) wave amplitude (V/m) frequency (Hz) angular frequency (rad/sec) time snapshot (sec) phase constant (degrees) phase constant (radians) speed of light in vacuum (m/s)

4-6 % % % %

lambda B E z

wavelength (m) phase constant (1/m) electric field intensity position

clc clear

%clears the command window %clears variables

% Initialize Variables atten=0.001 Eo=4; f=1e6; t=1; phi=0; phir=phi*pi/180; c=2.998e8; lambda=c/f; B=2*pi/lambda; omega=2*pi*f; % Perform Calculation z=0:4*lambda/100:4*lambda; E=Eo.*exp(-atten.*z).*cos(omega*t-B*z+phir); % Generate a Reference Frame plot(z,E) axis([0 4*lambda -2*Eo 2*Eo]) grid xlabel('z(m)') ylabel('E(V/m)') pause % Make the Movie t=0:1/(40*f):1/f; for n=1:40; E=Eo.*exp(-atten.*z).*cos(omega*t(n)-B*z+phir); plot(z,E) axis([0 4*lambda -2*Eo 2*Eo]) grid title('General Wave Equation'); xlabel('z(m)'); ylabel('E(V/m)');

4-7 M(:,1)=getframe; end

Fig. P4.8 (at end of animation)

3. Faraday’s Law and Transformer EMF P4.9: The magnetic flux density increases at the rate of 10 (Wb/m2)/sec in the z direction.

A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch) and magnitude of the induced current in the conducting loop.

dB Wb = 10 2 a z dt ms dB Wb Vemf = −  dS = −  10 2 a z dxdya z dt ms Wb Vs Vemf = −0.1 = −0.1V s Wb 0.1V I= I = = 10mA 10 I=10mA clockwise (when viewed From +z)

Fig. P4.9

P4.10: A bar magnet is dropped through a conductive ring. Indicate in a sketch the direction of the induced current when the falling magnet is just above the plane of the ring and when it is just below the plane of the ring, as shown in Figure 4.22.

Refer to Figure P4.10.

4-8 When the north pole first goes through the loop, flux is increasing and the current induced to oppose this change in flux is as shown.

When the south pole is exiting the loop, flux is decreasing and the current induced acts to oppose this change in flux.

Fig. P4.10

P4.11: Considering Figure 4.7, suppose the area of a single loop of the pair is 100 cm 2, and the magnetic flux density is constant over the area of the loops but changes with time as B = Boe−  t a z , where Bo = 4.0 mWb/m2 and  = 0.30 Np/sec. Determine VR at 1, 10, and 100 seconds.

Vemf = − N 

B dB dS; = −  Bo e−  t a z ; t dt

Vemf = 2  Bo e−  t S 2

 1m  −0.30t VR = 2  Bo Se −  t = 2 ( 0.30 ) ( 4 x10−3 )(100cm 2 )  = 24 x10−6 e −0.30 t  e  100cm  at t = 1 sec, VR = 17.8 V at t = 10 sec, VR = 1.20 V at t = 100 sec, VR = 2.25x10-18 V

P4.12: Sometimes a transformer is used as an impedance converter, where impedance is given by v/i. Find an expression for the impedance Z1 seen by the primary side of the transformer in Figure 4.11 that has a load impedance Z2 terminating the secondary.

We have i2 =

N1 N i1 and v2 = 2 v1 N2 N1

4-9

 N2  v1  2  v1 v2  N 1   N 2  Z1 = , Z 2 = = =  Z1 i1 i2  N1   N1   N i1   2  2

N  Z1 =  1  Z 2  N2 

P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It is placed in a plane normal to a magnetic field increasing with time as B = 1.0 t Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center of the loop resulting from the induced current, and compare this with the original magnetic flux density that generated the induced current at t = 1.0 sec.

We find the distributed resistance of the loop and work the problem assuming this resistance is lumped in one spot as shown in the figure. (a) The induced current is Vemf divided by the distributed resistance of the wire loop.

4 ( 0.04m) ) 1 l 1m R= = = 3.5m 7  A 5.8 x10  ( 0.0005m )2 0.04

0.04

dB Wb dB Wb = 1 2 a z ; Vemf = −  dS = −1 2  dx  dy = −1.6mV dt ms dt ms 0 0

I ind =

1.6mV = 0.46 A (note that this answer has no time dependence) 3.5m

Fig. P4.13 (b) The field at the center of the loop from a single arm of the loop is found from Eqn. (3.7): +a

 I  z I H= −a z ) =  2 2  ( 4   z +  2  2 −a

( 0.46 ) 1 1 A ( -a z ) = ( −a z ) = −2.59a z ; 2 m 2a 2 ( 0.02 )

4-10 So B = 4o H = −13

Wb m2

az .

P4.14: The mean length around a nickel core of a transformer like the one shown in Figure 4.11 is 16 cm, and its cross sectional area is 1 cm2. There are 30 turns on the primary side and 45 on the secondary side. If the current on the primary side is 1.0 sin20x106t mA, (a) calculate the amplitude of the magnetic flux in the core in the absence of the output windings. (b) With the output windings in place, calculate i2. (a)  =

=

Vm

R

; Vm = N1i1; R =

N1i1r o A

(b) i2 =

l

l  r o A

( 30)(1mA)( 600) ( 4 x10−7 )(1cm2 ) m2 = = 14 x10−9Wb 2 ( 0.16m ) (100cm )

(

)

N1 30 2 i1 = 1sin ( 20 x106 t ) mA = sin ( 20 x106 t ) mA N2 45 3

P4.15: A triangular wire loop has its vertices at the points (2, 0, 0), (0, 3, 0) and (0, 0, 4), with dimensions in meters. A time-varying magnetic field is given by B = 4t ay Wb/m2 (t in seconds). If the wire has a total distributed resistance of 2 , calculate the induced current and indicate its direction in a carefully drawn sketch.

B dS, t B Wb = 4a y 2 t ms 1 MN S = MN 2 MN

Vemf = − 

M = −2a x + 3a y , N = −2a x + 4a z

S = 6a x + 4a y + 3a z m2 Vemf = -(4)(4)=-16V Vemf 16V I= = = 8A R 2

Fig. P4.15

4. Faraday’s Law and Motional EMF P4.16: Referring to Figure 4.23, suppose a conductive bar of length h = 2.0 cm moves with velocity u = -1.0 m/s a towards an infinite length line of current I = 4.0 A. Find an expression for the voltage from one end of the bar to the other when  reaches 10 cm and indicate which end is positive.

4-11 In Figure P4.16, an imaginary circuit has been chosen. For the chosen circulation direction, we have the sign for Vemf as shown. Then, h  o I   m  o I Vemf =  ( u  B ) dL =   -1a   a  dza z =  −1   dz , 2   s  2 0 

Vemf = −160nV . Therefore, the bottom of the bar is positive.

Fig. P4.16

P4.17: Suppose we have a conductive bar moving along a pair of conductive rails as in Figure 4.12, only now the magnetic flux density is B = 4.0ax + 3.0az Wb/m2. If R = 10. , w = 20. cm, and uy = 3.0 m/s, calculate the current induced and indicate its direction.

 Wb  m  Vemf = − B u y dxa z = −  3 2  3  ( 0.2m ) = −1.8V  m  s  1.8V I= = 0.18 A (clockwise when viewed from the +z axis) 10

P4.18: The radius r of a perfectly conducting metal loop in free space, situated in the x-y plane, increases at the rate of (r)-1 m/sec. A break in the loop has a small 2.0 ohm resistor across it. Meanwhile, there exists a magnetic field B = 1.0 az T. Determine the current induced in the loop, and show in a sketch the direction of flow.

Here we’ve assumed dS = -dSaz to get iind and Vemf as shown. Our approach will be to find , then Vemf = -d/dt.

4-12

 =  B dS =  1a z ( −  d  d a z ) r

= −2   d  = − r 2 0

d dr  1  = −2 r = −2 r   = −2V dt dt r  Vemf = +2V I=

2V = 1A, clockwise as shown 2 Fig. P4.18

P4.19: Rederive Vemf for the rectangular loop of Figure 4.16 if the magnetic field is now B = Boaz.

We see in Figure P4.19a that ( u  B ) dL = 0 for the 1 → 2 and 3 → 4 line sections.

For the 2 → 3 section we have:

dl

3

 =  ( u  B ) dL; B = B a , dL = d a  , u = dt = o

z

2

0

 ( u  B ) dL =  Bo   d  = a

 da dt

= a

− Bo a 2 2

+ Bo a 2 , so for the 2 → 3 section, the contributions to Vemf 2 0 cancel. This will also be the case for the 4 → 1 section, and therefore Vemf = 0; no current is induced. a

 ( u  B ) dL =  Bo   d  =

Fig. P4.19a

Fig. P4.19b

4-13 P4.20: In Figure 4.16, replace the rectangular loop with a circular one of radius a and rederive Vemf.

 =  B dS =  Boa y dS = Boa y  a 2 ( − sin a x + cos a y ) = Bo a 2 cos  d flux = − Bo a 2 sin  ; dt

Vemf = −

d flux dt

=  Bo a 2 sin 

P4.21: A conductive rod, of length 6.0 cm, has one end fixed on a grounded origin and is free to rotate in the x-y plane. It rotates at 60 revolutions per second in a magnetic field B = 100. mT az. Find the voltage at the end of the bar.

rad  rev  2 rad    = 120 s  rev  s  Vemf =  ( u  B ) dL

 =  60

u=

d a =  a dt l

Vemf =  ( a  Boa z ) d  a  = 0

 Bo l 2 2

1 rad  Wb  2 Vs Vemf = 120  0.1 2  ( 0.06m ) 2 s  m  Wb = 68mV

Fig. P4.21

We can confirm the sign by observing that a positive charge placed in the middle of the bar would move to the ungrounded end by the Lorentz force equation.

P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make conductive contact with a ring to make one end of the electrical contact to R; the other contact is made to the center of the bar. Given Bo = 100. mWb/m2, a = 6.0 cm, and R = 50. , determine I if the bar rotates at 1.0 revolution per second.

d a =  a , dL = d  a  dt Figure P4.22 indicates one of the paths for the circulation integral.

Vemf =  ( u  B ) dL; u = 

4-14 a

a

0

0

Vemf =  ( a  Boa z ) d  a   =  Bo   d  =

I=

Vemf

=

 Bo a 2

R 2R I = 22.6 A

 Bo a 2 2

1  rev  rad  2  1  Vs A −3 Wb  = 1 0.06m )   2 100 x10  2 ( 2  s  rev  m   50  Wb V

Fig. P4.22

P4.23: A Faraday Disk Generator is similar to the rotating conductor of P4.22, only now the rotating element is a disk instead of a bar. Derive an expression of the Vemf produced by a Faraday Disk Generator, and using the parameters given in problem 4.22, find I. Worked exactly as P4.22.

P4.24: Consider a sliding rail problem where the conductive rails expand as they progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y direction, and uy = 2.0 m/sec, find the Vemf across a 100.  resistor at the instant when y = 10. cm if the field is Bo = 100. mT. First we modify the figure so that the top rail is horizontal and all the spreading occurs via the bottom rail. As before, our approach will be to find  and then d /dt. We have:  =  B dS =  Boa z dxdya z Now, notice that x and y are not independent and are in fact related: x=y+w So we have y y+w y 1   = Bo   dxdy = Bo  ( y + w)dy = Bo  y 2 + wy  2  y =0 x =0 0

4-15

d dy Wb    m = − Bo ( y + w ) = − Bo ( y + w ) u y =  −0.100 2  ( 0.1m + 0.1m )  2  dt dt m    s Vemf = 40mV

Vemf = −

Alternate Method: Vemf =  ( u  B ) dL =  ( u y a y  Boa z ) dxa x 1 − y 2

Vemf = u y Bo  dx = −u y Bo ( w + y ) 1 w+ y 2

Fig. P4.24

5. Displacement Current P4.25: Suppose a vector field is given as A = 3x 2 yz 3a x .

Verify that the divergence of the curl of this vector field is equal to zero.

verify that   A = 0

ax  A =  x 2 3x yz 3

ay az   =  ( 3x 2 yz 3 ) a −  ( 3x 2 yz 3 ) a y z y z z y 0 0

= 9 x 2 yz 2a y − 3x 2 z 3a z   A =

  9 x 2 yz 2 ) − ( 3x 2 z 3 ) = 9 x 2 z 2 − 9 x 2 z 2 = 0 ( y z

4-16 P4.26: Suppose a vector field is given by A =  2 cos az .

Verify that the divergence of the curl of this vector field is equal to zero.

1    2 cos  ) a  − (  2 cos  ) a = −  sin  a  − 2  cos  a (    1  1    A =  ( −  sin  ) ) + ( −2 cos  ) (     1 = − 2  sin  + 2sin  = 0  A =



P4.27: A pair of 60 cm2 area plates are separated by a 2.0 mm thick layer of ideal dielectric characterized by r = 9.0. If a voltage v(t) = 1.0 sin (2x103t) V is placed across the plates, determine the displacement current. −12 2 dv  S ( 9 ) (8.854 x10 F m )( 60cm ) 10mm 1m id = C ; C = = = 239 pF dt d cm 100cm ( 2mm )

dv V = 2 x103 cos ( 2 x103 t ) dt s 3 id = 1.5cos ( 2 x10 t )  A P4.28: Plot the loss tangent of seawater ( = 4 S/m and r = 81) versus log of frequency from 1 Hz to 1 GHz. At what frequency is the magnitude of the displacement current density equal to the magnitude of the conduction current density?

tan  =

( 4 S m)  889 x106 = =  ( 2 )( f ( Hz ) )( 81) (8.854 x10−12 F m ) f

(Plot this in Figure P4.28) Solving for f when tan = 1: f = 890MHz % MLP0428 clear clc for i=1:9 for j=1:10 m=(i-1)*10+j; f(m)=j*10^(i-1); tand(m)=889e6/f(m); end

4-17 end loglog(f,tand) xlabel('frequency (Hz)') ylabel('loss tangent') grid on

Fig. P4.28

P4.29: A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer conductor diameter 6.0 mm is filled with an ideal dielectric with r = 10.2. A voltage v(t) = 10.cos(6x106 t) mV is placed on the inner conductor and the outer conductor is grounded. Neglecting fringing fields at the ends of the coax, find the displacement current between the inner and outer conductors.

C=

Q , so Q = Cv(t ) V

for coax (from chapter 2): C=

2 lVo cos t 2 l so Q = ln ( b a ) ln ( b a )

 D dS =  D a  d dza = 2l D = Q

2 l Vo cos t Vo cos t = 2 l ln ( b a )  ln ( b a ) D −Vo sin t = a = Jd t  ln ( b a ) 

so D =

id =  J d dS =

and D =

Vo cos t a .  ln ( b a ) 

−Vo sin t 1 2 l −2 l Vo sin t   d  dz = ln ( b a )  0 ln ( b a ) 0

4-18 Now evaluate id with the given parameters: −2 ( 6 x106 rad s ) (10.2 ) ( 8.854 x10−12 F m ) (1m )( 0.010V sin t ) C As id = ln ( 0.006 0.002 ) FV C

id = −97 sin ( 6 x106 t )  A

7. Lossless TEM Waves P4.30: Suppose in free space that E(z,t) = 5.0e-2zt ax V/m. Is the wave lossless? Find H(z,t).

Since the wave has an attenuation term (e-2zt) it is clearly not lossless. ax ay az H   =  ( 5e−2 zt ) a = −10te−2 zt a   E = − o =  y y x y z z t 5e−2 zt 0 0 −10t −2 zt 10 dH = e dta y , H =  te−2 zt dta y − o o This integral is solved by parts

(  udv = uv −  vdu ) where we let u = t and dv = e dt. −2 zt

We arrive at:  −10t −2 zt 10 −2 zt  A H= e − e ay 2 4 o z m  2 o z 

P4.31: An electric field propagating in a lossless non-magnetic media is characterized by V E( y, t ) = 100.cos 4 106 t − 0.1257 y az . m Find the wave amplitude, frequency, propagation velocity, wavelength, and the relative permittivity of the media. (b) Find H(y,t).

(

)

V 2  m 1 , f = 2MHz,  = ,  = 50m, u p = = 100 x106 = ; r = 9 m   s  H  E =  Eo sin (t −  y ) a x = − o t  Eo E H=−  sin (t −  y )dt = o cos (t −  y ) a x Eo = 100

o

H( y, t ) =

 Eo cos (t −  y ) a x o

o

Inserting the given values we find: H( y, t ) = 0.796cos ( 4 x106 t − 0.126 y ) a x

A m

4-19 P4.32: A magnetic field propagating in free space is given by

(

)

H( z, t ) = 20.sin   108 t +  z ax

Find f, , , and E(z,t).

A m

.

1 2

 = 2 f =  x108 , f = x108 = 50MHz c 3x108 2  rad = = 6m,  = = 8 f 0.5 x10  3 m D  H = (no J c ) t

=

ax  H = x H o sin (t +  z )

ay az   =  ( H sin (t +  z ) ) a y y z z o 0 0

= H o  cos (t +  z ) a y =  o

E t

Ho 

Ho 

 dE = E =   cos (t +  z ) dta =  sin (t +  z ) a y

o

so E( z, t ) =

 Ho V sin (t +  z ) a y  o m

y

o

  kV  inserting the given values we find: E( z, t ) = 7.5sin   x108 t + z  a y 3  m  P4.33: Find the instantaneous expression for E for the magnetic field of problem P4.6.

From P4.6 we have H = 2.0e−0.14 y cos ( 20 x106 t − 0.2 y ) as H = H o e− y cos (t −  y ) First we have   H = 

A a x . We can write this wave m

A a x , and can reinsert the values at the end. m

c E , . Assuming a nonmagnetic medium, we have u p = t r

or r = 9. Now we evaluate the curl of H:  ax a y az         H = = −  H o e− y cos (t −  y )  a z .  x y z  y   − y 0   H o e cos (t −  y ) 0

4-20

  H o e − y cos (t −  y )  a z = − H o  − e − y cos (t −  y ) +  e − y sin (t −  y )  a z . y E So  H = H o e− y cos (t −  y ) a z − H o  e− y sin (t −  y ) a z =  . t Solving for E: H H  E = o e− y  cos (t −  y ) dta z − o e− y  sin (t −  y ) dta z −



E=

H o



e− y sin (t −  y ) a z +

Ho





e− y cos (t −  y ) a z .

Now we can evaluate the variables: ( 2 )( 0.14 ) H o = = 56, 6  ( 20 x10 ) ( 9 ) (8.854 x10−12 )

Ho 



=

( 2 )( 0.2 ) = 251. ( 20 x10 ) (9) (8.854 x10−12 ) 6

So E = 56e−0.14 y sin ( 20 x10−6 t − 0.2 y ) a z + 251e−0.14 y cos ( 20 x10 −6 t − 0.2 y ) a z

V . m

We can use a trigonometric identity to simplify this answer. Given that cos ( x  y ) = cos x cos y sin x sin y, we can let E = e− y a z  A ( b cos (t −  y ) + c sin (t −  y ) ) , where b=cosy, c=siny, and tany=c/b. Now, since b/c = 251/56, then y=12.6° and A=257.

Thus V E( y, t ) = 257e−0.14 y cos ( 20 x106 t − 0.2 y − 12.6 ) a z . m This result is easier to obtain using the phasor approach of the next section.

P4.34: Given, at some point distant from a point source at the origin in free space, V E(r , t ) = 8.0cos 9 106 t −  r a , m find frequency, phase constant and H(r,t).

(

 = 2 f , f = 4.5MHz,  = E = −

B H = − o t t

)

c 2 rad = 66.7m,  = = 0.094 f  m

4-21

E =

1  ( rE ) a , E = Eo cos (t −  r ) , r r

 ( Eo r cos (t −  r ) ) = Eo cos (t −  r ) + Eo  r sin (t −  r ) r E   E = o cos (t −  r ) +  r sin (t −  r )  a r E cos (t −  r ) = o a + Eo  sin (t −  r ) a r Assume we can ignore the first term in the far-field for large r, then we have H  E = Eo  sin (t −  r ) a = − o t  Eo E  dH = H = −  sin (t −  r ) dta = o cos (t −  r ) a

o

o

Inserting the given values we find mA H(r , t ) = 21cos (t −  r ) a m

P4.35: In a lossless, non-magnetic media, the magnetic field at some point distant from a source at the origin is given by A H(  , t ) = 6.0sin 3 108 t + 10  a , m find the relative permittivity of the media, the frequency and phase constant of the wave, and E(,t).

(

)

2 f = 3x108 ; f = 48MHz  m c u p = = 0.3x108 = ;  r = 100  s r

 r

rad c m D E H = = t t 1  1 H = H o  sin (t + 10  ) ) a z = ( H o sin (t +  ) +  H o  cos (t +  ) ) a z (    In the far-field, we will ignore the first term (has 1/) leaving us with  H =  H o cos (t +  ) a z

=

= 10

Now,  dE = E =

E=

 Ho cos (t +  ) dta z  

 Ho sin (t +  ) a z . 

4-22 Plugging in the given values we arrive at: E(  , t ) = 226sin(3x108 t + 10  )a z

P4.36: Suppose, in a non-magnetic medium of relative permittivity 3, that

E ( y, t ) = 4.0sin ( x107 t −  y ) a x + 9.0cos ( x107 t −  y ) a z

V . m

Determine  and H(y,t).

f = 5MHz, u p =

c

r

= 1.73x108

u m 2 rad ,  = p = 34.6m,  = = 0.181 s f  m

 E = 9 sin (t −  y ) a x + 4 cos (t −  y ) a z = − o

H=

9

o

cos (t −  y ) a x +

−4

o

H t

sin (t −  y ) a z

Inserting the given values we arrive at:

H( y, t ) = 41cos (t −  y ) a x − 18sin (t −  y ) a z

mA m

8. Time-Harmonic Fields and Phasors P4.37: Show that  − Re B s e jt  = − Re ( jB s ) e jt  . t

(

)

We first show that :    − Re B s e jt  = − Re  B s e jt  t  t  Expanding the left side we have:   − Re B s e jt  = − ( Re  B s cos(t ) + jB s sin(t )) t t  = − B s cos(t ) = B s sin(t ) t Now expanding the right side we have:   − Re  B s e jt  = − Re  jB s e jt  = − Re  jB s ( cos (t ) + j sin (t ) )   t 

(

)

(

)

= − Re  jB s cos (t ) + j 2 sin (t )  = B s sin (t ) So since we have :

V . m

4-23

−

   Re B s e jt  = − Re  B s e jt  = − Re  jB s e jt  , t  t 

(

)

then     − Re B s e jt  = − Re B s e jt  = − Re  jB s e jt  t  t 

(

)

P4.38: Derive the differential phasor form of (a) Gauss’s Law, and (b) Ampere’s Circuit Law.

 D = v ,  D ( x, y, z, t ) = v ( x, y, z, t )  Re  Ds e jt  = Re  vs e jt  ;

Re  Ds e jt  = Re  vs e jt  ;

 Ds = vs

H =

D E = t t

 Re ( H s e jt ) =   H s = j Es

 Re ( Es e jt ) ; t

Re (  H s e jt ) = Re ( j Es e jt )

P4.39: Find H(y,t) in problem P4.31(b) using phasors.

Es = Eo e− j y a z ,  Es = − j H s   Es = − j  Eo e− j y a x = − j H s  Hs =

 Eo − j y  Eo e a x and H( y, t ) = cos (t −  y ) a x . o o

Plugging in the values from P4.31, we find A H( y, t ) = 0.796cos ( 4 x106 t − 0.126 y ) a x . m

P4.40: Find E(z,t) in problem P4.32 using phasors.

H s = H o e j z a x ,  H s = j o Es ax   Hs =  x jH o e j z

ay az   =  jH e j  z a o y y z z 0 0

4-24

 jH o e j z a y = j 2  H o e j z a y = −  H o e j z a y = j o Es z − H o e j z a y  H o j z  Ho V Es = = j e a y ,  E( z, t ) = sin (t +  z ) a y j o  o  o m Plugging in the values from P4.32, we find   kV  E( z, t ) = 7.5sin   108 t + z  a y .

3  m   H s =

P4.41: In free space,

(

)

(

)

E( z, t ) = 10.cos  106 t −  z a x + 20.cos  106 t −  z a y Find H(z,t).

The phasor version of E is: Es = 10e− j z a x + 20e− j  z a y

ax   Es =  x 10e− j z

ay  y 20e− j z

az  = −  20e− j z a +  10e− j z a x y z z z 0

 Es = j  20e− j z a x − j  10e− j  z a y = − jo H s

Hs =

j  20 − j z − j 10 − j z −20 − j z 10 − j z e ax + e ay = e ax + e ay − jo − jo o o

   x106 rad up = ;  = = = 10.47 x10−3 8  u p 3x10 m Inserting the appropriate values we then find: mA H s = −53e− j  z a x + 27e− j  z a y , m

and H( z, t ) = −53cos (t −  z ) a x + 27 cos (t −  z ) a y

mA m

P4.42: Find H(y,t) in problem P4.36 using phasors.

In phasor form the field of P4.36 is: Es = j 4e− j y a x + 9e− j y a z Now apply  Es = − jo Hs

ax   Es =  x j 4e− j y

ay az   = − j9 e− j y a x + j 2 4 e− j y a z y z 0 9e− j y

− j9 e− j y a x − 4 e− j y a z = − jo H s

V . m

4-25

− j 9 − j  y −4 − j y 9 − j  y 4 − j  y e ax + e az = e ax − j e az − jo − jo o o Inserting the appropriate values we then find: c m  rad up = = 1.73x108 ,  = = 0.181 s up m r Hs =

H( y, t ) = 41cos (t −  y ) a x − 18sin (t −  y ) a z

mA m

P4.43: MATLAB: In MATLAB 4.4, a polar plot of the phasor corresponded to a location on a sine wave for a particular time, at a fixed position in space. You can also make a polar phasor plot for a snapshot in time, where you change position. Modify MATLAB 4.4 to provide an animation of the phasor versus sine wave as you change the position. % % % % % % % % % % % % % % % % % %

M-File: MLP0443 This program modifies ML0404, generating a position plot animation synchronized with a polar plot.

Wentworth, 1/19/03 Variables: Eo E E f theta theta1 T t t1

clc clear clf

field amplitude (V/m) electric field intensity (V/m) E-field for movie frequency (Hz) angle angle for movie period (1/f) time (s) time for movie

%clears the command window %clears variables

% Generate the reference frame % Initialize Variables c=3e8; Eo=1; f=1000; lambda=c/f;

4-26 beta=2*pi/lambda; T=1/f; % Perform Calculation z=0:lambda/100:2*lambda; theta=-beta*z; E=Eo*cos(theta); % Generate the Plot subplot(211),plot(z,E,0,Eo,'ro'); subplot(212),polar(0,Eo,'ro'); pause %Make the Movie z1=0:lambda/50:2*lambda; for n=1:100 theta1(n)=-beta*z1(n); E1(n)=Eo*cos(theta1(n)); subplot(211),plot(z,E,z1(n),E1(n),'ro'); subplot(212),polar(theta1(n),Eo,'ro'); M(:,1)=getframe; end

Fig. P4.43 (at start of movie) P4.44: MATLAB: Repeat problem P4.43, now accounting for attenuation. Run the program assuming an attenuation of 2 x 10-6 Np/m. %

M-File: MLP0444

4-27 % % % % % % % % % % % % % % % % % % %

Same as MLP0443 but add attenuation.

This program modifies ML0404, generating a position plot animation synchronized with a polar plot and accounting for attenuation.` Wentworth, 1/19/03 Variables: Eo E E f theta theta1 T t t1

clc clear clf

field amplitude (V/m) electric field intensity (V/m) E-field for movie frequency (Hz) angle angle for movie period (1/f) time (s) time for movie

%clears the command window %clears variables

% Generate the reference frame % Initialize Variables atten=0.000002 ; c=3e8; Eo=1; f=1000; lambda=c/f; beta=2*pi/lambda; T=1/f; % Perform Calculation z=0:lambda/100:2*lambda; theta=-beta*z; E=Eo.*exp(-atten.*z).*cos(theta); % Generate the Plot subplot(211),plot(z,E,0,Eo,'ro'); subplot(212),polar(0,Eo,'ro');

4-28 hold on pause %Make the Movie z1=0:lambda/50:2*lambda; for n=1:100 theta1(n)=-beta*z1(n); E2(n)=Eo.*exp(-atten.*z1(n)); E1(n)=E2(n).*cos(theta1(n)); E2(n)=Eo.*exp(-atten.*z1(n)); subplot(211),plot(z,E,z1(n),E1(n),'ro'); subplot(212),polar(theta1(n),E2(n),'ro'); M(:,1)=getframe; end

Fig. P4.44 (at end of movie)

5-1 Solutions for Chapter 5 Problems 1. General Wave Equations P5.1: Starting with Maxwell’s equations for simple, charge-free media, derive the Helmholtz equation for H.

E      (  H ) =    E +   =   E +    E t  t  2 H H = -  −  2 t t Using a vector identity we also have:  (  H ) =  H −  2 H But  H = 0 , leading to H 2H  2 H =  +  2 t t P5.2: Derive equation (5.10) by starting with the phasor point form of Maxwell’s equations for simple, charge-free media.

For charge-free media the phasor form of Maxwell’s equations are:  Ds = 0

 Bs = 0   Es = − j H s

  H s = ( + j ) Es Now we take the curl of both sides of Faraday’s Law,  (  Es ) =  ( − j H s ) = − j H s = − j ( + j ) Es Now since  (  Es ) =  Es −  2Es , and since  E s = 0 , we have

2Es = j ( + j ) Es P5.3: A wave with  = 6.0 cm in air is incident on a nonmagnetic, lossless liquid media.

In the liquid, the wavelength is measured as 1.0 cm. What is the wave’s frequency (a) in air? (b) in the liquid? (c) What is the liquid’s relative permittivity?

3x108 m s = 5GHz   0.06m (b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz (c) (a) f =

up

=

c

=

5-2

1 m c  u p =  f =  5 x109  ( 0.01m ) = 5 x107 = s s r  2

 3x108   r =  = 36 8   0.5 x10 

P5.4: Suppose Hs(z) = Hys(z) ay. Start with (5.14) and derive (5.29).

Since Hs is only a function of z, (5.14) becomes 2 H s −  2 H s = 0. (a) 2 z H s 2 H s z z =  Ae , and =  2 Ae z .

If we let H s = Ae , then 2 z z 2 2 So (a) becomes  −  = 0, or (  +  )(  −  ) = 0. This has two solutions: (1) for  +  = 0, we have  = − , H s = Ae− z , or H s = H o+ e− z .

(2) for  −  = 0, we have  =  , H s = Ae z , or H s = H o−e z .

The general solution is the linear superposition of the two, or H s = ( H o+ e− z + H o−e z ) a y .

P5.5: Given  = 1.0x10-5 S/m , r = 2.0, r = 50., and f = 10. MHz, find , , , and .

=

jr o ( + j r o ) =  + j 

=

j r o  + j r  o

jr o = j 2 (10 x106 ) ( 50 ) ( 4 x10−7 ) = j3948

 + j r o = 1x10−5 + j 2 (10 x106 ) ( 2 ) (8.854 x10−12 ) = 1x10−5 + j1.11x10−3 Inserting these into the expressions for  and ,  = 9.4 x10−3 + j 2.1 1 m ,  = 9.4 x10−3 Np m ,  = 2.1 rad m ,  = 1880e j 257  These results are confirmed by ML0501.

P5.6: MATLAB: In some material, the constitutive parameters are constant over a large frequency range and are given as  = .10 S/m , r = 4.0, and r = 600. Write a MATLAB routine that will plot , , and  (magnitude and phase) versus the log of frequency from 1 Hz up to 100 GHz. % %

M-File: MLP0506

5-3 % This program is a modification of ML0501. % For a given material, it will plot the attenuation, % phase constant and intrinsic impedance vs f. % % Wentworth, 1/23/03 % clc %clears the command window clear %clears variables % Initialize Variables uo=pi*4e-7; eo=8.854e-12; sig=0.10; er=4; ur=600; % Perform Calculation for i=1:10 for j=1:10 m=(i-1)*10+j; f(m)=j*10^(i-1); w(m)=2*pi*f(m); A(m)=i*(w(m)*ur*uo); B(m)=complex(sig,w(m)*er*eo); gamma(m)=sqrt(A(m)*B(m)); alpha(m)=real(gamma(m)); beta(m)=imag(gamma(m)); eta(m)=sqrt(A(m)/B(m)); meta(m)=abs(eta(m)); aeta(m)=180*angle(eta(m))/pi; end end subplot(3,1,1) plot(f,alpha,'-o',f,beta,'-*') ylabel('1/m') xlabel('frequency (Hz)')a legend('alpha','beta') subplot(3,1,2) semilogx(f,meta) ylabel('magnitude of eta (ohms)') subplot(3,1,3) semilogx(f,aeta) ylabel('phase of eta (degrees)') xlabel('frequency (Hz)')

5-4

Fig. P5.6

P5.7: Suppose E(x,y,t) = 5.0 cos(x106t – 3.0x + 2.0y) az V/m. Find the direction of propagation, ap, and H(x,y,t).

Es = 5e− j 3 x e j 2 y a z We assume nonmagnetic material and therefore have  Es = − j H s = j10e− j 3 x e j 2 y a x + j15e− j 3 x e j 2 y a y

j10 − j 3 x j 2 y j15 − j 3 x j 2 y e e ax + e e a y = −2.53e− j 3 x e j 2 y a x − 3.8e− j 3x e j 2 y a y − jo − j A H( x, y, t ) = −2.53cos ( x106 t − 3x + 2 y ) a x − 3.80cos ( x106 t − 3x + 2 y ) a y m To find the direction of propagation, E  Hs aP = s Es  H s Hs =

Es  H s = 19e− j 6 x e j 4 y a x − 12.65e− j 6 x e j 4 y a y And with the exponential terms canceling in the top and bottom of the equation for ap, we have: a P = 0.83a x − 0.55a y .

5-5

P5.8: Suppose in free space, H(x,t) = 100.cos(2x107t – x + /4) az mA/m. Find E(x,t).

H s = 0.100e− j  x e j a z , a P = a x ,

( =  4 )

Es = − a P  H s = −120 a x  0.100e− j  x e j a z = 12 e− j x e j a y

E = 12 cos (t −  x +  ) a y

Since free space is stated, 2 2 = = = 2 30 rad m  c f and then 2  V  E = 12 cos  2 x107 t − x + ay 30 4 m 

2. Propagation in Lossless, Charge-Free Media P5.9: Start with the Helmholtz equation (5.11), and using  = j, derive (5.41), the traveling wave equation.

 2E s −  2E s = 0, let E s = Exs ( z )a x , and with  =j we have Let Exs = Ae z , so

 2 Exs +  2 Exs = 0. z 2

Exs  2 Exs = A e x and = A 2e x 2 z z

Now we have A 2e z +  2 Ae z = 0, or  2 +  2 = 0 This can be factored:  2 +  2 = (  + j  )(  − j  ) = 0 , suggesting two solutions. The first solution uses  = − j  and

Exs = Ae− j z = Eo+ e− j z .

Likewise, the second solution uses  = + j  and Exs = Ae j z = Eo−e j z .

The complete solution is a linear superposition of these two solutions, or Es = Eo+ e− j z + Eo−e+ j z a x .

P5.10: A 100 MHz wave in free space propagates in the y direction with an amplitude of 1 V/m. If the electric field vector for this wave has only an az component, find the instantaneous expression for the electric and magnetic fields.

From the given information we have  = 2 f = 200 x106

 2 rad rad = , and  = up 3 m s

5-6

2  V  or E( y, t ) = 1cos  200 x106 t − y  az .

3  m  Now to find H. 1 1 1 − j y Es = 1e− j y a z , H s = a P  Es = a y 1e− j y a z = e ax  120 120 So 1 2  A  H ( y, t ) = cos  200 x106 t − y  ax 120 3  m  or 2  mA  H ( y, t ) = 2.7 cos  200 x106 t − y  ax .

3  m  P5.11: In a lossless, nonmagnetic material with r = 16, H = 100 cos(t – 10y) az mA/m.

Determine the propagation velocity, the angular frequency, and the instantaneous expression for the electric field intensity.

up =

 c 3x108 m = = = 0.75 x108  s r 16

 = u p  = ( 0.75 x108 ) (10 ) = 7.5 x108

rad s mA H( y, t ) = 100cos ( 7.5 x108 t − 10 y ) a z m − j y H s = 0.100e a z ,

Es = − a P  H s =

−120

r

a y  0.100e− j  y a z = −3 e− j  y a x

E( y, t ) = −9.4cos ( 7.5 x108 t − 10 y ) a x

V m

P5.12: Given E = 120 cos(xt – y) az V/m and H = 2.00cos(xt – y) ax A/m, find r and r.

Es = 120 e− j y a z , H s = 2e− j 0.080 y a x Hs =

1



a P  Es =

so we know

r =2 r

1



a y 120 e− j y a z =

120 120 r  r

e− j y a x =

 r − j y e ax r

5-7

1

Now, u p =

r  r =



c

=

r  r

=

 6 x106 = = 75 x106  0.080

c =4 75 x106

And now

r r  r = (2)(4) = 8 =  r r

r  r 4 = = 2 = r r  r 2

3. Propagation in Dielectrics P5.13: Work through the algebra to derive equation  and  equations (5.52) from equations (5.50) and (5.51).

 2 = − 2  + j = ( 2 −  2 ) + j 2 ; Comparing the imaginary parts, we see  = 2 , or  = and comparing the real parts,  2 −  2 +  2  = 0 .

Rearranging and inserting our value for :

 4 +  2  2 −

 2  2 2

=0 4 This is a quadratic expression (x2 + bx + c = 0), where here    x =  , b =   , c = −    2  Solving the quadratic: 2

x=

2

2

−b  b 2 − 4c 1 2 1 1  4c  = b − 4c − b = b  1 − 2 − 1 2 2 2 2  b 

Reinserting the a, b and c values:



1 2

 2 =  2   1 + 

2   4 2  2 2    2   − 1 =   1 + − 1    4 4  2 2        

2        =   1 +  − 1       

Now for : 2

    2  −  −   = 0,  = , so  2 −   −   = 0 2  2  2

2

2

Rearranging,

 , 2

5-8

    =0  2  Solving this quadratic we find 2

 4 −  2  2 −  

2     + 1      

 =    1 +   

P5.14: MATLAB: Write a routine to prompt the user for a material’s constitutive parameters and an operating frequency, and calculate the  and  from (5.52). Verify the program by running Drill 5.6. % MLP0514 % % Prompts user for material's constitutive % parameters and an operating frequency, then % calculates alpha(Np/m) and beta(rad/m). % % Wentworth, 1/24/03 % clc clear ur=input('relative permeability: '); erp=input('real part of rel permittivity: '); erdp=input('complex part of rel permittivity: '); s=input('conductivity (S/m): '); f=input('frequency (Hz): '); w=2*pi*f; uo=pi*4e-7; eo=8.854e-12; seff=s+w*erdp*eo; A=sqrt(1+(seff/(w*erp*eo))^2); B=ur*uo*erp*eo/2; alpha=w*sqrt(B*(A-1)) beta=w*sqrt(B*(A+1)) Now run the program for Drill 5.6: (a) relative permeability: 1 real part of rel permittivity: 10 complex part of rel permittivity: .01 conductivity (S/m): 1e-12 frequency (Hz): 100

5-9

alpha = 3.3730e-009

beta = 6.6268e-006 (b) relative permeability: 1 real part of rel permittivity: 10 complex part of rel permittivity: .01 conductivity (S/m): 1e-12 frequency (Hz): 1e6 alpha = 3.3134e-005

beta = 0.0663 These results agree with Drill 5.6.

P5.15: Given a material with  = 1.0x10-3 S/m, r = 1.0, and r’ = 3.0, r’’ = 0.015, compare a plot of  versus frequency from 1 Hz to 1 GHz using (5.52) to a similar plot using (5.54). At what frequency does the % error exceed 2%? % MLP0515 % % Compares alpha calculated using (5.52) to % that calculated using (5.54). % % Wentworth, 1/25/03 % clc clear % Initialize variables ur=1; erp=3; erdp=.015; s=1e-3; uo=pi*4e-7; eo=8.854e-12; B=ur*uo*erp*eo/2;

5-10 % Perform calculations for i=1:10 for j=1:10 m=(i-1)*10+j; f(m)=j*10^(i-1); w(m)=2*pi*f(m); seff(m)=s+w(m)*erdp*eo; A(m)=sqrt(1+(seff(m)/(w(m)*erp*eo))^2); alpha1(m)=w(m)*sqrt(B*(A(m)-1)); alpha2(m)=(seff(m)/2)*sqrt(ur*uo/(erp*eo)); diff(m)=abs(100*(alpha1(m)-alpha2(m))/alpha1(m)); C(m)=diff(m)<2; if diff(m)<2 if diff(m-1)>2 fdiff=f(m); Fstr=num2str(fdiff); end end end end % generate plot loglog(f,alpha1,'-o',f,alpha2,'-*') legend('(5.52)','(5.54)') xlabel('frequency (Hz)') ylabel('alpha(Np/m)') S=strcat('Error drops below 2% when frequency > ',Fstr); title(S) grid on

Fig. P5.15

5-11

P5.16: In a media with properties  = 0.00964 S/m , r = 1.0, r = 100., and f = 100.

MHz, a 1.0 mA/m amplitude magnetic field travels in the +x direction with its field vector in the z direction. Find the instantaneous form of the related electric field intensity.

 mA  − x − x − j  x H = 1  e cos (t −  x ) a z ; H s = H o e e a z  m  Es = −a P  H s = −a x  H o e− x e− j xa z =  H o e− x e− j xa y j 2 (100 x106 ) (100 ) ( 4 x10−7 ) j = = = 2664e j 30  6 −12  + j 0.00964 + j 2 (100 x10 )(8.854 x10 )

=

j ( + j ) = 14.8 + j 25.7

1 m

Finally,

E( x, t ) = 2.66e−15 x cos ( 200 x106 t − 26 x + 30 ) a y

V m

P5.17: MATLAB: Make a pair of plots similar to Figure 5.4 for the 3 materials of Table 5.1. Instead of loss tangent, one plot is to contain the magnitude of  and the other is to have the phase of . %ML P5.17 clc;clear %want to plot intrinsic impedance vs frequency for %the data listed in table 5.1 %Here, we'll plot the magnitude and phase of the %intrinsic impedance. %enter data from Table 5.1 sigC=5.8e7; %conductivity of copper in S/m sigS=4; % conductivity of seawater sigG=1e-12; % conductivity of glass er1C=1; %real part of rel perm for Copper er1S=72; %real part of rel perm for seawater er1G=10; %real part of rel perm for glass er2C=0; %imag part of rel perm for Copper er2S=12; %imag part of rel perm for seawater er2G=0.010; %imag part of rel perm for glass %enter constant values eo=8.854e-12; %free space permittivity, F/m uo=pi*4e-7; %free space permeability, H/m

5-12 %calculations n=2:.2:14; f=10.^n;w=2*pi*f; seffC=sigC+w*er2C*eo; seffS=sigS+w*er2S*eo; seffG=sigG+w*er2G*eo; etaC=sqrt(i*w*uo./(seffC+i*er1C*eo)) etaS=sqrt(i*w*uo./(seffS*er1S*eo)) etaG=sqrt(i*w*uo./(seffG+i*er1G*eo)) magC=abs(etaC); angC=180*angle(etaC)/pi; subplot(3,2,1) semilogx(f,magC) ylabel('mag, ohms') title('copper') subplot(3,2,2) semilogx(f,angC) ylabel('phase, deg') magS=abs(etaS); angS=180*angle(etaS)/pi; subplot(3,2,3) semilogx(f,magS) ylabel('mag, ohms') title('seawater') subplot(3,2,4) semilogx(f,angS) ylabel('phase, deg') magG=abs(etaG); angG=180*angle(etaG)/pi; subplot(3,2,5) semilogx(f,magG) ylabel('mag, ohms') xlabel('freq (Hz)') title('glass') subplot(3,2,6) semilogx(f,angG) xlabel('freq (Hz)') ylabel('phase, deg')

5-13

Fig. P5.17

4. Propagation in Conductors P5.18: Starting with (5.13), show that  =  for a good conductor.

=

j ( + j ) 

j

j=

1+ j ,  = (1 + j ) 2

j   = +j 2 2 2

= =

for a good conductor

 2

(Note: we get the same result starting with (5.52) and assuming 

  1.

P5.19: In seawater, a propagating electric field is given by E(z,t) = 20.e-z cos(xt – z + 0.5) ay V/m. Assuming ’’=0, find (a)  and , and (b) the instantaneous form of H. For seawater we have r = 72,  = 5, and r = 1.

5-14 So: jo = j7.896, j r o = j0.004

=

j = 1.257e j 44.98   + j

=

j ( + j ) = 4.441 + j 4.445 1 m

 =  = 4.4

1 m

V V = 20e − z e −  z e j 28.6 a y m m 1 1 − 20 A H s = a P  Es = a z  20e − z e −  z e j 28.6 a y = e − z e −  z e j 28.6 a x    m A H( z, t ) = −15.9e −4.4 z cos ( 2 x106 t − 4.4 z + 28.6 − 45 ) a x m or with appropriate significant digits: A H( z, t ) = −16e −4.4 z cos ( 2 x106 t − 4.4 z − 16 ) a x m Es = 20e − z e −  z e j 0.5radians a y

P5.20: Calculate the skin depth at 1.00 GHz for (a) copper, (b) silver, (c) gold, and (d) nickel.

=

1 ;as an example, for copper at 1 GHz:  f 

=

1 1 H 1  Vs A   1x109   4 x10−7   5.8 x107  s   Table P5.19 Cu Ag Au Ni

= 2.1x10−6 m = 2.1 m

m  HA V

m 

(S/m)

r

(m)

7

1 1 1 600

2.1 2.0 2.5 0.17

5.8x10 6.2x107 4.1x107 1.5x107

P5.21: For Nickel ( = 1.45 x 107, r = 600), make a table of , , , up, and  for 1Hz, 1kHz, 1MHz, and 1 GHz. For Ni we have  = 1.45x107S/m, r = 600

 =  =  f  =  f ( Hz )( 600 ) ( 4 x10−7 )(1.45x107 ) = 34.35x103 f ( Hz )

 = 1/

= 2

 j 45 e  = 18.08 x10−6 f ( Hz )e j 45  

5-15

up =

c

r r

= 12 x106

Table P5.21 f(Hz)= (Np/m) (rad/m)

 

up(m/s)

m s

103 5860 5860 570ej45º  170m 12x106

1 185 185 18ej45º 5.4mm 12x106

106 185x103 185x103 18ej45º m 5.3m 12x106

109 5.9x106 5.9x106 0.57ej45º  170nm 12x106

P5.22: A semi-infinite slab exists for z > 0 with  = 300 S/m, r = 10.2, and r = 1.0. At the surface (z = 0), E(0,t) = 1.0 cos( x 106t) ax V/m.

Find the instantaneous expressions for E and H anywhere in the slab.

The general expression for E is: E( z, t ) = 1.0e− z cos ( x106 t −  z ) a x

j = j ( x106 )( 4 x10−7 ) = j3.948

V m

j = j ( x106 ) (10.2 ) (8.854 x10−12 ) = j 284 x10−6 Here,    (i.e. it is a good conductor), so 1  =  f  = 24.3 =  m

= 2

 j 45 e = 0.115e j 45  

So now we have

E( z, t ) = 1.0e−24 z cos ( x106 t − 24 z ) a x

V m

To find B we’ll work in phasors.

Es = 1e− z e− j z a x , H s =

1



a P  Es =

1



a z 1e− z e− j  z a x =

1



e − z e − j  z a y

1 A e−24 z cos ( x106 t − 24 z − 45 ) a y 0.115 m A H( z, t ) = 8.7e−24 z cos ( x106 t − 24 z − 45 ) a y m H( z, t ) =

P5.23: In a nonmagnetic material, E(z,t) = 10.e-200z cos(2 x 109t - 200z) ax mV/m.

Find H(z,t).

5-16 Since  = , the media is a good metal. With r = 1 we have

( 200 ) 2 S  =  f o , or  = = = 10.13 9 −7  f o  (1x10 )( 4 x10 ) m 2

= 2

 j 45 e = 28e j 45  

E s = 10e− z e− j z a x , H s =

1



a P  Es =

1



a z 10e− z e− j  z a x =

H( z, t ) = 360e−200 z cos ( 2 x109 t − 200 z − 45 ) a y

10 − z − j  z e e ay



mA m

P5.24: A 0.1 m layer of copper is deposited atop a very thick slab of nickel. For a field incident on the copper surface, (a) calculate Rs at 1.0 GHz. Compare this with Rs at 1.0 GHz for (b) a semi-infinite slab of copper and (c) for a 0.1 m thickness of copper by itself.

Fig. P5.24 Refer to Figure P5.24..

In the copper portion the field is Ex = Exo e−Cu z

(

)

In the nickel portion, Ex = Exo e−Cut e− Ni ( z −t ) The current density in the copper is J xCu =  Cu Exoe−Cu z , and in the nickel is

(

)

J xNi =  Ni Exoe−Cut e− Ni ( z −t ) . The current is I =   Cu Exoe−Cu z dydz +   Ni Exoe−Cut e− Ni ( z −t ) dydz , or t

I = w Cu Exo  e o

− Cu z

dz + w Ni Exo e

−Cu t



e t

− Ni ( z −t )

dz , and upon evaluating

5-17

   I = wExo  Cu 1 − e−Cut + Ni e−Cut  , and with V=ExoL,  Ni   Cu 

(

)

−1

   L we have  R = Rs , where Rs =  Cu 1 − e−Cu t + Ni e −Cu t  . w  Ni   Cu  Now we’re ready to perform the calculations using the following data: S Np  Cu = 5.8 x107 , r = 1,  Cu = 479 x103 m m S Np  Ni = 1.5 x107 , r = 600,  Cu = 596 x106 m m (a) 0.1m Cu over Ni: Rs = 176 m (b) Semi-infinite Cu: Rs = 8.3 m (c) 0.1 m Cu: Rs = 177 m

(

)

P5.25: Calculate the DC resistance per meter length of a 4.0 mm diameter copper wire.

Now find the resistance at 1.0 GHz. DC:

R 1 1 1 1 m = = = 1.37 2 2 7 L  a m ( 5.8x10 )  ( 0.002 )

R R 1 = s = ;  = 1  f  = 2.09 x10−6 m L 2 a  2 a R 1  = = 0.66 7 −6 L ( 5.8 x10 )( 2.09 x10 ) 2 ( 0.002 ) m

1 GHz:

5. The Poynting Theorem and Power Transmission P5.26: In air, H(z,t) = 12.cos(x106t - z + /6) ax A/m. Determine the power density passing through a 1.0 square meter surface that is normal to the direction of propagation.

2

1 1 kW  A Pavg =  H xo 2a z = (120 ) 12  a z = 27 2 a z 2 2 m  m

P5.27: A 600 MHz uniform plane wave incident in the z direction on a thick slab of Teflon (r = 2.1, r = 1.0) imparts a 1.0 V/m amplitude y-polarized electric field intensity at the surface. Assuming  = 0 for Teflon, find in the Teflon (a) E(z,t), (b) H(z,t) and (c) Pav.

(

) Vm

E(0, t ) = 1cos 2 ( 600 x106 ) t −  z a y

5-18

E( z, t ) = 1e− z cos (t −  z ) a y Teflon:  = 0 so  = 0, and  =   =

 c

r =

V m

2 ( 600 x106 ) 8

3x10

2.1 = 18.2

rad m

(a) E( z, t ) = 1cos (1.2 x109 t − 18.2 z ) a y

V m 1 2.1 V (b) H s = a P  Es = a z 1e− j z a y ,  120 m mA H( z, t ) = −3.8cos (1.2 x109 t − 18.2 z ) a x m 2 1 (1) 2.1 mW a z = 1.9 2 a z (c) Pavg = 2 120 m P5.28: Assume distilled water ( = 10-4 S/m, r = 81, r = 1.0) fills the region z > 0. At the surface, we have E(0,t) = 8.0cos(2x108t) ax V/m. Determine, for z > 0, (a) E(z,t), (b) H(z,t), and (c) Pav at z = 1.0 m. (d) Find the power passing through a 10 square meter surface located at z = 1.0 m. (a) The general expression for E is: E( z, t ) = Eo e− z cos (t −  z +  ) a x and we can see from the given information that V rad Eo = 8 ,  = 2 x108 , f = 108 Hz,  = 0 . Also m s

 = ( 2 x108 ) (81) (8.854 x10−12 ) = 0.45,  = 10−4 , so

  1 (low loss dielectric). 



 10−4 1 Np = (120 ) = 0.0021 2  2 81 m  rad  =   =  r = 18.8

=

c

=

m

 1 = 120 = 41.9  81

so E( z, t ) = 8e−0.0021z cos ( 2 x108 t − 18.8 z ) a x

V m

(b)

Es = 8e−0.0021z e− j18.8 z a x Hs =

1



a P  Es =

V , m

V , m

8 −0.0021z − j18.8 z mA e e a y = 191e−0.0021z e − j18.8 z a y 41.9 m

5-19 so H( z, t ) = 191e−0.0021z cos ( 2 x108 t − 18.8 z ) a y

mA m

1 Exo 2 −2 z W e a z = 0.764e−2(0.0021)(1)a z = 0.761 2 2  m 2 (d) P = Pavg (10m ) = 7.6W (c) Pavg =

P5.29: The density of solar radiation is approximately 150 W/m2 at some locations on the earth’s surface. How much solar power is incident on a typical “100 Watt” solar panel (.6 m x 1.6 m area) if the panel is normal to the radiation propagation direction? How much power is incident if the panel is tilted 45 to the radiation propagation direction?

P = Pavg S = 144W , P = Pavg S cos 45 = 102W

P5.30: A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0 mV/m amplitude at the surface. How much power passes through a square meter at the surface? How much power passes through a square meter area 10. m beneath the surface? f = 200MHz , Eo = 1

Cu:  = 2

mV 1 Eo 2 , Pavg = m 2 

 j 45 Np e ,  =  f  = 214 x103 , so  = 5.22e j 45 m  m

−3 1 (10 ) W Pavg = = 96 2 ; P = Pavg S = 96W −3 2 5.22 x10 m Now at 10 m beneath the surface, we have 2

E ( z = 10 m) = Eo e− (10  m ) = 10−3 e− (214 x10 )(10  m) = 118x10−6 3

1 (118 x10 ) W Pavg = = 1.3 2 ; P = 1.3W −3 2 5.22 x10 m

V m

−6 2

6. Wave Polarization P5.31: Suppose E(z,t) = 10.cos(t-z)ax + 5.0cos(t-z)ay V/m. What is the wave polarization and tilt angle?

The figure indicates linear polarization.

The tilt angle is: 5  = tan −1   = 27  10 

5-20

Fig. P5.31

Fig. P5.32 P5.32: Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ay V/m, find the polarization and handedness.

The field can be rewritten as E(z,t) = 10.cos(t-z)ax + 20.cos(t-z-45-180°)ay or E(z,t) = 10.cos(t-z)ax + 20.cos(t-z+135°)ay Running ML0503: Polarization Plot enter x-amplitude: 10 enter x-phase angle (degrees): 0 enter y-amplitude: 20 enter y-phase angle (degrees): 135 To determine direction of polarization, move from the o to + along the plot. >> From the figure, we have left-hand elliptical polarization.

P5.33: Given H(z,t) = 2.0cos(t-z)ax + 6.0cos(t-z-120)ay A/m, find the polarization and handedness.

Convert to E(z,t):

5-21

(

)

Es = −a P  H s = −oa z  2e− j z a x + 6e− j z e− j120 a y = −2o e− j z a y + 6o e− j z e− j120 a x

(

E ( z, t ) = o 6cos (t −  z − 120 ) a x + 2cos (t −  z − 180 ) a y

)

With this we can run ML0503: Polarization Plot enter x-amplitude: 6 enter x-phase angle (degrees): -120 enter y-amplitude: 2 enter y-phase angle (degrees): -180 To determine direction of polarization, move from the o to + along the plot. >> From the figure, we have right-hand elliptical polarization.

Fig. P5.33

P5.34: Given

E( z, t ) = E xo cos ( t −  z ) ax + E yo cos ( t −  z +  ) a y , we say that Ey leads Ex for 0 <  < 180, and that Ey lags Ex when –180 <  < 0.

Determine the handedness for each of these two cases.

For 0 <  < 180°, we have LHP For 180° <  < 360°, we have RHP

5-22 P5.35: MATLAB: For a general elliptical polarization represented by E( z, t ) = E xo cos ( t −  z ) ax + E yo cos ( t −  z +  ) a y , the axial ratio and tilt angle can be found from the following formulas (from K. R. Demarest, Engineering Electromagnetics, Prentice-Hall, 1998, pp. 451-453): a=|Exo|, b=|Eyo| MAJ = length of majority-axis MIN = length of minority-axis 1 MAJ = 2  a 2 + b2 + a 4 + b4 + 2a 2b2 cos 2   2

1 2 a + b2 − a 4 + b4 + 2a 2b2 cos 2    2 axial ratio=MAJ/MIN 1  2ab   = tan −1  2 2 cos   .

2 a − b  Compose a program that not only draws a polarization plot like MATLAB 5.3, but that also calculates the axial ratio and tilt angle. Run the program on Drill 5.11.

MIN = 2

% M-File: MLP0535 % % This program modifies ML0503. As before, it will % trace polarization ellipses, given the amplitude % and phase of a pair of linearly polarized waves. % Now it will also calculate axial ratio and tilt %angle. % % Wentworth 1/28/03 % Variables: % Exo,Eyo amplitudes for the pair of waves % fxd,fyd phase angle for each wave % fx,fy phase (radians) for each wave % wtd ang freq * time, in degrees % wtr ang freq * time, in radians % x,y superposed position % x0,y0 position at wtd=0 degrees % x45,y45 position at wtd=45 degrees % a,b shorthand for Exo,Eyo % MAJ,MIN majority,minority axis length % AR,tiltangle axial ration, tilt angle % clc %clears the command window clear %clears variables % Prompt for input values disp('Polarization Plot') disp(' ') Exo=input('enter x-amplitude:

');

5-23 fxd=input('enter x-phase angle (degrees): '); fx=fxd*pi/180; Eyo=input('enter y-amplitude: '); fyd=input('enter y-phase angle (degrees): '); fy=fyd*pi/180; disp(' ') disp('To determine direction of polarization,') disp('move from the o to + along the plot.') disp(' ') %Perform calculations wtd=0:360; %wt in degrees wtr=wtd*pi/180; x=Exo*cos(wtr+fx); y=Eyo*cos(wtr+fy); x0=Exo*cos(fx); y0=Eyo*cos(fy); x45=Exo*cos(fx+pi/4); y45=Eyo*cos(fy+pi/4); fdiff=fy-fx; a=abs(Exo);b=abs(Eyo); temp=sqrt(a^4+b^4+2*a^2*b^2*cos(2*fdiff)); MAJ=2*sqrt(0.5*(a^2+b^2+temp)); MIN=2*sqrt(0.5*(a^2+b^2-temp)); AR=MAJ/MIN temp2=(2*a*b/(a^2-b^2))*cos(fdiff); tiltangle=(0.5*atan(temp2)*180/pi) %Make the plot plot(x,y,x0,y0,'ok',x45,y45,'+k') xlabel('x') ylabel('y') title('Polarization Plot') axis('equal') Now we run the program for Drill 5.11.

Polarization Plot enter x-amplitude: 3 enter x-phase angle (degrees): -30 enter y-amplitude: 8 enter y-phase angle (degrees): 90 To determine direction of polarization,

Fig. P5.35

5-24 move from the o to + along the plot.

AR = 3.1997 tiltangle = 11.7874 7. Reflection and Transmission at Normal Incidence P5.36: Starting with (5.107) and (5.109), derive (5.110) and (5.111).

(1) Eoi + Eor = Eot  (2) Eoi − Eor = 1 Eot 2 Add (1) and (2): 2 Eoi = Eot +

1 t  1  t 22 Eo = 1 +  Eo , so Eot = Eoi 2 1 + 2  2 

Now subtract (2) from (1):       22  − 2 Eor = 1 − 1  Eot = 1 − 1  Eoi , Eor = 2 1 Eoi 2 + 1  2   2  1 + 2 P5.37: A UPW is normally incident from media 1 (z < 0,  = 0, r = 1.0, r = 4.0) to media 2 (z > 0,  = 0, r = 8.0, r = 2.0). Calculate the reflection and transmission coefficients seen by this wave.

2 − 1  120  8 ; 1 = = = 60, 2 = = 120 = 240 2 + 1   2 4 240 − 60 3 = = = 0.60 240 + 60 5  = 1 +  = 1.60 =

P5.38: Suppose media 1 (z < 0) is air and media 2 (z > 0) has r = 16. The transmitted magnetic field intensity is known to be Ht = 12 cos (t-2z)ay mA/m. (a) Determine the instantaneous value of the incident electric field. (b) Find the reflected average power density.

Hts = 12e− j2 z a y

mA Eot − j2 z mA = e ay m 2 m Eot

mA t V V , E o = 0.36 , and  Est = 1.13e− j2 z a x 2 m m m  − 3 2 Eot =  Eoi = (1 +  ) Eoi ;  = 2 1 = − ,  = 1 +  = 2 + 1 5 5

2 = 30, so

= 12

5-25

E = i o

Eot



= 2.83, so  Eis = 2.83e− j1z a x

V . m Eor = Eoi = −1.70, so Ers = −1.70e+ j1z a x 1 1 mA H rs = a P  Ers = ( −a z )  ( −1.70e+ j1za x ) = 4.5e+ j1za y  120 m 1 mW r Pavg = (1.70 ) ( 4.5 x10−3 ) ( −a z ) = 3.8 2 ( -a z ) 2 m

 E( z, t ) = 2.83cos (t − 1 z ) a x

P5.39: Suppose a UPW in air carrying an average power density of 100 mW/m 2 is normally incident on a nonmagnetic material with r = 11. What is the time-averaged power density of the reflected and transmitted waves?

 1  −1   120 11   1 = o = 120; 2 = ; = = −0.537  1  +1 11   11    = 1 +  = 0.463 mW 2 i r Pavg =  Pavg = 28.8 2 m 2 1 Exo mW t i Pavg = =  2 Pavg 11 = 71.2 2 2 2 m P5.40: A UPW in a lossless nonmagnetic r = 16 media (for z < 0) is given by E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m.

This is incident on a lossless media characterized by r = 12, r = 6.0 (for z > 0). Find the instantaneous expressions for the reflected and transmitted electric field intensities.

Eis = 10e− j1z a x + 20e− j1z e j 3a y

Ers = 10e j1z a x + 20e j 1z e j 3a y 120 12 = 30; 2 = 120 = 120 2 6 16  −  = 2 1 = 0.700;  = 1 +  = 1.70 2 + 1

1 =

Ers = 7e j1z a x + 14e j 1z e j 3a y

 V  E(rz ,t ) = 7 cos (t + 1 z ) a x + 14cos  t + 1 z +  a y 3 m 

5-26

Ets = 10 e− j2 z a x + 20 e− j 2 z e j 3a y , or Ets = 17e− j2 z a x + 34e− j2 z e j 3a y , so

 V  .

Et( z ,t ) = 17 cos (t −  2 z ) a x + 34cos  t −  2 z +  a y 3 m  P5.41: The wave Ei = 100 cos( x 106t - 1z + /4) ax V/m is incident from air onto a perfect conductor. Find Er and Et. For the perfect conductor, 2 = 0. So  = -1 and Er = -100 cos( x 106t + 1z + /4) ax V/m Et = 0 P5.42: A UPW given by E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m is incident from air (for z < 0) onto a perfect conductor (for z > 0). Find the instantaneous expression for the reflected electric field intensity and the SWR. As in the previous problem,  = -1. We then have E(z,t) = -10.cos(t+1z)ax - 20.cos(t+1z+/3)ay V/m SWR =

1+  = 1− 

P5.43: The wave Ei = 10.cos(2 x 108t - 1z) ax V/m is incident from air onto a copper conductor. Find Er, Et and the time-averaged power density transmitted at the surface.

For copper we have

2 = 2

 2 j 45 e  2

where  2 =  f 2 2 =  (108 )( 4 x10−7 )( 5.8 x107 ) = 151x103

Np = 2 m

so 2 = 3.7e j 45 m

22 2  2 = 19.6 x10−6 e j 45 1 + 2 1 r 8 So E = -10.cos(2 x 10 t + 1z) ax V/m V V Ets = (196 ) e−2 z e− j2 z e j 45 a x , and Et = 196e−2 z cos (t −  2 z + 45 ) a x m m We find   −1, and  =

1 (196 x10 V m ) W = cos ( 45 ) a z = 3.7 2 a z . −3 2 ( 3.7 x10  ) m −6

t avg

P

2

5-27

P5.44: Given a UPW incident from medium 1 ( = 0, r = 1.0, r = 25.) to medium 2 ( = 0.0080, r = 1.0, r = 81.), calculate  SWR and  at 1 kHz, 1 MHz, and 1 GHz.

120 j j 7.896 x10−6 f ( Hz ) 1 = = 24; 2 = =  + j 0.008 + j 4.506 x10−9 f ( Hz ) 25 Table P5.44 f 1kHz 1MHz 1GHz

2() 0.994ej44.98° 29.3ej30.3° 41.9ej0.05°

=

2 − 1 2 + 1

SWR =

0.9815ej178.9° 0.513ej155° 0.286ej179.9°

1+  1− 

107.3 3.11 1.80

P5.45: MATLAB: Write a program that prompts the user for the constitutive parameters in medium 1 and medium 2 separated by a planar surface. You are to assume a wave is normally incident from media 1 to media 2. The program is to plot  and  versus a frequency range supplied by the user. Use this program to plot  and  from 100 Hz to 10 GHz for the pair of media specified in the previous problem. %ML P0545 clear clc %prompt user for constit parameters of media 1 & 2 %then plot ref & trans coeff over a freq range. %We'll plot mag and angle of each. %enter constant values eo=8.854e-12; %free space permittivity, F/m uo=pi*4e-7; %free space permeability, H/m %enter media 1 values er1=input('enter er1: '); ur1=input('enter ur1: '); s1=input('enter s1: '); %enter media 2 values er2=input('enter er2: '); ur2=input('enter ur2: '); s2=input('enter s2: '); %calculations n=2:.5:10; f=10.^n;

5-28 w=2*pi.*f; eta1=sqrt(i*w*ur1*uo./(s1+i*w*er1*eo)); eta2=sqrt(i*w*ur2*uo./(s2+i*w*er2*eo)); Gamma=(eta2-eta1)./(eta2+eta1); Gmag=abs(Gamma); Gang=180*angle(Gamma)/pi; Tau=1+Gamma; Tmag=abs(Tau); Tang=180*angle(Tau)/pi; subplot(2,1,1) semilogx(f,Gmag,'-o',f,Tmag,'-*') xlabel('frequency (Hz)') ylabel('magnitude') legend('reflection','transmission') subplot(2,1,2) semilogx(f,Gang,'-o',f,Tang,'-*') xlabel('frequency (Hz)') ylabel('phase angle (degrees)') legend('reflection','transmission') Run the program: enter er1: 25 enter ur1: 1 enter s1: 0 enter er2: 81 enter ur2: 1 enter s2: .008 >>

Fig. P5.45

P5.46: A wave specified by Ei = 100.cos(x107t-1z)ax V/m is incident from air (at z < 0) to a nonmagnetic media (z > 0,  = 0.050 S/m, r = 9.0). Find Er, Et and SWR. Also find the average power densities for the incident, reflected and transmitted waves.

5-29

rad 2  rad so f = 5 x106 Hz, 1 = = = 0.105 s 1 c m In this problem we find in medium 2 (z > 0) that  = 0.0025 and  = 0.05. These values are too close to allow for simplifying assumptions. Using (5.13) and (5.31), we calculate: Np rad  2 = 0.969 ,  2 = 1.019 , 2 = 28.1e j 43.6  . m m Then, 1+   −  = 2 1 = 0.898e j174 , SWR = = 18.6,  = 1 +  = 0.141e j 40.8 2 + 1 1− 

1 = 120,  =  x107

V m V V Ers = 100e+ j1z a x = 89.8e+ j1z e j174 a x , m m Eis = 100e− j1z a x

so Er ( z, t ) = 89.8cos ( x107 t + 0.105 z + 174 ) a x

Ets = 100 e− j2 z a x

V . m

V V = 14.1e− j2 z e j 40.8 a x , m m

so Et ( z, t ) = 14.1cos ( x107 t − 1.02 z + 40.8 ) a x

V . m

14.1 − j 43.6 − j2 z j 40.8 A A e e e a y = 0.502e− j2 z e− j 2.8 a y 28.1 m m 1 W Pt = (14.1)( 0.502 ) cos ( 40.8 + 2.8 ) a z = 2.6 2 a z 2 m 2 (100 ) = 13.3 W a Pi = z 2 (120 ) m2 Hts =

− ( 89.8 ) W P = = 10.7 2 ( -a z ) 2 (120 ) m 2 (check: 13.3 W/m = 10.7 W/m2 + 2.6 W/m2) 2

r

P5.47: A wave specified by Ei = 12 cos(2x107t-1z+/4)ax V/m is incident from a nonmagnetic, lossless, r = 9.0 media (at z < 0) to a media (z > 0) with  = 0.020 S/m, r = 2.0, and r = 16.). Find Hi, Er, Hr, Et, Ht, and the average power densities for the incident, reflected and transmitted waves.

We use ML0501 in each media to find: rad 1 = 0; 1 = 0.628 ; 1 = 40 m Np rad  2 = 1.01 ;  2 = 1.56 ; 2 = 84.9e j 33  m m We also will need reflection and transmission coefficients:

5-30

=

2 −1 = 0.353e j126 ;  = 1 +  = 0.84e j19.8 2 + 1

Incident:

V m 12 − j1z j 4 A A His = e e a y = 0.300e− j1z e j 4a y , 1 m m  A  Hi ( z, t ) = 0.300cos  2 x107 t − 0.628 z +  a y .

4 m 

Eis = 12 e− j1z e j 4a x

1 (12 ) W P = a z = 5.655 2 a z 2 1 m Reflected: V V V Ers =  (12 ) e+ j1z e j 4a x = 13.3e+ j1z e j 45 e j126 a x = 13.3e+ j 1z e j171 a x m m m V Er ( z, t ) = 13.3cos ( 2 x107 t + 0.628 z + 171 ) a x . m − 13.3 A A H rs = e+ j1z e j171 a y = 0.106e+ j 1z e j171 a y 40 m m A H r ( z, t ) = −0.106cos ( 2 x107 t + 0.628 z + 171 ) a y . m 2 1 (13.3) W r Pavg =− a z = 0.704 2 ( -a z ) 2 40 m Transmitted: V V Ets =  (12 ) e− j1z e j 4a x = 31.67e− j2 z e j 64.8 a x , m m V Et ( z, t ) = 31.7 cos ( 2 x107 t − 1.56 z + 64.8 ) a x . m 31.67 − j 33 − j2 z j 64.8 A A Hts = e e e a y = 0.373e− j2 z e j 31.8 a y , 84.9 m m A Ht ( z, t ) = 0.373cos ( 2 x107 t − 1.56 z + 31.8 ) a y . m 31.67 0.373 ( )( ) cos 64.8 − 31.8 a = 4.954 W a t Pavg = ( ) z z 2 m2 2

i avg

(Check: 5.655W/m2 = 0.704W/m2 + 4.954W/m2)

8. Reflection and Transmission at Oblique Incidence P5.48: A 100 MHz TE polarized wave with amplitude 1.0 V/m is obliquely incident from air (z < 0) onto a slab of lossless, nonmagnetic material with r = 25 (z > 0). The angle of

5-31 incidence is 40. Calculate (a) the angle of transmission, (b) the reflection and transmission coefficients, and (c) the incident, reflected and transmitted fields. (a) 1 =

 c 1

=

2 (100 x106 ) 8

3x10

= 2.09

 r rad rad , 2 = = 10.45 . m c m

1 1 1 = = ; sin t = sin 40 ; t = 7.4 2 r2 5 5 (b) Now we need to calculate the reflection and transmission coefficients.

120 1 = 120; 2 =  = 24 25  cos i −1 cos t TE = 2 = −0.732;  TE = 1 + TE = 0.268 2 cos i + 1 cos t (c) The fields, Incident: − j 2.09( x sin 40 + z cos 40 ) V Eis = 1e a y = 1e− j1.34 x e− j1.60 z a y m V Ei ( z, t ) = 1cos (t − 1.34 x − 1.60 z ) a y m 1 His = e− j1.34 x e− j1.60 z ( − cos 40 a x + sin 40 a z ) 120 mA His = ( −2.03a x + 1.71a z ) e− j1.34 x e− j1.60 z m mA Hi ( z, t ) = ( −2.03a x + 1.71a z ) cos (t − 1.34 x − 1.60 z ) m Reflected: Eor = TE Eoi = −0.732 V Ers = ( −0.732 ) e− j1.34 x e+ j1.60 z a y m V Er ( z, t ) = −0.732cos (t − 1.34 x + 1.60 z ) a y m − 0.732 ( ) e− j1.34 x e+ j1.60 z cos 40 a + sin 40 a A H rs = ( x z) 120 m mA H rs = ( −1.49a x − 1.25a z ) e− j1.34 x e+ j1.60 z m mA H r ( z, t ) = ( −1.49a x − 1.25a z ) cos (t − 1.34 x + 1.60 z ) m transmitted: Eot =  TE Eoi = 0.268 V − j  x sin  + z cost ) Ets = 0.268e 2 ( t a y = 0.268e− j1.35 x e− j10.4 z a y m

5-32

0.268 − j1.35 x − j10.4 z A e e − cos 7.4 a x + sin 7.4 a z ) ( 24 m mA Hts = ( −3.5a x + 0.46a z ) e− j1.35 x e− j10.4 z m mA Ht ( z, t ) = ( −3.5a x + 0.46a z ) cos (t − 1.35 x − 10.4 z ) m Hts =

P5.49: A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incident from air (z < 0) onto a slab of lossless, nonmagnetic material with r = 25 (z > 0). The angle of incidence is 40. Calculate (a) the angle of transmission, (b) the reflection and transmission coefficients, and (c) the incident, reflected and transmitted fields. (a) The material parameters in this problem are the same as for P5.48. So, once again we have t = 7.4°. Also, 1 = 2.09 rad/m and 2 = 10.45 rad/m. (b)  cos t −1 cos i TM = 2 = −0.589 2 cos t + 1 cos i 22 cos i  TM = = 0.318 2 cos t + 1 cos i (c) Incident: Eis = 1e− j1.34 x e− j1.60 z ( cos 40 a x − sin 40 a z )

Ei ( z, t ) = ( 0.766a x − 0.643a z ) cos (t − 1.34 x − 1.60 z )

His =

V m

1 − j1.34 x − j1.60 z A e e ay 120 m

Hi ( z, t ) = 2.65cos (t − 1.34 x − 1.60 z ) a y

mA m

Reflected: Ers = −0.589e− j1.34 x e+ j1.60 z ( cos 40 a x + sin 40 a z )

Er ( z, t ) = ( −0.452a x − 0.379a z ) cos (t − 1.34 x + 1.60 z ) H rs =

−0.589 − j1.34 x + j1.60 z A e e ay 120 m

H r ( z, t ) = −1.56cos (t − 1.34 x + 1.60 z ) a y

V m

mA m

transmitted: Ets = 0.318e− j1.35 x e− j10.4 z ( cos 7.4 a x − sin 7.4 a z )

Et ( z, t ) = ( 0.315a x − 0.041a z ) cos (t − 1.35 x − 10.4 z )

V m

5-33

Ht ( z, t ) = 4.22cos (t − 1.35 x − 10.4 z ) a y

mA m

P5.50: A randomly polarized UPW at 200 MHz is incident at the Brewster’s angle from air (z < 0) onto a thick slab of lossless, nonmagnetic material with r = 16 (z > 0). The wave can be decomposed into equal TE and TM parts, each with an incident electric field amplitude of 10. V/m. Find expressions for the instantaneous value of the incident, reflected and transmitted electric fields.

First we calculate the Brewster’s angle: sin  BA =

1 ;  BA = 76 1 + 1 16

Also, we calculate 1 = 4.19 rad/m, 2 = 16.8 rad/m, 1 = 120 , and 2 = 30  .

TE

V m At the Brewster’s angle of incidence, we have from Snell’s Law:   t = sin −1  1 sin i  = 14  2   cos i −1 cos t TE = 2 = −0.883;  TE = 1 + TE = 0.117 2 cos i + 1 cos t V V Eor = TE Eoi =-8.83 ; Eot =  TE Eoi =1.17 m m Eis = 10e

− j 1 ( x sin i + z cosi )

a y = 10e− j 4.06 x e− j1.01z a y

Ers = −8.83e− j 4.06 x e+ j1.01z a y

V m

Ets =  TE (10 ) e

a y = 1.17e− j 4.06 x e− j16.3 z a y

− j  2 ( x sin t + z cost )

TM: At the Brewster’s angle, TM = 0 and Ers = 0.

Eis = 10e− j 4.06 x e− j1.01z ( cos i a x − sin i a z ) ,

V . m Ets = 10e− j 4.06 x e− j16.3 z ( cos t a x − sin t a z ) , Eis = ( 2.42a x − 9.70a z ) e− j 4.06 x e− j1.01z

Ets = ( 9.70a x − 2.40a z ) e− j 4.06 x e− j16.3 z Combining the results we arrive at:

V . m

V m

5-34

Ei ( z, t ) = ( 2.4a x + 10a y − 9.7a z ) cos (t − 4.06 x − 1.01z ) Er ( z, t ) = −8.83cos (t − 4.06 x + 1.01z ) a y

V m

V m

Et ( z, t ) = ( 9.7a x + 1.2a y − 2.4a z ) cos (t − 4.06 x − 16.3z )

V m

6-1 Solutions for Chapter 6 Problems 1. Distributed Parameters Model P6.1: RG-223/U coax has an inner conductor radius a = 0.47 mm and inner radius of the outer conductor b = 1.435 mm. The conductor is copper, and polyethylene is the dielectric. Calculate the distributed parameters at 800 MHz.

for copper:  Cu = 5.8 x107

S m

for polyethylene:  r = 2.26,  = 10−16 R' =

S m

1 1 1  f  +  2  a b   c

6 −7 1  1 1    ( 800 x10 )( 4 x10 ) = + = 3.32  −3 −3  7 2  0.47 x10 1.435 x10  m ( 5.8x10 )

L' = G' = C'=

  b  4 x10−7  1.435  nH ln   = ln   = 223 2  a  2 m  0.47  2 (10−16 ) 2 S −18 ln ( b a )

=

ln (1.435 0.47 )

= 560 x10

2 ( 2.26 ) (8.854 x10 2 = ln ( b a ) ln (1.435 0.47 )

−12

m0

) = 112 pF m

P6.2: MATLAB: Modify MATLAB 6.1 to account for a magnetic conductive material.

Apply this program to problem P6.1 if the copper conductor is replaced with nickel.

S and r = 600. m Note that this program has also been modified for P6.04 as well. for Nickel we have  Ni = 1.5 x107

%Coax distributed parameters % % Modified: P0602 % add rel permeability % also modified for P0604 % clear clc disp('Calc Coax Distributed Parameters')

6-2 %Some constant values muo=pi*4e-7; eo=1e-9/(36*pi); %Prompt for input values a=input('inner radius, in mm, = '); b=input('outer radius, in mm, = '); er=input('relative permittivity, er= '); sigd=input('dielectric conductivity, in S/m, = '); sigc=input('conductor conductivity, in S/m, = '); ur=input('conductor rel. permeability, = '); f=input('input frequency, in Hz, = '); %Perform calulations G=2*pi*sigd/log(b/a); C=2*pi*er*eo/log(b/a); L=muo*log(b/a)/(2*pi); Rs=sqrt(pi*f*ur*muo/sigc); R=(1000*((1/a)+(1/b))*Rs)/(2*pi); omega=2*pi*f; RL=R+i*omega*L; GC=G+i*omega*C; Gamma=sqrt(RL*GC); Zo=sqrt(RL/GC); alpha=real(Gamma); beta=imag(Gamma); loss=exp(-2*alpha*1); lossdb=-10*log10(loss); %Display results disp(['G/h = ' num2str(G) ' S/m']) disp(['C/h = ' num2str(C) ' F/m']) disp(['L/h = ' num2str(L) ' H/m']) disp(['R/h = ' num2str(R) ' ohm/m']) disp(['Gamma= ' num2str(Gamma) ' /m']) disp(['alpha= ' num2str(alpha) 'Np/m']) disp(['beta= ' num2str(beta) 'rad/m']) disp(['Zo = ' num2str(Zo) ' ohms']) disp(['loss=' num2str(loss) ' /m']) disp(['lossdb=' num2str(lossdb) ' dB/m']) Now run the program for Nickel: Calc Coax Distributed Parameters

6-3 inner radius, in mm, = 0.47 outer radius, in mm, = 1.435 relative permittivity, er= 2.26 dielectric conductivity, in S/m, = 1e-16 conductor conductivity, in S/m, = 1.5e7 conductor rel. permeability, = 600 input frequency, in Hz, = 800e6 G/h = 5.6291e-016 S/m C/h = 1.1249e-010 F/m L/h = 2.2324e-007 H/m R/h = 159.7792 ohm/m Gamma= 1.78881+25.252i /m alpha= 1.7888Np/m beta= 25.252rad/m Zo = 44.6608-3.1637i ohms loss=0.027942 /m lossdb=15.5374 dB/m >> Summarizing the distributed parameter data from this routine we have: R ' = 160  , L ' = 223 nH , G ' = 560 x10−18 S , C ' = 112 pF m m m m

P6.3: Modify (6.3) to include internal inductance of the conductors. To simplify the calculation, assume current is evenly distributed across the conductors. Find the new value of L’ for the coax of Drill 6.1.

From Ampere’s Circuit Law we can find H versus :

I for   a 2 a 2 I H = for a    b 2

H =

c2 −  2 for b    c 2 c 2 − b 2 H = 0 for   c H =

I

Using the energy approach, Wm =

1 2 o LI = H 2 dv , we find  2 2

 o b  o  o  c 2   c   c 2  1  c 2 + b 2   L' = ln + +   ln   −  +   2 a 8 2  c 2 − b 2   b   c 2 − b 2  4  c 2 − b 2     2

Inserting the given values we find

6-4

nH nH = 328 m m With two significant digits we therefore have L’ = 330 nH/m.

L ' = ( 237 + 50 + 41.2 )

2. Time Harmonic Waves on Transmission Line P6.4: MATLAB: Modify MATLAB 6.1 to also calculate    and Zo. Confirm the program using Drill 6.2.

See the solution for P6.2.

Calc Coax Distributed Parameters inner radius, in mm, = 0.45 outer radius, in mm, = 1.47 relative permittivity, er= 2.26 dielectric conductivity, in S/m, = 1e-16 conductor conductivity, in S/m, = 5.8e7 conductor rel. permeability, = 1 input frequency, in Hz, = 1e9 G/h = 5.3078e-016 S/m C/h = 1.0606e-010 F/m L/h = 2.3675e-007 H/m R/h = 3.8112 ohm/m Gamma= 0.0403332+31.4857i /m alpha= 0.040333Np/m beta= 31.4857rad/m Zo = 47.246-0.0605221i ohms loss=0.9225 /m lossdb=0.35033 dB/m >> This agrees with the results of Drill 6.2.

P6.5: The impedance and propagation constant at 100 MHz for a T-Line are determined to be Zo = 18.6 – j0.253  and  = 0.0638 + j4.68 /m. Calculate the distributed parameters.

R '+ j L ' ,  = ( R '+ j L ' )( G '+ jC ' ) G '+ jC ' Z o = R '+ j L ' = 2.37 + j87.0 Zo =

 R ' = 2.37

 nH ,  L ' = 87.0 so L ' = 139 m m

6-5

 Zo

= G '+ jC ' = 7.63x10−6 + j 0.252,

 G ' = 7.63

S m

, and C ' = 0.252 so C ' = 401

pF m

P6.6: The specifications for RG-214 coaxial cable are as follows: • 2.21 mm diameter copper inner conductor • 7.24 mm inner diameter of outer conductor • 9.14 mm outer diameter of outer conductor • Teflon dielectric (r = 2.10) Calculate the characteristic impedance and the propagation velocity for this cable.

60 b  3.62  ln   = ln   = 49.1 r  a  2.1  1.105  c m up = = 2.07 x108 s r Zo =

60

P6.7: For the RG-214 coax of problem P6.6 operating at 1 GHz, how long is this T-line in terms of wavelengths if its physical length is 50 cm?

2.07 x108 up =  f ,  = = = 0.207m f 1x109 up

   1m    = 2.4  0.207m  100cm 

l ( ) = ( 50cm ) 

P6.8: If 1 watt of power is inserted into a coaxial cable, and 1 microwatt of power is measured 100 m down the line, what is the line’s attenuation in dB/m?

 1W  A = −10 log   = +60dB  1W  60dB dB A' = = 0.6 100m m P6.9: Starting with a 1 mm diameter solid copper wire, you are to design a 75  coaxial T-Line using mica as the dielectric. Determine (a) the inner diameter of the outer copper conductor, (b) the propagation velocity on the line and (c) the approximate attenuation, in dB/m, at 1 MHz.

6-6

((

) )

((

b ln   , b=a exp Zo  r 60 = ( 0.5mm ) exp 75 5.4 r  a  So the inner diameter of the outer conductor is 18 mm. c 2.998 x108 m m up = = = 1.29 x108 , so u p = 1.3x108 s s r 5.4 Zo =

60

) 60) = 9.1mm

To calculate , will need . Therefore we calculate R’, L’, G’ and C’.

6 −7 1  1 1 m   (1x10 )( 4 x10 ) R' = + = 87.6  −3 −3  7 2  0.5 x10 9.1x10  5.8 x10 m

L' = G' =

C'=

4 x10−7  9.1  nH ln   = 580 2 m  0.5  2 (10−15 )

ln ( 9.1 0.5 )

= 2.17 x10−15

2 ( 5.4 ) (8.854 x10−12 )

( 0.5)

ln 9.1

S m

= 103.5

pF m

Now, with  = 2f,

1 m Np 8.686 dB dB   Finally,  =  585 x10−6 = 5.1x10−3  m  Np m  This is confirmed using MLP0602.

=

( R '+ j L ')( G '+ jC ') = 585x10−6 + j 0.049

P6.10: MATLAB: A coaxial cable has a solid copper inner conductor of radius a = 1mm and a copper outer conductor of inner radius b. The outer conductor is much thicker than a skin depth. The dielectric has r = 2.26 and eff = 0.0002 at 1 GHz. Letting the ratio b/a vary from 1.5 to 10, generate a plot of the attenuation (in dB/m) versus the line impedance. Use the lossless assumption to calculate impedance. % MLP0610 % % Plot of alpha vs Zo for a particular coax clear clc %Some constant values muo=pi*4e-7; eo=8.854e-12; a=1; er=2.26; sigd=0.0002; sigc=5.8e7;

6-7 f=1e9; %Perform calulations b=1.5:.1:10; G=2*pi*sigd./log(b./a); C=2*pi*er*eo./log(b./a); L=muo*log(b./a)/(2*pi); Rs=sqrt(pi*f*muo/sigc); R=(1000*((1./a)+(1./b))*Rs)/(2*pi); w=2*pi*f; RL=R+i*w*L; GC=G+i*w*C; Gamma=sqrt(RL.*GC); Zo=abs(sqrt(RL./GC)); alpha=real(Gamma); loss=exp(-2*alpha*1); lossdb=-10*log10(loss); plot(Zo,lossdb) xlabel('Characteristic Impedance (ohms)') ylabel('attenuation (dB/m)') grid on

Fig. P6.10

3. Terminated T-Lines P6.11: Start with equation (6.54) and derive (6.55).

6-8

Vo+ e+ l + Vo− e− l Zin = + + l Zo Vo e − Vo− e− l With Vo− =  LVo+ , we then have

(e Z = (e in

+ l + l

+  L e− l ) −  L e− l )

Zo

We also know that Z − Zo L = L , Z L + Zo So now we have  Z − Z o  − l e+ l +  L e Z L + Zo  ( Z + Z o ) e +  l + ( Z L − Z o ) e − l Z  Zin = Zo = L o Z L + Z o ) e+  l − ( Z L − Z o ) e− l  Z L − Z o  − l ( + l e − e  Z L + Zo  and with rearranging, Z L ( e +  l + e − l ) + Z o ( e +  l − e − l ) Zin = Zo .

Z L ( e +  l − e − l ) + Z o ( e +  l + e − l ) We can convert the exponential terms into hyperbolic functions, given 1 1 sinh(x) sinh( x) = ( e x − e − x ) , cosh( x) = ( e x + e − x ) , and tanh(x)= .

2 2 cosh(x) This leads to 2Z cosh (  l ) + 2 Z o sinh (  l ) Z in = Z o L , 2Z L sinh (  l ) + 2 Z o cosh (  l ) or finally Z + Z o tanh (  l ) Z in = Z o L .

Z o + Z L tanh (  l )

P6.12: Derive (6.56) from (6.55) for a lossless line.

Z L + Z o tanh (  l ) , and tanh (  l ) = tanh ( l + j  l ) = tanh ( j  l ) since  = 0 for Z o + Z L tanh (  l ) lossless line. Using the hyperbolic definitions, we have + j l − e− j l ) sinh ( j  l ) ( e tanh ( j  l ) = = . cosh ( j  l ) ( e+ j l + e− j l ) Z in = Z o

Now using Euler’s formula,

6-9 cos (  l ) + j sin(  l ) - cos ( −  l ) − j sin(−  l ) j 2sin (  l ) = = j tan(  l ) cos (  l ) + j sin(  l ) + cos ( −  l ) + j sin(  l ) 2 cos (  l ) Plugging this in, we find, Z + jZ o tan (  l ) Z in = Z o L .

Z o + jZ L tan (  l ) tanh ( j  l ) =

P6.13: A 2.4 GHz signal is launched on a 1.5 m length of T-Line terminated in a matched load. It takes 6.25 ns to reach the load and suffers 1.2 dB of loss. Find the propagation constant.

 =  + j 1.2dB 1Np Np = 0.092 1.5m 8.686dB m  l 1.5m m  : up = = = = 2.4 x108  t 6.25ns s

=

=

 up

=

2 ( 2.4 x109 ) 8

2.4 x10

= 62.8

rad m

So

 = 0.092 + j 62.8

1 m

P6.14: A source with 50  source impedance drives a 50  T-Line that is 1/8 of a wavelength long, terminated in a load ZL = 50 – j25 . Calculate L, VSWR, and the input impedance seen by the source.

Z L − Z o 50 − j 25 − 50 = = 0.242e− j 76 Z L + Z o 50 − j 25 + 50 1+ L VSWR = = 1.64 1 − L L =

2     = , tan   = 1  8 4 4 Z + jZ o tan (  l ) Z in = Z o L Z o + jZ L tan (  l )

l =

50 − j 25 + j 50 50 + j 50 + 25 = 30.8 − j 3.8 

= 50

Fig. P6.14

6-10 P6.15: A 1 m long T-Line has the following distributed parameters: R’ = 0.10 /m, L’ = 1.0 H/m, G’ = 10.0 S/m, and C’ = 1.0 nF/m. If the line is terminated in a 25  resistor in series with a 1 nH inductor, calculate, at 200 MHz, L and Zin.

Z L = 25 + j 2 ( 200 x106 )(10−9 ) = 25 + j1.257  Now, MLP0615 is used to solve the problem. % MLP0615 % % calculate gamma and char impedance % given the distributed parameters % Then, calculate gammaL and Zin % % define variables clc clear R=0.1; L=1.0e-6; G=10e-6; C=1.0e-9; f=200e6; w=2*pi*f; length=1; ZL=25+j*1.257; % Perform calcuations A=R+i*w*L; B=G+i*w*C; gamma=sqrt(A*B) %Propagation Constant Zo=sqrt(A/B) gammaL=(ZL-Zo)/(ZL+Zo) %Reflection coefficient TGL=tanh(gamma*length); Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL)) Running the program, Gamma = 0.0017 +39.7384i Zo = 31.6228 - 0.0011i gammaL = -0.1164 + 0.0248i Zin = 34.0192 - 7.4618i >> So the answers are, with the appropriate significant digits,  L = 0.12e j168 and Zin = 34 − j 7.5 

6-11 P6.16: The reflection coefficient at the load for a 50  line is measured as L = 0.516ej8.2 at f = 1 GHz. Find the equivalent circuit for ZL. Z L − Zo 1 + L , we find Z L = Zo = 150 + j30  .

Z L + Zo 1 − L This is a resistor in series with an inductor. The inductor is found by considering 30 j L = j 30, or L = = 4.8nH , 2 (1x109 )

Rearranging  L =

So the load is a 150  resistor in series with a 4.8 nH inductor.

P6.17: The input impedance for a 30 cm length of lossless 100  impedance T-line operating at 2 GHz is Zin = 92.3 – j67.5 . The propagation velocity is 0.7c. Determine the load impedance.

Rearranging Z in = Z o

=

 0.7c

=

Z L + jZ o tan (  l ) Z − jZ o tan (  l ) , we find Z L = Z o in Z o + jZ L tan (  l ) Z o − jZ in tan (  l )

2 ( 2 x109 )

0.7 ( 3x108 )

= 59.84

rad ; m

 rad   tan (  l ) = tan   59.84  ( 0.3m )  = −1.254 m   

Evaluating, we have Z L = 50 + j 0.016  = 50 + j 2 2 x109 L, or L = 1.3 pH.

(

)

This is a very small inductance, so we have Z L  50 .

P6.18: For the lossless T-Line circuit shown in Figure 6.51, determine the input impedance Zin and the instantaneous voltage at the load end vL.

25 − 50 1 2  = − , l = =  , tan  = 0 25 + 50 3  2 Z +0 Z in = Z o L = Z L = 25 ZL + 0 25 Vin = 8V = 2V = Vo+ e− j z + Vo− e+ j z 25 + 75 2 = Vo+ ( e j l +  L e− j l ) L =

e j = cos  + j sin  = −1, e− j = −1, 1   −2 Vo+  −1 − ( −1)  = Vo+ = 2; Vo+ = −3V 3   3  1 VL = Vo+ (1 +  L ) = −3 1 −  = −2V , so vL = 2cos (t + 180 )V  3

6-12

P6.19: Referring to Figure 6.10, a lossless 75  T-Line has up = 0.8c and is 30 cm long.

The supply voltage is vs = 6.0 cos(t) V with Zs = 75 . If ZL = 100 + j125  at 600 MHz, find (a) Zin, (b) the voltage at the load end of the T-Line, and (c) the voltage at the sending end of the T-Line.

up =

  rad ,= = 15.7 ,  l = 4.71, tan  l = 418.6  up m

Zin = 75

100 + j125 + j 75 ( 418.6 ) 75 + j (100 + j125 )( 418.6 )

= 22 − j 28  Referring to Fig P6.19, Zin Vin = 6 = 2.1e− j 36 V Zin + 75

 vin = 2.1cos (t − 36 )V L =

Fig. P6.19

Z L − Zo = 0.593e j 43 Z L + Zo

Vin = Vo+ ( e+ j l +  L e− j l ) = 0.70e− j126 Vo+ = 2.1e− j 36 V Vo+ =

2.1e− j 36 − j126

= 3e j 90 V

0.70e VL = Vo+ (1 +  L ) = 4.47e j105.8 V vL = 4.5cos (t + 106 )V

P6.20: Suppose the T-Line for Figure 6.10 is characterized by the following distributed parameters at 100 MHz: R’ = 5.0 /m, L’ = 0.010 H/m, G’ = 0.010 S/m, and C’ = 0.020 nF/m. If ZL = 50 – j25 ,vs = 10cos(t)V, Zs = 50, and the line length is 1.0 m, find the voltage at each end of the T-line.

The following MATLAB routine was used to find the required parameters. % MLP0620 % % calculate gamma and char impedance % given the distributed parameters % Then, calculate gammaL and Zin % % define variables clc clear

6-13 R=5; L=.010e-6; G=.01; C=.020e-9; f=100e6; w=2*pi*f; length=1; ZL=50-j*25; % Perform calcuations A=R+i*w*L; B=G+i*w*C; gamma=sqrt(A*B) Zo=sqrt(A/B) gammaL=(ZL-Zo)/(ZL+Zo) TGL=tanh(gamma*length); Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL)) Running the program, gamma = 0.2236 + 0.2810i Zo = 22.3607 gammaL = 0.4479 - 0.1908i Zin = 27.2079 -15.4134i >>

Vin = VSS

Zin = 3.97e− j18.2 V ,  vin = 4.0cos (t − 18.2 )V Zin + Z S

(

Vin = Vo+ ( e l +  L e− l ) = Vo+ (1.504 + j 0.101) = Vo+ 1.507e j 3.84 so Vo+ =

3.97e− j18.2 j 3.84

)

= 2.63e− j 22

1.507e VL = V (1 +  L ) = 3.85e− j 29.6 ,  vL = 3.9cos (t − 30 )V + o

4. The Smith Chart P6.21: Locate on a Smith Chart the following load impedances terminating a 50  TLine. (a) ZL = 200  , (b) ZL = j25 , (c) ZL = 50 + j50 , and (d) ZL = 25 – j200 .

6-14

Fig. P6.21

P6.22: Repeat problem P6.14 using the Smith Chart.

First we locate the normalized load, zL = 1 – j0.5 (point a). By inspection of the Smith Chart, we see that this point corresponds to  L = 0.245e j −76 . Also, after drawing the constant  circle we can see VSWR = 1.66. Finally, we move from point a, at 0.356 on the WTG scale, clockwise (towards the generator) a distance 0.125  to point b, at 0.481 . At this point we see zin = 0.62 – j0.07. Denormalizing we find: Zin = 31 – j3.5 .

Fig. P6.22a

6-15

Fig. P6.22b

P6.23: A 0.690 long lossless Zo = 75  T-Line is terminated in a load ZL = 15 + j67 .

Use the Smith Chart to find (a) L, (b) VSWR, (c) Zin and (d) the distance between the input end of the line and the first voltage maximum from the input end.

After normalizing ZL and locating it on the chart (point a), we see  L = 0.80e j 95 .

After drawing the constant  circle, we see that VSWR = 9 (point c). We locate the input impedance by moving from the load (point a at WTG = 0.118) clockwise towards the generator to the input point (point b at WTG = 0.118  + 0.690  – 0.500  = 0.308 ) At this point, zin = 0.8 – j2.4, so Zin = 60 – j180 . Finally, the distance from the input end of the line (point b) to the first voltage maximum (point c) is simply 0.308  – 0.250  = 0.058 . Or, using the WTL scale, it is 0.250  – 0.192  = 0.058 .

Fig. P6.23

6-16 P6.24: A 0.269 long lossless Zo = 100  T-Line is terminated in a load ZL = 60 + j40 .

Use the Smith Chart to find (a) L, (b) VSWR, (c) Zin and (d) the distance from the load to the first voltage maximum. (a) zL = 0.6 + j0.4 located at WTG=0.082.

We read off the Smith Chart that this point corresponds to:  L = 0.34e j121 .

After drawing the constant  circle we notice the VSWR = 2.05 (point c).

Moving from this point a distance 0.269  (clockwise, towards generator), we find the input point (point b at WTG = 0.351 ). At this point we have zin = 0.96j0.72, or Zin = 96-j72 . Finally, we move from point a towards the generator at point c to reach the voltage maximum, a distance 0.168.

Fig. P6.24

P6.25: The input impedance for a 100  lossless T-Line of length 1.162 is measured as 12 + j42 . Determine the load impedance.

We first locate the normalized input impedance, zin = 0.12 + j0.42, at point a (WTL=0.436). Then we move a distance 1.162  towards the load to point b, at WTL = 0.436  + 1.162  =1.598 ; 1.598  – 1.500  = 0.098 . At this point, we read zL = 0.15-j0.7, or ZL = 15 – j70 .

Fig. P6.25

6-17 P6.26: On a 50  lossless T-Line, the VSWR is measured as 3.4. A voltage maximum is located 0.079 away from the load. Determine the load.

We can use the given VSWR to draw a constant  circle as shown in the figure.

Then we move from Vmax at WTG = 0.250 to point a at WTG = 0.250  0.079  = 0.171 . At this point we have zL = 1 +j1.3, or ZL = 50 + j65 .

Fig. P6.26 P6.27: Figure 6.52 is generated for a 50  slotted coaxial air line terminated in a short circuit and then in an unknown load. Determine (a) the measurement frequency, (b) the VSWR when the load is attached and (c) the load impedance.

From the locations of minima on the shorted line we find :  = 2 ( 7.55cm − 1.25cm ) = 12.6cm

(a) f =

c



= 2.4GHz

(b) From the voltage maxima and voltage minimum on the loaded line, we have 4 VSWR = = 2 2 Using VSWR=2 we draw the constant || circle on the Smith Chart. Point a on the circle Fig. this P6.28 represents the 1.9 cm minimum. We move from point towards the load at the 1.25 cm reference location, a move of  1.9cm − 1.25cm    = 0.0516  12.6cm   At this point (point b on the circle) we have zL = 0.55 – j0.25, and upon denormalizing we have (c) ZL = 28 – j12 .

P6.28: Figure 6.53 is generated for a 50  slotted coaxial air line terminated in a short circuit and then in an unknown load.

Fig. P6.27

6-18 Determine (a) the measurement frequency, (b) the VSWR when the load is attached and (c) the load impedance.

From the location of the maxima on the shorted line, we find :  = 2 ( 9.3cm − 1.7cm ) = 15cm

(a) f =

c



= 2.0GHz

(b) From the load line, 10 VSWR = = 2.5 4 Using VSWR=2.5 we draw the constant || circle on the Smith Chart. Point a on the circle represents the minimum at 7.9 cm. We move from this point towards the load at the 5.5 cm reference location, a move of  7.9cm − 5.5cm    = 0.16  15cm   At this point (point b on the circle) we have zL = 1 – j0.95, and upon denormalizing we have (c) ZL = 50 – j48 .

P6.29: Referring to Figure 6.20, suppose we measure Zinsc = +j25  and ZinL = 35 + j85 . What is the actual load impedance? Assume Zo = 50 .

We normalize the short circuit impedance to zinsc = 0+j0.5 and locate this on the Smith Chart to determine the length of the T-Line is 0.074. Then we normalize ZinL to zinL=0.70+j1.70, locate this on the chart at 0.326 (WTL scale) and draw a constant || circle. We then move towards the load, or to 0.336 + 0.074  = 0.400 , and find this point on the Smith Chart (zL = 0.25+j0.7). Denormalizing, we find ZL = 12+j35 .

P6.30: MATLAB: Modify MATLAB 6.3 to draw the normalized load point and the constant Γ L circle, given Zo and ZL. Demonstrate your program with the values from Drill 6.11.

Add this to the end of the Matlab 6.3 program: %now add constant gamma circles ZL=50; fudge=0.001+i*0.001; newZL=ZL+fudge; Zo=50; zL=newZL/Zo; gamma=(zL-1)/(zL+1); plot(gamma,'-o');

6-19 constgamma(zL); You must change the value of ZL for each load point. Notice the addition of a ‘fudge factor’. This ensures that gamma has both a nonzero and finite real and imaginary part to work with in the plot.

You’ll also need to add an additional function: function [h]=constgamma(zL) %constgamma(zL) draws the constant gamma circle; phi=1:1:360; theta=phi*pi/180; a=abs((zL-1)/(zL+1)); Re=a*cos(theta); Im=a*sin(theta); z=Re+i*Im; h=plot(z,'--k'); axis('equal') axis('off') The program is run for each point of Drill 6.11 by changing the ZL value.

Since the MATLAB routine has the ‘hold on’, each new point is added to the plot.

Fig. P6.30 5. Impedance Matching P6.31: A matching network, using a reactive element in series with a length d of T-Line, is to be used to match a 35 – j50  load to a 100  T-Line. Find the through line length d and the value of the reactive element if (a) a series capacitor is used, and (b) a series inductor is used.

First we normalize the load and locate it on the Smith Chart (point a, at zL = 0.35-j0.5, WTG = 0.419). (a) need to move to point b, at z = 1+j1.4, so that a capacitive element of value jx = -j1.4 can be added to provide an impedance match. Moving to this point b gives d = 0.500+0.173  -0.419  = 0.254 . The capacitance is

6-20

Fig. P6.31a

−j = − j1.4, CZ o C=

1

2 (1x10 ) (100 )(1.4 ) 9

= 1.14 pF

Fig. P6.31b

(b) Now we need to move to point c, at z = 1-j1.4, so that an inductive element of value jx = +j1.4 can be added. Moving to this point c gives d = 0.500  + 0.327  – 0.419  = 0.408 . The inductance is (1.4 )(100 ) = 22.3nH j L = j1.4, L = Zo 2 (1x109 )

P6.32: A matching network consists of a length of T-Line in series with a capacitor.

Determine the length (in wavelengths) required of the T-Line section and the capacitor value needed (at 1.0 GHz) to match a 10 – j35  load impedance to the 50  line.

We find the normalized load, zL = 0.2 – j0.7, located at point a (WTG = 0.400). Now we move from point a clockwise (towards the generator) until we reach point b, where we have z = 1 + j2.4. Moving from a to b corresponds to d = 0.500+0.194-0.400 = 0.294. For the series capacitance we have −j − j 2.4 = , CZ o

or C =

1

2 (1x10 ) ( 50 )( 2.4 )

Fig. P6.32a

9

= 1.33 pF

Fig. P6.32b

6-21 P6.33: You would like to match a 170  load to a 50  T-Line. (a) Determine the characteristic impedance required for a quarter-wave transformer. (b) What through-line length and stub length are required for a shorted shunt stub matching network? (a) Z s = Z o RL = 92 (b) (1)Normalize the load (point a, zL = 3.4 + j0).

(2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle (d = 0.170) (4) move from the shorted end of the stub (normalized admittance point c) to the point y = 0 – jb. (l = 0.354  – 0.250  = 0.104 .)

Note in step 3 we could have gone to the point y = 1-jb. This would have resulted in d = 0.329  and l = 0.396 .

Fig. P6.33a

Fig. P6.33b

P6.34: A load impedance ZL = 200 + j160  is to be matched to a 100  line using a shorted shunt stub tuner. Find the solution that minimizes the length of the shorted stub.

Refer to Figure P6.33a for the shunt stub circuit.

(1)Normalize the load (point a, zL = 2.0 + j1.6).

(2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.170 -0.458 = 0.212) (4) move from the shorted end of the stub (normalized admittance point) to the point y = 0 – jb. (l = 0.354  – 0.250  = 0.104 .)

Fig. P6.34

6-22 P6.35: Repeat P6.34 for an open-ended shunt stub tuner.

(1)Normalize the load (point a, zL = 2.0 + j1.6).

(2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1-jb circle(0.500 + 0.330 -0.458 = 0.372).

We choose this point for c so as to minimize the length of the shunt stub.

(4) move from the open end of the stub (normalized admittance point) to the point y = 0 + jb. (l = 0.146 )

Fig. P6.35a

Fig. P6.35b

P6.36: A load impedance ZL = 25 + j90  is to be matched to a 50  line using a shorted shunt stub tuner. Find the solution that minimizes the length of the shorted stub.

Refer to Figure P6.33a for the shunt stub circuit.

(1)Normalize the load (point a, zL = 0.5 + j1.8).

(2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.198 -0.423 = 0.275) (4) move from the shorted end of the stub (normalized admittance point) to the point y = 0 – jb. (l = 0.308  – 0.250  = 0.058 .)

Fig. P6.36

6-23 P6.37: Repeat P6.36 for an open-ended shunt stub tuner.

Refer to Figure P6.35a.

(1)Normalize the load (point a, zL = 0.5 + j1.8).

(2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.392 -0.423 = 0.379) (4) move from the open end of the stub (normalized admittance point) to the point y = 0 + jb. (l = 0.191 )

Fig. P6.37

P6.38: (a) Design an open-ended shunt stub matching network to match a load ZL = 70 + j110  to a 50  impedance T-Line. Choose the solution that minimizes the length of the through line. (b) Now suppose the load turns out to be ZL = 40 + j100 . Determine the reflection coefficient seen looking into the matching network. (a) Refer to Figure P6.35a.

(1)Normalize the load (point a, zL = 1.4 + 2.2).

(2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.185 -0.448 = 0.237) (4) move from the open end of the stub (normalized admittance point) to the point y = 0 jb. (l = 0.328 ) (b)

Fig. P6.38a

Fig. P6.38b

6-24 (1) Normalize the load (point a: zL = 0.8 + j2.0) (2) locate yL (point b) (3) Move a distance 0.237 to point c (0.434  + 0.237  = 0.671 ; or WTG = 0.171 ) (4) Move from yopen to 0.328  (point d) (5) add admittances of point c and d to get ytot = 0.6 – j0.2.

(6) locate the corresponding ztot (point f) and read the reflection coefficient as:  = 0.28e j 34

6. Microstrip P6.39: A 6.00 cm long microstrip transmission line is terminated in a 100.  resistive load. The signal line is 0.692 mm wide atop a 0.500 mm thick polyethylene substrate.

What is the input impedance of this line at 1.0 GHz? What is the maximum frequency at which this microstrip can operate?

This can be solved using either the Smith Chart or ML0604 in conjunction with the Zinput function from Matlab 6.2. Using the latter approach we have: Microstrip Analysis enter width & thickness in the same units enter the line width: 0.692e-3 enter the substrate thickness: 0.500e-3 enter substrate rel permittivity: 2.26 eeff = 1.8326 up = 221461941.7986m/s Zo = 80.2454ohms To run the Zinput routine, we also need the propagation constant. Assuming lossless line, we have  = j  = j

 up

= j

2 (1x109 ) 8

2.215 x10

= j 28.4

>> Zinput(80.2,100,j*28.4,0.06) ans = 64.7278 + 3.7906i >> So we have Zin = 65 + j3.8  To find fmax, we have c 3x108 m s f max = = = 100GHz 4h  r 4 ( 0.5 x10−3 m ) 2.26

1 . m

6-25 P6.40: A 75  impedance microstrip line is to be designed on a 2.0 mm thick Teflon substrate using copper metallization. What is the maximum operating frequency for this microstrip? Now determine w, and the physical length of a quarter wave section of line at 800. MHz.

f max =

c 4h  r

=

3x108 m s = 26GHz 4 ( 2 x10−3 m ) 2.1

Using ML0605: Microstrip Design width & thickness will be in the same units enter the desired impedance: 75 enter the substrate thickness: 2 enter substrate rel permittivity: 2.1 w = 3.2929 eeff = 1.741 up = 227209857.0703m/s >> so w = 3.29mm and eff = 1.741.

The guide wavelength is:  c f 3x108 800 x106 G = o = = = 0.284m  eff  eff 1.741 The quarter wave section length is then: l =

G 4

= 0.071m = 7.1cm

P6.41: Analysis of a 2.56 cm long microstrip line reveals that it has a 50  characteristic impedance and an effective relative permittivity of 5.49. It is terminated in a 60  resistor in series with a 1.42 pF capacitor. Determine the input impedance looking into this terminated line at 1.60 GHz. This problem may be solved analytically or with the Smith Chart. For the analytical solution we have: 8 u p c  eff ( 3x10 ) 5.49 .0256m = 0.320. = = = = 0.080m, l = 9 0.080 m  f f 1.6 x10 2 l = ( 0.320 ) , tan  l = −2.125



The load capacitance has an impedance:

6-26

−j = − j 70 C so the total load impedance is ZL = 60-j70 .

Then, the input impedance is 60 − j 70 + j 50 ( −2.125) Z in = 50 = 31.8 + j 48.2  50 + j ( 60 − j 70 )( −2.125) Zc =

With the Smith Chart, the answer is Zin  32 + j 48  P6.42: A 100  impedance microstrip line is to be designed using copper metallization on a 0.127 cm thick dielectric of relative permittivity 3.8. Determine (a) w, (b) fmax, and at 2 GHz find (c) up and G. From ML0605: Microstrip Design width & thickness will be in the same units enter the desired impedance: 100 enter the substrate thickness: 0.127 enter substrate rel permittivity: 3.8 w = 0.066625 eeff = 2.6865 up = 182909468.597m/s >> (a) So we have w = 0.0666 cm = 0.67 mm. (b) c 3x108 m s f max = = = 30GHz 4h  r 4 (1.27 x10−3 m ) 3.8 (c) We know up, from the program (up = 1.83x108 m/s), so at 2 GHz u G = p = 0.0915m = 9.15cm f

P6.43: MATLAB: Modify MATLAB 6.4 to calculate attenuation. Try out your program using the parameters of Drill 6.21 and Drill 6.22. % % % % %

M-File: MLP0640 Microstrip Analysis Given the physical dimensions and er, this

6-27 % % % % % % % % % % % % % % % % % % % % %

will calculate eeff, Zo and up for microstrip.

Wentworth, 8/3/02 modified ML0604 on 9/5 to calculate attenuation Variables: w h t sigc ur er eeff up Zo ad ac atot ds Rs tand

clc clear

line width (m) substrate thickness (m) conductor thickness (m) conductor conductivity (S/m) conductor rel permeability substrate relative permittivity effective relative permittivity propagation velocity (m/s) characteristic impedance (ohms) dielectric attenuation(dB/m) conductor attenuation dB/m) total attenuation (dB/m) skin depth (m) skin effect resistance (ohms/square) dielectric loss tangent

%clears the command window %clears variables

disp('Microstrip Analysis') disp(' ') % Prompt for input values w=input('enter the line width (m): '); h=input('enter the substrate thickness(m): '); er=input('enter substrate rel permittivity: '); t=input('enter conductor thickness (m): '); sigc=input('enter conductor conductivity (S/m): '); ur=input('enter conductor relative permeability: '); tand=input('enter dielectric loss tangent: '); f=input('enter frequency (Hz): '); uo=pi*4e-7; eo=8.854e-12; c=2.998e8; u=ur*uo; e=er*eo;

6-28

% Perform Calculations eeff=((er+1)/2)+(er-1)/(2*sqrt(1+12*h/w)); up=2.998e8/sqrt(eeff); if w/h<=1 Zo=(60/sqrt(eeff))*log((8*h/w)+(w/(4*h))); else if w/h>1 Zo=120*pi/(sqrt(eeff)*((w/h)+1.393+0.667*log((w/h)+1.444))) ; end end ds=1/sqrt(pi*f*u*sigc); Rs=1/(sigc*ds*(1-exp(-t/ds))); ac=8.686*Rs/(Zo*w); ad=8.686*2*pi*f*er*(eeff-1)*tand/(c*2*sqrt(eeff)*(er-1)); atot=ac+ad; % Display results disp(['eeff = ' num2str(eeff) ]) disp(['up = ' num2str(up) 'm/s']) disp(['Zo = ' num2str(Zo) 'ohms']) disp(['ac = ' num2str(ac) 'dB/m']) disp(['ad = ' num2str(ad) 'dB/m'])

Now we run the program using the information from Drill 6.21 and Drill 6.22. Note that we’ve changed the dimensions to metric units.

Microstrip Analysis enter the line width (m): 9.8e-4 enter the substrate thickness(m): 1.016e-3 enter substrate rel permittivity: 9.9 enter conductor thickness (m): 6e-6 enter conductor conductivity (S/m): 5.8e7 enter conductor relative permeability: 1 enter dielectric loss tangent: 0.0001 enter frequency (Hz): 1e9 eeff = 6.6638 up = 116137011.5308m/s Zo = 49.8369ohms

6-29 ac = 1.5554dB/m ad = 0.022214dB/m >>

P6.44: A 50  impedance microstrip line is desired for operation at 2.4 GHz. It is to be built on a 20 mil thick mica substrate using a 10 m thick copper conductor. Calculate (a) w, (b) c, (c) d, and (d) tot at this frequency.

Mica has r = 5.4 and tan = 0.0003. Using ML0605: Microstrip Design width & thickness will be in the same units enter the desired impedance: 50 enter the substrate thickness: 20 enter substrate rel permittivity: 5.4 w = 32.6859 eeff = 3.9617 up = 150623255.8531m/s >> To use this information in ML0640, we need to convert h and w to metric units. We find h = 5.08x10-4m and w = 8.3x10-4m.

Microstrip Analysis enter the line width (m): 8.3e-4 enter the substrate thickness(m): 5.08e-4 enter substrate rel permittivity: 5.4 enter conductor thickness (m): 10e-6 enter conductor conductivity (S/m): 5.8e7 enter conductor relative permeability: 1 enter dielectric loss tangent: 0.0003 enter frequency (Hz): 2.4e9 eeff = 3.9616 up = 150624957.4337m/s Zo = 50.1512ohms ac = 2.6687dB/m ad = 0.11967dB/m >> With more appropriate significant digits we find: w = 33 mils

6-30 c = 2.67 dB/m d = 0.12 dB/m tot = 2.79 dB/m

P6.45: One type of board routinely used to build microwave circuits is 50 mils thick Rogers Corporation RT/Duroid©, with r = 10.8 and tan = 0.0028. It is coated on both sides by “1/4 oz copper”. This translates to a 0.35 mil thickness of copper. Find w and up for a 50  line. Then determine the c, d and tot at three frequencies: 1, 10 and 20 GHz. What is the maximum frequency of operation for this microstrip?

Using ML0605: Microstrip Design width & thickness will be in the same units enter the desired impedance: 50 enter the substrate thickness: 50 enter substrate rel permittivity: 10.8 w = 44.3241 eeff = 7.1852 up = 111844030.4942m/s >> Now convert w and h and t to metric (1.125x10-3m, 1.27x10-3m, and 8.89x10-6m, respectively), and run MLP0640 at each frequency. For instance, at 1 GHz we have: Microstrip Analysis enter the line width (m): 1.125e-3 enter the substrate thickness(m): 1.27e-3 enter substrate rel permittivity: 10.8 enter conductor thickness (m): 8.89e-6 enter conductor conductivity (S/m): 5.8e7 enter conductor relative permeability: 1 enter dielectric loss tangent: 0.0028 enter frequency (Hz): 1e9 eeff = 7.1847 up = 111847474.32m/s Zo = 49.8031ohms ac = 1.2975dB/m ad = 0.64805dB/m >>

6-31 Tabulating the results for each frequency: 1 GHz  in dB/m 1.3 c= 0.65 d= 2.0 tot=

10 GHz 4.0 6.5 10.5

20 GHz 5.7 13.0 18.7

The maximum frequency is c 3x108 m s 1mil f max = = = 18GHz −6 4h  r 4 ( 50mils ) 10.8 25.2 x10 m

P6.46: A 1.5 inch length of microstrip line of width 48.86 mils sits atop a 50 mil thick substrate with dielectric constant 4. Determine the impedance looking into this circuit at 2 GHz if it is terminated in a 300  resistor. Assume ideal conductors and lossless dielectric.

From ML0604 we find: Microstrip Analysis enter width & thickness in the same units enter the line width: 48.86 enter the substrate thickness: 50 enter substrate rel permittivity: 4 eeff = 2.9116 up = 175697087.6994m/s Zo = 74.9641ohms >> So we have Zo = 75 . Also, we find up  0.0254m  l = (1.5in )  = 0.0381 m ,  = = 0.08785m, G  in f   0.0381m so l = = 0.434G 0.08785m G  1 = 71.5 .

We can also calculate :  = up m Now either a Smith Chart or the Zinput equation may be used to evaluate Z in. From the Zinput function we defined earlier in MATLAB, >> Zo=75; >> ZL=300; >> G=j*71.5; >> L=0.0381; >> Zinput(Zo,ZL,G,L)

6-32

ans = 8.6567e+001 +1.2031e+002i >> So Zin = 87 + j120 .

A Smith Chart approach is more approximate, yielding Zin  90 + j120 .

P6.47: The top-down view of a microstrip circuit is shown in Figure 6.54. If the microstrip is supported by a 40 mil thick alumina substrate, (a) determine the line width required to achieve a 50  impedance line. (b) What is the guide wavelength on this microstrip line at 2 GHz? (c) Suppose at this frequency the load impedance is ZL = 150 j100 . Determine the length of the stubs (dthru and lstub) required to impedance match the load to the line.

We first apply ML0605: Microstrip Design width & thickness will be in the same units enter the desired impedance: 50 enter the substrate thickness: 40 enter substrate rel permittivity: 9.9 w = 38.6273 eeff = 6.6644 up = 116131354.2486m/s >> From this we have (a) w = 38.6 mils.

Also, u 1.161x108 (b) G = p = = 0.058m = 2.29in f 2 x109 (c) Now we use a Smith Chart to determine the open-ended shunt stub matching network.

Fig. P6.47

(1) Normalize the load (point a: zL = 3.0 - j2.0) (2) locate yL (point b) (3) Move to point c (0.180 - 0.025 = 0.155; or dthru = 0.155 = 9 mm (354 mils)) (4) Move from yopen to 0.336, so lstub = 0.336 = 19.5 mm (768 mils)

P6.48: Suppose the microstrip circuit shown in Figure 6.54 is realized atop the RT/Duroid© board of problem P6.45. Assuming the board material is lossless, (a) determine the line width required to achieve a 75  impedance line. (b) Now suppose at

6-33 1.0 GHz the load impedance is ZL = 150 + j150 . Find the length of the stubs (dthru and lstub) required to impedance match the load to the line.

Applying ML0605: Microstrip Design width & thickness will be in the same units enter the desired impedance: 75 enter the substrate thickness: 50 enter substrate rel permittivity: 10.8 w = 15.7817 eeff = 6.6844 up = 115957584.2884m/s >> So we have (a) w = 15.8 mils. u We also have: G = p = 0.116m f (b) Now we employ a Smith Chart to determine the open-ended shunt stub matching network.

Fig. P6.48

(1) Normalize the load (point a: zL = 2.0 + j2.0) (2) locate yL (point b) (3) Move to point c (0.178 + 0.500 - 0.459 = 0.219; or dthru = 0.219 = 2.54cm (1 in)) (4) Move from yopen to 0.339, so lstub = 0.339 = 3.93cm (1.55 in)

7. Transients P6.49: Consider Figure 6.33 with the following values: Vs = 10 V, Zs = 30 , Zo = 50 , up = 0.666c, ZL = 150 , l = 10 cm. Plot, out to 2 ns, (a) the voltage at the source end, (b) the voltage at the middle, and (c) the voltage at the load end of the T-Line.

0.1m = 0.5ns  8 m 0.666  3x10  s  150 − 50 1 30 − 50 1 L = = ; S = =− 150 + 50 2 30 + 50 4 50 Vo = 10 = 6.25V 50 + 30 tl =

6-34

Fig. P6.49a

Fig. P6.49b

6-35 P6.50: Repeat problem P6.49 for a 10 V pulse of duration 0.4 ns.

Fig. P6.50a

Fig. P6.50b

6-36 P6.51: MATLAB: Consider a 12 cm long 50  transmission line terminated in a 25  load and having a matched source impedance (Zs = 50 ). Propagation velocity on the TLine is 0.67c. The source is a 0.4 ns square pulse of amplitude 6 V. Modify MATLAB 6.6 to plot v(z,t) at two points: z = 2 cm and z = 10 cm. % % % % % % % % % % % % %

M-File: ML0651 modifies ML0606 replaces triangular pulse with rectangular pulse ML0606: Analysis of a triangular pulse (matched source impedance) traveling down a T-Line and reflecting off a resistive load. We want to be able to trace the voltage at an arbitrary point along the line.

Wentworth, 4/25/03

% Variables % Vo % t1 % t2 % L % T % z % tau % up % Zo,ZL % N % GL clc clear

pulse height (V) pulse start (ns) pulse end (ns) line length (cm) transit time (ns) location to find pulse (cm) time "location" to find pulse (ns) propagation velocity (m/s) line,load impedance (ohms) number of points load reflection coefficient

%enter variables Vo=6; t1=0; t2=.4; L=12; z=2; up=2e8; Zo=50; ZL=25; T=1e9*(L/up)/100; tau=1e9*(z/up)/100; N=500; GL=(ZL-Zo)/(ZL+Zo); %initialize array

6-37 for i=1:N+1 v(i)=0; end dt=2*T/N; %enter rectangular pulse function t=0:dt:2*T; vo=0.5*Vo*(step(t,t1)-step(t,t2)); %Note that “step” is a function defined %in ML0606. %Generate + wave data for i=1:N+1 ta=i*dt; if ta>tau j=ceil((ta-tau)/dt); vplus(i)=vo(j); end end %Generate - wave data for i=1:N+1 ta=i*dt; tb=2*T-tau; if ta>tb j=ceil((ta-tb)/dt); vmin(i)=GL*vo(j); end end %Sum the data for i=1:N+1 v(i)=vplus(i)+vmin(i); end plot(t,v) xlabel('time (ns)') ylabel('voltage') AXIS([0 2*T -Vo Vo]) grid on

Fig. P6.51 (plot at z = 2 cm)

P6.52: MATLAB: Modify MATLAB 6.6 to plot v(z,t) at z = 4.5 cm if the source pulse is as indicated in Figure 6.55. % % % %

M-File: MLP0652 Analysis of a triangular pulse (matched source impedance) traveling down a T-Line and reflecting off a

6-38 % % % % %

resistive load. We want to be able to trace the voltage at an arbitrary point along the line.

Wentworth, 4/25/03

% Variables % Vo % t1 % t2 % t3 % L % T % z % tau % up % Zo,ZL % N % GL clc clear

pulse height (V) pulse start (ns) pulse middle (ns) pulse end (ns) line length (cm) transit time (ns) location to find pulse (cm) time "location" to find pulse (ns) propagation velocity (m/s) line,load impedance (ohms) number of points load reflection coefficient

%enter variables Vo=10; t1=0; t2=1; t3=2; L=6; z=4.5; up=3e7; Zo=50; ZL=0; T=1e9*(L/up)/100; tau=1e9*(z/up)/100; N=500; GL=(ZL-Zo)/(ZL+Zo); %initialize array for i=1:N+1 v(i)=0; end dt=2*T/N; %enter triangular pulse function m1=0.5*Vo/(t2-t1); b1=0.5*Vo-m1*t2; m2=0.5*Vo/(t2-t3); b2=0.5*Vo-m2*t2; for i=1:N+1 t(i)=i*dt; if t(i)<t1

6-39 vo(i)=0; end if and(t(i)>t1,t(i)<=t2) vo(i)=m1*t(i)+b1; end if and(t(i)>t2,t(i)<=t3) vo(i)=m2*t(i)+b2; end if t(i)>t3 vo(i)=0; end end %Generate + wave data for i=1:N+1 ta=i*dt; if ta>tau j=ceil((ta-tau)/dt); vplus(i)=vo(j); end end %Generate - wave data for i=1:N+1 ta=i*dt; tb=2*T-tau; if ta>tb j=ceil((ta-tb)/dt); vmin(i)=GL*vo(j); end end %Sum the data for i=1:N+1 v(i)=vplus(i)+vmin(i); end plot(t,v) xlabel('time (ns)') ylabel('voltage') AXIS([0 2*T -Vo Vo]) grid on

Fig P6.52

P6.53: The expression for iL(t) and vL(t) of equations (6.106) and (6.107) were derived for a T-Line terminated in an inductor. Find similar expressions for a T-Line terminated in a capacitor.

iL (t ) = C

dvL (t ) , vL (t ) = (Voi + Vor )U ( ), ZoiL (t ) = (Voi − Vor )U ( ) dt

6-40 Suppressing U(), we have

dvL (t ) dt i 2V 2V i dvL (t ) dv (t ) 1 −1 + vL (t ) = o , or L = vL (t ) + o . dt Z oC Z oC dt Z oC Z oC This is in the form: 2Voi dv −1 = Av + B, where A = , and B = dt Z oC Z oC solving the integral dv  Av + B =  dt we have 1 ln ( Av + B ) = t + C. A 1 Now since at t = 0, v = 0, we have C = ln( B), and A 1  Av + B  t = ln  .

A  B  −t B Solving for v: v = ( e At − 1) = −2Voi  e ZoC − 1 A   −t    vL (t ) = 2Voi 1 − e ZoC U ( ).   Now for iL(t), −t  1  2V i −t dv (t ) iL (t ) = C L = C  2Voi e ZoC  = o e ZoC dt  Z oC  Zo 2Voi = vL (t ) + Z oiL (t ) = vL (t ) + Z oC

 iL (t ) =

2Voi −t ZoC e U ( ).

Zo

P6.54: For Figure 6.42, Zo = 100  and up = 0.1c. Estimate L if the VL vs t is given in Figure 6.56.  − Zo   L 

Using (6.107), vL (t ) = 2Voi e

U ( ), here we have  = t – 2 ns. Choosing the voltage

at 2 ns we have 0.9V = 2V , or Voi = 0.45V . Then, at approximately 2.2 ns we have 0.2 V, so i o

 −100( 0.2 ns )  L  i  o

0.2V = 2V e

Solving for L:

, or

0.2 = e −20 ns L .

0.45

6-41

L=

−20ns = 25nH . ln ( 0.2 0.45 )

P6.55: A 50  T-Line with up = 0.5c is terminated in some load such that the TDR is given by Figure 6.57. Determine the location and the value of the load.

From the figure we see the two-way travel time is 7 ns, so therefore one-way is 3.5 ns and we have m  l = ( 3.5ns )( 0.5)  3x108  = 0.525m. s  Also from the figure we can calculate : V tot − V + 0.4 − 1 = = = −0.6 V+ 1 and then 1 − 0.6 R = 50 = 12.5.

1 + 0.6 So we have a 12.5  terminating resistor located 0.525 m along the line.

P6.56: The TDR plot for a 75  T-Line with up = 0.2c is given in Figure 6.58. What type components terminate the line? Estimate the component values.

By inspection we see it is a series combination of resistance and inductance. The location is: 1 l = (10 x10−9 sec ) ( 0.2 ) ( 3x108 m sec ) = 0.30m 2 The resistance is found by determining the reflection coefficient: 1.75 − 1 = = 0.75, 1 1.75 and then the resistance is: R = 50 = 350.

0.25 Only a very rough estimate can be found for the inductance. At  = 10 ns – 10 ns = 0,. we have 2Voi = 2V . Then, at (very roughly)  = 11 ns – 10 ns = 1ns,, vL(t) has dropped to 1.75 V, so −75ns 1.75 = 2e−75(1ns ) L , or L = = 560nH . ln (1.75 2 ) So our rough estimate is an inductance between 500 and 600 nH.

6-42 8. Dispersion P6.57: MATLAB: Use Fourier Series to construct a 5 V pulse of duration 5 ns that repeats every 10 ns. % % % % % % % % % % % % % % %

M-File: MLP0657 This program assembles a pulse using Fourier series. It modifies ML0607.

Wentworth, 8/3/02 Variables: N aO T fo wo t ftot

clc clear

number of Fourier coefficients avg value of the waveform (volts) period (s) fundamental frequency (Hz) fund angular freq (rad/s) time (sec) fourier sum at a particular time(volts)

%clears the command window %clears variables%MstripDesign

% Initialize variables clear N=1000; a0=2.5; T=10e-9; fo=1/T; wo=2*pi*fo; % Evaluate Fourier Series Coefficients for n=1:N a(n)=(10/(pi*n))*sin(n*pi/2); end % Generate data and plot for i=1:180 t(i)=i*T/90; for n=1:N f(n)=a(n)*cos(n*wo*t(i)); end ftot(i)=a0+sum(f); end plot(t,ftot) xlabel('time(s)')

6-43 ylabel('volts') grid on

Fig. P6.57

P6.58: MATLAB: Actual pulses have some slope to the leading and trailing edge.

Suppose a symmetrical pulse is 5 V from –2 ns to +2ns, and has a linear slope to 0 V on each edge of duration 0.2 ns. The pulse repeats every 20 ns. Construct this pulse using Fourier Series for N = 10, 100 and 1000. Comment on how this pulse compares to the one of Figure 6.48.

We must first find the Fourier coefficients: ao = 1.05 10  n  −50 an = sin  ( 2.2sin ( 0.22n ) − 2.0sin ( 0.20n ) ) + n  5  n 500 110 − cos ( 0.22n ) − cos ( 0.20n ) ) + (sin ( 0.22n ) − sin ( 0.20n ) ) 2 ( n ( n ) % % % % % % % % % %

M-File: MLP0658 This program assembles a pulse using Fourier series. It modifies ML0607.

Wentworth, 8/3/02 Variables: N aO

number of Fourier coefficients avg value of the waveform (volts)

6-44 % % % % %

T fo wo t ftot

period (ns) fundamental frequency (Hz) fund angular freq (rad/s) time (ns) fourier sum at a particular time(volts)

clc clear

%clears the command window %clears variables%MstripDesign

% Initialize variables clear N=1000; a0=1.05; T=20; fo=1/T; wo=2*pi*fo; % Evaluate Fourier Series Coefficients for n=1:N a1(n)=(10/(pi*n))*sin(n*pi/5); a2a(n)=(-50/(pi*n))*(2.2*sin(.22*pi*n)2.0*sin(.20*pi*n)); a2b(n)=(-500/((pi*n)^2))*(cos(.22*pi*n)-cos(.20*pi*n)); a3(n)=(110/(pi*n))*(sin(.22*pi*n)-sin(.2*pi*n)); a(n)=a1(n)+a2a(n)+a2b(n)+a3(n); end % Generate data and plot for i=1:180 t(i)=i*T/90; for n=1:N f(n)=a(n)*cos(n*wo*t(i)); end ftot(i)=a0+sum(f); end plot(t,ftot) xlabel('time(ns)') ylabel('volts') grid on save 'Fourier1000' t ftot -ascii

6-45

6 5

Volts

4 3 2 1 0 -1 0

5

10

15

20

25

30

35

40

time(ns)

Fig. P6.58a (the N=100 and N=100 cases are indistinguishable)

6 N=100 5 N=1000

4 3

N=10

2 1 0 -1 17.5

17.6

17.7

17.8

17.9

18

18.1

18.2

18.3

18.4

18.5

time (ns)

Fig. P6.58b To see the difference between the N = 100 and N = 1000 cases requires looking at the plot over a reduced portion of time, as shown in Figure P6.59b.

P6.59: MATLAB: A material has a constant r = 4 from DC up to 20 GHz. Then  f − 20 x109   r = 4 cos  , 9  60 x10  for 20 GHz < f < 50 GHz. Show the pulse from problem P6.58 after it has traveled along a coaxial T-Line with this dielectric.

6-46 % % % %

M-File: MLP0659 This program modifies ML0607.

Wentworth, 2/2/03

clc clear

%clears the command window %clears variables%MstripDesign

% Initialize variables clear N=1000; a0=1.2; T=20e-9; fo=1/T; wo=2*pi*fo; z=10; %evaluate Fourier Series Coefficients for n=1:N a(n)=(12/(pi*n))*sin(n*pi/5); end %Generate data for i=1:180 t(i)=i*T/90; for n=1:N f(n)=n*50e6; er(n)=4; if f(n)>20e9 er(n)=4*cos((f(n)-20e9)/60e9); end beta(n)=2*pi*f(n)*sqrt(er(n))/3e8; V(n)=a(n)*cos(n*wo*t(i)-beta(n)*z); end Vtot(i)=a0+sum(V); end plot(t,Vtot) xlabel('time(ns)') ylabel('volts') grid on save 'dispoff' t Vtot -ascii

6-47

Fig. P6.59

7-1 Problems 1. Rectangular Waveguide Fundamentals P7.1: Find the cutoff frequency for the first 8 modes of WR430. a = 4.3 in = 0.1092 m, b = 2.15 in = 0.0546 m For air-filled guide we have: 2

2

c m n fcmn =   +  2  a  b Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find Mode fcmn (GHz) TE10 1.374 TE01 2.747 TE20 2.747 TE11 3.07 TM11 3.07 TE21 3.885 TM21 3.885 TE30 4.121

P7.2: Calculate the cutoff frequency for the first 8 modes of a waveguide that has a = 0.900 inches and b = 0.600 inches. a = 0.900 in = 0.02286 m, b = 0.600in = 0.01524 m For air-filled guide we have: 2

2

c m n   +  2  a  b Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find Mode fcmn (GHz) TE10 6.56 TE01 9.84 TE11 11.83 TM11 11.83 TE20 13.12 TE21 16.40 TM30 19.69 TE02 19.69 fcmn =

P7.3: Calculate the cutoff frequency for the first 8 modes of a waveguide that has a = 0.900 inches and b = 0.300 inches.

7-2

a = 0.900 in = 0.02286 m, b = 0.300 in = 0.00762 m For air-filled guide we have: 2

2

c m n fcmn =   +  2  a  b Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find Mode fcmn (GHz) TE10 6.56 TE20 13.12 TE30 19.68 TE01 19.68 TE11 20.75 TM11 20.75 TE21 23.66 TM21 23.66

P7.4: Calculate uG, the wavelength in the guide and the wave impedance at 10 GHz for WR90.

From Table 7.1 for WR90 we have fc10 = 6.56 GHz. So 2

uG = uU

=

 fc   6.56  8 m 1 −   = 3x108 1 −   = 2.26 x10 s  10   f 

U 2

=

2

3x108 10 x109 2

= 0.0397m,  = 4cm

 fc   6.56  1−  1−     10   f  Since fc10 = 6.56 GHz, at 10 GHz only TE10 is present and therefore we only have the Z10TE impedance. U 120 Z10TE = = = 500 2 2  fc   6.56  1−  1−     10   f 

P7.5: Consider WR975 is filled with polyethylene. Find (a) uu, (b) up and (c) uG at 600 MHz. From Table 7.1 for WR975 we have a = 9.75 in and b = 4.875 in. Then c 3x108 m s 1  1in  fc10 = =   = 403MHz 2 r a 2 2.26 9.75in  0.0254m 

7-3 2

 fc   403  F = 1−   = 1−   = 0.741  600   f  Now, c 3x108 m uU = = = 2 x108 s r 2.26 uP =

2

uU m = 2.7 x108 F s

uG = uU F = 1.48 x108

m s

P7.6: MATLAB: Plot up and wavelength in the guide as a function of frequency over the cited useful frequency range for WR90. % MLP0706 % % Plot propagation velocity and guide wavelength % over the cited useful freq range of WR90. % % 2/2/03 Wentworth % c=3e8; a=0.900;b=0.450; fc=(c/(2*.0254*a)); flo=8.2e9; fhi=12.4e9; N=100; df=(fhi-flo)/N; f=flo:df:fhi; A=sqrt(1-(fc./f).^2); Lu=c./f; LG=Lu./A; up=c./A; fG=f./1e9; Fig. P7.6 subplot(2,1,1) plot(fG,LG) ylabel('guide wavelength (m)') grid on subplot(2,1,2) plot(fG,up)

7-4 xlabel('frequency(GHz)') ylabel('propagation velocity (m/s)') grid on

P7.7: WR90 waveguide is to be operated at 16 GHz. Tabulate the values of the guide wavelength, phase velocity, group velocity and impedance for each supported mode.

For the TE10 mode we have c fc10 = , where a = 0.900 in = .02286m, so fc10 = 6.562GHz. 2a Then u c 3x108 = , where u = = = 1.88cm 2 f 16 x109 fc 1− f

( )

.0188m

=

(

1 − 6.562

16

uu

up =

)

= 0.0206m

2

c

=

3x108

=

= 3.3x108

( f ) 1 − ( fc f ) 1 − (6.56216) m u = u 1 − ( fc ) = 3x10 1 − ( 6.562 ) = 2.74 x10 f 16 s 2

1 − fc

2

2

2

2

8

G

m s

8

u

u

TE Z mn =

( f)

1 − fc

2

120

=

(

1 − 6.562

16

)

2

= 413

Likewise values are found for the TE20 and TE11 mode. expression for impedance is used:

( f)

TM Z mn = u 1 − fc

Mode TE10 TE20 TE11 TM11

For the TM11 mode, a different

2

fc(GHz) 6.56 13.1 14.7 14.7

(m) 0.0206 0.0328 0.0470 0.0470

up(m/s) 3.3x108 5.2x108 7.5x108 7.5x108

uG(m/s) 2.7x108 1.7x108 1.2x108 1.2x108

Z() 413 659 945 150

P7.8: MATLAB: Modify MATLAB 7.1 by plotting uG and up versus frequency for the same guide over the same frequency range.

7-5

% M-File: MLP0708 % % Waveguide Velocity Plot % Plots uG and uP for TE11 and TM11 % vs freq. for air-filled waveguide % (modifies ML0701) % % Wentworth, 11/26/02 % clc %clears command window clear%clears variables % Initialize variables c=2.998e8; %speed of light Zo=120*pi; ainches=0.900; binches=0.450;

Fig. P7.8

% convert to metric a=ainches*0.0254; b=binches*0.0254; % calc fc11 fc=c*sqrt((1/a)^2+(1/b)^2)/2; % Perform calculations f=15e9:.1e9:25e9; fghz=f/1e9; Factor=sqrt(1-(fc./f).^2); uu=c.*Factor./Factor; %just filling array with c up=c./Factor; ug=c.*Factor; % Display results plot(fghz,up,'-.k',fghz,uu,'--k',fghz,ug,'-k') legend('up','c','uG') xlabel('frequency, (GHz)') ylabel('velocity (m/s)') grid on

P7.9: MATLAB: Plot the TE10 wave impedance for WR430 waveguide versus frequency if the guide is filled with Teflon. Choose a suitable frequency range for your plot.

7-6

% MLP0709 % % Plot TE10 wave impedance for teflon filled % WR430 guide over a suitable frequency range. % % 2/2/03 Wentworth % c=3e8; er=2.1; uu=c/sqrt(er); a=4.30;b=2.150; fc=(uu/(2*.0254*a)); flo=1.7e9/sqrt(er); fhi=2.6e9/sqrt(er); N=100; df=(fhi-flo)/N; f=flo:df:fhi; Fig. P7.9 A=sqrt(1-(fc./f).^2); ZTE=(120*pi/sqrt(er))./A; fG=f./1e9; plot(fG,ZTE) xlabel('frequency(GHz)') ylabel('TE10 mode impedance (ohms)') grid on

P7.10: Suppose a length of WR137 waveguide operated at 7.0 GHz is terminated in a short circuit. At what distance from this short circuit does the input impedance appear infinite?

From our study of T-Lines, we know that looking into a quarter guide-wavelength section of waveguide terminated in a short circuit, the input impedance appears infinite. The cutoff frequency for the TE10 mode is 4.29 GHz. Then, U c f 3x108 7 x109 = = = = 0.0542m 2 2 2 1 − 4.29 1 − fc 1 − fc 7 f f

( )

( )

(

)

So the quarter wave length is 0.0542m/4 = 0.0136 m. Therefore a distance 1.4 cm away from the short circuit, the input impedance appears infinite.

2. Waveguide Field Equations P7.11: Manipulate (7.41) to get (7.1).

7-7 Rearranging (7.41), we have  m   n   − =  +  ,  a   b  Also from (7.41) we have 2

2 u

2

2

( ).

 = u 1 − fc f So



2

( )

u2 −  2 =, u 2 1-1+ fc f 

2

( )

2  2 fc =  u  f 

and 2 2  fc   m   n   2    =   +   ,  f   a   b   2

2 u

2 2  2   fc   m   n   2     =   +      u   f   a   b   2

2

2

2

2 fc m n =   +  u f  a  b Solving for fc, where we have uu = uf, 2

2

1 1 m n fc = uu   +   = 2 2   a  b

2

m n   +   a  b

2

P7.12: Find expressions for the phasor field components of the TE01 mode.

With m = 0 and n = 1, the nonzero field components in equations (7.67) - (7.71)are   y  − j z H zs = H o cos  e  b  j    y  − j z Exs = 2 H o sin  e 2 u −  b  b 

H ys =

j    y  − j z H o sin  e 2  − b  b  2 u

P7.13: Find an expression for the magnetic field of the TE11 mode.

7-8

H= +

− j  x   y  H o sin   cos   cos(t −  z )a x 2 2 u −  a  a   b 

j  x   y  H o cos   sin   cos(t −  z )a y 2  − b  a   b  2 u

x   y  + H o cos   cos   cos(t −  z )a z  a   b 

P7.14: Modify MATLAB 7.2 to look at the Hz field for the TE20 mode. % % % % % % % % % % % % % % %

M-File: MLP0714 TE02 Hz Field Pattern Generates contour and surface plots Wentworth, 11/26/02 Variables m,n mode indicators a,b unitless guide dimensions betax x component of phase constant betay y component of phase constant Ez Ez for contour plot Ezc Ez for conventional plot Ezs Ez for surface plot

clc clear

%clears the command window %clears variables

% Initialize variables m=0;n=2; a=40;b=20; betax=m*pi/a; betay=n*pi/b; % Generate data for contour plot for i=1:a/40:a x(i)=(i/40)*a; for j=1:b/20:b y(j)=(j/20)*b; Hz(j,i)=cos(betax*x(i))*cos(betay*y(j)); end

7-9 end % Generate data for conventional plot at y=b/2 xc=a/2; yc=1:b/20:b; Hzc=20*cos(betax*xc)*cos(betay*yc); % Generate data for surface plot [X,Y]=meshgrid(0:a,0:b); Hzs=10*cos(betax.*X).*cos(betay.*Y); subplot(3,1,1) contour(x,y,Hz,4) title('Hz') ylabel('y') axis('equal') axis([0 a 0 b]) subplot(3,1,2) plot(Hzc,yc) ylabel('y') xlabel('10*Hz, (at y=b/2)') axis('equal') axis([-20 20 0 b]) subplot(3,1,3) surf(Hzs) axis([0 40 0 20 0 10]) axis('equal') xlabel('x') ylabel('y') zlabel('Hzs*10')

Fig. P7.14

P7.15: MATLAB: You are to create a movie showing how a surface plot of Hz over a cross section of rectangular waveguide changes with position. Use WR284 waveguide operating at 5 GHz and animate Hz for the TE11 mode. % % % % %

M-File: MLP0715 Movie of the Hz contour field pattern in TE11 mode as a function of z-position along the line.

Here we'll take a time snapshot at t = 0.

7-10 % % Wentworth, 2/5/03 % clc %clears the command window clear %clears variables % Initialize variables m=1;n=1; f=5e9; c=3e8; a=2.84*.0254; b=1.34*.0254; fc=(c/2)*sqrt((m/a)^2+(n/b)^2); betau=2*pi*f/c; betax=m*pi/a; betay=n*pi/b; beta=betau*sqrt(1-(fc/f)^2); M=40; N=20; Lguide=2*pi/beta; %note: as movie runs the pattern shape remains the same, %but the intensity varies as indicated by the change in % color as the position z changes. % Generate reference frame [X,Y]=meshgrid(0:a/40:a,0:b/20:b); Hzs=10*cos(betax.*X).*cos(betay.*Y); surf(Hzs); axis([0 40 0 20 -20 20]) axis('equal') xlabel('x') ylabel('y') zlabel('Hzs*10') pause % Generate data for surface plot M=Moviein(1); [X,Y]=meshgrid(0:a/40:a,0:b/20:b); for n=1:60 z(n)=(n/30)*Lguide; Hzs=10*cos(betax.*X).*cos(betay.*Y).*cos(-beta.*z(n));

7-11 surf(Hzs); axis([0 40 0 20 -20 20]) axis('equal') xlabel('x') ylabel('y') zlabel('Hzs*10') M(n)=getframe; end movie(M);

Fig. P7.15 (Snapshot at end of movie)

3. Dielectric Waveguide P7.16: Start with (7.83) and derive (7.84) and (7.85).

n1 cos i − n2 cos t n1 cos i + n2 cos t Now, from the definition of index of refraction, we have n1 sin t n = , or sin t = 1 sin i , n2 sin i n2 (7.83)  =

2

n  cos t = 1 −  1 sin i   n2  Substituting this into (7.83) we have

=

n  n1 cos i − n2 1 −  1 sin i   n2 

2

n  n1 cos i + n2 1 −  1 sin i   n2 

2

=

n  n cos i − 2 1 −  1 sin i  n1  n2 

2

n  n cos i + 2 1 −  1 sin i  n1  n2 

2

( n2 n1 ) − ( sin i ) cos i + j 2 ( n2 n1 ) − ( sin i ) = = 2 2 2 2 cos i + ( n2 n1 ) − ( sin i ) cos i − j 2 ( n2 n1 ) − ( sin i ) cos i −

2

2

2

( sin i ) − ( n2 n1 ) = 2 2 cos i − j ( sin i ) − ( n2 n1 ) cos i + j

2

2

(7.84)

2

7-12 To arrive at the phase, we must break (7.84) into its real and imaginary parts. Multiplying both numerator and denominator by the conjugate of the denominator results in:

=

(

cos i + j

( sin i ) − ( n2 n1 ) 2

cos 2 i + sin 2 i − ( n2 n1 )

2

2

), 2

or

=

cos i + j

( sin i ) − ( n2 n1 ) , 2 1 − ( n2 n1 ) 2

2

So we can find the phase angle for the square root of ,  sin 2  − ( n n )2  i 2 1 −1  .   = tan   cos i   Now, since  = 2  ,

 sin 2  − ( n n )2  i 2 1  . (7.85)  = 2 tan    cos i   −1

P7.17: MATLAB: Compose a program that will plot the left and right side of (7.86) versus all possible values of i for m = 0. Test the program using the following values: a = 1 mm, f = 100 GHz, n1 = 3 and n2 = 1. % MLP0717 % plot the left and right sides of (7.86) % vs incident angle. clc clear n1=3; n2=1; m=0; a=.001; f=100e9; c=3e8; N=100; b1=2*pi*f*n1/c; thc=asin(n2/n1); %critical angle in radians thcd=thc*180/pi; %critical angle in degrees dth=round(90-thcd)/N;

7-13

thmin=acos(pi/(a*b1)); %This corresponds to the tangent % argument of (7.86) going to infinity. thmind=thmin*180/pi; thi1=thmind+dth:dth:90; thir1=thi1*pi/180; phi=a*b1*cos(thir1)/2; left=tan(phi); thi2=thcd+dth:dth:90-2*dth; % Note that we avoid thi2 getting too close to 90 degrees % where the function goes to infinity. thir2=thi2*pi/180; right=sqrt(sin(thir2).^2-(n2/n1)^2)./cos(thir2); plot(thi1,left,'-*',thi2,right,'-o') xlabel('incident angle (degrees)') ylabel('equation (7.86)') legend('left','right') grid on

Fig. P7.17

P7.18: MATLAB: Compose a program that will plot the left and right side of (7.89) versus all possible values of i for m = 0. Test the program using the following values: a = 1 mm, f = 100 GHz, n1 = 3 and n2 = 1.

7-14

% MLP0718 % plot the left and right sides of (7.89) % vs incident angle. clc clear n1=3; n2=1; m=0; a=.001; f=100e9; c=3e8; N=100; b1=2*pi*f*n1/c; thc=asin(n2/n1); %critical angle in radians thcd=thc*180/pi; %critical angle in degrees dth=round(90-thcd)/N; thmin=acos(pi/(a*b1)); thmind=thmin*180/pi; thi1=thmind+dth:dth:90; thir1=thi1*pi/180; phi=a*b1*cos(thir1)/2; left=tan(phi); thi2=thcd+dth:dth:90-10*dth; thir2=thi2*pi/180; rightnum=sqrt(sin(thir2).^2-(n2/n1)^2); rightden=((n2/n1)^2)*cos(thir2); right=rightnum./rightden; plot(thi1,left,'-*',thi2,right,'-o') xlabel('incident angle (degrees)') ylabel('equation (7.89)') legend('left','right') grid on

7-15

Fig. P7.18

Fig. P7.19

P7.19: MATLAB: Devise a Newton-Raphson iterative technique to solve for i from equation (7.86). Test the program for m = 0, 1 and 2 using the following values of Figure 7.16: a = 50 mm, f = 4.5 GHz, and r = 4.

The Newton-Raphson iterative approach is explained with Figure P7.19, a plot of a function F(x). We start by ‘guessing’ the value of x such that F(x) = 0. We take the slope of the function at this point, or F’(xi0), and for improved value of x we then have F ( xi0 ) 1 0 xi = xi − ' 0 .

F ( xi ) We then calculate F and F’ at this better point for x in order to arrive at a better still value for x.

Such an iterative procedure can usually quickly arrive at a solution.

Applying this approach to equation (7.86), we have

sin 2 i − ( n2 n1 )  a cos i m  F = tan  1 − − .  2 2  cos i  The first derivative is sin i cos 2 i du F'= − + v sin i , cos 2 u v where a cos i m 2 u= 1 − , v = sin 2 i − ( n2 n1 ) , 2 2 and −a1 sin i du = .

2 2

7-16 We can also consider that the minimum possible i occurs where tan(u) = ∞, at cos u = 0. And the maximum possible i occurs where tan(u) = 0, at sin u = 0. Solving, we find  (m + 1)  −1  m  i min = cos−1   , and i max = cos  .  a1   a1  % % % % % % % % % % % % % % % % % % % % %

MLP0719 Use a Newton-Raphson iterative approach to solve for data in Equation 7.86.

2/3/03

Wentworth

Variables a dielectric thickness (m) er relative permittivity f operating frequency (Hz) m mode n1,n2 indices of refraction n21 ratio of n2 to n1 qi angle theta (rad) qimin minimum qi (rad) qimax maxiumum qi (rad) qicrit critical angle for qi (rad) b1 beta (rad/m) N number of iterations F,Fp function and 1st derivative

clc clear % Initialize Variables a=0.050; er=4; f=4.5e9; m=0; n1=sqrt(er); n2=1; n21=n2/n1; c=3e8; b1=2*pi*f*sqrt(er)/c; N=100;

7-17 % Determine range for qi qicrit=asin(n21); qimin=acos((m+1)*pi/(a*b1)); if qimin<qicrit qimin=qicrit; end qimax=acos(m*pi/(a*b1)); qi(1)=(qimin+qimax)/2;

%initial guess for qi

for n=1:N u=(a*b1*cos(qi(n))-m*pi)/2; du=-a*b1*sin(qi(n))/2; v=sqrt(sin(qi(n))^2-(n21)^2); F(n)=tan(u)-v/cos(qi(n)); Fp(n)=(du/cos(qi(n))^2)(sin(qi(n))*cos(qi(n))^2/v)+v*sin(qi(n)); qi(n+1)=qi(n)-F(n)/Fp(n); end qidegrees=180*qi(N+1)/pi

Now run the program: qidegrees = 74.5224 >> The program also works with m = 1 and m = 2. For m = 3, where there is no solution, a complex number is calculated.

P7.20: Find (i)critical for a wave incident from distilled water into air.

Distilled water has r = 81, or n = 9. Then n 1 sin icrit = air = , so icrit = 6.4 nwater 9

P7.21: Suppose a Teflon slab of thickness 60 mm exists in air. What is the maximum frequency at which this slab will support only one mode?

7-18

For Teflon we have r = 2.1, or n = 1.45. Then, a 1 1 a 1 1  , or f  o 2 n12 − n22 c 2 n12 − n22 For a single mode we have c 1 3x108 m s 1 f = = 2.38GHz − 3 2a n12 − n22 2 ( 60 x10 m ) 1.452 − 12 So fmax = 2.4 GHz.

P7.22: Suppose a polystyrene dielectric slab is sandwiched between thick slabs of polyethylene.

How thin must the polystyrene slab be such that only one propagating mode is supported at 1 GHz? We have for polyethylene: For polystyrene: a

o



a

 1 1 , so a  o 2 n12 − n22 2

1 c 2 f

r = 2.26, n2 = 1.503 r = 2.56, n1 = 1.600 1 n12 − n22

.

1 n − n22 2 1

1 3x108 m s 1 a = 0.273m 9 2 2 1x10 / s 1.6 − 1.5032 So a < 273 mm.

P7.23: MATLAB: Modify MATLAB 7.3 to find the odd TE mode field patterns. Use the example information and duplicate the m = 1 plot for Figure 7.19. % % % % % % % % % %

M-File: MLP0723 This modifies ML0703 to find the odd TE mode field patterns at z = 0.

The theta angle must be entered for a particular mode. The "hold on" function allows the results of multiple runs to be placed on one plot.

Wentworth, 2/2/03

7-19 % % % % % % % % % % % % %

Variables a dielectric thickness (m) b phase constant thdeg angle theta in degrees th angle theta in radians n1,n2 indices of refraction n21 the ratio n2/n1 f frequency (Hz) c speed of light (m/s) w radian frequency (rad/s) Eo initial amplitude (V/m) alpha

Fig P7.23

clc %clears the command window clear %clears variables % initialize variables a=50e-3; thdeg=57.9; %corresponds to m = 1 th=pi*thdeg/180; n2=1; n1=2; n21=n2/n1; f=4.5e9; w=2*pi*f; c=2.998e8; Eo=1; b=(w/c)*n1; alpha=b*sqrt(sin(th)^2-n21^2); %m=0 x=-a/2:a/40:a/2; Ey=Eo*sin(b*cos(th)*x); hold on plot(x,Ey,'k') grid on xlow=-2*a/2:a/40:-a/2; Eylow=-Eo*sin(b*cos(th)*a/2)*exp(alpha*(xlow+a/2)); plot(xlow,Eylow,'k')

7-20

xhi=a/2:a/40:2*a/2; Eyhi=Eo*sin(b*cos(th)*a/2)*exp(-alpha*(xhi-a/2)); plot(xhi,Eyhi,'k')

P7.24: Generate a figure similar to Fig. 7.17(b) for the m=0 mode of 10 mm thick r1 = 9 dielectric for n1/n2 = 1.5, 2, and 3.

This is a relatively straightforward application of equation (7.89). The minimum critical angle occurs for n1/n2 = 3; from (7.78) we find icrit = 19.5°. The m = 0 plot is calculated assuming f = 4.5 GHz (as in Figure 7.17), and

1 =

  r1 c

.

P07.24 50 m=0 40 n1/n2=3 30

20 n1/n2=2 10 n1/n2=1.5 0 90

80

70

60

50

40

30

20

incidence angle (degrees)

Fig. P7.24

4. Optical Fiber P7.25: A 100/240 silica optical fiber has a core index of 1.460 and a cladding index of 1.450.

Estimate the number of propagating modes for a source wavelength (a) 850 nm, (b) 1300 nm and (c) 1550 nm.

For a 100 micrometer diameter fiber we have a = 50 x 10-6 m. Then,

7-21 2

  (50 x10−6 )  a  2 2 N = 2 (1.46)2 − (1.45)2 ) (  ( n f − nc ) = 2         2

N=

1.436 x10−9

2

.

(a) for 850 nm we have 1.436 x10−9 N= = 1990 (850 x10−9 )2 Likewise, (b) N = 850 and (c) N = 598.

P7.26: Given a fiber of core index 1.478 and cladding index 1.445, find the numerical aperture for a source of light incident from (a) air, (b) distilled water. (a) for air, no = 1 so we have

NA = sin  a =

n 2f − nc2 n0

1.4782 − 1.4452 = 0.310,( a = 18 ) 1

=

(b) for distilled water, n0 = 9, so

NA = sin  a =

n 2f − nc2 n0

1.4782 − 1.4452 = 0.034,( a = 2 ) 9

=

P7.27: Suppose nf = 1.475 and nc = 1.470. Determine the numerical aperture for a source of light incident from air. What is the maximum core diameter allowed to support only a single propagating mode if the source wavelength is 1300 nm?

NA = sin  a =

amax =

n 2f − nc2 n0

 k01 2 n2f − nc2

1.4752 − 1.4702 = 0.121 1

=

(1300 x10 ) ( 2.405) = 4.1 m, −9

=

2 ( 0.121)

diameter = 2amax = 8.2 m.

P7.28: Given a step-index fiber with nf = 1.480 and a source wavelength of 1550 nm, determine the minimum value of nc that will allow only one propagating mode for a core radius of 2 m.

There is single mode propagation so long as

7-22



2 a n 2f − nc2 k01

, or

n 2f − nc2 

 k01 .

2 a

Rearranging,

 k   k  nc  n −  01  , so nc min = n 2f −  01   2 a   2 a  2

2

2 f

 (1550 x10−9 ) ( 2.405)  2 nc min = (1.480 ) −   = 1.450 −6   2  2 x 10 ( )   2

P7.29: At a source wavelength of 1550 nm for a 5/125 silica fiber with nf = 1.470, what is the minimum value nc can be and only allow one propagating mode?

From the given information we have a = 2.5 m. Then, rearranging



2 a n 2f − nc2 k01

, we arrive at

 k  nc  n −  01  =  2 a  2

2 f

 (1550 x10−9 ) ( 2.405)  1.470 −  ( )  −6  2  2.5 x 10 ( )  

2

2

nc  1.451

6. Optical Link Design P7.30: A 10. km optical link is established between a typical LED and a typical PIN photodiode using a 1300 nm graded-index multimode fiber. Find the power margin and the maximum frequency analog signal that can be supported by this link. Assume 2 connectors and 4 splices.

Power budget: Source Source-to-fiber Connector Fiber (1dB/km * 10km) 4 splices (.05dB each) Connector Fiber-to-detector

0dBm -12dB -0.7dB -10dB -0.2dB -0.7dB -1.5dB -25.1dB Power Margin=-25.1dB-(-35dBm)= 9.9dB Frequency budget:

(from Table 7.5) (from Table 7.5) (from Table 7.2) (from Table 7.5) (from Table 7.5) (from Table 7.5)

7-23 For an LED with  = 50 nm, ns nm   tchrom =  0.003  (50nm )(10km ) = 1.5ns (see Tables 7.2, 7.3) km   also ns   tinterm =  0.5  (10km ) = 5ns (see Table 7.2) km   Combining, 2 2 t f = tchrom + tinterm = 1.5ns 2 + 0.5ns 2 = 5.22ns

and using LED risetime of 10 ns from Table 7.3, and PIN diode risetime of 0.3 ns from Table 7.4,

ts = tt2 + t 2f + tr2 =

(10ns ) + ( 5.22ns ) + ( 0.3ns ) = 11.28ns.

2

2

2

The rise-time budget is satisfied for t < T/2, so T > 2t, or Tmin = 22.6 ns. Since T=1/f, we have fmax =1/Tmin=44 MHz.

P7.31: Consider coupling efficiency from a typical LED to a graded-index multimode fiber can be approximated as (NA)2. What is the power received by a typical PIN diode if 2 km of 850 nm fiber is used? Repeat for 2.0 km of 1300 nm fiber.

Using the data available in Tables 7.2 – 7.5, and with source-to-fiber loss = 10log(NA2), we find: 850nm GRIN MMF 1300nm GRIN MMF NA 0.24 0.20 Atten 4dB/km 1dB/km PIN = Source-to-fiber Connector Fiber(2km*atten) Connector Fiber-to-detector Prec(dB)

(

Prec = (1mW ) 10 rec ( dB ) P

10

)=

0dB -12.4dB -0.7dB -8dB -0.7dB -1.5dB -23.3dBm

0dB -14dB -0.7dB -2dB -0.7dB -1.5dB -18.9dB

4.7W

13W

P7.32: Calculate the maximum data rate, in bits per second, that could be supported for each case of the previous problem.

7-24

For a typical LED we have (from Table 7.3)  = 50 nm and tt = 10 ns. The detector has (from Table 7.4) tr = 0.3 ns. Data from Table 7.2 is also used.

chromatic dispersion intermodal dispersion tt tchrom (see(1)) tintermodal(see(2)) tf (see(3)) tr ts (see(4)) bps (see(5))

850nm GRIN MMF 0.10 (ns/nm)/km 3 ns/km 10 ns 10 ns 6 ns 11.7 ns 0.3 ns 15.4 ns <32MHz

1300nm GRIN MMF 0.003 (ns/nm)/km 0.5 ns/km 10 ns 0.3 ns 1 ns 1.04 ns 0.3 ns 10.1 ns <49.5MHz

ns nm   () tchrom =  0.10  ( 50nm )( 2km ) = 10ns km    ns  () tintermodal =  3  ( 2km ) = 6ns  km  2 2 (3) t f = tchrom + tintermodal = 11.7ns

(4) ts = tt2 + t 2f + tr2 =

(10ns ) + (11.7ns ) + ( 0.3ns ) = 15.4ns 2

2

2

(5)ts<T/2, and T = 1/bps, so bps<1/(2ts)

P7.33: A 10. km optical link is established between a typical laser diode and a typical PIN photodiode using a 1300 nm step-index single-mode fiber. Find the power margin and the maximum frequency analog signal that can be supported by this link. Assume 2 connectors and 4 splices. Compare your answer with that of problem P7.30.

Using the data from Tables 7.2 – 7.5, Power budget: Source 0dBm Source-to-fiber -2dB 2 Connectors -1.4dB Fiber (0.6 dB/km * 10km) -6dB 4 splices (.05dB each) -0.2dB Connector -0.7dB Fiber-to-detector -1.5dB -11.1dB Power Margin=-11.1dB-(-35dBm)= 23.9dB

7-25

Frequency budget: For an LED with  = 50 nm, ns nm   tchrom =  0.003  ( 3nm )(10km ) = .09ns km   and 2 ts = tt2 + tchrom + tr2 =

( 0.4ns ) + ( 0.09ns ) + ( 0.3ns ) = 0.508ns.

2

2

2

Tmin = 2ts = 1.016 ns, and fmax=1/Tmin=984MHz. So, the laser diode source and the SMF:SI results in a much larger power margin and much higher maximum frequency.

8-1 Solutions for Chapter 8 Problems (11/15/04) 1. General Properties P8.1: In free space, a wave propagating radially away from an antenna at the origin has Hs =

−Is r

cos2  a ,

where the driving current phasor Is = Ioej. Determine (a) Es, (b) P(r,) and (c) Rrad.

 −I  Es = −a P  H s = −oa r   s cos 2  a  ,  r  I (a)Es = o s cos 2  a r  − I e − j 1 1  I e j P = Re E s  H*s  = Re  o o cos 2  a  o cos 2  a  2 2  r r  2

1 I  (b) P ( r , ,  ) = o  o  cos 4  a r 2 r  Now to find Rrad: 1 Prad =  P(r , ,  ) dS = I o2 Rrad , 2  2 4 1 cos  1 Prad = o I o2  2 a r r 2 sin  d d a r = o I o2  cos 4  sin  d  d 2 r 2 0 0 Prad =

o I o2 5

( − cos  ) = 52  I = 12 I R 5

 0

2 o o

2 o

rad

Solving: 2  (120 ) I o2 Rrad = 5 = 96 2 1 2 Io 2 (c) Rrad = 950

P8.2: What is the pattern solid angle and the directivity for an isotropic antenna? How about for a semi-isotropic antenna, that radiates equally in all directions above  = /2, but is zero otherwise? 

2

0

0

 P =  Pn ( ,  ) d  =  (1) sin  d  d = 4 4 = 1. P For a semi-isotropic antenna:  P = 4 , and Dmax =

8-2  2

2

0

0

 P =  (1)sin  d  d = 2 , and Dmax = 2.

P8.3: Sketch an appropriate cross-section of the radiation pattern and determine the beamwidth, pattern solid angle and directivity for the following normalized radiation intensities: (a ) Pn ( , ) = cos for 0     2, 0 otherwise.

(b) Pn ( ,  ) = cos2  for 0     2, 0 otherwise. (c) Pn ( ,  ) = cos3  for 0     2, 0 otherwise.

The patterns are drawn using MLP0803: % MLP0803 % Polar Plots for P8.03 % % The polar plot function doesn't allow multiple % plots. So we have to let the angle theta loop % around several times, changing the rho function % each time. % for i=1:100 theta(i)=-pi/2+i*pi/100; rho(i)=cos(theta(i)); end for j=101:200 theta(j)=(-pi/2)+j*pi/100; rho(j)=0; end for i=201:300 theta(i)=-pi/2+i*pi/100; rho(i)=(cos(theta(i)))^2; end for j=301:400 theta(j)=-pi/2+j*pi/100; rho(j)=0; end for i=401:500 theta(i)=-pi/2+i*pi/100; rho(i)=(cos(theta(i)))^3; Fig. P8.3 end for j=501:600

8-3 theta(j)=-pi/2+j*pi/100; rho(j)=0; end polar(theta,rho) (a) Beamwidth: solving  for cos  = 0.5, we find  = 60°, and beamwidth =2  = 120°.

Pattern solid angle:  2

 P =  Pn ( ,  )d  =  cos  sin  d d = 2  cos  sin  d 0

u = cos  , du = − sin  d , so  P = −2  udu = − u 2 = − cos 2 

Dmax =

4 =4 P

 2 0

=  sr.

(b) Beamwidth: solving  for cos2 = 0.5, we find  = 45°, and beamwidth =2  = 90°.

Pattern solid angle:  2

 P =  cos  sin  d d = 2  cos 2  sin  d 2

u = cos  , du = − sin  d , so  P = −2  u 2 du = −

Dmax =

4 =6 P

0

 2

2 3 2 2 u =− cos3  = sr 3 3 3 0

(c) Beamwidth: solving  for cos3 = 0.5, we find  = 37.5°, and beamwidth =2  = 75°.

Pattern solid angle:  P =  cos3  sin  d d = −2  u 3du

where u = cos  and du = -sin  d ,  2

2  cos 4  = sr 4 2 0 4 Dmax = =8 P

P = −

8-4 P8.4: Sketch an appropriate cross-section of the radiation pattern and determine the beamwidth, pattern solid angle and directivity for the following normalized radiation intensities: (a ) Pn ( ,  ) = sin 

(b) Pn ( ,  ) = sin 2  (c) Pn ( ,  ) = sin 3 

1 1 = sin  ,  = sin −1   = 30 .

2 2 Now, since the beam is maximum at  = 90°, the beamwidth is from 30° to 150°, or BW = 120°. (a) Pn =



2

 P =  Pn d  =  sin  sin  d d =  sin  d  d 2

0

0



1 4 (1 − cos 2 ) d =  2 sr , Dmax = = 1.27 2 P 0

so  P = 2  (b) Pn =

 1 1 o = sin 2  ,  = sin −1   = 45 , BW = 135 − 45 = 90 2  2 

 P =  Pn d  =  sin 2  sin  d d = 2  (1 − cos 2  ) sin  d 0





0

0

 P = 2  sin  d − 2  cos 2  sin  d 

= 2 ( − cos  ) 0 + so Dmax =

 2 4 8 cos3  = 4 − = sr 0 3 3 3

4 = 1.5 P

(

)

1 1 = sin 3  ,  = sin −1 (1 2 ) 3 = 52.5 , BW = (180o − 52.5o ) − 52.5o = 75o 2   1   1 2   P = 2  sin 4  d = 2  (1 − cos 2 ) d =  1 − 2 cos 2 + (1 + cos 4 ) d 4 2 0 2  0 0

(c) Pn =

 



   1 1  P =   d − 2 cos 2 d +  d +  cos 4 d  2 0 20 20 0 

P =

3 2 4 , and Dmax = = 1.7 4 P

8-5 Plots are generated using MLP0804: % MLP0804 % % Generate polar plots % clc clear for i=1:100 theta(i)=i*pi/50; rho(i)=abs(sin(theta(i))); end for i=101:200 theta(i)=i*pi/50; Fig. P8.4 rho(i)=(sin(theta(i)))^2; end for i=201:300 theta(i)=i*pi/50; rho(i)=abs(sin(theta(i))^3); end polar(theta,rho)

P8.5: (JustAsk): You are given the following normalized radiation intensity: Pn ( , ) = sin 2  sin 3  for 0     ,

0 otherwise.

Find the beamwidth, pattern solid angle, and directivity.

The beam is pointing in the ay direction, and we have BW = To find BW, we fix  = /2 and set sin2 equal to ½. Then,  1  = sin −1   = 45 , so BW = (180 − 45 ) − 45 = 90 .  2 To find BW, we fix  = /2, and set sin3 = ½, giving us

(

)

1 ( BW + BW ) .

2

 = sin −1 (1 2 ) 3 = 52.5 , so BW = (180 − 52.5 ) − 52.5 = 75 .

1

1 (90 + 75 ) = 82.5 .

2 The pattern solid angle is Finally, BW =

8-6

 P =  Pn d  =  ( sin 2  sin 3  ) sin  d d , 



 P =  sin  d  sin 3 d , (note limits on  ) 3

0

0

Each integral is solved as follows: 







0

0

0

0

y =  sin 3 xdx =  (1 − cos 2 x ) sin xdx =  sin xdx −  cos 2 x sin xdx. 



 sin xdx = − cos x = 2 0

0



1

 cos x sin xdx = − u du = − 3 u , where u = cos x, du = − sin xdx.

2

2

3

0



 1 1 2 so  cos 2 x sin xdx = − cos3 x = − (−1 − 1) = .

0 3 3 3 0 So we have  2 4 y =  sin 3 xdx = 2 − = , 3 3 0 



 4  4  16 and  P =  sin  d  sin 3 d =    = = 1.78sr.  3  3  9 0 0 4 4 Dmax = = = 7.1  P 1.78 3

P8.6: You are given the following normalized radiation intensity:



Pn ( , ) = sin 2  sin .

2 Determine the beamwidth, direction of maximum radiation, pattern solid angle and directivity.

1 ( BW + BW ) , 2 BW: Fix  = , sin2 = 1/2,  = 45°, BW = (180° – 45°) – 45° =90°.

BW: Fix  = /2, sin(/2)=1/2, BW = (360° – 60°) – 60° = 240° BW = (90°+240°)/2=165° BW =

By inspection, the direction of maximum radiation is at  =  and  = /2. (i.e. the –ax direction).



2





 P =  sin 2  sin sin  d d =  sin d  sin 3  d 2 2 0 0 Do each integral separately:

8-7 2





2

0 sin 2 d = −2cos 2 0 = −2(−1 − 1) = 4









 sin  d =  (1 − cos  ) sin  dx =  sin  d −  cos  sin  d 3

0

2

0

2

0

0

 1 1 2 4  = − cos  0 + cos3  = −(−1 − 1) + (−1 − 1) = 2 − = 0 3 3 3 3 4 3  4  16 So  P = (4)   = , and Dmax = = = 2.4 P 4 3 3

2. Electrically Short Antennas P8.7: Use the phasor form of Ampere’s Circuit Law, 1 Eos =   Hos , j o to find Eos from (8.46) without assuming the far-field condition. Then, show that this value of Eos reduces to (8.50) in the far-field.

To make the calculations easier, we’ll let I l 2 A= s .

4   j 1   1   sin  Ae − j r   ar   HOS = + sin     r (  r )2   r sin       

  j 1   1   rAe − j r   a + sin     r (  r )2   r r      We’ll break these up and do them separately: ar derivative:  e− j r   j Ae − j r  j 1   1  2 + sin  a = 2 A + cos ar .    ( ) r  r  r 2  r sin    r (  r )2   (  r )    a derivative: A sin    je − j r e − j r  A sin   − j r e − j r j  e − j r  − + a = − e − 2 2− a r r    2 r   r   r  2r   −

=−

Ae− j r  1 j  − 2  sin  a 1 − 2 r  (  r )  r 

  HOS = Now,

  Ae − j r  j 2 2  1 j  + cos  a + − 1 + + sin  a     r  2 2   r   r (  r )2   r  r ( )    

8-8

EOS =

1 j o

  HOS , so

  A e − j r  j 2 2  1 j  + cos  a + − 1 + + sin  a     r .

2 2   j o r   r (  r )2   r  r ( )     This is EOS without assuming the far-field condition. For far-field, we only need to consider the single 1/r term (1/r2<<1/r). So we have − A e − j r − I l  2 e − j r EOS = sin  a = s sin  a j o r j 4 o r upon reinserting our value for A from above. Rearranging, we have    I sl  e− j r EOS = j  sin  a .  r   o  4 The term in parenthesis can be manipulated: EOS =

 o o  o = = = o , so  o  o o I sl  e − j r sin  a , 4 r which is equation (8.50) for the far-field.

EOS = jo

P8.8: Suppose, for a particular antenna in free space, Aos = o I oe− j y a z .

Find Hos, Eos, and the time-averaged power density vector P.

BOS =   A OS =

HOS =

BOS

 o I o e − j y ) a x = − j o I o e − j y a x . ( y

= − j  I oe− j y a x .

o EOS = −oa y  HOS = −oa y  − j  I oe − j y a x = − jo  I oe − j y a z .

1 Re  EOS  H*OS  , so 2 1 1 2 P = Re  − jo  I oe − j y a z  + j  I o e + j y a x  = o (  I o ) a y .

2 2 Note that this problem is not very realistic, but is good to teach the mechanics of going from AOS to P. P=

8-9 P8.9: (JustAsk): Suppose a Hertzian dipole antenna is 1.0 cm long and is excited by a 10. mA amplitude current source at 100. MHz. What is the maximum power density radiated by this antenna at a 1.0 km distance? What is the antenna’s radiation resistance?

c 3x108 m s c=f,  = = = 3m. f 100 x106 1 s

  2 I 2l 2 120 ( 2 ) ( 0.010 ) ( 0.010 ) pW Pmax = o 2o 2 = = 0.052 2 2 2 2 32 r 32 3  1000 m 2 2 l   0.01  Rrad = 80 2   = 80 2   = 8.8m   3  2

2

2

P8.10: A 1.0 cm long, 1.0 mm diameter copper wire is used as a Hertzian dipole radiator at 1.0 GHz. (a) Find Rrad. (b) Estimate Rdiss by considering the skin effect resistance of the wire. (c) Find efficiency, e. (d) Find the maximum power gain Gmax.

=

c 3x108 m s = = 0.3m f 1x109 1 s

l   0.01  Rrad = 80   = 80 2   = 0.877   .3  From Example 8.2 we have  Cu 1GHz = 2.09 x10−6 m 2

2

2

S =  d Cu =  ( 0.001m ) ( 2.09 x10−6 m ) = 6.57 x10−9 m2 Rdiss =

e=

1l 1 0.01 = = 0.026 7  S ( 5.8 x10 ) 6.57 x10−9

Rrad 0.877 = = 0.97 Rrad + Rdiss 0.877 + 0.026

Gmax = eDmax = 0.97 (1.5) = 1.46

P8.11: Evaluate the curl of Aos (equation (8.59)) to find Hos. Now apply a far-field approximation to verify (8.60).

Eqn. (8.59): AOS =

o I s S 1 + j  r ) e− j r sin  a 2 ( 4 r

Since AOS only has an A component, we have 1  1    AOS = sin  A ) ar − ( ( rA ) a r sin   r r A A jA − j r We’ll let A = 2 (1 + j  r ) e− j r sin  = 2 e− j r sin  + e sin  , r r r

8-10 where A =

o I S S .

4

We break up the derivative into two parts:  jA − j r   sin 2   A − j r jA − j r  A 1. sin  A ) =  2 e− j r + e = 2 e + e (   2sin  cos   r r r   r  or  2A j 2 A − j  r sin  A ) = 2 e− j r sin  cos  + e sin  cos  (  r r 1  j 2 A  2A  sin  A ) ar =  3 e− j r cos  + 2 e− j  r cos   ar ( r sin   r r 

  A  rA ) =  e− j r sin  + jA e− j r sin   , ( r r  r  the left-most derivative is  −e− j r j  e− j  r  − A − j r   e− j  r  jA − j r A sin   = A sin  sin  − e sin    2 − = 2 e r  r  r r  r  r and the right derivative is  jA sin  ( e− j r ) = jA sin  ( − j  e− j r ) = A 2e− j r sin  r So  A − j r  1  jA − j r A 2 − j  r − rA a = e sin  + e sin  − e sin   a (  3 )  2 r r r r r  2.

Putting this all together j 2 A  2A    AOS =  3 e− j r cos  + 2 e− j  r cos   a r r r 

A  jA A 2 − j  r +  3 e− j r sin  + 2 e− j  r sin  − e sin   a r r r  Now, reinserting our value for A, we can find HOS as  1 1  2  I S j 2   I S  HOS =   AOS =  3  o s  e− j r cos  + 2  o s  e− j r cos   a r o o  r  4  r  4  

 1  I S  j   I S   2  o I s S  − j  r +  3  o s  e− j  r sin  + 2  o s  e− j  r sin  − sin   a  e r  4  r  4   r  4   and reducing we find j  I s S − j r  IS  HOS =  s 3 e− j r cos  + e cos   a r 2 2 r  2 r   IS  j  I s S − j r  2 I s S − j r +  s 3 e− j r sin  + e sin  − e sin   a 2 4 r 4 r  4 r  With the far-field approximation, only the 1/r term will be significant and we have

8-11

 2 I s S − j r H OS = − e sin  a 4 r Now, making use of  =  o o and o = o  o , we find

 2 I s S o I s S  = , 4 r 40 r  I S  − j r so HOS = − o s e sin  a , which is Equation (8.60).

40 r P8.12: Neglecting resistive losses in the wire, how much current must drive a loop antenna of radius 2.0 cm at 60 MHz to radiate 1.0 W of power? Repeat for a 20 turn loop.

At 60 MHz,  =

c 3x108 = = 5m . So at 2 cm radius, we have the small loop situation. f 60 x106

4o 3 I o2  S  For a small loop: Prad =  2 , 3   Here, 2 S =  a 2 =  ( 0.02m ) = 1.257 x10−3 m2 .

2

Solving for Io, 2

3Prad   2  (a) I o =   = 159 A 4o 3  S  with N=20, (b) I o = 159 A

1 159 = = 8A 2 N N

P8.13: Suppose in the far-field for an antenna at the origin,  I e − j r Hos = s sin  cos  a 4 r where Is = Ioej. What is the radiation resistance of this antenna at 100 MHz?

o  I s e − j  r sin  cos  a 4 r 2 1 1   Io  *   P(r , ,  ) = Re EOS  H OS  = o  sin 2  cos 2  a r  2 2  4 r  Note also that P(r, ,  ) = Pmax Pn ( ,  )ar , where here EOS = −oa r  H OS =

8-12

1  I  Pmax = o  o  , and Pn ( ,  ) = sin 2  cos 2  .

2  4 r  Then, P =  Pn ( ,  )d .

2



Referring to P8.5,  sin 3  d = 2 − 0

2

2 4 = , and 3 3

1

 cos  d = 2   d +  cos 2 d  =  .

2

0

So,  P =  Pn ( ,  )d  =

4 sr 3

 1   I o 2   4  Prad = r Pmax  P = r  o  .  2  4 r    3    Using o = 120 and  = 2  , we find 2

Prad =

2

20 2 I o2

2

Finally, Prad =

.

2P 1 2 40 2 I o Rrad , so Rrad = rad = , 2 I o2 2

and since for this problem,  = c f = 3m, Rrad = ( 40 2 9 ) = 44

P8.14: Suppose in the far-field for a particular antenna at the origin, the electric field is e − j r Eos = o I o sin  a . r What is the radiation resistance of this antenna?

1 2 I o Rrad = r 2 Pmax  P , so we must find Pmax and p.

2 1 e− j r e− j r H S = a r  o I o sin  a = I o sin  a o r r 1 1 1 P = Re EOS  H*OS  = o I o2 sin 2  a r , 2 2 2 ( r )

We’ll use: Prad =

so 1 1 Pmax = o I o2 , and Pn = sin 2  .

2 2 ( r ) 

2



0

0

 P =  Pn d  =  sin  d  d = 2  (1 − cos 2  ) sin  d 3

0





   1 8    P = 2   sin  d −  cos 2  sin  d  = 2  − cos  0 + cos 2   = sr 0 3 3   0 0  

8-13 Rrad =

2 2 2 1 1 8 8 (120 )8 r Pmax  P = 2 r 2 o I o2 = o = = 320 2 2 Io Io 2 3 3 ( r ) 3

P8.15: Derive the expressions for radiated power (equation (8.64)) and radiation resistance (equation (8.65)) for a small loop antenna.

1 2 I o Rrad = r 2 Pmax  P 2  2 o 2 I o2 S 2  2 From (8.63) we have Pmax = 32o 2 r 2 and  2 8  P =  sin 2  d  =  sin 3  d  d = sr (see integral solution of P8.14) 3 0 0 Now,   2 o 2 I o2 S 2  2   8  Prad = r 2 Pmax  P = r 2    2 2  32o r   3  We’ll use: Prad =

Using the conversions:  =  o o ,  = 2  , and o =

o

o

we arrive at: 2

4  S  1 Prad = o 3 I o2  2  = I o2 Rrad 3 2   Solving for Rrad, 2

 S  Rrad = 320  2     4

3. Dipole Antennas P8.16: MATLAB: Develop a routine to calculate the beamwidth for a dipole antenna of arbitrary length between 0.1 and 1 . % % % % % % % % % %

MLP0816 Determine beamwidth for an arbirary length dipole antenna. Equation (8.74) is used.

2/10/03 Wentworth Variables L th(i)

dipole length in wavelengths theta angle (degrees)

8-14 % num,den calculation variables % F(i) function F from (8.74) % Fmax maximum F(i) % thmax angle where Fmax occurs % diff(i) calculation variable used to find BW % diffmin calculation variable used to find BW % thhalf theta at the half-power point % clc clear % dipole length L=1; %dipole length in wavelengths % initialize settings diffmin=.1; % perform calculations bL=2*pi*L; % Find Fmax for i=1:1:180 th(i)=i*pi/360; num=cos((bL/2)*cos(th(i)))-cos(bL/2); den=sin(th(i)); F(i)=(num/den)^2; end Fmax=max(F); % Find thmax for i=1:1:180 if F(i)>=Fmax Fmax=F(i); thmax=180*th(i)/pi; end end for i=1:1:180 Pn(i)=F(i)/Fmax; diff(i)=abs(Pn(i)-0.5); if diff(i)<diffmin diffmin=diff(i); thhalf=180*th(i)/pi; end end

8-15

BW=2*abs(thhalf-thmax) Running the program for several values of L: L = 0.1 0 BW = 90° L = 0.25 BW = 87° L = 0.50 BW = 78° L = 0.75 BW = 64° L = 1.0 BW = 48°

P8.17: How long is a 1.5 long dipole antenna at 1.0 GHz? Suppose this antenna is constructed using AWG#20 (0.406 mm radius) copper wire. Determine Rdiss, e, and Gmax.

c 3x108 = = 0.3m, L = 1.5 = 0.45m f 1x109 1l Rdiss =  S From example 8.2, the skin depth for this wire at 1 GHz is 2.09x10-6m. Then, the crosssectional surface over which we consider the current to be conducted is: S = 2 r Cu = 5.33x10−9 m2 Then: 1 0.45m Rdiss = = 1.456 7 1 5.33x10−9 m2 5.8 x10 m 30 Now we need radiation resistance, Rrad = F ( )max  P , and we use Matlab 0804 to find

=



P = 8.08 (and Dmax = 1.55), and Fmax = 1.366. Therefore, Rrad = 105.

The efficiency is Rrad e= = 0.986 Rrad + Rdiss Finally, Gmax = e Dmax = 1.53.

P8.18: Find the half-power beamwidth of a /2 dipole antenna.

One approach is to carefully plot the pattern and then estimate the beamwidth (see Figure P8.18). A more exact method is as follows.   cos 2  cos   2  = 1 , or 2cos 2   cos   = sin 2  Here we have Pn =   2 sin  2 2  We can define a function

8-16

  F ( ) = 2cos 2  cos   − sin 2  , and then 2  dF ( )       = −4cos  cos   sin  cos   − sin   − 2sin  cos  . d 2  2  2  Rearranging, we have dF     F '( ) = = 2 sin  cos  cos   sin  cos   − 2sin  cos  . d 2  2  Now we can apply the Newton Raphson routine to converge onto a solution: F ( i ) i +1 = i − .

F '( i ) Newton-Raphson routine: % MLP0818 % % Use Newton-Raphson to solve for BW % on half wavelength dipole antenna. % % 2/10/03 Wentworth clc clear N=20; th(1)=45; %initial wild guess for th thr(1)=th(1)*pi/180; for i=1:N F=2*cos((pi/2)*cos(thr(i)))^2-sin(thr(i))^2; dF=2*pi*sin(thr(i))*cos((pi/2)*cos(thr(i)))*sin((pi/2)*sin( thr(i)))-2*sin(thr(i))*cos(thr(i)); thr(i+1)=thr(i)-F/dF; end th=180*thr(N+1)/pi; BW=2*(90-th) Running the program: BW = 78.0777 >> So BW = 78°

Fig. P8.18

8-17 P8.19: A 2.45 GHz /2 dipole antenna is driven by a 2.0 A amplitude current source.

Find the maximum power density at a distance of 1.0 km.

Pmax =

15I o2 15(2)2 W = = 19 2 2 2 r  (1000) m

P8.20: Given a z-polarized half-wave dipole antenna at the origin, and a driving current i(t) = 10cos(2x109t) A, find the instantaneous electric and magnetic fields at a point 2.0 km distant and angle =60.

We have  =  c = (2 )(1x109 ) 3x108 = 20.944 / m To find Hos we modify equation (8.71) by considering (L/2)=/2:    cos    − j  r  cos  jI e 2 a  H os = o  2 r  sin        cos 60   − j (20.944)(2000)  cos  j (10) e 2   a = 6.5 x10−4 e − j156 a A  =   2 2000 sin 60 m    

V m Converting to instantaneous form: V E = 0.245cos (t − 156 ) a m mA H = 0.65cos (t − 156 ) a m These equations can also be written in terms of sin as: V E = 0.245sin (t + 114 ) a m mA H = 0.65sin (t + 114 ) a m Eos = −oar  H os = 0.245e− j156 a

P8.21: MATLAB: Modify MATLAB 8.4 to calculate directivity and radiation resistance for an arbitrary length dipole antenna. Evaluate these properties for a 0.75 dipole antenna. % M-File: MLP0821 % Modify ML0804 to calculate D and radiation resistance. % All that is needed is a line to calculate Rrad

8-18 % using Equation (8.79). % clc %clears the command window clear %clears variables % Initialize variables L=.75; bL2=pi*L; N=90; % Perform calculations i=1:1:N; dth=pi/N; th(i)=i*pi/N; num(i)=cos(bL2.*cos(th(i)))-cos(bL2); den(i)=sin(th(i)); F(i)=((num(i)).^2)./den(i); Fmax=max(F); Pn=F./Fmax; omegaP=2*pi*dth*sum(Pn) Dmax=4*pi/omegaP Fmax % Calculate Rrad Rrad=(30/pi)*Fmax*omegaP Running the program: omegaP = 6.6769 Dmax = 1.8821 Fmax = 2.9142 Rrad = 185.8086 >> So we have Dmax = 1.88 and Rrad = 186 .

8-19 P8.22: (JustAsk): Find a 3.0 m long dipole antenna’s directivity and radiation resistance if it is operated at (a) 250 MHz, (b) 500 MHz, and (c) 750 MHz.

c 3x108 m s 3m = = 1.2m, L = = 2.5 6 f 250 x10 / s 1.2m  Now we use this information in MLP0821. Plugging in L = 2.5, we have omegaP = 7.4529 (a)  =

D= 1.6861 Fmax = 1.6969 Rrad = 120.7662 >> Following the same approach for the other two frequencies, we arrive at the following table of results: f(MHz) Dmax L() Rrad() 250 2.5 1.69 121 500 5.0 2.37 342 750 7.5 2.23 154 P8.23: A 50  impedance line is terminated in a 3.0 m long dipole antenna at 50 MHz. What is the VSWR looking into this antenna? Design a shorted shunt stub network to impedance match the antenna to the 50  line.

At 50 MHz,  = 6 m so the antenna is a /2 dipole and Zant = 73.2+j42.5 . We can find the VSWR with the Smith chart. Or we can calculate it as follows: Z − Zo 73.2 + j 42.5 − 50  L = ant = = 0.3715e j 42.3 Z ant + Zo 73.2 + j 42.5 − 50 1 + L VSWR = = 2.18 1 − L The stub matching solution uses the approach of chapter 6. We first locate the normalized load (z = 1.46+j0.85) at point a, convert it to a normalized admittance at point b, and move along the constant gamma circle to point c (distance traveled is 0.215) where the admittance is y = 1+j1.8. Then we move from a short in the admittance chart to the point 0-j1.8, moving a distance 0.142.

8-20

(b) Fig. P8.23

P8.24: MATLAB: Use MATLAB 8.2 to generate plots like those of Figure 8.19 for a dipole antenna of length 3.

Inserting L = 3 in the Matlab routine generates the following figure.

Fig. P8.24

8-21 P8.25: A 0.485 dipole antenna is constructed for operation at 4.0 GHz. (a) How long is the antenna? (b) What impedance is required of a quarter-wave transformer to match this antenna to a 50  impedance line?

c 3x108  0.075m  = = 0.075m, L = ( 0.485 )   = 0.0364m, L = 3.6cm 9 f 4 x10    From section 8.3 describing the half-wave dipole, we know that a 0.485 dipole has Zant = 73. A quarter wavelength matching section will have an impedance: Z = (73)(50) = 60

=

P8.26: MATLAB: Modify MATLAB 8.3 to run the movie from 0.1 up to 4. % M-File: MLP0826 % % Modifies ML0803 to extend movie for L % up to 4 wavelengths % clc %clears the command window clear %clears variables % Initialize variables N=360; th=1:1:N; thr=th*pi./180; % Generate Reference Frame L=0.1; polar(0,10); %sets scale for polar plot T=num2str(L); S=strvcat('Length',T,'wavelengths'); text(10,10,S) axis manual title('Linear Antenna Radiation Pattern') hold on pause % Make the Movie L=0.1:0.02:4; for n=1:196 polar(0,10) axis manual title('Linear Antenna Radiation Pattern')

8-22 T=num2str(L(n)); S=strvcat('Length',T,'wavelengths'); text(10,10,S) hold on num=cos(pi*L(n)*cos(thr))-cos(pi*L(n)); den=sin(thr); F=(num./den).^2; polar(thr,F) hold off M(:,1)=getframe; end The figure shows a snapshot of the movie when the angle reaches 4.

Fig. P8.26

P8.27: MATLAB: Using MATLAB 8.4, generate data of the pattern solid angle versus number of increments N to see the function convergence. Consider a 1.25 dipole. Try N = 2, 4, 8, 16, 32, 64, 128.

The following data is generated: N WP 2 9.87 4 5.182 6 3.868 8 3.83 16 3.828 32 3.828

8-23

10

Pattern Solid Angle (sr)

9 8 7 6 5 4 3 2 1 0 0

5

10

15

20

25

30

35

N iterations

Fig. P8.27 The function converges above N = 8, so data from the N = 64 and 128 runs was omitted.

4. Monopole Antennas P8.28: Consider a 1.0 nC charge at (0.0, 0.0, 5.0m) above a conductive sheet occupying the x-y plane at z = 0. Use image theory to find the electric field intensity at the point (0.0, 5.0m, 5.0m).

Letting Q2 be the image charge, the total field at P is Q1 Q2 EP = a + aR2 2 R1 4 o R1 4 o R22 where R1 = 5ax and R2 = 5ax + 10az.

Also, R2 = 125m.

Evaluating the field from Q1: 1x10−9 C E1 = a −9 2 x 4 10 F 5 m 36 m ( ) V = 0.36a x . m from Q2: −1x10−9 C ( 5a x + 10a z ) E2 = 3 −9 4 10 F 125m 36 m Fig. P8.28 V = −0.032a x − 0.064a z . m The total field is then EP = 0.33ax – 0.064az V/m.

(

)

(

)(

)

8-24

P8.29: Find the half-power beamwidth for a quarter-wave monopole antenna.

In problem P8.18, we found the bandwidth for a  /2 dipole antenna was 78°. For the /4 monopole, the bandwidth will be half that of the  /2 dipole, or 39°.

P8.30: MATLAB: Devise a routine to give a polar plot of the normalized power radiated for an arbitrary length monopole antenna. Use your program to generate the polar plot for a half-wave monopole. % MLP0830 % Modify ML0802 to plot the normalized power radiated % for an arbitrary length dipole. % We remove the current distribution plot, and plot for % theta from -90° to +90°. % clc %clears the command window clear %clears variables clf %clear figure % Initialize variables Lmono=0.5; L=2*Lmono; bL2=pi*L; N=180; % Calculate normalized power function th=1:.1:N; thr=(th-90)*pi./180; F=((cos(bL2.*cos(thr))-cos(bL2))./sin(thr)).^2; Fmax=max(F); Pn=F./Fmax; % Generate Plots polar(0,1) hold on polar(thr,Pn) T=num2str(Lmono); S=strvcat('Monopole length',T,'wavelengths'); text(1.0,.8,S)

8-25

Fig. P8.30

P8.31: Determine the pattern solid angle, directivity and radiation resistance for a halfwave monopole antenna.

From Example 8.3 we found the following for a 1 dipole: P = 5.21 sr Dmax = 2.41 Rrad = 200  Now, for a /2 monopole, 1 1 p = p , Dmax monopole = 2 Dmax dipole , Rrad monopole = Rrad dipole , monopole dipole 2 2 So, P = 2.6 sr Dmax = 4.8 Rrad = 100 

P8.32: How long is a 0.75 monopole antenna at 1.0 GHz? Suppose this antenna is constructed using AWG#20 (0.406 mm radius) copper wire. Determine Rdiss, e and Gmax.

Compare your results with the 1.5 dipole antenna of problem P8.17.

c 3x108 m s 3 = = = 0.30m, L = = 0.225m = 22.5cm 9 f 1x10 / s 4

8-26

(from P8.17 solution) 1l Rdiss =  S From example 8.2, the skin depth for this wire at 1 GHz is 2.09x10-6m. Then, the crosssectional surface over which we consider the current to be conducted is: S = 2 r Cu = 5.33x10−9 m2 Then, for the monopole, 1 0.225m Rdiss = = 0.73 7 1 5.33x10−9 m2 5.8 x10 m To find e and Gmax, we need P, Dmax and Rrad. We can find these by running Matlab 8.4 for a 1. dipole, and then use 1 1 p = p , Dmax monopole = 2 Dmax dipole , Rrad monopole = Rrad dipole . monopole dipole 2 2 We find P = 4.04 sr Dmax = 3.10 Rrad = 52.7  (half that of a 1.5 dipole) So

e=

Rrad 52.7 = = 0.986 (same as a 1.5 dipole) Rrad + Rdiss 52.7 + 0.73

and Gmax = eDmax = 3.04 (twice that of a 1.5 dipole)

P8.33: What is the VSWR looking into a quarter-wave monopole antenna if the feed line has a 50  impedance? Design an open-ended shunt stub matching network to match this antenna to the line.

For a /4 monopole, we have Zant = 36.6 + j21.25 .

Z − Z o 36.6 + j 21.5 − 50  = ant = = 0.282e j108 Z ant + Z o 36.6 + j 21.5 + 50

1 +  1 + 0.282 = = 1.79 1 −  1 − 0.282 The Smith Chart solution for the open-ended stub matching network is shown in the figure. After locating zL, we find yL and notice it is almost on the the 1 ± jb circle, at about 1 – j0.6. So we don’t need a through line. The stub needs to give us y = 0 + j0.6, or a length 0.086 is required.

VSWR =

8-27

Fig. P8.33

P8.34: Given a 1 GHz quarter-wave monopole antenna at the origin, excited by a 1.0 A amplitude current, find the amplitudes for the electric and magnetic field intensities at a point 1.0 km distant at an angle  = 80.

H os =

=

 c

H os =

− j r

jI o e 2 r

=

  cos  cos   2  a and E = j H a  os o os  sin 

2 (1x109 / s ) 3x108 m s

= 20.94 rad m , r = 1000m,

(1A) 0.978 = 156  A , E = 120 H 2 1000m

m

os

os

= 59

  cos  cos80  2  = 0.978 sin 80

mV m

5. Antenna Arrays P8.35: (JustAsk): Find and plot the far-field radiation pattern at  = /2 for a two element dipole antenna array given the following: 1. the dipoles are driven in-phase 2. each dipole is 1 in length oriented in the z-direction 3. the pair of dipoles are 1 apart on the x axis.

Also find the maximum time-averaged power density, in W/m2, 1.0 km away from the array if each antenna is driven by a 1.0 A amplitude current source at 1.0 GHz. We’ll first use ML0804 to find p, D and Fmax. % %

M-File: MLP0835a application of ML0804

8-28 % clc clear

%clears the command window %clears variables

% Initialize variables L=1.0; bL2=pi*L; N=90; % Perform calculations i=1:1:N; dth=pi/N; th(i)=i*pi/N; num(i)=cos(bL2.*cos(th(i)))-cos(bL2); den(i)=sin(th(i)); F(i)=((num(i)).^2)./den(i); Fmax=max(F); Pn=F./Fmax; omegaP=2*pi*dth*sum(Pn) D=4*pi/omegaP Fmax Running the program: omegaP = 5.2121 D = 2.4110 Fmax = 4 >> Using Eqn. (8.76) for a dipole of length L =  at an angle  = /2, we have 15I o2 15I o2 60 I o2 Pmax = F  = 4 = = Funit ( ) ( ) max  r2  r2  r2 Then, for the array we have  Farray = 4cos 2   , where  =  d cos  +  = 2 cos  2 2     60 I o   P  r, ,  =  4cos 2 ( cos  ) , 2   2   r  or

240 (1)  60 I 2  240 I o2 W Pmax =  2o  4 = = = 76.4 2 2 2 r m  (1000 )  r  2

8-29 The far-field radiation pattern (plot of cos2(cos)): %MLP0835b clc clear phi=.5:.5:360; phir=phi*pi./180; Psi=2*pi*cos(phir); Pn=(cos(Psi./2)).^2; polar(phir,Pn)

Fig. P8.35

P8.36: Repeat problem P8.35 if the dipoles are 180 out of phase.

From P8.35 we have p=5.212, D =2.411 and Fmax=4, and 60 I o2 Funit = .  r2 Then, for the array we have  Farray = 4cos 2   , where  =  d cos  +  = 2 cos  + 2 2     240 I o   P  r, ,  = cos 2   cos  +  , 2 2  2  r  or W Pmax = 76.4 2 m The far-field radiation pattern (plot of cos2(cos+)): %MLP0836 clc clear phi=.5:.5:360; phir=phi*pi./180; Psi=2*pi*cos(phir)+pi; Pn=(cos(Psi./2)).^2; polar(phir,Pn)

8-30

Fig. P8.36

P8.37: Repeat P8.35 for the case where the dipoles are 90 out of phase, 1.5 in length, and separated by /2.

We’ll first use ML0804 to find p, D and Fmax. % M-File: MLP0837a % modify ML0804 clc %clears the command window clear %clears variables % Initialize variables L=1.5; bL2=pi*L; N=90; % Perform calculations i=1:1:N; dth=pi/N; th(i)=i*pi/N; num(i)=cos(bL2.*cos(th(i)))-cos(bL2); den(i)=sin(th(i)); F(i)=((num(i)).^2)./den(i); Fmax=max(F); Pn=F./Fmax; omegaP=2*pi*dth*sum(Pn) D=4*pi/omegaP

8-31 Fmax Executing the program we find p= 8.087, D =1.554 and Fmax= 1.366.

Now: 15I o2 15I o2 Pmax = F  = ( ) (1.366 ) = Funit max  r2  r2 Then, for the array we have   Farray = 4cos2   , where  =  d cos  +  =  cos  + 2 2 2     15I o   P  r, ,  = (1.366)4cos 2  cos  +  2 4  2  r 2 or

Pmax =

15 (1)

2

 (1000 )

2

(1.366)4 = 26

W m2

.

Then we plot:   Pn = cos2  cos  +  , 4 2

Fig. P8.37

P8.38: Two z-polarized /2 dipole antennas are spaced /4 apart, centered at the origin on the x-axis. (a) If the dipole located at x = - /8 is driven by Is1 = Ioej0º, what phase shift  would you employ on the other dipole (Is2 = Ioej) to get maximum power at a far-field point on the +x axis? (b) If the dipole antennas are each driven by 1.0 A amplitude currents at 500 MHz, with the phase shift from part (a), find the time-averaged power density vector at 2.0 km on the x-axis. (a)

 Farray = 4cos 2   , where  =  d cos  +  . 2 2   On the x-axis,  = 0 so  = + = +  4 2

 We want maximum Farray, or Farray = 4, when cos 2   = 1. 2

8-32

   = + = 0,  , 2 ..., 2 4 2 satisfied when  = -/2 or 3/2. So we employ a -90° or a +270° phase shift.

This occurs when

2

   cos    2  cos  15I o 2  a .  (b) P(r , ) unit = r 2 r  sin       2 15I o .

At  = /2, we then have Funit =  r2 The radiated power vector is then

15 (1)  15I o2  P(r , ) array =  2  4a r = 4a r = 4.78Wa r 2  ( 2000 )  r  2

P8.39: Two small loop antennas, each oriented in the x-y plane, are centered at ±/2 on the x-axis. They each have a 1.0 cm radius and are driven in-phase by a 10. mA current source at 500. MHz. Find and plot the radiation pattern at  = /2 and determine the maximum time-averaged power density at a distance 100. m from the array.

2  Farray = 4cos 2   ,  =  d cos  +  =  cos  + 0 = 2 cos   2 so Farray = 4cos2 ( cos  ) , Farray =4 max

For magnetic dipoles we have:

1  o I o  S  Pmax 1 loop =   , 32o   r  where 2 c 3x108 2 = , = = = 0.6m, S =  a 2 =  ( 0.01) 8  f 5 x10 so 2 2 6 −7   2  500 x 10 4  x 10 0.01 2  0.6  0.01 ( ) ( )( ) ( ) ( )( ) 1   Pmax 1 loop =  32 (120 )   (100 )   pW Pmax 1 loop = 14.2 m2 Pmax = Pmax 1 loop 4 = 57 pW 2 m 2 A plot of cos (cos) gives the same result at P8.35.

2

(

)

8-33 P8.40: Given a pair of dipole antennas separated by /4 and driven in-phase, determine, for  = /2, (a) the values for  at the nulls in the radiation pattern, and (b) the values of  where the radiated power is maximum.

2  cos  + 0 =  cos   4    Pn = cos   = cos  cos   , 2 2  and maximum Pn occurs at  = /2, 3  /2. (a)To find the location for the minimum Pn, we must take a derivative of the Pn function: (b)  =  d cos  +  =

dPn    = sin  sin  cos   = 0 at  = 0,  (for Pn min ) d 2 2 

P8.41: MATLAB: Create a movie to plot the radiated power pattern in the x-y plane for the pair of dipoles in Example 8.7 as the separation distance varies from /10 to 4. % MLP0841 % Show in a movie how the radiated power pattern % in the x-y plane varies with separation distance % between a pair of in-phase dipoles. % % Wentworth, 2/11/03 % clc %clears the command window clear %clears variables % Initialize variables N=360; phi=1:1:N; phir=phi*pi./180; % Generate Reference Frame L=0.1; polar(0,1); T=num2str(L); S=strvcat('Separation distance:',T,'wavelengths'); text(1.0,1.0,S) axis manual title('dual element dipole array radiation pattern') hold on pause

8-34

% Make the Movie L=0.1:0.01:4.0; for n=1:391 polar(0,1) axis manual title('dual element dipole array radiation pattern') T=num2str(L(n)); S=strvcat('Separation distance:',T,'wavelengths'); text(1.0,1.0,S) hold on P=(cos(pi*L(n)*cos(phir))).^2; polar(phir,P) hold off M(:,1)=getframe; end

Fig. P8.41: snapshot at end of movie

P8.42: Plot the normalized radiation pattern at  = /2 for 3 dipole antenna elements spaced /2 apart with progressive phase steps of 90º.

We will use Equation (8.103). %MLP0842 clc clear

8-35 phi=.5:.5:360; phir=phi*pi./180; Psi=pi*cos(phir)+pi/2; N=3; num=(sin(N.*Psi./2)).^2; den=(sin(Psi./2)).^2; Pn=(num./den)./(N^2); polar(phir,Pn)

Fig. P8.42 P8.43: A particular broadside antenna array consists of 10 /2 dipole antenna elements spaced /2 apart with all currents driven at the same phase. Plot the radiation pattern, and find the maximum broadside power density (i.e. at  =  = /2) at a distance of 10. km if the antenna is driven by 10.A current sources at 2.45 GHz. The broadside array has 2  15I o2  2  15 (10 )  2 W Pmax = Funit Farray =  2 ( N ) =   10 ) = 480 2 2 ( max   m  r    (10, 000 )  2 N 1 sin 2 ,  =  d cos  +  =  cos  Pn = 2 N sin 2  2 So we plot 2 1 sin ( 5 cos  ) Pn = 100   sin 2  cos   2   %MLP0843 clc clear phi=.5:.5:360; phir=phi*pi./180; Psi=pi*cos(phir); N=10; num=(sin(N.*Psi./2)).^2; den=(sin(Psi./2)).^2; Pn=(num./den)./(N^2); Fig. 8.43

(

( )

)

8-36 polar(phir,Pn) P8.44: A particular endfire antenna array consists of 10 /2 dipole antenna elements spaced /2 apart with a progressive phase shift of 90º to each antenna. Plot the radiation pattern, and find the maximum endfire power density (i.e. at  = /2 and  = 0º) at a distance of 10. km if the antenna is driven by 10A current sources at 2.45 GHz. This has the same Pmax as P8.43 (480 W/m2). In the MATLAB routine of P8.43, we use

 =  cos  +



2

.

6. The Friis Transmission Equation P8.45: Consider a pair of half-wave dipole antennas operating at 2.45 GHz, separated by 50. m and aligned for maximum power transfer.

If the output power must be at least -35 dBm to be detectable, calculate how much power is required to drive the transmitting antenna.

Assume the antennas are 100% efficient.

Fig. P8.44

c 3x108 At f = 2.45 GHz we have  = = = 0.122m f 2.45 x109 Also, for half-wave dipoles the maximum directivity is 1.64.

Applying the Friis Equation: 2

2 Prec 2  0.122     −9 W = Dmax1Dmax 2   = 102 x10  = (1.64 )  Prad W  4 R   4 ( 50 ) 

or -70dB. Now, with Prec = -35 dBm, we have Prad = -35 dBm + 70 dB = 35 dBm, or 3.2 W. P8.46: (JustAsk): A half-wave dipole transmitting antenna is centered on the z-axis oriented in the z direction. Show in a sketch where would you place a small loop antenna, 100 m distant, to receive the maximum power. (Hint: consider both radiation pattern and polarization to achieve maximum power transfer.) Calculate the power transfer ratio for the maximum power transfer case at 800 MHz if the small loop antenna has a 2.0 cm diameter.

8-37 Maximum radiation occurs at  = 90°, so we choose, for instance, a point on the y-axis.

Then, we know for a small loop antenna the polarization will be in a plane containing the loop. For maximum power transfer the polarizations must match. Therefore the loop can be parallel to the y-z plane or to the x-z plane, but not to the x-y plane. Figure P8.46 shows a suggested orientation.

Fig. P8.46 Now, to find the power transfer ratio we apply the Friis equation. We first find c 3x108 = = = 0.375m f 0.800 x109 and we also know for a half-wavelength dipole Dmax = 1.64. Since the loop diameter is small compared to the wavelength, we have for the small loop Dmax = 1.5. So 2

2  0.375  Prec    −9 W = Dmax1Dmax 2   = 219 x10  = (1.64 )(1.5)  Prad W  4 R   4 (100 ) 

So the power transfer ratio is Prec/Prad = -67 dB.

P8.47: A pair of z-polarized dipole antenna with lengths indicated is shown in Figure 8.50. If the 3.0 m dipole is driven by a 50. MHz source, calculate the power transfer ratio.

Prec    = D ( ,  ) D ( ,  )   Prad  4 R  where  = c = 6m and R = 4000m. f P ( ,  ) P ( ,  ) 4 D ( ,  ) = n = n = Pn ( ,  ) = Dmax Pn ( ,  ) Pn ( ,  )avg  p 4  p 2

For the transmitter,

8-38

  cos 2  cos   2  = 0.667 Pnt ( = 60 ,  ) = 2 sin  and Dmax = 1.64 for the /2 dipole.

For the receiver, a 4 cm length dipole is much smaller than the wavelength, so we can consider this antenna to be a Hertzian dipole. Then, Pnr ( = 120 ,  ) = sin 2  = 0.750 and Dmax = 1.5 for the Hertzian dipole.

So, 2

  Prec 6 −9 W = (1.64 )( 0.667 )(1.50 )( 0.750 )   = 17.5 x10 Prad W  4 ( 4000 )  or Prec = −77 dB Prad

P8.48: Consider a pair of half-wave dipole antennas operating at 1.0 GHz and separated by 100. m on the y-axis. Initially, both antennas are aligned in the z-direction for maximum power transfer. Now to test the effect of polarization, the antenna at the origin is allowed to rotate an angle  in the x-z plane as shown in Figure 8.51. Plot the power transfer ratio versus  from  =0º (maximum transfer case) to  = 90º.

Pout    = e p D1 ( ,  ) D2max   Pin  4 R 

2

  cos 2  cos   2  D1 ( ,  ) = Dmax Pn ( ,  ) , Pn ( ,  ) = 2 sin  Inspecting the problem’s geometry, we see that the angle  in our Pn equation is referenced to the axis of the dipole. For simplification, we’ll let  be the angle the dipole makes with the z-axis and then let  be the angle from the dipole axis to a line drawn to the second antenna. Figure P8.48a shows this situation. Then we have    cos 2  cos  −    2  2 Pn ( ,  ) =   sin 2  −   2 

8-39

Fig. P8.48a Fig. P8.48b The polarization efficiency term is seen to be ep = cos2.

At 1 GHz, we have l = 0.3 m. And for half-wave dipoles we have Dmax = D2max = 1.64. % MLP0848 clc clear D=1.64; lambda=0.3; R=100; B=(lambda/(4*pi*R))^2; alpha=1:1:88; alphar=alpha*pi./180; A=cos((pi/2)*cos((pi/2)-alphar)); Pn=A.^2./(sin((pi/2)-alphar)).^2; D1=D.*Pn; ep=(cos(alphar)).^2; Prat=ep.*D1.*D.*B; PdB=10*log10(Prat); plot(alpha,PdB) xlabel('angle(degrees)') ylabel('(Pout/Pin) dB') grid on

8-40 P8.49: Design an open-ended shunt stub matching network to match a half-wave dipole transmitting antenna to a source with 50  impedance. Now suppose this antenna network is to be used as a receiver. Use a Smith Chart to determine the impedance looking into the matching network from the antenna.

The impedance matching network is solved similar to approach used in P8.23, only now the shunt stub is open-ended instead of shorted. The solution is shown in FigP8.49a.

(a)

(b)

Fig P8.49 a & b

Now we need to find the input impedance looking into the matching network from the antenna, as indicated in Fig P8.49c.

Fig P8.49 b & c

8-41

First, looking into the open-ended stub of length 0.392 we see normalized admittance ya = -j0.8. Adding this to the load admittance (yload = 1+j0) we have point b: yb = 1-j0.8.

We move from this point (at 0.344  on the WTG scale) a distance 0.215  towards the generator to the point 0.0559 . Then we move to the impedance chart and see zin = 1.5 – j0.8. De-normalizing, we find: Zin = 75 – j40 . This is close the the theoretical Zin = Zant* = 73 – j42 .

P8.50: Referring to Figure 8.52, suppose a source voltage with amplitude 12.V and source resistance 50  drives a half-wave dipole transmitting antenna at 500 MHz. An identical receiving antenna, 100. meters away and aligned for maximum power transfer, is coupled to a 50  load resistance. Clearly neither antenna is impedance-matched to the transmit and receive circuitry. Calculate the voltage amplitude across this load resistor.

Fig. P8.50 a&b Using FigP8.50a we see that vs 1 12V Prad = i 2 Rrad , i = = = 0.092 A, 2 Z o + Z ant ( 50 + 73.2 + j 42.5) 

 Prad =

1 2 (.092 ) ( 73.2 ) = 310mW 2

Now we find the transfer ratio by applying the Friis equation where  = 2

2  0.6  Prec    −9 = Dmax t Dmax r   = 613x10  = (1.64 )(1.64 )  Prad  4 R   4 (100 ) 

So Prec = 190 nW. Now we calculate Voc by assuming a matched load:

c = 0.6 m : f

8-42

Voc2 Prec = , Voc = 4 Rrad Prec = 7.46mV 4 Rrad Finally, we calculate vL (see Figure P8.50b): Zo 50 vL = Voc = ( 7.46mV ) = 2.9mV Zo + Z ant 50 + 73.2 + j 42.5

P8.51: Design open-ended shunt stub matching networks for both the transmitter and receiver of problem P8.50. Now recalculate the voltage amplitude across the load resistor.

The matching networks are the same as for P8.49.

Now we calculate Prad: v 1 Prad = i 2 Zo , i = s = 120mA,  Prad = 360mW 2 2Z o The power transfer ratio is: Prec = 613x10−9 , from P8.50.

Prad Therefore, Prec = 220.7nW. Now, since the receiver is matched, half the power must be dissipated in the load, or 1 1 vL2 Prec = , vL = Prec Z o = 3.3mV 2 2 Zo

7. Radar P8.52: Manipulate (8.125) using (8.113) to arrive at (8.126).

Rearranging (8.113) we find 4 D ( ,  ) = 2 Ae ( ,  )



This can be inserted into (8.125) for D(), or 2 2 Prec1  s  2  4 s  = A  ,  = A  ,  . ( ) ( ) ( ) e e   3 2 4 2 Prad 1 ( 4 ) R 4   4 R  

P8.53: (JustAsk): Suppose a 2 GHz radar antenna of effective area 6.0 m 2 transmits 100 kW. If a target with a 12 m2 radar cross section is 100 km away, (a) what is the roundtrip travel time for the radar pulse? (b) What is the received power? (c) What is the maximum detectable range if the radar system has a minimum detectable power of 2.0 pW? The given information is:

8-43 f = 2GHz, so  = 0.15m Ae = 6.0m2 Prad1 = 100 kW s = 12m2 R=100 km (a) t = 2

R = 0.67ms c

(b) Prec1 = Prad 1

s

A = (100kW ) 2 e

12m2

( 6m ) = 1.53 pW 2

4 (100km ) ( 0.15m ) (c) To find Rmax if (Prec)min = 2.0pW, we rearrange the radar equation as P  100kW 12m2 2 R 4 = rad 1 s 2 Ae2 = 6m ) = 76.4 x1018 2 ( Prec1 4 2 pW 4 ( 0.15m ) 4 R  4

2

4

2

R = 93 km.

P8.54: A half-wave dipole antenna is used in a radar system to determine range to a target that has a 1.0 m2 radar cross section. Consider that 1.0 kW is available to drive the antenna at 300 MHz. What power is received if the target is (a) 100 m distant? (b) 1.0 km distant?

At 300 MHz we have  = 1m.

Now we use (8.125): 2 1m2 ) (1m ) ( Prec1  s 2 2 2 = D ( ,  ) = (1.64 ) 3 3 4 4 Prad 1 ( 4 ) R ( 4 ) R With Prad1 = 1 kW, we then have 1.355 Prec1 = R4 (a) Prec1 = 13.6 nW (b) Prec1 = 1.36 pW P8.55: Suppose a 10 GHz radar antenna of effective area 100 m2 is to be used to determine the distance to the moon. The moon, with radius 1.74 x 106 m, has a measured radar cross section of 6.64 x 1011 m2. A 27 pW echo signal is received 2.56 seconds after transmission. (a) What is the distance to the moon, and (b) approximately how much power was radiated?

The given information is: f = 10 GHz (and therefore  = 0.030 m) Ae = 100m2 s = 6.64x1011 m2 Prec = 27 pW

8-44 t = 2.56 sec (a) we know that t = 2R/c, so R = tc/2 = 384x106 m (b) Prec1 s 6.64 x1011 m2 2 2 = Ae = (100m ) = 27 x10−18 4 4 2 2 6 Prad 1 4 R  4 ( 384 x10 m ) ( 0.030m )

Prad 1 =

Prec1 27 x10−12 = = 1MW 27 x10−18 27 x10−18

9-1 Solutions for Chapter 9 Problems 2. Passive Circuit Elements P9.1: Given a 2.0 cm length of AWG20 copper wire, (a) calculate Rdc, (b) calculate Rac at 800 MHz, (c) estimate L.

31.96mils 25.4 m m = 406 x10−6 m 2 mil 106  m l 0.02 Rdc = = = 666 2  a ( 5.8 x107 )  ( 406 x10−6 )2

a=

At 800 MHz:

Rac =

l , where  = 1 = 2.336 x10−6 m  f   2 a

So Rac =

0.02 = 58m ( 5.8x10 ) ( 2 ) ( 406 x10−6 )( 2.336 x10−6 ) 7

  2(0.02)   L  ( 2 x10−7 ) ( 0.02 ) ln  − 1 = 14.4nH , −6    406 x10   so L = 14 nH P9.2: MATLAB: Repeat MATLAB 9.1 for a typical 200  chip resistor (LL = 0.40 nH, Cx = 50 fF). Compare the resulting Bode plot with that of Figure 9.7. %Resistor Equivalent Circuit %P0902: modify ML0901 for typical chip resistor %Enter equivalent circuit values R=200; %resistance in ohms LR=0; %element inductance in nH LLcc=13.4; %lead inductance in nH for carbon comp res CRcc=2; %element capacitance, in pF for carb comp res LLcr=0.4; %lead inductance in nH for chip res CRcr=.05; %elem cap, in pF, for chip res f=10e6:10e6:10e11; w=2*pi*f; XLR=complex(0,w*LR*1e-9); XLLcc=complex(0,w*LLcc*1e-9); XCRcc=complex(0,-1./(w*CRcc*1e-12)); XLLcr=complex(0,w*LLcr*1e-9); XCRcr=complex(0,-1./(w*CRcr*1e-12));

9-2

Zcc=XLLcc+parallel(R+XLR,XCRcc); Zcr=XLLcr+parallel(R+XLR,XCRcr); Zmagcc=abs(Zcc); Zmagcr=abs(Zcr); loglog(f,Zmagcc,'--k',f,Zmagcr,'k') legend('carbon composite','chip resistor') grid on xlabel('frequency (Hz)') ylabel('Z magnitude (ohms)')

Fig. P9.2 We see the chip resistor may be operated almost two orders of magnitude higher in frequency than the carbon composite resistor.

P9.3: Recalculate L, Cx and fSRF if the AWG30 wire for the coil of Example 9.2 is replaced with AWG40 wire.

For this wire we have t =3.145 mil = 79.9 m.

L hasn’t changed from the Example, so L = 1.3 H.

d=

h − Nt 300mil − 20(3.145mil ) = = 12.5mils N −1 19

S = t (2 a)( N − 1) = (3.145mils)(2 )(0.25cm)(19)

1000mils = 37 x103 mil 2 2.54cm

9-3

(8.854 x10 C =

−12

x

f srf =

2

F m )( 37 x103 mil 2 ) 25.4 x10−6 m

12.5mil 1

(1.3x10−6 )( 666 x10−15 )

mil

= 666 x10−15 F

= 171MHz ,

so fsrf = 170 MHz.

P9.4: Estimate L and the SRF if a 99.8% iron core is inserted inside the coil of Example 9.2.

From Example 9.2 we had  N 2 a 2 L= o = 1.3 H . h Now, we have the iron core with r = 5000, so L = (5000)(1.3 H ) = 6.5mH .

For the self resonance frequency, we use Cx = 5.2 pF as before and find 1 f srf = = 866kHz, 2 ( 6.5mH )( 5.2 pF ) or fsrf = 870kHz.

P9.5: Consider a 99.8% iron toroidal core of inner diameter 0.50 cm and outer diameter 1.0 cm wrapped with 20 turns of evenly spaced AWG26 copper wire. Estimate inductance and the self-resonance frequency of this toroidal inductor.

For inductance, from P3.56 we found: L =



=

N2

(b − a )

2

I 2b where b = (a+c)/2 = 0.375 cm.

For the capacitance we have S C= , where S = t(2ac)(N-1), and ac = (c-a)/2 = 0.125 cm, and t = 405 m. d Also, h − Nt for a toroid, where h = core length = 2b. d= N Plugging in the numbers we have 2 2 ( 5000 ) ( 4 x10−7 ) ( 20 ) ( 0.00375 − 0.0025 ) L= = 524 H 2 ( 0.00375 )

S = ( 405x10−6 ) ( 2 )( 0.00125)(19 ) = 60.4 x10−6 m2

9-4

d=

2 ( 0.00375) − 20 ( 405 x10−6 )

C=

20 )( 60.4 x10−6 )

(8.854 x10 ( 773x10 ) −12

−6

f srf =

1 2 LC

=

= 773x10−6 m

= 0.69 fF 1

2

(524 x10 )( 0.69 x10 ) −6

−15

= 8.4MHz.

P9.6: Calculate the self-resonance frequency for a 47 nF mica capacitor with a pair of 1.0 cm long AWG 24 copper leads.

We can apply the equivalent circuit of Figure 9.13(b), where we can neglect the very small resistance Rx. Then,  H    2l   Lx = 2 x10−7   l ln   − 1 , where each lead is l = 1 cm, and m   a  

a=

 25.4 x10−6 m  1 −6 ( 20.1mils )   = 255 x10 m, so 2 mil  

  2 ( 0.01)   H Lx = 2 x10−7   ( 0.01) ln  − 1 = 6.7nH . −6  m   255 x10   With the two leads in series we then have Ltot = 13.4 nH, and then 1 1 f srf = = = 6.3MHz , 2 Ltot C 2 (13.4 x10−9 )( 47 x10−9 ) so fsrf = 6 MHz. P9.7: A thin film capacitor is made by sandwiching a 0.10 m thick layer of Teflon between copper conductive layers. Determine the capacitance per unit area and the maximum voltage that can be applied across such a capacitor.

For Teflon, from the appendix we have r = 2.1 and Ebr = 60x106 V/m.

So, −12 C  2.1( 8.854 x10 ) F = = = 186 2 , −6 S d 0.1x10 m and V Ebr = max , or Vmax = Ebr d = ( 60 x106 )( 0.1x10−6 ) = 6V . d

9-5 P9.8: If the 2.2 nF capacitor of Example 9.3 has an area of 20. mm 2, what thickness mica is used? What is the maximum voltage that can be applied across this capacitor?

For mica, from the appendix we have r = 5.4 and Ebr = 200x106 V/m.

F  5.4 )  8.854 x10−12  ( 20mm2 ) ( 2  r o S S m  1m   C= , d= =   = 435nm d C 2.2 x10−9 F  1000mm  Vmax = Ebr d = ( 200 x106 )( 435x10−9 ) = 87V .

P9.9: (JustAsk): Suppose a standard 300.  twin-lead T-line is constructed with AWG 24 wire separated by a center-to-center spacing of 0.800 cm. If this line is terminated in a short circuit realized using the shortest possible length of AWG 24 wire, calculate the reflection coefficient looking into this “short” at 100 MHz, 1 GHz and 10 GHz. AWG24 has a radius a = 255 m. Using Equation (9.4),   2 ( 0.008 )    H    2l   Lx = 2 x10−7   l ln   − 1 = 2 x10−7 ( 0.008 ) ln  − 1 = 5.02nH −6  m   a     255 x10   The impedance looking into the line, neglecting the very small resistance, is ZL = jL. At 100 MHz, we then have Z L = j ( 2 ) (100 x106 )( 5.02 x10−9 ) = j3.16.

The reflection coefficient is found by Z − Zo j3.16 − 300 L = L = = 1e j179 .

Z L + Z o j3.16 + 300 Repeating the calculation at 1 GHz and at 10 GHz, we have: Frequency 100 MHz 1 GHz 10 GHz

ZL() j3.16 j31.5 j315

L 1ej179° 1ej168° 1ej87°

3. Digital Signals P9.10: (JustAsk): What is the spectral bandwidth for a 4.0 ns risetime signal, using equation (9.13)? What risetime is required to achieve a 1 GHz bandwidth? (a) The bandwidth is approximated by Equation (9.13), 1 1 BW  = = 250MHz. tr 4 x10−9 s (b) Modifying Equation (9.13), 1 1 tr  = = 1ns.

BW 1x109

9-6

P9.11: Suppose a 1.0 GHz clock rate is assumed. bandwidth calculated using Eqn. (9.13).

What is the minimum spectral

The minimum bandwidth occurs for the maximum possible rise time, or when the signal is a sawtooth function. At 1 GHz, the period is T = 1/f = 1 ns. For a sawtooth wave, then, the risetime would be half the period, or tr = 0.5 ns. Then we have 1 1 BW  = = 2GHz. tr 0.5 x10−9 s

P9.12: MATLAB: Modify MATLAB 9.3 to look at a 1.0 GHz clock rate signal.

Minimize the rise and fall times by letting the signal be a sawtooth wave. % MLP0912 % Modify ML0903 to look at a 1 GHz % clock rate signal, with minimum % rise and fall times of a sawtooth wave. % clc %clears the command window clear %clears variables % Initialize variables N=40; fo=1e9; wo=2*pi*fo; To=1/fo; Vo=1; t1=0; t2=0.50*To; tf=t2-t1; % Determine the coefficients a0=(Vo/To)*(t1+t2); for i=1:N n(i)=i; a(i)=((2*Vo)/(pi*wo*tf*i^2))*(cos(i*wo*t1)cos(i*wo*t2)); end % Determine the function components for j=1:1000 t(j)=j*To/1000; for i=1:N

9-7 Fn(i)=a(i)*cos(i*wo*t(j)); end F(j)=a0+sum(Fn); end % Generate the plot subplot(2,1,1) plot(t,F) axis([0 1e-9 0 1]) xlabel('time(sec)') ylabel('voltage(V)') grid on subplot(2,1,2) m=1:40; b=a(m); bar(m,b,'-k') axis([0 40 0 0.5]) xlabel('n') ylabel('an')

Fig. P9.12

4. Grounds P9.13: Repeat Example 9.4 using AWG22 wire and 200 MHz current.

AWG22 wire has a =

1 25.4 m  ( 25.35mils )   = 322 m, so we can use equation (9.4) to 2  mil 

find

  2 ( 0.04 )    H    2l   Lx = 2 x10−7   l ln   − 1 = 2 x10−7 ( 0.04 ) ln  − 1 = 36nH −6  m   a     322 x10   At 200 MHz, the impedance is Z = j L = j ( 2 ) ( 200 x106 )( 36 x10−9 ) = j 45 Figure P9.13 indicates our situation. By circuit analysis we then have VA = ( 3mA)( 45 ) = 135mV ,

VB = ( 2mA)( 45 ) + 135mV = 225mV , VC = (1mA )( 45 ) + 225mV = 270mV .

9-8

Fig. P9.13

P9.14: Repeat Example 9.5 using AWG22 wire and 200 MHz current.

As in P9.13, we have for AWG22 wire a = 322 m, and for a 4 cm length equation (9.4) gives   2 ( 0.04 )    H    2l   Lx = 2 x10−7   l ln   − 1 = 2 x10−7 ( 0.04 ) ln  − 1 = 36nH −6  m   a     322 x10   At 200 MHz, the impedance of the 4 cm length impedance is Z = j L = j ( 2 ) ( 200 x106 )( 36 x10−9 ) = j 45 .

Likewise, for the 8 cm length wire we have Lx = 83nH and Z = j105.

For the 12 cm length wire we have Lx = 135nH and Z = j170.

Now, VA = (1mA)( 45 ) = 45mV ,

VB = (1mA)(105 ) = 105mV , VC = (1mA)(170 ) = 170mV .

5. Shields P9.15: The field within a shielded enclosure is 12 kV/m. What shielding effectiveness is required such that the field outside the shield is no more than 1.0 nV/m?

 E   12 kV m  SE = 20 log  ns  = 20 log   = 262dB  1 nV m   Es  P9.16: (JustAsk): Compare the attenuation in dB at 1.0 GHz for 20 m thick layers of (a) copper, (b) aluminum and (c) nickel.

=

1



=  f  =  (1x109 ) r ( 4 x10−7 ) 

9-9

(

) (

)(

)

For copper,  =  1x109 (1) 4 x10−7 5.8 x107 = 479 x103 Np m,

 dB   Np  −6 And the attenuation is Atten(dB) =    , which for copper is  ( 20 x10 m )  8.686 Np   m  

Atten(dB) = ( 479 x103 )( 20 x10−6 m ) (8.686 ) = 83dB.

Using this approach the following table is generated: r (S/m) (a) copper 1 5.8x107 (b) aluminum 1 3.8x107 (c) nickel 600 1.5 x107

(Np/m) 479x103 387x103 5.96x106

Atten(dB) 83 67 1040

P9.17: A particular silver-filled paint is to be used as an absorptive layer. It has  = 1.0x106 S/m with r and r assumed equal to one. Calculate the attenuation of a 100 MHz wave propagating through a 50. m thick layer and compare with the attenuation through a pure silver layer of the same thickness.

Atten = t, where  =  f  .

For the silver paint we have

 paint =  (100 x106 )( 4 x10−7 )(106 ) = 19.9 x103

Np . m

Then,

 Np  dB   −6 Attenpaint = 19.9 x103  = 8.6dB.  ( 50 x10 m )  8.686 m  Np    Repeating these calculations for pure silver, we find silver = 156x103 Np/m and Attensilver = 68dB.

P9.18: Shielding low frequency magnetic fields often requires a magnetic shield. What thickness of 99.8% iron is required to give 20 dB attenuation of a 1.0 kHz magnetic field?

For 99.8% iron we have r = 5000 and  = 1x107 S/m. So we have Np  =  f  =  (1000 )( 5000 ) ( 4 x10−7 )(1x107 ) = 14 x103 . m In terms of dB, we have Np   dB   3 dB  (dB) = 14 x103 .  = 122 x10   8.686 m  Np  m  Since atten = t, then we have Atten 20dB t= = = 163 m.  122 x103 dB m

9-10 P9.19: Find the shielding effectiveness for the silver-filled paint shield of P9.17 and compare the result with that of pure silver.

Using the program ML0904, we find SEpaint = 80 dB SEsilver = 156 dB P9.20: MATLAB: Consider a 10.0 m thick copper shield. Plot the contributions to shielding effectiveness (and the total shielding effectiveness) from each of the reflective terms and from the absorption term from 1 MHz up to 1 GHz. Repeat for the same thickness nickel shield. % % % % % % % % % % % % % % % % % % % % % % % % % % % %

used for P9.20 Modify ML0904 Here we want to plot the SE contributions vs frequency.

Wentworth, 2/12/03 Variables d shield thickness (m) s shield conductivity (S/m) ur rel permeability uo free space permeability er rel permittivity eo free space permittivity f,w freq. and ang. freq. (1/s) c speed of light (m/s) Zo free space impedance (ohms) A,B,C calculation variables prop propagation constant (1/m) Z1 impedance (ohms) taud transmission coeff at z = -d tao0 trans coeff at z = 0 ratio power ratio SErefd SE from reflection at front face SEabs SE from atten in shield SEref0 SE from reflection at back face SEtot total shielding effectiveness

clc clear

%clears the command window %clears variables

9-11

% Initialize variables d=10e-6; s=5.8e7; ur=1; er=1; f=100e6; eo=8.854e-12; uo=pi*4e-7; c=2.998e8; Zo=120*pi; % Perform calculations for i=1:1:4 f(i)=10^(i+5); w=2*pi*f(i); A=i*w*ur*uo; B=s+i*w*er*eo; prop=sqrt(A*B); Z1=sqrt(A/B); C=tanh(prop*d); Zin=(Z1*(Zo+Z1*C))/(Z1+Zo*C); taud=2*Zin/(Zin+Zo); tau0=2*Zo/(Zo+Z1); ratio=abs(taud*tau0*exp(-prop*d)); SErefd(i)=-20*log10(abs(taud)); SEabs(i)=-20*log10(abs(exp(-prop*d))); SEref0(i)=-20*log10(abs(tau0)); SEtot(i)=-20*log10(ratio); end semilogx(f,SErefd,'-o',f,SEabs,'-+',f,SEref0,'*',f,SEtot,'-^') legend('front face reflection','absorption','back reflection','total') title('10 micron thick copper shield') xlabel('frequency (Hz)') ylabel('loss (dB)') grid on This is repeated for nickel and the results plotted in Figure P9.20.

face

9-12

Fig. P9.20

6. Filters P9.21: Derive the insertion loss expression Equation (9.20) for the series inductor circuit Figure 9.25(b).

Removing the inductor from the circuit of Figure 9.25(b) (that is, replacing it with a short), we have the voltage across the load without the filter element 1 vL = vS .

2 Now, with the filter in place, we have R vLf = vS .

2 R + j L

9-13 Therefore, vLf 2R 1 1 = = = . vL 2 R + j L 1 + j L 1 + j fL 2R R We find the magnitude of this ratio, vLf 1 = .

2 vL  fL   1+   R   The insertion loss is then 2    v  vLf    fL 2   fL   L IL = 20log  = 20log  1 +  =   = 10log 1 +    .  vLf  vL  R R          

P9.22: (JustAsk): Suppose an L = 100. nH inductor is used in the series inductance filter of Figure 9.25(b). Determine the insertion loss at 200 MHz if (a) R = 10.  and (b )R = 10. k.

   fL 2  We apply the equation IL = 10 log 1 +  .   R        200 x106 100 x10−9 2  ( )( )   = 16dB. (a) IL = 10log 1 +      10    

   200 x106 100 x10−9 2  ( )( )   = 0.17 x10−3 dB. (b) IL = 10log 1 +      10 x103    

P9.23: Suppose a C = 47. pF capacitor is used in the shunt capacitance filter of Figure 9.25(a). Determine the insertion loss at 200. MHz if (a) R = 10.  and (b) R = 10. k.

(

)

We apply the equation IL = 10log 1 + ( fRC ) .

2

) ) = 0.36dB. ( ( (b) IL = 10log (1 + ( ( 200 x10 )(10 x10 )( 47 x10 ) ) ) = 49dB. (a) IL = 10log 1 +  ( 200 x106 ) (10 ) ( 47 x10−12 ) 6

3

2

−12

2

P9.24: Determine the insertion loss at 1.0 GHz for a T-filter inserted between a 10.  source impedance and a 10.  load impedance. Consider L = 10. nH and C = 47. pF.

9-14

Fig. P9.24 (a) We can analyzer the circuit shown in Figure 9.24(a) by using the models in Figures 9.24 (b) & (c): Z1 = R + j L, and Fig. 9.24(b) & (c) − j Z 2 = Z1 C .

Then we have Z2 R R R Z2 v2 = vS , vLf = v2 = v2 ,  vLf = vS .

Z1 + Z 2 R + j L Z1 Z1 Z1 + Z 2 1 vL = vS , 2  v  and we then wish to calculate insertion loss as IL = 20 log  L  .  vLf    This is calculated using MLP0924.

(

)

% MLP0924 % % A T-Filter Problem % clc clear % Variables R=10; L=10e-9; C=47e-12; f=1e9; % Run Program w=2*pi*f; XL=i*w*L; XC=-i/(w*C);

9-15 Z1=R+XL; Z2=parallel(Z1,XC); VLF=R*Z2/(Z1*(Z1+Z2)); ratio=1/(2*abs(VLF)); IL=20*Log10(ratio) Now we run the program: IL = 34.5662 So the insertion loss is IL = 34.6 dB. P9.25: Determine the insertion loss at 40 MHz for a -filter inserted between a 10. k source impedance and a 10. k load impedance. Consider L = 10. nH and C = 47. pF. Referring to Figure P9.25, we have −j Z1 = RL , C and Z2 = j L + Z1.

Then, −j Z3 = Z 2 . C Using these values we find Z3 v1 = vS , Z1 + Rs

v2 =

Z1 v1 = vLf , Z1 + j L

 vLf =

Z3 Z1 vS Z1 + j L Z1 + Rs

and RL .

RL + RS Finally,  v  IL = 20 log  L  .  vLf    This is calculated using MLP0925. vL =

% MLP0925 % % This is a pi filter. %

Fig. P9.25

9-16 clc clear % enter variables RS=10e3; RL=10e3; L=10e-9; C=47e-12; f=40e6; % Perform calculations w=2*pi*f; XC=-i/(w*C); XL=i*w*L; Z1=parallel(RL,XC); Z2=XL+Z1; Z3=parallel(XC,Z2); vLf=(Z1*Z3)/(Z2*(Z3+RS)); vL=RL/(RL+RS); IL=20*Log10(vL/abs(vLf)) Now running this program: IL = 41.3 So IL = 41 dB.

10-1 Solutions for Chapter 10 Problems 1. Lumped Element Matching Networks P10.1: A matching network consists of a length of a T-Line through section in series with a capacitor. Determine the length (in wavelengths) required of the through section and the capacitor value needed (at 1.0 GHz) to match a 10 - j35  load impedance to the 50  line.

Fig. P6.32a This problem is identical to P6.32: We find the normalized load, zL = 0.2 – j0.7, located at point a (WTG = 0.400).

Now we move from point a clockwise (towards the generator) until we reach Fig. P6.32b point b, where we have z = 1 + j2.4.

Moving from a to b corresponds to d = 0.500+0.194-0.400 = 0.294. For the series capacitance we have −j − j 2.4 = , CZ o

or C =

1

2 (1x10 ) ( 50 )( 2.4 ) 9

= 1.33 pF

P10.2: Design an L-section matching network to match a 10 + j15  load to a 50  line. Determine specific values of the lumped elements at a 1.0 GHz operating frequency.

Referring to the figure, we normalize the load (point a) at zL = 0.2 + j0.3. We then add series +j0.1 to move along the real = 0.2 circle to the point zb = 0.2+j0.4 (point b). We next jump to the admittance chart, at yb = 1.0-j2.0 (point b’). Finally, we add shunt +j2.0 to reach the center of the chart at point c’.

Fig. P10.2(a)

10-2 Realizing these components, we have for the series element: ( 0.1)( 50 ) = 0.796nH . j L j 0.1 = , or L = Zo 2 (1x109 ) So L = 0.80 nH. Then, the shunt element is: j 2.0 = jCZ o ,

or C =

2.0 = 6.37 pF .

2 (1x109 ) 50

Fig. P10.2b

So C = 6.4 pF. The final circuit is shown in Fig P10.2 (b). An alternate solution is also shown.

P10.3: Design an L-section matching network to match a 80 - j50  load to a 50  line.

Determine specific values of the lumped elements at a 1.0 GHz operating frequency.

We first normalize and locate the load, zL = 1.6 – j1.0 (point a). Then we move to the admittance chart (point a’) at yL = 0.44+j0.28.

We add +j0.22 (shunt element) to get to the rotated circle (point b’) where we have yb = 0.44 + j0.50.

Next, we move back to the impedance chart, at zb = 1 – j1.1 (point b). Finally, adding a series +j1.1 brings us to the matched condition.

Realizing these components, we have for the shunt +j0.22: j 0.22 = jCZ o ,

or C =

0.22 = 0.7 pF .

2 (1x109 ) ( 50 )

For the series +j1.1 element: j L j1.1 = , Zo

or L =

(1.1)( 50 ) = 8.7nH .

2 (1x109 )

Fig. P10.3

The final circuit is shown in Fig P10.3. An alternate solution replaces the shunt capacitor with a shunt inductor of value 10 nH, and the series inductor is replaced with a series capacitor of value 2.9 pF.

10-3 P10.4: Design an L-section matching network to match a 30 + j70  load to a 50  line.

Determine specific values of the lumped elements at a 2.5 GHz operating frequency.

The normalized load is zL = 0.6+j1.4, at point a. We move to the rotated circle by adding –j0.9 (series element) to get to zb = 0.6+j0.5 (point b). We then move to the admittance chart (point b’) at yb = 1 – j0.8.

Finally, adding a shunt element of value +j0.8 brings us to the matched condition.

Realizing these components, for the series –j0.9 we have: −j − j 0.9 = , or CZ o

C=

1 = 1.4 pF .

2 ( 2.5 x109 ) ( 50 )( 0.9 )

For the shunt +j0.8 we have: j 0.8 = jCZ o , or

C=

( 0.8)

2 ( 2.5 x109 ) ( 50 )

= 1.02 pF .

The figure shows the final matching Fig. P10.4 network. An alternate solution replaces the series capacitance with another series capacitance of value 0.68pF, and the shunt capacitance with a shunt inductor of value 4.1 nH. P10.5: Suppose you want to match a 20 + j50  load to a 50  line. For the design of an L-section matching network, you notice the normalized load impedance lies outside both the 1 ± jx circle and the rotated 1 ± jx circle. Find all four possible solutions, and for each one determine specific values of the lumped elements at a 2.5 GHz operating frequency.

In each solution, we proceed from a to b to c, with primes denoting points in the admittance chart.

Conversion of normalized element values to components values shown in the figures is accomplished using Table 10.2.

10-4

Fig. P10.5a

Fig. P10.5b

Fig. P10.5c

Fig. P10.5d

10-5

P10.6: Suppose you want to match a 100  line to a load ZL = 200 - j100  (a resistor in series with a capacitor) at a frequency of 500 MHz. (a) Determine the element values for the load. (b) Design a shorted shunt stub matching network. (c) Design an L-section matching network.

Fig. P10.6b

Fig. P10.6a

The load is a series resistance/capacitance combination, where the capacitance is given by: −j 1 − j100 = , or C = = 3.18 pF . C 2 ( 500 x106 ) (100 ) For the stub matching network, we first locate the normalized load zL on the Smith Chart, move to the normalized admittance yL, and then proceed along the constant || circle to the point y = 1+j1, traveling a through-line length of 0.125G. Then, we travel from the normalized admittance of a short circuit (at 0.250 G on the WTG scale) to y = 0 - j1, traveling a stub line length of 0.125 G. So for this particular problem the through and stub lengths are the same.

Next, we turn to the lumped element matching problem. We locate the load at zL = 2 – j1 (point a), and then move to yL = 0.4 + j0.2 (point a’). Then we move a distance +j0.3

10-6 (corresponding to a shunt element) to yb = 0.4 + j0.5 (point b’). Switching to the impedance chart, or zb = 1 – j1.2, we add a series +j1.2 to reach the matched condition (point c).

Realizing these components, we have for the shunt j0.3: ( 0.3) j 0.3 = jCZ o , or C = = 0.95 pF .

2 ( 500 x106 ) (100 ) For the series j1.2: (1.2 )(100 ) = 38nH . j L j1.2 = , or L = Zo 2 ( 500 x106 )

P10.7: (JustAsk): Design an L-section matching network to match a load ZL = 100 + j80  to a 50  line. Find the lumped element values at an operating frequency of 11.18 GHz. Further, your design should allow for DC biasing the load element through the matching network.

Allowing DC bias through the network requires use of a shunt capacitor and a series inductor. Referring to the figure, we first normalize the load, zL = 2 + j1.6 (point a), and find the corresponding point in the normalized admittance chart, yL = .31 – j.25 (point a’). To get a shunt capacitance, we must add a positive admittance. So we move to the point yb = .31 + j.47 (point b’) by adding shunt +j0.72. Then we move to the impedance chart, at zb = 1 – j1.45. Adding a series +j1.45 gives us the necessary series inductance. The element value for the shunt capacitance is: j 0.72 = jCZ o ,

or C =

( 0.72 )

2 (11.18 x109 ) ( 50 )

= 0.20 pF .E

The series inductance is: j L j1.45 = , Zo

or L =

(1.45)( 50 )

2 (11.18 x109 )

= 1.03nH .

Fig. P10.7

10-7 P10.8: Design an L-section matching network for a load that has a 25.0  resistor in series with a 1.061 pF capacitor.

Assume a 50  system impedance at 3.0 GHz. We normalize the load to find zL = 0.5 – j1.0 (point a), and then add series j0.5 to reach zb = 0.5 –j0.5 (point b). Moving to the admittance chart, yb = 1 + j1, so we add a shunt –j1.0 to reach the matched condition.

The element values are: j L j 0.5 = , or Zo

L=

( 0.5)( 50 ) = 1.33nH .

2 ( 3x109 )

and

− jZ o , or L ( 50 ) = 2.65nH .

L= 2 ( 3 x109 )

− j1 =

Fig. P10.8

P10.9: MATLAB: There are two fundamental solutions for the L-section matching network of P10.2. Develop a routine to plot the || versus frequency for both solutions from 0.1 GHz to 10 GHz. (Hint: this is somewhat similar to MATLAB 9.1).

The two fundamental solutions are shown in Figure P10.9a. For the top solution, we have: −j Zin1 = ( j Lm1 + Z L ) Cm1 And for the bottom solution:  −j  Z in 2 = j Lm 2  + ZL    Cm 2  Once Zin is found, then

Fig. P10.9a

10-8 Z in − Z o , Z in + Z o and we can plot || versus frequency.

From P10.2 we have ZL = 10 + j15 , Lm1 = 0.80 nH, Cm1 = 6.4 pF, Lm2 = 4.0 nH and Cm2 = 4.5 pF. The following MATLAB routine generates the plot. =

% MLP1009 % % Plot magnitude of reflection coefficient % from 0.1GHz to 10 GHz looking into each % matching network from P10.02. % clc clear ZL=10+j*15; Zo=50; f=0.1:0.01:10; w=2*pi.*f*1e9; Lm1=0.80e-9; XL1=j*w.*Lm1; Cm1=6.4e-12; XC1=-j./(w.*Cm1); Zin1=parallel(XC1,ZL+XL1); Ref1=abs((Zin1-Zo)./(Zin1+Zo)); Lm2=4.0e-9; Cm2=4.5e-12; XL2=j*w.*Lm2; XC2=-j./(w.*Cm2); Zin2=parallel(XL2,ZL+XC2); Ref2=abs((Zin2-Zo)./(Zin2+Zo)); semilogx(f,Ref1,'-*',f,Ref2,'-o') Legend('L=.8nH,C=6.4pF','C=4.5pF,L=4nH') xlabel('frequency(GHz)') ylabel('magnitude of reflection coefficient') grid on

10-9

Fig P10.9b

P10.10: MATLAB: There are two fundamental solutions for the L-section matching network of P10.3. Develop a routine to plot the || versus frequency for both solutions from 0.1 GHz to 10 GHz. The two fundamental solutions are shown in Figure P10.10a. For the top solution, we have: −j Zin1 = j Lm1 + Z Cm1 L And for the bottom solution: −j Zin 2 = + j Lm 2 Z L Cm 2 Once Zin is found, then Z − Zo  = in , Z in + Z o Fig. P10.10a and we can plot || versus frequency.

From P10.3 we have ZL = 80 – j50 , Lm1 = 8.7 nH, Cm1 = 0.7 pF, Lm2 = 10 nH and Cm2 = 2.9 pF. The following MATLAB routine generates the plot. % % %

MLP1010 Plot magnitude of reflection coefficient from 0.1GHz to 10 GHz looking into each

10-10 % matching network from P10.03. clc clear ZL=80-j*50; Zo=50; f=0.1:0.01:10; w=2*pi.*f*1e9; Lm1=8.7e-9; XL1=j*w.*Lm1; Cm1=0.7e-12; XC1=-j./(w.*Cm1); Zin1=XL1+parallel(ZL,XC1); Ref1=abs((Zin1-Zo)./(Zin1+Zo)); Lm2=10e-9; Cm2=2.9e-12; XL2=j*w.*Lm2; XC2=-j./(w.*Cm2); Zin2=XC2+parallel(ZL,XL2); Ref2=abs((Zin2-Zo)./(Zin2+Zo)); semilogx(f,Ref1,'-*',f,Ref2,'-o') Legend('L=8.7nH,C=.6pF','C=2.9pF,L=10nH') xlabel('frequency(GHz)') ylabel('magnitude of reflection coefficient') grid on

Fig. P10.10b

10-11 P10.11: MATLAB: Suppose the L-section matching network of Example 10.1 is realized with a capacitor that can be characterized by the circuit model of Figure 9.13(b), where Rx = 0.010  and Lx = 7.2 nH, and an inductor that can be characterized by the circuit model of Figure 9.10 where Rx = 0.10  and Cx = 5.2 pF. Compare plots of || versus frequency for the ideal case to the case where parasitics are included. The frequency range is from 0.1 GHz to 10 GHz. Replacing the shunt capacitance with its circuit model we get the circuit of Fig p10.11a.

We now have: j   Zin = j Lm +  Rx − + j Lx  Z L C   Then, Z − Zo  = in , Z in + Z o and we can plot || versus frequency. % MLP1011 % % Plot magnitude of reflection coefficient % from 0.1GHz to 10 GHz looking into each % matching network of Ex10.1. Compare with % plot including parasitic effects. % clc clear ZL=250-j*250; Zo=50; f=0.1:0.01:10; w=2*pi.*f*1e9; Lm1=30e-9; XL1=j*w.*Lm1; Cm1=0.8e-12; XC1=-j./(w.*Cm1); Zin1=XL1+parallel(ZL,XC1); Ref1=abs((Zin1-Zo)./(Zin1+Zo)); Rx=.010; Lx=7.2e-9; XC2=XC1+j*w*Lx+Rx; Zin2=XL1+parallel(ZL,XC2); Ref2=abs((Zin2-Zo)./(Zin2+Zo));

10-12

semilogx(f,Ref1,'-*',f,Ref2,'-o') Legend('ideal cap','cap with parasitics') xlabel('frequency(GHz)') ylabel('magnitude of reflection coefficient') grid on

Fig. P10.11a In this problem, the parasitics introduce only a slight change in the resonance frequency.

Fig. P10.11b

2. Scattering Parameters P10.12: Find the scattering matrices for the simple two-port networks shown in Figure 10.50.

A port terminated in a short circuit will have a reflection coefficient of -1. An open termination will have a +1 reflection coefficient. The ideal through connections will have no reflection. By inspection, then, the scattering matrices for the circuits shown are: 0 1  1 0  (a)  S  =  (b)  S  =    1 0  0 1  (a)  −1 0  1 0  (c )  S  =  (d )  S  =     0 −1 0 −1 P10.13: Cut a 50  T-Line and insert a series 50  resistor followed by a shunt 50  resistor. Determine the scattering matrix for this 2-port network. Is the network lossless?

Is it reciprocal? Calculate the insertion loss.

10-13 The circuit (Figure P10.13(a)) is redrawn in (b) to calculate S11 and S21. Here, 75 − 50 1 S11 =  = = = 0.2 75 + 50 5 V−  25   1  1   2  S21 = 2+ =    =  + 1  =   V1  75   5  3   5  S21 = 0.40 Figure P10.13(c) is used to calculate S22 and S12: Z L = 50 100 = 33.3,

33.3 − 50 = −0.2 33.3 + 50 V−  50  1 S12 = 1+ =    = 0.8   = 0.4 V2  100  2 Fig. P10.13 0.20 0.40  So:  S  =   , and we see by 0.40 −0.20  inspection that the scattering matrix is reciprocal. It is also not lossless (summing the squares of any column yields a number less than one). Finally, to calculate the insertion loss: IL21 = −20log S21 = −20log 0.4 = 8dB S22 =  =

P10.14: In a 50  system, a two-port network consists of a 25  series resistor followed by a 50  shunt resistor (see Figure 10.11(a)). Calculate the return loss looking into port 1 of this network if port 2 is terminated in a 100  resistor.

The difficult approach: We have the scattering matrix from example 10.5 for the circuit of Fig P10.14(a): 1  0 2  S  =  1 −1   2 4 By staring with V1− = S11V1+ + S12V2+

V2− = S21V1+ + S22V2+ Fig. P10.14 and considering that 100 − 50 − 1 − V2+ = V2− = V2 = V2 , 100 + 50 3 we can manipulate the scattering parameters to arrive at:

10-14

( )( ) ( )

1 1 V1− S12 S21 2 2 = 0.07692 = S + = 0 + 11 + − 1 V1 3 − S22 3− 4 The return loss is then: RL = −20log 0.07692 = 22.3dB. The easier approach: All we need is the reflection coefficient looking into the circuit of Fig P10.14(b). We have Z L = 25 + 50 100 = 58.33,

Z L − Zo 58.33 − 50 = = 0.07693, Z L + Z o 58.33 + 50 and then RL = −20log 0.07693 = 22.3dB L =

P10.15: A series capacitor of value C = 2.0 pF is inserted in a 50  T-Line. At 1.0 GHz, determine [S], the return loss and the insertion loss.

The impedance of the capacitor at 1 GHz is −j −j Zc = = = − j80 9 C 2 (1x10 )( 2 x10−12 ) To find S11, 50 − j80 − 50 S11 =  = 50 − j80 + 50 − j80 = = 0.625e − j 51 100 − j80 And for S21, V− 50 S21 = 2+ =  V2 50 − j80

(

= 1 + 0.625e− j 51

) 50 −50j80 = 0.78e

So

0.625e− j 51 0.780e j 39  S = ,     0.780e j 39 0.625e− j 51  RL = −20 log S11 = 4.1dB IL = −20 log S21 = 2.2dB

Fig. P10.15

j 39

10-15 P10.16: A series inductor of value L = 3.5 nH is inserted in a 50  T-Line. At 1.0 GHz, determine [S], the return loss and the insertion loss.

The impedance of the inductor at 1 GHz is Z L = j L = j 2 (1x109 )( 3.5x10−9 ) = j 22 To find S11, 50 + j 22 − 50 S11 =  = 50 + j 22 + 50

= 0.215e j 77.6 And for S21, V2− 50 S21 = + =  V2 50 + j 22

(

= 1 + 0.215e

j 77.6

)

Fig. P10.16

50 = 0.977e− j12.4 50 + j 22

So

0.215e j 78 S  =   0.98e − j12

RL = −20 log S11 = 13.4dB 0.98e− j12   , and IL = −20 log S21 = 0.18dB 0.215e j 78 

P10.17: The scattering matrix for a 3-port network is 0 j 0.80  0.60   0 .  S  =  0 1.0e j 90  j 0.80 0 0.60    (a) Is this network reciprocal? (b) Is it lossless? (c) Determine the return loss at port 1 if ports 2 and 3 are connected together by a matched T-Line of electrical length 45.

(a) Fig. P10.17

(b)

10-16 (a) by inspection, we see the network is reciprocal. (b) Each column’s values, when squared, add to one, so the network appears to be t * lossless. This is confirmed by evaluation of  S   S  which indeed results in a unitary matrix. (c) To determine the return loss, we need to determine V1− V1+ .

We start with: (1)V1− = 0.6V1+ + j 0.8V3+

(2)V2− = 1e j 90 V2+ = jV2+ (3)V3− = j 0.8V1+ + 0.6V3+ Examining Figure P10.17(b), we have (4)V2+ = V3−e − j 45

(5)V3+ = V2− e − j 45 .

Inserting (5) and (2) into (3), we have

(

)

(6)V3− = j 0.8V1+ + 0.6 V2−e− j 45 = j 0.8V1+ + 0.6e− j 45 jV2+ Now inserting (4) into (6),

(

)

(7)V3− = j 0.8V1+ + 0.6e− j 45 j V3−e− j 45 = j 0.8V1+ + 0.6V3− and solving (7) for V3−

(8)V3− (1 − 0.6 ) = j 0.8V1+ , so V3− = j 2V1+ .

Next, inserting (8) into (4) we have (9)V2+ = ( j 2V1+ ) e− j 45 = j 2e− j 45 V1+ , and plugging this into (2),

(

)

(10)V2− = j j 2e− j 45 V1+ = −2e− j 45 V1+ Finally, inserting this value into (5):

(

)

(11)V3+ = −2e− j 45 V1+ e− j 45 = j 2V1+ .

Inserting (11) into (1) leads us to the desired ratio: (12)V1− = 0.6V1+ + j 0.8 ( j 2V1+ ) = −1V1+ or

V1− = −1 , and RL = 0 dB. V1+

P10.18: (JustAsk): The scattering matrix (assuming a 50  impedance system) for a 2port network is:  0.5 0.5e j 45  .  S  =  j 45 0.5e 0.5e j 90 

10-17 (a) Is this network reciprocal? (b) Is this network lossless? (c) Determine the return loss looking into port 1 if port 2 is terminated in an open-ended Zo stub of electrical length 45.

By inspection it is reciprocal but not lossless. For the scattering matrix: (1)V1− = S11V1+ + S12V2+ = 0.5V1+ + 0.5e j 45 V2+

(2)V2− = S21V1+ + S22V2+ = 0.5e j 45 V1+ + 0.5e j 90 V2+ Finding the return loss we will need the relationship between V2− and V2+ resulting from the 45 line. There is a round trip delay of 90, and at the open end the reflected wave equals the incident wave. So we have (3)V2+ = V2− e− j 90 .

Inserting (3) into (2) we can solve for Fig. P10.18 V2− : S21V1+ 1 − S22e − j 90 (4) is now inserted into (3) to find S e − j 90 V1+ (5)V2+ = 21 .

1 − S22e − j 90 Finally, inserting (5) into (1) we can find the ratio needed: V1− S12 S21e − j 90 = S + 11 V1+ 1 − S22e − j 90 (4)V2− =

(0.5e )(0.5e ) e = 0.5 + 1 − ( 0.5e ) e j 45

j 45

j 90

− j 90

− j 90

=1

 RL = −20log (1) = 0dB

This is an interesting result. Note that terminating port 2 in a short circuit results in a 7dB return loss.

P10.19: Three T-Lines with the same characteristic impedance Zo are connected as shown in Figure 10.51. Determine the scattering matrix that represents this 3-port network. Is this network reciprocal? Is it lossless?

The load seen by each line is a pair of Zo impedances in parallel. So

10-18

1 −1 1 2 S11 =  L = =− , 1 3 +1 2 2 S21 =  = 1 +  L = .

3 So the scattering matrix is:  −1 2 2  1  S  =  2 −1 2  .

3  2 2 −1

3. Couplers and Dividers P10.20: Consider a 3-port network that is matched at all ports (S11 = S22 = S33 = 0).

Show that it is impossible to construct a reciprocal network that is lossless for this case.

For a 3-port scattering matrix matched at all ports and reciprocal we have  0 S21 S31   S  =  S21 0 S32  .  S31 S32 0  If it is also lossless, then  S   S  = U  , which gives the following equations: t

*

 S21 2 +  S31 2 = 1  S21 2 +  S32 2 = 1  S31 2 +  S32 2 = 1 and * S31S32 =0 * S21S32 =0 * S21S31 = 0.

If we choose any pair of S31, S21 and S32 to be zero to satisfy the bottom set of equations, then one of the top set of equations will be zero. Therefore, the network cannot be lossless, reciprocal and matched at all ports.

P10.21: A circulator referenced to a 50  impedance is characterized by  0.50 0.050e j 60 0.75e j 60    0.50 0.050e j 60  .  S  =  0.75e j 60   j 60 0.75e j 60 0.50  0.050e 

10-19 Is this network (a) reciprocal? (b) Lossless? Calculate (c) insertion loss, (d) isolation and (e) VSWR. (a) not reciprocal, since (for instance) S12  S21. (b) not lossless, since summing the squared column 1 elements gives a value less than one. (c) IL = −20log S21 = −20log 0.75 = 2.5dB (d) I = −20log S31 = −20log 0.05 = 26dB (e) VSWR =

1 +  1 + S11 1 + 0.5 = = =3 1 −  1 − S11 1 − 0.5

P10.22: Calculate the insertion loss and the VSWR for the previous problem if the isolated port is terminated in a short circuit.

The insertion loss will be V− IL = −20log 2+ , V1 and 1+  VSWR = , 1−  Fig. P10.22

V1− where  = + .

V1

For a 3 port scattering matrix we have V3− = S31V1+ + S32V2+ + S33V3+ , and since port 2 is terminated in a matched load we therefore have (1)V3− = S31V1+ + S33V3+ .

With port 3 terminated in a short circuit, V3+ = −V3− , and using (1) we find

− S31V1+ .

1 + S33 Now, for a 3 port scattering matrix matched at port 2 we have (3)V1− = S11V1+ + S12V2+ + S13V3+ = S11V1+ + S13V3+ Inserting (2) into (3), we find (2)V3+ =

(

)(

)

0.75e j 60 0.05e j 60 V1− S13S31 = S11 − = ( 0.5) − = 0.513e− j 2.2 , V1+ 1 + S33 1 + 0.5 1 +  1 + 0.513 = = 3.1 so VSWR = 1 −  1 − 0.513

10-20 Finally, for a 3 port scattering matrix matched at port 2 we also have (4)V2− = S21V1+ + S22V2+ + S23V3+ = S21V1+ + S23V3+ , Inserting (2) into (4), we find

(

)

0.05e ( 0.5) V2− S31S33 j 60 = S − = 0.75 e − = 0.733e + j 60 , 21 V1+ 1 + S33 1 + 0.5

(

and IL = −20log

)

j 60

V2− = −20log 0.733 = 2.7dB V1+

P10.23: The following information is supplied for a commercial L-Band circulator: ILmax = 0.60 dB, Imin = 18 dB, and VSWRmax = 1.35. Calculate the worst-case magnitudes for the scattering matrix. Assume a symmetrical circulator.

ILmax = −20log S21 = 0.6dB, I min = −20log S31 = 18dB, VSWRmax =

1 + S11 , or 1 − S11

S21 = 10−0.6 20 = 0.933 S31 = 10−18 20 = 0.126

S11 =

VSWR − 1 1.35 − 1 = = 0.15 VSWR + 1 1.35 + 1

So, for the symmetrical network we have the following magnitudes for the S-matrix: 0.15 0.13 0.93  S  =  0.93 0.15 0.13    0.13 0.93 0.15

P10.24: Verify the scattering matrix (equation (10.27)) for the resistive power divider of Figure 10.16.

The impedance seen looking into, say, port 1 is Zo/3 plus the series combination of a pair of 4Zo/3 impedances. So, with a total load impedance of Zo, there is no reflection.

S21 is found using a pair of resistive divider circuits: 4 4    Zo Zo   Z  23 1 1 3 3 S21 =     o  = (1) = 1 43 2  4 Z 4 Z + 1 Z  4 Z   3 o 3 o 3 o  3 o    Here, the transmission coefficient  is simply 1+ = 1. The first ratio in parenthesis determines how much of the voltage at port 1 is dropped across the pair of parallel 4Zo/3 impedances, and the second ratio determines how much of this voltage drops across the port 2 termination.

So,

10-21

0   S  =  1 2 1  2

1

1  2 1 .

2 0 

2

0 1

2

P10.25: Suppose 10.0 mW of microwave power is fed into port 1 of the resistive divider shown in Figure 10.16. With ports 2 and 3 terminated in matched loads, determine how much power is transmitted to each port and how much is dissipated in the divider.

(

P2 = P3 = 10mW S21

2

)

2

1 = 10mW   = 2.5mW 2 Pdiss = 10mW − 2(2.5mW ) = 5mW

Fig. P10.25

P10.26: Repeat problem P10.25 for a Wilkinson power divider.

(

P2 = P3 = 10mW S21

2

)

2

 1  = 10mW   = 5mW  2 Pdiss = 10mW − 2(5mW ) = 0mW P10.27: A 4-port “20-dB Coupler” is specified as having 20 dB coupling, 50 dB isolation, and 0.25 dB of insertion loss. If 100 mW of power is input, calculate the power out of the other three ports. Assume all ports are terminated in matched loads.

P  C = 20dB = −10log  3  , so P3 = ( P1 ) (10−20 10 ) = 1mW  P1 

10-22

I = 50dB = −10log ( S41 ) , so

S41 = 10−50 10 = 0.00316

P4 = S41 P = 1W 2

IL = 0.25dB = −10 log ( S21 ) , so

S21 = 10−0.25 10 = 0.9716

P2 = S21 P = 94.4mW 2

P10.28: Suppose the coupling for an ideal symmetrical 4-port coupler is 3 dB. Find the scattering matrix and determine the insertion loss.

In the ideal coupler, all the power entering port 1 is split and exits ports 2 and 3 with none reflected and none to the isolated port, and none lost to dissipation.

C = 3dB = −20log S31 , or S31 = 10−3 20 = 0.707,

S21 = S31 = 0.707 and IL = −20log S21 = 3dB. The scattering matrix is: 0   0 .707 .707 .707 0 0 .707    S  = 0 0 .707  .707   0   0 .707 .707

P10.29: (JustAsk): Suppose to port 1 of an ideal ring hybrid coupler we apply the appropriate frequency voltage V1+ = 1.0e j 0 V .

If port 2 is terminated in a short circuit, determine the voltage exiting ports 1, 3 and 4.

For the ideal ring hybrid, we have 0 1 1 0    − j 1 0 0 −1 . S  =  2 1 0 0 1    0 −1 1 0  Assuming ports 3 and 4 are terminated in matched loads ( V3+ = V4+ = 0 ), then we have

V1− = S12V2+ V2− = S21V1+ V3− = S31V1+ V4− = S42V2+ Now, at port 2 terminated in a short circuit, V2+ = −V2− so

10-23

V1− = S12V2+ = − S12V2− = − S12 S21V1+

(

)

 − j   − j  j0 V1− = −  = 0.5e j 0 V   1e  2  2   − j  j0 V3− = S31V1+ =  = 0.707e − j 90 V  1e  2 − + V4 = S42V2 = − S42V2− = − S42 S 21V1+

( )

( )

 + j  − j  j 0 V4− = −  = −0.5e j 0 V   1e  2  2 

P10.30: Consider an ideal ring hybrid coupler, with all ports terminated in matched loads.

A signal 5.0ej30V is injected into port 2, and 3.0ej30V is injected into port 3. Determine the signals exiting ports 1 and 4.

For the ideal ring hybrid, we have 0 1 1 0    − j 1 0 0 −1 . S  = 2 1 0 0 1    0 −1 1 0  With all ports terminated in matched loads, all the incident voltages ( Vk+ ) will be zero, except, of course, for the injected signals into ports 2 and 3. So we have V1− = S12V2+ + S13V3+ V4− = S42V2+ + S43V3+ . −j = 0.707e − j 90 , we have With 2 − V1 = S12V2+ + S13V3+

(

= 0.707e − j 90

)(5e ) + (0.707e )(3e ) j 30

− j 90

j 30

= 3.535e − j 60 + 2.121e − j 60 = 5.66e − j 60 V and V4− = S42V2+ + S43V3+

(

= −0.707e − j 90

)(5e ) + (0.707e )(3e ) j 30

− j 90

j 30

= −1.414e − j 60 = −1.414e j120 V

P10.31: Suppose to port 1 of a quadrature hybrid coupler we apply the appropriate frequency voltage V1+ = 1.0e j 0 V .

If port 2 is terminated in a short circuit, determine the voltage exiting ports 1, 3 and 4.

10-24

The scattering matrix for the quadrature hybrid coupler is 0 j 1 0    −1 j 0 0 1  .  S  =  2 1 0 0 j   0 1 j 0 With ports 3 and 4 terminated in matched loads V3+ = V4+ = 0 and port 2 terminated in a short (Vv+ = −V2− ) ,we then have

(

)

V1− = − S12 S21V1+ V2− = S21V1+ V3− = S31V1+ V4− = − S42 S21V1+ Evaluating,  − j  − j  j 0 V1− = −  = 0.5e j 0 V   1e  2  2 

( )

( )

 −1  j 0 V3− = S31V1+ =  = −0.707e j 0 V  1e  2  −1  − j  j 0 V4− = −  = 0.5e − j 90 V   1e  2  2 

( )

P10.32: Given a 50.0 mil thick Teflon substrate, design a quadrature hybrid coupler for 2.50 GHz operation.

For h = 50 mils and r = 2.1 for Teflon, and assuming a copper conductor where we may neglect the conductor thickness, we must design quarter wave microstrip at 50  and at 50/ 2 . We use ML0605 to find: Zo = 50  w=159 mils  = 0.0893 m = 3520 mils  /4 = 879 mils The resulting design is shown in Figure P10.32.

Zo = 35.4  w=258 mils  = 3470 mils  /4 = 867 mils

10-25

Fig. P10.32

P10.33: Suppose you join a pair of quadrature hybrid couplers (port 2 of coupler 1 attached to port 1 of coupler 2, port 3 of coupler 1 attached to port 2 of coupler 2). The resulting network will have 4 ports: ports 1 and 3 from the first coupler, and ports 2 and 4 from the second coupler. Determine the overall scattering matrix.

Each individual coupler in Figure P10.33 has the matrix: 0 j 1 0    −1  j 0 0 1  . S  =  2 1 0 0 j   0 1 j 0 We can expand the matrix to yield the following equations, where we’ve also indicated the equivalent terms VA−2 = VB+1 , VA+2 = VB−1 , VA−3 = VB+4 , and VA+3 = VB−4 :

Fig. P10.33

10-26 For the A matrix: VA−1 = S12VA+2 + S13VA+3 = S21VB−1 + S13VB−4

VA−2 = S21VA+1 + S24VA+4 VA−3 = S31VA+1 + S34VA+4 VA−4 = S42VA+2 + S43VA+3 = S42VB−1 + S43VB−4 For the B matrix: VB−1 = S12VB+2 + S13VB+3 VB−2 = S21VB+1 + S24VB+4 = S21VA−2 + S24VA−3 VB−3 = S31VB+1 + S34VB+4 = S31VA−2 + S34VA−3 VB−4 = S42VB+2 + S43VB+3 Note that for the overall matrix, we are only interested in VA−1 , VA−4 , VB−2 , and VB−3. So we insert reflected values into those equations. For example, VA−1 = S21VB−1 + S13VB−4 = S21 ( S21VB+2 + S13VB+3 ) + S13 ( S42VB+2 + S43VB+3 ) , which expanding and rearranging becomes VA−1 = ( S21S21 + S13 S42 )VB+2 + ( S21S13 + S13S 43 )VB+3 Inserting the scattering parameter values, we find   − j  − j   −1  −1   +   − j  −1   −1  − j   + + VA−1 =    +   VB 2 +    +    VB 3 = − jVB 3   2  2   2  2     2  2   2  2   Likewise, we find VA−4 = jVB+2

VB−2 = jVA+4 VB−3 = − jVA+1.

The overall scattering matrix is then: 0 −j 0  0 0 0 0 + j    Soverall  =  −j 0 0 0   0  0 +j 0

4. Filters P10.34: Derive an insertion loss expression for a high-pass filter realized using a shunt inductor inserted in a Zo = Ro system. What inductance value is needed for a 1.0 GHz cutoff frequency if Ro = 50 ?

1 With no filter present, we have vL = vs . With the filter in place, 2 Ro j L j LRo vLf = vs , where Ro j L = .

Ro + j L Ro j L + Ro

10-27 Expanding and rearranging: 1 vLf = . jRo 2− L jR  jR v 1 Now we have L =  2 − o  = 1 − o , vLf 2  L  2 L 2   Ro 2  vL  Ro  = 1+   , so IL = 10log 1 +    . vLf  2 L    2 L  

Now, for a 1 GHz cutoff frequency, Ro R 50 = 1, or L = o = = 3.98nH .

2 ( 2 f c ) L 4 f c 4 (1x109 ) So L = 4.0 nH.

P10.35: Derive an insertion loss expression for a high-pass filter realized using a series capacitor inserted in a Zo = Ro system. What capacitance value is needed for a 1.0 GHz cutoff frequency if Ro = 50 ?

1 With no filter present, we have vL = vs . With the filter in place, 2 Ro vLf = vs .  j  2 Ro −    C  j   2−  vL 1  j C  = = 1− , vLf 2 1 2 RoC 2 2       1 vL 1 = 1+   .  , and IL = 10 log 1 +  4 fRoC   vLf   4 fRoC    For a 1 GHz cutoff frequency, 1 1 1 = 1, so C = = = 1.6 pF .

4 f c RoC 4 f c Ro 4 (1x109 ) ( 50 )

P10.36: Design a 5th order low-pass filter with fc = 2.0 GHz for a 50  system where we will allow only 0.1 dB of ripple. Use the Figure 10.28(b) circuit configuration, and compare your insertion loss plot with that of Figure 10.31.

For N = 5 and 0.1 dB ripple, we read the appropriate values from Table 10.3 as indicated in Figure P10.36(a).

10-28 Impedance Transformation: L1 ' = g1Ro = (1.1468)( 50 ) = 57.3H

L5 ' = L1 ' g 1.3712 C2 ' = 2 = = 27.4mF Ro 50 C4 ' = C2 '

L3 ' = g3 Ro = (1.975)( 50 ) = 98.75H

(a)

Frequency Transformation: L' 57.3 L1 '' = 1 = = 4.6nH c 2 ( 2 x109 )

L5 '' = L1 '' C ' 27.4mF C2 '' = 2 = = 2.18 pF , C4 '' = C2 '' c 2 ( 2 x109 )

(b) Fig. P10.36 a&b

(c)

(d)

(e) Fig. P10.36 c,d & e

L3 '' =

L3 '

c

=

98.75 = 7.86nH .

2 ( 2 x109 )

These values are indicated on the filter design shown in Figure P10.36(b).

For generating the insertion loss plot, we make use of the equivalent circuits shown in Figure P10.36(c)-(e). We have

10-29

Z1 =

  −j R ( R + j L1 ) , vLf = v1   C1  R + j L1 

Z2 =

  Z1 −j ( Z1 + j L2 ) , v1 = v2   C1  Z1 + j L2 

  Z2 v2 =   vs  Z 2 + R + j L1  So finally we have, vLf 1     Z1 Z2 R = ,=    , vs 2  R + j L1  Z1 + j L2  Z 2 + R + j L1  and with  v  vLf v vL 1 v = L to arrive at IL = 20log  L  . = , we divide the two ratios L  vLf  vs vs vLf vs 2  

These equations are used in MLP1036 to generate the insertion loss plot of Figure P10.36(f). % MLP1036 % Design N=5 LPF, 0.1 dB ripple % Compare with N=3 LPF of Example 10.14. % clear;clc;clf

R=50; f=0.1:0.1:10; w=2*pi*f*1e9; % N=3 L31=4.1e-9; C31=1.83e-12; ZL31=i*w*L31; Fig. P10.36(f) ZC31=-i./(w*C31); Z131=R+ZL31; Z132=parallel(Z131,ZC31); Z133=Z132+R+ZL31; sqrtPLR3=Z131.*Z133./(2*R*Z132); IL3=10*LOG10(abs(sqrtPLR3)); %

N=5

10-30 L51=4.6e-9; C52=2.2e-12; L53=7.9e-9; XL51=j*w*L51; XC52=-j./(w*C52); XL53=j*w*L53; Z51=parallel(XC52,R+XL51); Z52=parallel(XC52,Z51+XL53); A=(R./(R+XL51)).*(Z51./(Z51+XL53)).*(Z52./(R+Z52+XL51)); IL5=20*log10(abs(1./(2*A))); semilogx(f,IL3,'--b',f,IL5,'.-k') Legend('N=3','N=5') grid on xlabel('frequency(GHz)') ylabel('IL(dB)') axis([0.1 10 0 40])

P10.37: Design a 3rd order low-pass filter with fc = 1.0 GHz for a 50  system starting with the Figure 10.28(a) circuit configuration. Determine component values for each amount of ripple (0.1, 1 and 3 dB) and compare the three insertion loss responses.

Referring to Figure 10.28(a), we have the following equations based on values of g1, g2 and g3 read from the appropriate portion of Table 10.3, after applying both impedance and frequency transformations: g gR C1 '' = 1 = C3 '', L2 '' = 2 . c R c For the MATLAB routine used to compare the insertion loss responses, we let −j XC = and XL = j L. C Considering v1 is the voltage across C1, we define Z1 as the impedance seen looking to the right of C1 and Z2 as the parallel impedance of Z1 and C1. Then,  R XC  vLf = Z1v1 , where Z1 =   ,  R XC + XL 

 Z2  and v1 =   vs , where Z 2 = Z1 XC.  Z2 + R  Then, with vL = vs/2, we arrive at vL R + Z 2 = . vLf 2Z1Z 2

10-31 % % % %

MLP1037 Compare N=3 LPF with different ripples

R=50; fc=1e9; wc=2*pi*fc; % 0.1 dB ripple g1=1.0315; g2=1.1474; C=g1/(wc*R); L=g2*R/wc; f=0.1:0.01:10; w=2*pi*f*1e9; XC=-j./(w.*C); XL=j.*w.*L; Z1=parallel(R,XC)./(parallel(R,XC)+XL); Z2=parallel(Z1,XC); A=abs((R+Z2)./(2.*Z1.*Z2)); ILa=20*log10(A); % 1 dB ripple g1=2.0236; g2=.9941; C=g1/(wc*R); L=g2*R/wc; f=0.1:0.01:10; w=2*pi*f*1e9; XC=-j./(w.*C); XL=j.*w.*L; Z1=parallel(R,XC)./ (parallel(R,XC)+XL) ; Z2=parallel(Z1,XC); A=abs((R+Z2)./(2.*Z 1.*Z2)); ILb=20*log10(A);

Fig. P10.37

10-32

% 3 dB ripple g1=3.3487; g2=.7117; C=g1/(wc*R); L=g2*R/wc; f=0.1:0.01:10; w=2*pi*f*1e9; XC=-j./(w.*C); XL=j.*w.*L; Z1=parallel(R,XC)./(parallel(R,XC)+XL); Z2=parallel(Z1,XC); A=abs((R+Z2)./(2.*Z1.*Z2)); ILc=20*log10(A); semilogx(f,ILa,'--',f,ILb,'-',f,ILc) legend('0.1 dB','1 dB','3 dB') grid on xlabel('frequency (GHz)') ylabel('IL (dB)')

P10.38: (JustAsk): Starting with the Figure 10.28(b) circuit configuration, design a 3rd order high-pass filter with fc = 2.4 GHz for a 50  system where we will allow only 1 dB of ripple. Plot the insertion loss.

For the 3rd order, 1 dB ripple, high-pass filter, we first select the component values from Table 10.3: g1 = g3 = 2.0236, and g2 = 0.9941, as shown in Figure P10.38(a).

Impedance Transformation: g L1 ' = g1Ro = 101.2, C2 ' = 2 = 0.0199.

Ro Frequency Transformation: (using Table 10.4) 1 C1 '' = c L1 '

=

1 = 0.655 pF 2 ( 2.4 x109 ) (101.2 )

10-33

L2 '' = =

1 cC2 '

1 = 3.33nH .

2 ( 2.4 x109 ) ( 0.0199 )

See Figure P10.38(b). (a)

To develop an expression for the insertion, we consider v2 to be the voltage across the inductor L2’’, and also we let Z1 = XL ( R + XC ) , where XL = jwL and

XC =

−j . C

Then,

(b) Fig. P10.38a&b

v Z1 Z1 R R v1 , v1 = vs , so Lf = .

R + XC Z1 + R + XC vs R + XC Z1 + R + XC Now, with vL=vs/2, we have vL ( R + XC )( Z1 + R + XC ) = . vLf 2 RZ1 We then generate the insertion loss plot via ML1038: % MLP1038 % Design N=3 high pass filter.% fc=2.4; %corner frequency in GHz wc=2*pi*fc*1e9; g1=2.0236; g2=.9941; R=50; vLf =

L=g1*R; C=g2/R; C1=1/(wc*L); L2=1/(wc*C); f=0.1:0.1:10; w=2*pi*f*1e9; XC1=-j./(w.*C1); XL2=j.*w.*L2;

Fig. P10.38c

10-34 Z1=parallel(XL2,R+XC1); A=((R+XC1).*(R+XC1+Z1))./(2.*R.*Z1); IL=20*log10(abs(A)); semilogx(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('2.4 GHz High Pass Filter')

P10.39: Starting with the Figure 10.28(a) circuit configuration, design a 5th order highpass filter with fc = 1.0 GHz for a 50  system where we will allow 3 dB of ripple. Plot the insertion loss.

Assuming 3dB of ripple, from Table 10.3 we have: g1 = g5 = 3.4817, g2 = g4 = 0.7618, and g3 = 4.5381 (see Figure P10.39(a)) Now we perform the impedance transformation: g g C1 ' = 1 = 0.06963, L2 ' = g 2 R = 38.09, and C3 ' = 3 = 0.09076 .

R R The frequency transformation is performed using Table 10.4: 1 1 1 L1 '' = = 2.29nH , C2 '' = = 4.18 pF , and L3 '' = = 1.75nH . cC1 ' c L2 ' cC3 ' The final circuit is shown in Figure P10.39(b).

The MATLAB insertion loss plot is based on analysis of the Fig P10.39(b) circuit.

Assuming the reactances −j XL1 = j L1 '', XC2 = , and XL3 = j L3 '', C2 '' the impedance to the right of L3 is:

(a) Fig. P10.39a&b

(b)

10-35

Z1 = ( R XL1 ) + XC2 .

We’ll also define Z2 to be the impedance of Z1 in parallel with XL3, and Z3 to be: Z3 = ( Z 2 + XC2 ) XL1.

Now we can find the voltage ratio relations:  R XL1   Z3   Z2  vLF =   v1 , v1 =   vs .  v2 , and v2 =   Z 2 + XL2   Z3 + R   Z1  Combining, and realizing vL = vs/2, we arrive at: vL Z1 ( Z 2 + XC2 )( Z3 + R ) = . vLf 2Z 2 Z3 ( R XL1 ) The insertion loss is then:  v  IL = 20log  L  .  vLf    This is plotted by ML1039 in Figure P10.39(c). % MLP1039 % % Design N=5 high pass filter, 3 dB ripple. % fc=1; %corner frequency in GHz wc=2*pi*fc*1e9; g1=3.4817; g2=.7618; g3=4.5381; R=50; Ca=g1/R; L=g2*R; Cb=g3/R; L1=1/(wc*Ca); C2=1/(wc*L); L3=1/(wc*Cb); f=0.1:0.01:10; w=2*pi*f*1e9; XL1=j.*w.*L1; XC2=-j./(w.*C2); XL3=j.*w.*L3; Z1=parallel(R,XL1)+XC2;

Fig. P10.39(c)

10-36 Z2=parallel(Z1,XL3); Z3=parallel(Z2+XC2,XL1); A=(Z1.*(Z2+XC2).*(Z3+R))./(2.*parallel(R,XL1).*Z2.*Z3); IL=20*log10(abs(A)); semilogx(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('1 GHz High Pass Filter: N=5')

P10.40: Starting with the Figure 10.28(b) circuit configuration, design a 3rd order Chebyshev bandpass filter for a 50  system. The passband is to be from 900 MHz to 1100 MHz with only 1 dB of ripple allowed.

Assuming 1dB of ripple, from Table 10.3 we have: g1 = g3 = 2.0236, and g2 = 0.9941. (see Figure P10.28(b)) Now we perform the impedance transformation: g L1 ' = L3 ' = g1R = 101.2, and C2 ' = 2 = 0.01988 R The frequency transformation is performed using Table 10.4, where L1 ' is transformed into a series circuit containing L1 '' and C1 '' . C2 ' is transformed into a parallel circuit containing L2 '' and C2 '' . This is shown in Figure P10.40(a). To perform the transformation, we calculate BW and wo: BW = 2 ( 200MHz )

o = 2

( 900MHz )(1100MHz ) = 2 ( 995MHz )

We then have:

(a) Fig. P10.40a&b

(b)

10-37 L1 '' =

L1 ' 101.2 = = 80.5nH BW 2 ( 200 x106 )

C1 '' =

2 ( 200MHz ) BW = = 0.317 pF 2 2 o L1 ' ( 2 ) ( 995MHz )2 (101.2 )

L2 '' =

2 ( 200MHz ) BW = = 1.617nH 2 2 o C2 ' ( 2 ) ( 995MHz )2 ( 0.01988)

C2 '' =

C2 ' 0.01988 = = 15.8 pF BW 2 ( 200 x106 )

To plot the insertion loss using ML1040, we use the impedances shown in Figure P10.40(b), given as j −j Z1 = j L1 ''− , Z 2 = j L2 '' C1 '' C2 '' We also define Z3 as the impedance seen by the voltage v1: Z3 = Z 2 ( R + Z1 ) .

Now we can find the voltage ratio relations:    R  Z3 vLF =   vs .  v1 , and v1 =   R + Z1   Z3 + R + Z1  Combining, and realizing vL = vs/2, we arrive at: vL ( R + Z1 )( R + Z1 + Z 3 ) = . vLf 2 RZ 3 The insertion loss is then:  v  IL = 20log  L  .  vLf    This is plotted by ML1040 in Figure P10.40(c). % MLP1040 % Design N=3 band pass filter, 1 dB ripple. % clc clear R=50; flo=900e6;fhi=1100e6; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); g1=2.0236;g2=.9941;

10-38

Fig. P10.40(c) L=g1*R;C=g2/R; L1=L/BW; C1=BW/((wo^2)*L); L2=BW/((wo^2)*C); C2=C/BW; f=0.1:0.01:10; %f in GHz w=2*pi.*f*1e9; Z1=j.*(w.*L1-1./(w.*C1)); Z2=parallel(j.*w.*L2,-j./(w.*C2)); Z3=parallel(Z2,R+Z1); A=((R+Z1).*(R+Z1+Z3))./(2.*R.*Z3); IL=20*log10(abs(A)); semilogx(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=3 BPF: 900-1100 MHz')

P10.41: Starting with the Figure 10.28(b) circuit configuration, design a 3rd order Chebyshev bandpass filter for a 50  system. The passband is to be from 900 MHz to 1100 MHz with 3 dB of ripple allowed. Plot the insertion loss and calculate the shape factor.

10-39

(a) Fig. P10.41a&b

(b)

Assuming 3 dB of ripple, from Table 10.3 we have: g1 = g3 = 3.3487, and g2 = 0.7117. (see Figure P10.28(b)) Now we perform the impedance transformation: g L1 ' = L3 ' = g1R = 167.4, and C2 ' = 2 = 0.01423 R The frequency transformation is performed using Table 10.4, where L1 ' is transformed into a series circuit containing L1 '' and C1 '' . C2 ' is transformed into a parallel circuit containing L2 '' and C2 '' . This is shown in Figure P10.41(a). To perform the transformation, we calculate BW and wo: BW = 2 ( 200MHz )

o = 2

( 900MHz )(1100MHz ) = 2 ( 995MHz )

We then have: L' 167.4 L1 '' = 1 = = 133.2nH BW 2 ( 200 x106 )

C1 '' =

2 ( 200MHz ) BW = = 0.1921 pF 2 2 o L1 ' ( 2 ) ( 995MHz )2 (167.4 )

L2 '' =

2 ( 200MHz ) BW = = 2.259nH 2 2 o C2 ' ( 2 ) ( 995MHz )2 ( 0.01423)

C2 '' =

C2 ' 0.01423 = = 11.32 pF BW 2 ( 200 x106 )

To plot the insertion loss using ML1041, we use the impedances shown in Figure P10.41(b), given as j −j Z1 = j L1 ''− , Z 2 = j L2 '' C1 '' C2 '' We also define Z3 as the impedance seen by the voltage v1:

10-40

Z3 = Z 2 ( R + Z1 ) .

Now we can find the voltage ratio relations:    R  Z3 vLF =   vs .  v1 , and v1 =  R + Z Z + R + Z  1  1   3 Combining, and realizing vL = vs/2, we arrive at: vL ( R + Z1 )( R + Z1 + Z 3 ) = . vLf 2 RZ 3 The insertion loss is then:  v  IL = 20log  L  .  vLf    This is plotted by ML1041 in Figure P10.40(c). % MLP1041 % % Design N=3 BPF, 3 dB ripple. % clc clear R=50; flo=900e6;fhi=1100e6; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); g1=3.3487;g2=.9941; L=g1*R;C=g2/R; L1=L/BW; C1=BW/((wo^2)*L); L2=BW/((wo^2)*C); C2=C/BW; f=0.5e9:0.005e9:2e9; %f in GHz w=2*pi.*f; Z1=j.*(w.*L1-1./(w.*C1)); Z2=parallel(j.*w.*L2,-j./(w.*C2)); Z3=parallel(Z2,R+Z1); A=((R+Z1).*(R+Z1+Z3))./(2.*R.*Z3); IL=20*log10(abs(A));

Fig. P10.41 (c)

10-41

plot(f,IL) axis([.5e9 1.8e9 0 70]) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=3 BPF: 900-1100 MHz') Figure 10.41(c) also shows lines marking the frequencies at 60 dB insertion loss for calculation of shape factor. We have BW60 dB (1.7 − 0.57 ) GHz SF = = = 5.6 BW3dB 0.2GHz

P10.42: Starting with the Figure 10.28(b) circuit configuration, design a 3rd order Chebyshev bandpass filter for a 50  system. The passband is to be from 2.2 GHz to 2.6 GHz with only 1 dB of ripple allowed.

Assuming 1dB of ripple, from Table 10.3 we have: g1 = g3 = 2.0236, and g2 = 0.9941. (see Figure P10.28(b)) Now we perform the impedance transformation: g L1 ' = L3 ' = g1R = 101.2, and C2 ' = 2 = 0.01988 R The frequency transformation is performed using Table 10.4, where L1 ' is transformed into a series circuit containing L1 '' and C1 '' . C2 ' is transformed into a parallel circuit containing L2 '' and C2 '' . This is shown in Figure P10.42(a). To perform the transformation, we calculate BW and wo: BW = 2 ( 400MHz )

o = 2

( 2200MHz )( 2600MHz ) = 2 ( 2392MHz )

(a) Fig. P10.42a&b

(b)

10-42 We then have: L' 101.2 L1 '' = 1 = = 40.2nH BW 2 ( 400 x106 )

C1 '' =

2 ( 400MHz ) BW = = 0.110 pF 2 2 o L1 ' ( 2 ) ( 2392MHz )2 (101.2 )

L2 '' =

2 ( 400MHz ) BW = = 0.560nH 2 2 o C2 ' ( 2 ) ( 2392MHz )2 ( 0.01988 )

C2 '' =

C2 ' 0.01988 = = 7.93 pF BW 2 ( 400 x106 )

To plot the insertion loss using ML1043, we use the impedances shown in Figure P10.42(b), given as j −j Z1 = j L1 ''− , Z 2 = j L2 '' C1 '' C2 '' We also define Z3 as the impedance seen by the voltage v1: Z3 = Z 2 ( R + Z1 ) .

Now we can find the voltage ratio relations:    R  Z3 vLF =   vs .  v1 , and v1 =   R + Z1   Z3 + R + Z1  Combining, and realizing vL = vs/2, we arrive at: vL ( R + Z1 )( R + Z1 + Z 3 ) = . vLf 2 RZ 3 The insertion loss is then:  v  IL = 20log  L  .  vLf    This is plotted by ML1042 in Figure P10.42(c). % MLP1042% % Design N=3 BPF, 1 dB ripple. % clc clear R=50; flo=2.2e9;fhi=2.6e9; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi);

10-43

Fig. P10.42(c) g1=2.0236;g2=.9941; L=g1*R;C=g2/R; L1=L/BW; C1=BW/((wo^2)*L); L2=BW/((wo^2)*C); C2=C/BW; f=1:0.01:5; %f in GHz w=2*pi.*f*1e9; Z1=j.*(w.*L1-1./(w.*C1)); Z2=parallel(j.*w.*L2,-j./(w.*C2)); Z3=parallel(Z2,R+Z1); A=((R+Z1).*(R+Z1+Z3))./(2.*R.*Z3); IL=20*log10(abs(A)); plot(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=3 BPF: 900-1100 MHz')

10-44 P10.43: Starting with the Figure 10.28(b) circuit configuration, design a 3rd order Chebyshev bandstop filter for a 50  system. The stopband is to be from 900 MHz to 1100 MHz with 3 dB of ripple allowed. Plot the insertion loss.

(a) Fig. P10.43a&b

(b)

Assuming 3 dB of ripple, from Table 10.3 we have: g1 = g3 = 3.3487, and g2 = 0.7117. (see Figure P10.28(b)) Now we perform the impedance transformation: g L1 ' = L3 ' = g1R = 167.4, and C2 ' = 2 = 0.01423 R The frequency transformation is performed using Table 10.4, where L1 ' is transformed into a parallel circuit containing L1 '' and C1 '' . C2 ' is transformed into a series circuit containing L2 '' and C2 '' . This is shown in Figure P10.43(a). To perform the transformation, we calculate BW and wo: BW = 2 ( 200MHz )

o = 2

( 900MHz )(1100MHz ) = 2 ( 995MHz )

We then have: 1 1 C1 '' = = = 4.75 pF BW L1 ' 2 ( 200 x106 ) (167.4 )

2 ( 200MHz )(167.4 )

L1 '' =

BW L1 '

L2 '' =

1 1 = = 55.9nH BW C2 ' 2 ( 200MHz )( 0.01423)

C2 '' =

BW C2 '



2 o

o2

=

=

( 2 ) ( 995MHz ) 2

2

= 5.39nH

2 ( 200 x106 ) ( 0.01423)

( 2 ) ( 995MHz ) 2

2

= 0.458 pF

10-45 To plot the insertion loss using ML1043, we use the impedances shown in Figure P10.43(b), given as j −j Z 2 = j L2 ''− , Z1 = j L1 '' C2 '' C1 '' We also define Z3 as the impedance seen by the voltage v1: Z3 = Z 2 ( R + Z1 ) .

Now we can find the voltage ratio relations:    R  Z3 vLF =   vs .  v1 , and v1 =  Z + R + Z  R + Z1  3 1   Combining, and realizing vL = vs/2, we arrive at: vL ( R + Z1 )( R + Z1 + Z 3 ) = . vLf 2 RZ 3 The insertion loss is then:  v  IL = 20log  L  .  vLf    This is plotted by ML1043 in Figure P10.43(c). % MLP1043 % % Design N=3 band-stop filter, 3 dB ripple. % clc;clear R=50; flo=900e6;fhi=1100e6; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); g1=3.3487;g2=.7117; L=g1*R;C=g2/R; C1=1/(BW*L); L1=BW*L/((wo^2)); C2=BW*C/((wo^2)); L2=1/(BW*C);

Fig. P10.43c

f=.8:0.001:1.2; %f in GHz w=2*pi.*f*1e9; Z2=j.*(w.*L2-1./(w.*C2)); Z1=parallel(j.*w.*L1,-j./(w.*C1));

10-46 Z3=parallel(Z2,R+Z1); A=((R+Z1).*(R+Z1+Z3))./(2.*R.*Z3); IL=20*log10(abs(A)); plot(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=3 Band-Stop Filter: 900-1100 MHz') axis([0.8 1.2 0 50])

P10.44: Starting with the Figure 10.28(a) circuit configuration, design a 5th order Chebyshev bandstop filter for a 50  system. The stopband is to be from 2.3 GHz to 2.5 GHz with 1 dB of ripple allowed. Plot the insertion loss.

From Table 10.3 for 1 dB of ripple and N = 5 we find: g1 = g5 = 2.1349

g 2 = g 4 = 1.0911 g3 = 3.0009 We first perform transformation: g C1 ' = 1 = 0.0427, Ro

the

impedance

(a)

L2 ' = g 2 Ro = 54.56, C3 ' =

g3 = 0.0602 Ro

Figure P10.44(b) shows the filter portion of the circuit after transforming based on Table 10.4.

Fig. P10.44a&b To perform the frequency transformation, we first calculate: BW = 2 ( 2.5GHz − 2.3GHz ) = 2 ( 200MHz )

o = 2

( 2.3GHz )( 2.5GHz ) = 2 ( 2.398GHz )

Then, L1 '' =

1 1 = = 18.6nH BW C1 ' 2 ( 200 x106 ) ( 0.0427 )

(b)

10-47

2 ( 200MHz )( 0.0427 )

C1 '' =

BW C1 '

C2 '' =

1 1 = = 14.6nH BW L2 ' 2 ( 200MHz )( 54.56 )

L2 '' =

BW L2 '

L3 '' =

1 1 = = 13.3nH BW C3 ' 2 ( 200MHz )( 0.06002 )

C3 '' =

BW C3 '



2 o

o2

o2

=

=

=

( 2 ) ( 2.398GHz ) 2

2

2 ( 200 x106 ) ( 54.56 )

( 2 ) ( 2.398GHz ) 2

2

= 0.236 pF

= 0.302nH

2 ( 200 x106 ) ( 0.06002 )

( 2 ) ( 2.398GHz ) 2

2

= 0.332 pF

The voltage ratios needed to determine insertion loss are found with the aid of Fig P10.44(c). Here we have j Z1 = j L1 '− , C1 '

Z 2 = j L2 '

−j  C2 '

j C3 ' Fig. P10.44(c) Some of these impedances are also lumped together: Z a = Z3 ( Z 2 + Z1 R ) and Zb = Z1 ( Z 2 + Z a ) .

Z 3 = j L3 '−

The voltage ratios are then:  Z1 R   Za   Zb  vLf =   v1 , v1 =   v2 , v2 =   vs ,  Za + Z2   R + Zb   Z1 R + Z 2  so vLf

 Z1 R   Z a  Zb  =    . vs  Z1 R + Z 2   Z a + Z 2  R + Z b  Now, since vL=vs/2, then after manipulation we have vL ( Z1 R + Z 2 ) ( Z a + Z 2 )( R + Z b ) = vLf 2Z a Z b ( Z1 R ) and this is used to find IL = 20 log

vL using MLP1044: vLf

10-48 % MLP1044 % Design N=5 Band-Stop filter, 1 dB ripple. % clc;clear R=50; flo=2.3e9;fhi=2.5e9; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); g1=2.1349;g2=1.0911;g3=3.0009; Ca=g1/R;L=g2*R;Cb=g3/R; L1=1/(BW*Ca); C1=BW*Ca/((wo^2)); L2=BW*L/((wo^2)); C2=1/(BW*L); L3=1/(BW*Cb); C3=BW*Cb/((wo^2)); f=1.8:0.0001:3; w=2*pi.*f*1e9;

%f in GHz

Z1=j.*w.*L1-j./(w.*C1); Z2=parallel(j.*w.*L2,-j./(w.*C2)); Z3=j.*w.*L3-j./(w.*C3); Za=parallel(Z1,R); Zb=parallel(Z3,Z2+Za); Zc=parallel(Z1,Z2+Zb); A=(Za+Z2).*(Zb+Z2).*(Zc+R); B=2.*Za.*Zb.*Zc; IL=20*log10(abs(A./B)); plot(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=5 Band-Stop Filter: 2.3 - 2.5 GHz') axis([1.8 3 0 50])

10-49

Fig. P10.44(d)

5. Amplifiers P10.45: (JustAsk): The following S-Parameters were measured at 2.0 GHz in a 50  system: S11 = 0.68e j125 S12 = 0

S21 = 3.6e j 40 S22 = 0.86e− j 74 . (a) Determine the gain, in dB, without any matching networks. (b) Determine the maximum gain, assuming optimized matching networks. (a) With out matching networks, we have S = L = 0 as indicated in Figure P10.45. So our transducer gain term is:

GT =

1 − S

2

S21

2

1 − L

2

= S21 = 12.96 2

1 −  IN  S 1 − S22 L So GT (dB) = 10 log(12.96) = 11.1 dB. 2

2

(b) Now assuming optimized matching, the transducer gain term reduces to:

Fig. P10.45

10-50 GT =

1 1 − S21

2

S21

2

1 1 − S22

2

= 92.6

and GT (dB) = 10 log(92.6) = 19.7dB.

P10.46: For P10.45, (a) design open-ended shunt stub matching networks. (b) You are to realize the matching networks in microstrip constructed on 25.0 mil thick Teflon.

Determine the required microstrip width, and provide a labeled sketch of your network similar to Figure 10.42.

Referring to Figure P10.46(a) we find the open-ended shunt stub matching network to * = 0.86e j 74 . We move from the open-ended stub in the admittance chart achieve  L = S22 (point a’) a distance 0.206 to reach the point b’. The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’. From here, we move a distance (0.397 – 0.209 = 0.188 ) along the constant || circle to point d’.

Transforming this to the impedance chart, we reach our desired impedance point d.

The same procedure is used for the source matching network. Referring to Figure P10.46(b) we find the open-ended shunt stub matching network to achieve  S = S11* = 0.68e− j125 . We move from the open-ended stub in the admittance chart (point a’) a distance 0.173 to reach the point b’. The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’. From here, we move a distance (0.500 + 0.174 – 0.185 = 0.489 ) along the constant || circle to point d’. Transforming this to the impedance chart, we reach our desired impedance point d.

The final stub matching network is indicated in Figure P10.46(c).

To realize the stub matching network in microstrip, we employ ML0605 for microstrip design: Microstrip Design width & thickness will be in the same units enter the desired impedance: 50 enter the substrate thickness: 25 enter substrate rel permittivity: 2.1 w = 79.3268 eeff = 1.8015 up = 223363676.0028m/s >> The length of a guide wavelength is

10-51

(a)

(b)

(c)

(d) Fig. P10.46

G =

c f  eff

=

3x108 m / s  1mil    = 4400mils −6 9 2 x10 1.8015  25.4 x10 m 

This is used to find the physical lengths required of the stubs, as shown in Figure 10.46(d).

10-52 P10.47: For P10.45, design a matching network using lumped elements. In the sketch of your solution, indicate line lengths in terms of wavelength.

Fig P10.47(a)

We first design the lumped element matching network to achieve *  L = S22 = 0.86e j 74 .

For this type of problem we actually work backwards, starting from point c and working back to point a, to identify the location of the points. Then we work the matching network forwards to determine the component values. From the matched condition at point a, we add a shunt inductance to point b’ (-j2.0). Then, transforming to the impedance chart, we add a series inductance to reach point c (j1.3 – j0.4 = j0.9).

For the shunt inductance we have: − jZ o − j 2.0 = , or L 50 L= = 2nH 2 ( 2 ) ( 2 x109 )

For the series inductance, j L j 0.9 = , or Zo

L=

( 0.9 )( 50 ) = 3.6nH .

2 ( 2 x109 )

Now we design the lumped element matching network to achieve  S = S11* = 0.68e− j125 .

From the matched condition at point a, we add a shunt capacitance to point b’ (j1.8).

Then, transforming to the impedance chart, we add a series capacitance to reach point c (-j0.49 + j0.42 = -j0.07).

For the shunt capacitance we have: j1.8 = jCZ o , or

C= Fig. P10.47(b)

1.8 = 2.9 pF 2 ( 2 x109 ) ( 50 )

For the series capacitance,

10-53

Fig. P10.47(c)

− j 0.07 = C=

−j , or CZ o

1 = 23 pF . ( 0.07 ) 2 ( 2 x109 ) ( 50 )

The overall circuit is given in Figure P10.47(c).

P10.48: The following S-Parameters were measured at 10 GHz in a 50  system: S11 = 0.72e j 76 S12 = 0

S21 = 4.4e j125 S22 = 0.58e− j 30 . (a) Determine the gain, in dB, without any matching networks. (b) Determine the maximum gain, assuming optimized matching networks. (a) With out matching networks, we have S = L = 0. So our transducer gain term is: 2 2 1 − S 2 1 − L 2 GT = S21 = S21 = 19.36 2 2 1 −  IN  S 1 − S22  L So GT (dB) = 10 log(19.36) = 12.9 dB. (b) Now assuming optimized matching, the transducer gain term reduces to: 1 1 2 GT = S21 = 60.58 2 2 1 − S21 1 − S22 and GT (dB) = 10 log(60.59) = 17.8dB.

P10.49: For P10.48, design shorted shunt stub matching networks with the overall line lengths minimized. In the sketch of your solution, indicate line lengths in terms of wavelength.

Referring to Figure P10.49(a) we find the shorted shunt stub matching network to achieve *  L = S22 = 0.58e j 30 . We move from the open-ended stub in the admittance chart (point a’) a distance 0.099 to reach the point b’. The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’. From here, we

10-54

(a)

(b)

(c) Fig. P10.49 move a distance (0.458 – 0.326 = 0.132 ) along the constant || circle to point d’.

Transforming this to the impedance chart, we reach our desired impedance point d.

The same procedure is used for the source matching network. Referring to Figure P10.49(b) we find the shorted shunt stub matching network to achieve  S = S11* = 0.72e − j 76 . We move from the open-ended stub in the admittance chart (point a’) a distance 0.072 to reach the point b’. The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’. From here, we move a distance (0.500 + 0.106 – 0.311 = 0.295 ) along the constant || circle to point d’. Transforming this to the impedance chart, we reach our desired impedance point d.

Figure P10.49(c) shows the completed circuit. Note that this pair of matching networks results in the smallest overall sum of stub lengths.

10-55 P10.50: For P10.48, design the matching networks using lumped elements.

We first design the lumped element matching network to achieve *  L = S22 = 0.58e j 30 .

For this type of problem we actually work backwards, starting from point c and working back to point a, to identify the location of the points. Then we work the matching network forwards to determine the component values. From the matched condition at point a, we add a series inductance to point b (+j1.6). Then, transforming to the admittance chart, we add a shunt capacitance to reach point c’ (-j.25 –- j.44 = j0.19).

For the series inductance we have: 50 (1.6 ) − j L j1.6 = , or L = = 1.27nH Zo ( 2 ) (10 x109 ) For the shunt capacitance,

j 0.19 = jCZo , or C =

(a)

Fig. P10.50

( 0.19 )

2 (10 x109 ) ( 50 )

= .060 pF

(b)

10-56 Now we design the lumped element matching network to achieve  S = S11* = 0.72e − j 76 .

From the matched condition at point a, we add a shunt capacitance to point b’ (+j1.2).

Then, transforming to the impedance chart, we add a series capacitance to reach point c (j1.2 -- j.48 = -j0.72).

For the shunt capacitance we have: (1.25) j1.25 = jCZo , or C = = 0.40 pF ( 2 ) (10 x109 ) (50 ) For the series capacitance, −j 1 − j 0.72 = , or C = = .44 pF 9  CZ o 2 (10 x10 ) ( 50 )( 0.72 )

6. Receiver Design P10.51: Determine the IF power, in watts, exiting a mixer that has a 6.0 dB conversion loss if 0 dBm of RF power and of LO power enters the mixer.

P  CL = 10log  RF  = 6dB,  PIF  PRF = 100.6 = 3.981 PIF PIF =

PRF 1mW = = 0.25mW 3.981 3.981

P10.52: (JustAsk): Referring to Example 10.21 and Figure 10.48, suppose you require a 100 W output power level and the antenna receives -80 dBm. If you have several of each amplifier available, design the receiver. You are also allowed to insert a fixed value attenuator.

One design is shown in Figure P10.52.

Fig. P10.52