Fundamentals of electromagnetics with engineering applications 1st edition wentworth solutions manua by vincent.hasty918

Fundamentals of Electromagnetics with Engineering Applications 1st Edition

Wentworth

Full download at: Solution Manual:

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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

Using a vector identity we also have:

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:

Now we take the curl of both sides of Faraday’s Law,

Now since

EEE , and since 0 s  E , we have

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?

5-1

  2 2 tt tt         E HEE+E HH =-

  2HHH

0  H , leading to 2 2 2tt     HH H=

But

  0 0 s s ss ss j j       D B EH HE

      ssss jjjj  EHHE

  2

  2 ss jj EE

sss

(b) the frequency doesn’t change with the media (the wavelength does) so f =

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

If we let ,

So (a) becomes

then

has two solutions:

(2) for 0,wehave , ,or .

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

5-2

8310 5 0.06 p u cxms fGHz m  

(a)

GHz (c)   97 2 8 8 1 5100.01510 310 36 0.510 p r r mc ufxmx ss x x           

2 2 2 0. s s H H z     (a)

 

2 2 2 ,and . sszz HH AeAe zz   

z s HAe

  

  This

22 0,or 0.

zz sso HAeHHe   

(1) for 0,wehave , ,or .

HAeHHe   

zz sso

  . zz sooy HeHe  Ha P5.5: Given = 1.0x10-5 S/m , r = 2.0, r = 50., and f = 10. MHz, find 

.   roro jjj   ro ro j j          67 21010504103948 ro jjxxj      561253 1102101028.854101101.1110 ro jxjxxxjx 

into the



332579.4102.11,

j xjmxNpmradme 

The general solution is the linear superposition of the two, or

, 



and

Inserting these

expressions for

and ,

9.410, 2.1 , 1880

routine that will plot , , and  (magnitude and phase) versus the log of frequency from 1 Hz up to 100 GHz.

% M-File: MLP0506

% 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)

5-3

semilogx(f,aeta) ylabel('phase of eta (degrees)') xlabel('frequency (Hz)') P5.7: Suppose E(x,y,t) =

We assume nonmagnetic material and therefore have

To find the direction of propagation,

5-4

x106t – 3.0x

2.0

V/m.

32 5 jxjy sz ee  Ea

5.0 cos(

y) az

Find the direction of propagation, ap, and H(x,y,t).

32321015jxjyjxjy ssxy jjeejee  EHaa 323232321015 2.533.8 jxjyjxjyjxjyjxjy sxyxy o jj eeeeeeee jj   Haaaa     66 A (,,)2.53cos10323.80cos1032 m xyxytxtxyxtxy   Haa

ss P ss    EH a EH Fig. P5.6

And with the exponential terms canceling in the top and bottom of the equation for ap, we have:

Since free space is stated, 22

Now we have 2222 0,or 0

This can be factored: 

22 0 jj

, suggesting two solutions. The first solution uses j

Likewise, the second solution uses j   and jzjz xso EAeEe

The complete solution is a linear superposition of these two solutions, or . jzjz soox EeEe Ea

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.

5-5 64641912.65jxjyjxjy ssxy eeee EHaa

0.830.55. Pxy  aaa

in free space, H(x,t) = 100.cos(2x107t – x + /4) az mA/m. Find E(x,t).   0.100, , 4 1200.10012 jxj szPx jxjjxj sPsxzy ee eeee       Haaa EaHaaa   12cos y tx  Ea

P5.8: Suppose

230radm cf     and then 7 2 12cos210 304 y V xtx m      Ea

Start with

Helmholtz equation (5.11), and using  = j, derive (5.41), the

2 222 2 0,let (),andwith=jwehave 0. xs sssxsxxs E EzE z     EEEa Let 2 2 2 ,so and zxx xsxs xs EE EAeAeAe zz   

2. Propagation in Lossless, Charge-Free Media

P5.9:

the

traveling wave equation.

AeAe 

 

 



 

and . jzjz xso EAeEe

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

5-6

6220010 rad fx s   and 2 , 3 p rad um    or 6 2 (,)1cos20010 3 z V ytxty m      Ea Now to find H. 111 1, 1 120120 jyjyjy szsPsyzx eee    EaHaEaaa So   6 12 ,cos20010 1203 x A ytxty m       Ha or   6 2 ,2.7cos20010. 3 x mA ytxty m      Ha

From the given information we have

–

8 8 310 0.7510 16 p r cxm ux s        88 0.7510107.510 p rad uxx s    8 (,)100cos7.51010 z mA ytxty m  Ha 0.100, 120 0.1003 jy sz jyjy sPsyzx r e ee        Ha EaHaaa   8 (,)9.4cos7.51010 x V ytxty m  Ea 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. 0.080 120, 2 jyjy szsx ee   EaHa 11120 120 120 jyjyjy r sPsyzxx r rr eee        HaEaaaa so we know

intensity.

3.

Propagation in Dielectrics

P5.13: Work through the algebra to derive equation  and  equations (5.52) from equations (5.50) and (5.51).

Comparing the imaginary parts, we see

comparing the real parts, 222 0

Rearranging and inserting our value for

This is a quadratic expression (x2 + bx + c = 0), where here

5-7 2 r r    Now, 6 6 1610 7510 0.080 p rr cx ux     6 4 7510 rr c x   And now (2)(4)8 r rrr r     4 2 2 rr r rr    

  2222 2;jj  

2,or , 2     and

 .

: 222 422 0 4  

2 22 , , 2 xbc       Solving the quadratic: 2 2 2 41114 411 2222 xbcbbbbcc b      Reinserting the a, b and c values: 2 222 222 422 14 1111 24               2 11           Now for :

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

5-8 2 22222 0,=,so 0 22        Rearranging, 2 422 0 2       Solving this quadratic we find 2 11          

complex part of rel permittivity: .01

conductivity (S/m): 1e-12

frequency (Hz): 100

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;

5-9

eo=8.854e-12; B=ur*uo*erp*eo/2;

% 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

5-10

Fig.

P5.15

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.

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

5-11

  1cos;xxjx zsoz mA etxHee m       HaHa xjxxjx sPsxozoy HeeHee  EaHaaa        67 30 612 210010100410 2664 0.009642100108.85410 j jxx j e j jxx           1 14.825.7jjj m 

  156 (,)2.66cos200102630 x y V xtextx m   Ea

Finally,

%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')

5-12

P5.18: Starting with (5.13), show that = for a good conductor.

(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-

cos(

+ 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.

5-13

4. Propagation in Conductors

  foragoodconductor jjj     1 ,1222 2 2 jj jjj      





xt –



Fig P5.17

P5.20: Calculate the skin depth at 1.00 GHz for (a) copper, (b) silver, (c) gold, and (d) nickel.

5-14 So: 7.896, 0.004 oro jjjj  44.98 1.257 j j e j        4.4414.4451mjjj  1 4.4 m   0.528.6

syy

eeeeee mm  Eaa 28.628.61120 20 zzjzzj sPszyx A eeeeee m    HaEaaa   4.46 (,)15.9cos2104.428.645

m   Ha or with

  4.46 (,)16cos2104.416 z x A

 Ha

2020zzjradianszzj

z x

ztextz

appropriate significant digits:

ztextz m 

6 977 1 ;asanexample,forcopperat1GHz: 1 2.1102.1 1104105.81011 f xmm HVsA xxx smmHAV            Table P5.19 (S/m) r (m) Cu 5.8x107 1 2.1 Ag 6.2x107 1 2.0 Au 4.1x107 1 2.5 Ni 1.5x107 600 0.17 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      773 6004101.451034.3510() ffHzxxxfHz  = 1/ 45645218.0810() jj exfHze    

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:

So now we have

246 (,)1.0cos1024 z x V ztextz m   Ea

To find B we’ll work in phasors. 111 1, 1 zjzzjzzjz sxsPszxy eeeeee  

1 (,)cos102445 0.115 z y A ztextz m   Ha

246 (,)8.7cos102445 z y A ztextz m   Ha

P5.23: In a nonmagnetic material, E(z,t) = 10.e-200z cos(2 x 109t - 200z) ax mV/m. Find H(z,t).

5-15 61210 p rr cm ux s   Table P5.21 f(Hz)= 1 103 106 109 (Np/m) 185 5860 185x103 5.9x106 (rad/m) 185 5860 185x103 5.9x106  18ej45º 570ej45º 18ej45º m 0.57ej45º  5.4mm 170m 5.3m 170nm up(m/s) 12x106 12x106 12x106 12x106

  6(,)1.0cos10 z x V ztextz m   Ea   

jjxxj      6126

jjxxjx 



  4545

ee    

67 104103.948

1010.28.8541028410

Here,

(i.e. it is a good conductor), so 1 24.3 f m

20.115jj

 



 



EaHaEaaa

246



Since

, the media is a good metal. With r = 1 we have

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. Refer

5-16



     2 2 97 200 ,or 10.13 110410 o o S f fm xx      4545228jj ee     1110 10, 10 zjzzjzzjz sxsPszxy eeeeee    EaHaEaaa   2009 (,)360cos21020045 z y mA ztextz m   Ha

to Figure P5.24.. In the copper portion the field is Cu z xxo EEe   In the nickel portion,     Ni Cu zt t xxo EEee    The current density in the copper is , Cu z xCuCuxo JEe    and in the nickel is     . Ni Cu zt t xNiNixo JEee     The current is () CuCuNi ztzt CuxoNixo IEedydzEeedydz   , or () , CuCuNi t ztzt CuxoNixo ot IwEedzwEeedz     and upon evaluating Fig. P5.24

Now we’re ready to perform the calculations using the following data:

(a)

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.

5. The Poynting Theorem and Power Transmission P5.26:

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)

5-17   1, CuCu CuNitt xo CuNi IwEee     and with V=ExoL, we have   1 ,where= 1. CuCu CuNitt ss CuNi L RRRee w     

73 Cu 5.810,

Cur SNp xx mm  76 Cu 1.510, 600,

Nir SNp xx mm 

1, 47910

59610

0.1m Cu



over Ni: Rs = 176 m

(b) Semi-infinite Cu: Rs = 8.3 m





m Cu: Rs = 177 m

   2 2 7 1111 DC: 1.37 5.8100.002 Rm Lam x     1 GHz: 6 1 ; 12.0910 22 s R R fxm Laa         76 1 0.66 5.8102.091020.002 R Lm xx   

In air, H(z,t) = 12.cos(x106t - z + /6) ax A/m. Determine the power

1.0

meter surface that is normal to the direction of propagation.   2 2 2 11 1201227 22 avgxozzz AkW H mm      Paaa

density passing through a

square

Pav.     6(0,)1cos260010 y V txtz m  Ea

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:

5-18   (,)1cos z y V ztetz m   Ea

= 0 so = 0, and   6 8 260010 2.118.2 310 r x rad cxm    (a)   9 (,)1cos1.21018.2 y V ztxtz m   Ea (b) 12.1 1, 120 jz sPszy V e m     HaEaa   9 (,)3.8cos1.21018.2 x mA ztxtz m   Ha (c)  2 2 112.1 1.9 2120 avgzz mW m   Paa

Teflon:

  (,)cos, z ox V ztEetz m   Ea and we can see from the

information that 88 8, 210, 10, 0 o Vrad ExfHz ms   . Also     8124210818.854100.45, 10,so 1(lowlossdielectric). xx       4 101 1200.0021 2281 Np m     18.8 r rad cm   1 12041.9 81     so   0.00218 (,)8cos21018.8 z x V ztextz m   Ea (b) 0.002118.8 0.002118.80.002118.8 8, 18 191 41.9 zjz sx zjzzjz sPsyy V ee m mA eeee m    Ea HaEaa

given

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?

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?

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

5-19 so   0.00218 (,)191cos21018.8 z y mA ztextz m   Ha (c) 2 22(0.0021)(1) 2 1 0.7640.761 2 z xo avgzz E W ee m    Paa (d) 2 (10)7.6 avg PPmW 

144, cos45102 avgavg PPSWPPSW 

2 1 200,1, 2 o oavg E mV fMHzEP m   Cu: 453452, 21410,so 5.22 jj Np efxem m      2 3 32 110 96; 96 25.2210 avgavg W PPPSW xm    Now at 10 m

3 (10)3(21410)(10)6 (10)1011810 mxm o V EzmEeex m     2 6 32 111810 1.3; 1.3 25.2210 avg x W PPW xm   

beneath the surface, we have

1 5 tan27 10     

angle is:

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.

find the polarization and handedness.

Convert to E(z,t):

5-20

Fig. P5.31 P5.32: Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ay V/m,

P5.33: Given H(z,t) = 2.0cos(t-z)ax + 6.0cos(t-z-120 )ay A/m, Fig. P5.32

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.

P5.34: Given

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

5-21   1201202626 jzjzjjzjzj sPsozxyoyox eeeeee    EaHaaaaa       (,)6cos1202cos180 oxy Ezttztz  aa

>> From the figure, we have right-hand elliptical polarization.

    (,)coscos, xoyo ztEtzEtz  xy Eaa

–



Fig P5.33

P5.35: MATLAB: For a general elliptical polarization represented by

xy Eaa 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=|E

MAJ = length of majority-axis

MIN = length of minority-axis

    

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.

% 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: ');

5-22

   



(,)coscos, xoyo ztEtzEtz xo|, b=|Eyo| 224422

2 1

224422    

22cos2 22cos2 MAJababab MINababab   axial ratio=MAJ/MIN

1 22 12 tancos 2 ab ab

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,

5-23

Fig. P5.35

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).

P5.37: A UPW is normally incident from media 1 (z < 0,  = 0, r = 1.0, r = 4.0) to media 2 (

0,  = 0, r = 8.0, r = 2.0). Calculate the reflection and transmission coefficients seen by this wave.

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.

5-24

irt ooo EEE  (2) 1 2 irt ooo EEE   

(1) and (2): 112 2212 2 21, so itttti oooooo EEEEEE       

subtract (2) from (1): 11221 221221 2 211, rtiri ooooo EEEEE       

(1)

Add

Now

21 12 21 1208 ; 60, 120240 2 4       240603 0.60 240605     11.60  

z >

22 2 12 t jzjz t o syy E mAmA ee mm    Haa 2 tt 2os 2 30,so 12,E0.36,and 1.13 t jz o x E mAVV e mmm    Ea   21 21 32 1; , 1 55 tii ooo EEE     

P5.39: Suppose a UPW in air carrying an average power density of 100 mW/m2 is normally incident on a nonmagnetic material with r = 11. What is the time-averaged power density of the reflected and transmitted waves?

P5.40: A UPW in a lossless nonmagnetic r = 16 media (for z < 0) is given by

(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.

5-25 1 2.83,so 2.83 t jz ii o osx E Ee   Ea   1 (,)2.83cos. x V zttz m  Ea 1 1.70,so 1.70 jz rir oosx EEe   Ea     11 11 1.704.5 120 jzjz rr sPszxy mA ee m     HaEaaa       3 2 1 1.704.5103.8 2 r avgzz mW x m  Pa-a

12 1 1 120 11 120; ; 0.537 11 1 1 11 o            10.463   2 2 PP28.8 ri avgavg mW m  2 2 2 2 1 PP1171.2 2 xo ti avgavg E mW m   

sxy

 Eaa

1020jzjz ij

eee  

sxy eee    Eaa

12012 30; 1201202 6 16    21 21 0.700; 11.70      11 3 714jzjz rj sxy eee    Eaa   (,)11 7cos14cos 3 r ztxy V tztz m     Eaa

11 3 1020jzjz rj

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

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

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.

5-26 22 3 1020, jzjz tj sxy eee    Eaa or 22 3 1734jzjz tj sxy eee    Eaa , so   (,)22 17cos34cos 3 t ztxy V tztz m     Eaa .

= 0

1 1 SWR   

For copper we have     45 2 2 2 8773 2222 2 where 104105.81015110 j e Np fxxx m       so 45 2 3.7 j em   We find 645 22 121 22 1,and=19.610 j xe      So Er = -10.cos(2x 108t + 1z) ax V/m   22 45 196, zjz tj sx V eee m    Ea and   2 2 196cos45 z t x V etz m    Ea       2 6 2 3 119610 cos453.7. 2 3.710 t avgzz xVm W m x    Paa

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

and 1 GHz.

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;

5-27

MHz,

6 12 9 1207.89610() 24; 250.0084.50610() jjxfHz jjxfHz      Table P5.44 f 2() 21 21     1 1 SWR    1kHz 0.994ej44.98° 0.9815ej178.9° 107.3 1MHz 29.3ej30.3° 0.513ej155° 3.11 1GHz 41.9ej0.05° 0.286ej179.9° 1.80

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

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.

5-28

Fig P5.45

16.).

the average power densities for the incident, reflected and transmitted waves.

We use ML0501 in each media to find:

We also will need reflection and transmission coefficients:

5-29 76 11 1 2 120, 10 so 510, 0.105 xfxHzradrad scm    

this



0.0025

 = 0.05.

(5.13) and (5.31), we calculate: 43.6 222 0.969, 1.019, 28.1 j Nprad e mm   . Then, 17440.8 21 21 1 0.898, 18.6,10.141 1 jj eSWRe       1 100 jz i sx V e m   Ea 11 174 10089.8jzjz rj sxx VV eee mm   Eaa , so   7 (,)89.8cos100.105174. r x V ztxtz m   Ea 22 40.8 10014.1jzjz tj sxx VV eee mm    Eaa , so   7 (,)14.1cos101.0240.8. t x V ztxtz m   Ea 22 43.640.82.8 14.1 0.502 28.1 jzjz tjjj syy AA eeeee mm  Haa      2 1 14.10.502cos40.82.82.6 2 t zz W m  Paa     2 2 100 13.3 2120 i z W m   Pa       2 2 89.8 10.7 2120 r z W m   P-a (check: 13.3 W/m2 = 10.7 W/m2 + 2.6 W/m2) P5.47:

Ei = 12 cos(2x107t-1z+/4)ax V/m

lossless, r

9.0 media (at z < 0) to a media (z > 0) with  = 0.020 S/m, r



Er , Hr , Et , Ht

problem we find in medium 2 (z >

that

and

These values are too close to allow for simplifying assumptions. Using

A wave specified by

is incident from

nonmagnetic,

= 2.0, and

r =

Find H

, and

111 0; 0.628; 40 rad m   33 222 1.01; 1.56; 84.9 j Nprad e mm 

(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

5-30 12619.8 21 21 0.353; 10.84 jj ee      Incident: 1 4 12 jz ij sx V ee m     Ea 1144 1 12 0.300 jzjz ijj syy AA eeee mm     Haa , 7 (,)0.300cos2100.628. 4 i y A ztxtz m      Ha  2 2 1 112 5.655 2 i avgzz W m    Paa Reflected:   1114451261711213.313.3 jzjzjz rjjjj sxxx VVV eeeeeee mmm     Eaaa   7 (,)13.3cos2100.628171. r x V ztxtz m   Ea 11171171 13.3 0.106 40 jzjz rjj syy AA eeee mm     Haa   7 (,)0.106cos2100.628171. r y A ztxtz m   Ha     2 2 113.3 0.704 240 r avgzz W m   Pa-a Transmitted:   12464.81231.67jzjz tjj sxx VV eeee mm     Eaa ,   7 (,)31.7cos2101.5664.8. t x V ztxtz m   Ea 22 3364.831.8 31.67 0.373 84.9 jzjz tjjj syy AA eeeee mm  Haa ,   7 (,)0.373cos2101.5631.8. t y A ztxtz m   Ha      2 31.670.373 cos64.831.84.954 2 t avgzz W m  Paa

incidence is 40. Calculate (a) the angle of transmission, (b) the reflection and transmission coefficients, and (c) the incident, reflected and transmitted fields.

(a)

(b) Now we need to calculate the reflection and transmission coefficients.

5-31

  6 12 8 210010 2.09, 10.45. 310 r x radrad cxmcm      1 22 111 ; sinsin40;7.4 55tt r    

12 120 120; 24 25    21 21 coscos 0.732; 10.268 coscos it TETETE it     

The fields, Incident:   2.09sin40cos40 1.341.60 11 jxz ijxjz syy V eee m   Eaa   (,)1cos1.341.60 i y V zttxz m   Ea   1.341.60 1 cos40sin40 120 ijxjz sxz ee   Haa   1.341.60 2.031.71 ijxjz sxz mA ee m  Haa     (,)2.031.71cos1.341.60 i xz mA zttxz m  Haa Reflected: 0.732 ri oTEo EE   1.341.60 0.732 rjxjz sy V ee m   Ea   (,)0.732cos1.341.60 r y V zttxz m   Ea       1.341.60 1.341.60 0.732 cos40sin40 120 1.491.25 rjxjz sxz rjxjz sxz A ee m mA ee m      Haa Haa     (,)1.491.25cos1.341.60 r xz mA zttxz m  Haa transmitted: 0.268 ti oTEo EE     2 sincos 1.3510.4 0.2680.268 ttjxz tjxjz syy V eee m   Eaa

(c)

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.

rjxjz sxz ee   Eaa

transmitted:

0.318cos7.4sin7.4tjxjz sxz ee  Eaa

t xz

5-32   1.3510.4 0.268 cos7.4sin7.4 24 tjxjz sxz A ee m   Haa   1.3510.4 3.50.46 tjxjz sxz mA ee m  Haa     (,)3.50.46cos1.3510.4 t xz mA zttxz m  Haa

(b) 21 21 coscos 0.589 coscos ti TM ti     2 21 2cos 0.318 coscos i TM ti      (c) Incident:   1.341.60 1cos40sin40ijxjz sxz ee  Eaa     (,)0.7660.643cos1.341.60 i xz V zttxz m  Eaa 1.341.60 1 120 ijxjz sy A ee m   Ha   (,)2.65cos1.341.60 i y mA zttxz m   Ha Reflected:   1.341.60 0.589cos40sin40

    (,)0.4520.379cos1.341.60

xz V zttxz m  Eaa

  

 

1.341.60 0.589 120 rjxjz sy A ee m

(,)1.56cos1.341.60 r y mA zttxz m   Ha

  1.3510.4

    (,)0.3150.041cos1.3510.4

zttxz m  Eaa

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: 1 sin;76

Also, we calculate 1 = 4.19 rad/m, 2 = 16.8 rad/m, 1 = 120, and

At the Brewster’s angle of incidence, we have from Snell’s Law:

Combining the results we arrive at:

5-33   (,)4.22cos1.3510.4

zttxz m   Ha

t y mA

 

1116 BABA

2

. TE   1 sincos 4.061.01 1010 iijxz ijxjz syy V eee m   Eaa

= 30

1 1 2 sinsin14ti        21 21 coscos 0.883; 10.117 coscos it TETETE it      VV =-8.83; =1.17 mm riti oTEooTEo EEEE   4.061.01 8.83 rjxjz sy V ee m   Ea     2 sincos 4.0616.3 101.17 ttjxz tjxjz sTEyy V eee m     Eaa TM: At the Brewster’s angle, TM = 0 and 0. r s  E   4.061.01 10cossinijxjz sixiz ee  Eaa ,   4.061.01 2.429.70.ijxjz sxz V ee m  Eaa   4.0616.3 10cossintjxjz stxtz ee  Eaa ,   4.0616.3 9.702.40.tjxjz sxz V ee m  Eaa

5-34     (,)2.4109.7cos4.061.01 i xyz V zttxz m  Eaaa   (,)8.83cos4.061.01 r y V zttxz m   Ea     (,)9.71.22.4cos4.0616.3 t xyz V zttxz m  Eaaa