Fundamentals of Electromagnetics with Engineering Applications Solutions Manual.pdf (33)

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     15I2 At  = /2, we then have Funit = o .  r2 The radiated power vector is then 2 15 I 2  15(1) P(r,) arary =  2o  4 ra =(20002 4a= 4.78Wa r r

)

r 

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=4cos2  cos+0=2cos   , =dcos  + = 4  2 = so Farray=4cos2(cos), array F For magnetic dipoles we have:

max

2

1   o oI S  Pmax 1 loop =  ,  32o   r  where c 3 2 2 = = 0.6m, S =  a2 =  ( 0.01) , = = 108 x f  5 so 2 2 x108   4 2 0.01 10−7 x  2 ( (  ) 0.6) ) 0.01) ( ( ) ( (500x106 ) 1   Pmax1 loop =   (100  ) ) 32 (120  pW Pmax1 loop = 14.2 m 2 pW max=(Pmax1 mop)4=57

P

lo

2

A plot of cos2(cos) gives the same result at P8.35.

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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.

(b)=dcos  + =

2  cos + 0 =  c o s   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: dPn

d

=

 2

  sinsin cos=0 at =0, (for P n min) 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 % clc in the x-y plane varies with separation distance clear between a pair of in-phase dipoles.

Wentworth, 2/11/03 %clears the command window %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

Pmax = FuitnF array

 15Io2  2  1510 ( )2 (10)=480 W 2 2  = ( N ) =  2 max  m2  r  )   (10, 000

N( 2) 1 sin Pn = ,  =  d cos +  = cos N2 sin2(  2) 2

So we plot 2 1 sin (5cos ) Pn = sin2  100   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 Fig. P8.44 antennas are 100% efficient.

c

3x108

=0.122m f 2.45x109 Also, for half-wave dipoles the maximum directivity is 1.64.

Applying the Friis Equation:

At f = 2.45 GHz we have = =

2

2

Prec W    2  0.122 = 102x 10−9 W = Dmax1D max2  = 1.64  ( )  4R Prad 4  ( )  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 yaxis. 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 3x108 c  = f = 0.800x109=0.375m 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

   Prec = Dmax1Dmax2  rad P  4 R 

2

−9 W = ( 1 . 6 40.375 ) ( 1 .=219x10 5)W   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.

2

Prec  = D (  ,  ) D (  ,  )  rad P 4R =6m and R = 4000m. where =c f P ( , ) P  (, ) = 4 P = n D (  ,  )n = n(,)=DmaxPn(,) p Pn ( , ) p 4 avg

For the transmitter,

8-38

  cos2  cos  2  Pn t (  = 6 0 , 2 ) == 0.667 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 ,) = sin2= 0.750 and Dmax = 1.5 for the Hertzian dipole.

So, 2   Prec 6 W −9 = (1.64 ( ) 1.50 ( ) 0.750 ( ) 0.667  = 17.5 x10 W  )  4 ( 4000 Prad )    or Prec =−77dB 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.

2 Pout    = ep D 1 ( 2max ,  ) D Pin  4 R 



cos2  cos  2  D1 (  ,  ) = D m a x P n (  , 2 ) , P n (  ,  ) = 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     cos2  cos −   2 2  Pn (  = ,  )  2 −  sin   2

8-39

Fig. P8.48a Fig. P8.48b 2 p= The polarization efficiency term is seen to be e cos.

At 1 GHz, we have l = 0.3 m. And for half-wave dipoles we have max D = D 2max = 1.64.

% clc clear MLP0848 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 = 1j0.8.

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 = 73 – j42 Z *ant zin = 1.5 – j0.8. De-normalizing, we find: Zin = 75 – j40 . This is close the the theoretical Zin = 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 = 0.092A, = Prad = 2i2Rrad,i = +j 42.5 ) 73.2 Zo + Zant (50+

1 2  Prad = (.092) (73.2=)310mW 2

= Now we find the transfer ratio by applying the Friis equation where 2

2  0.6  Prec    = DmaxtDmaxr  (  ) 1.64)  = 613 x10 −9  = (1.64  4(100) Prad  4R     So Prec = 190 nW. Now we calculate Voc by assuming a matched load:

c = 0.6 m: f

8-42

Voc2 , Voc= 4RradPrec=7.46mV rec= P 4Rrad Finally, we calculate vL (see Figure P8.50b): Zo 50 Voc = vL = (7.46mV)=2.9mV 50+73.2+ j42.5 Zo Z+ant 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: vs 1 2 i= =120mA, Prad=360mW Prad= iZ, o 2 2Zo The power transfer ratio is: Prec −9 =613x10, from P8.50.

Prad Therefore, Prec = 220.7nW. Now, since the receiver is matched, half the power must be dissipated in the load, or 1v2 1 PrecZo=3.3mV 2 Prec = L , vL= 2 Zo 7. Radar P8.52: Manipulate (8.125) using (8.113) to arrive at (8.126).

Rearranging (8.113) we find

4

D (  ,  )2A =e ( ) ,  This can be inserted into (8.125) for D(), or 2= 4 R4 2s   4  Prec1 2 s = . ,  A e (   ,  ) 2 ( A e )( )  2 ) 3 ( 4 r a d 1 R4  P   P8.53: (JustAsk): Suppose a 2 GHz radar antenna of effective area 6.0 m2 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 A e = 6.0m2 Prad1 = 100 kW  s = 12m2 R=100 km

(a) t=2

R =0.67ms c

rad1 (b) P rec=P 1

s

4

4 R 

12m2

2(6m)2 =1.53pW 4(100km)(0.15m)

A2 2 e =(100kW)

4

recmin = 2.0pW, we rearrange the radar equation as (c) To find Rmax if (P ) 4 2 P d1  s 2 100kW 12m R = ra Ae= 2(6m)2 =76.4x1018 rec1 4 2 P 2pW 4(0.15m) 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):

Pre1c Prad1

=

 s 2

( m) (m 12)1 D(,)= (1.64 )

(4)R4 3

2

2

2

(4)R4 3

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) A e= 100m2  s= 6.64x1011m2 Prec = 27 pW

8-44 t = 2.56 sec (a) we know that t = 2R/c, so R = tc/2 = 384x106 m (b) 2 Prec1 s 6.64x1011m = = A2 (100m )2 = 27x10−18 e 4 2 Prad1 4R42 4 (384x106 m )0.030 ( m)

Prad1 =

12 27 x10− Prec1 = = 1MW 18 27x10−18 27 x10−

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. a=

31.96mils25.4m m 2

=406x10−6m mil 106m 0.02

l =   a2 (5.8x107

Rdc=

=666

406 ) x10( − 6 2

)

At 800 MHz:

Rac =

l −6 = 2.336 x10 , where  = 1 m f 2a

So Rac =

0.02

=58m

( x10−6 (5.8x107 )(2 )( 406 x10−6 ) 2.336 )

  2(0.02)  L (2x10−7)(0.02) ln − 6   406x10 so L = 14 nH

 =44nH, 1. 

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 N−1

=

300mil−20(3.145 mil) = 12.5mils 19

1000mils =37x103mil2 2.54cm

S=t(2a)(N−1)= (3.145mils)(2)(0.25 cm)(19)

9-3

Cx=

) ( F m 37x103mil2 (8.854x10−12 ) 25.4x10−6m =666x10−15F

fsrf=

12.5mil 1

mil

=171MHz,

2(1.3x10−6)(666x10−15)

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 L=

 oN 2  a 2 =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 = 866kHz, fsrf =

2 (6.5mH)(5.2pF)

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.

 N2 For inductance, from P3.56 we found: L= = (b−a) where b = (a+c)/2 = 0.375 cm.

I 2b For the capacitance we have 2

S

, where S = t(2ac)(N-1), and ac = (c-a)/2 = 0.125 cm, and t = 405 m. d Also, − Nt d= h for a toroid, where h = core length = 2 b.

N Plugging in the numbers we have 2 −)0.0025)2 10  (x 20) (− 0.00375 (5000) ( 4 7 = 524 H L= ( 0.00375) 2 = m2 = 60.4 x10−6 S ( 405x10−6 ( ) ( 19) )(2 ) 0.00125 C=

9-4

2 ( 0.00375) − 20 405x10−6 ) =(773 x10m −6 20 x −12) 60.4 ( x10− =6 )fF (8.854 10 C= 0.69 x 10−6 ) (773 1 1 =8.4MHz. fsrf= 2 LC = 2 524(x10 −0.69 6 x10) − ( 15 ) d=

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   each lead is l = 1 cm, and Lx=2x10−7  lln   1, − where   m   a     25.4x10− 6m 1 −6 mils ) a = 2 ( 20.1 , so  =255 x10m mil     2(0.01)   H  −7 −1=6.7nH. Lx=2x10  (0.01) ln    255  x10− 6    m With the two leads in series we then have Ltot = 13.4 nH, and then 1 1 = =6.3MHz, fsrf= 2L ttC o 2(13.4x10−9 ) 47x0− ( 1 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 = 2.1 Ebr and= 60x106V/m. r So,

C  2.1(8.854x10 ) F = = =186 2 , − 6 0.1x10 S d m and V Ebr = max ( 6 )= 6V. ) 0.1x10−6 d , or Vmax = Ebrd= ( 60 x10 −12

9-5 P9.8: If the 2.2 nF capacitor of Example 9.3 has an area of 20. mm2, what thickness mica is used? What is the maximum voltage that can be applied across this capacitor?

6 For mica, from the appendix we have  and Ebr = 200x10 r = 5.4 V/m.

F  8.854x10  12 2 − (5.4) 2)  ( 2 0 m 1m   S m    S, d = r o m  C= =  = 435nm 2.2x10− 9 1000mm F d C V = Ebrd=(200x106)435x10−9 ( )=87V. max

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(.008) 0  − 1  = 5 . 0 2 n H  H    l   x0−7 l ln 2  − 1  = 2 −x7(0.008)ln 10 Lx= 21    2 5−5 x 1 0 6     m    a   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)(100x106 )(5.02x10−9 ) = j3.16.

The reflection coefficient is found by − j3.16−30 ZL L 0  L= ZZo =1ej179 . + Zo = j3.16+30 0 Repeating the calculation at 1 GHz and at 10 GHz, we have: Frequency 100 MHz 1 GHz 10 GHz

ZL()

L

j3.16 j31.5 j315

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 0 − 9 s BW = 4 x 1=250MHz. r t (b) Modifying Equation (9.13), 1 1 =1ns. tr BW= x 9 110

9-6

P9.11: Suppose a 1.0 GHz clock rate is assumed. What is the minimum spectral bandwidth calculated using Eqn. (9.13).

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, 1 1 x 1 0 − 9 s BW = 0 . 5 =2GHz. then, the tr risetime would be half the period, or tr = 0.5 ns. Then we have 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 clc clock rate signal, with minimum clear rise and fall times of a sawtooth wave. %clears the command window %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.

1 25.4m AWG22 wire has a= (25.35mils) =322m, so we can use equation (9.4) to 2  mil  find   2 (0.04 )    H   2l   Lx=2x10−7  lln   1− 2x10−7 (0.04) ln =  1− 36nH =  322x10− 6    m   a      At 200 MHz, the impedance is Z = j  L 20010 =x 6 )j 36x10−9)=j45 (2)( Figure P9.13 indicates our situation. By circuit analysis we then have VA = (3m ) 45 ( ) 135mV, VB = (A ) 45 (  ) =135mV= 225mV,

2m ) 45 VC = (1mA ( ) +225mV= 270mV.  + A

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=2x10−7  lln  a  1− 2x10−7(0.04) ln =  1− 36nH =  322x10− 6    m        impedance is At 200 MHz, the impedance of the 4 cm length

Z=jL=j(2)(200x106)(36x10−9)=j45.

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?

 Ens  12 kV m  SE=20log =20log  =262dB    1nV 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   = r((14x1x 01 90 )−7 ) 

9-9

For copper, =(1x109)(1)(4x10−7 ( ) 5.8x107)=479x103Npm, dB  Np −6 ( ) And the attenuation isAttendB= , which for copper is 20x10m)8.686 ( Np  m  

Atten(dB)=(479x103)(20x10−6m)(8.686)=83dB.

Using this approach the following table is generated: (S/m) r 5.8x107 1 (a) copper 3.8x107 1 (b) aluminum 1.5 x107 600 (c) nickel

(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

= 100x106 ( 4x10−7 () ) ( 106)=19.9x103 paint Then,

Np . m

 d  = 8 . 6 d B .  − 3 Np paint=19.9x10 Atten (50x106m)8.686 m B    Nwe find Repeating these calculations for pure silver,  silver = 156x103Np/m and Attensilver = 68dB. p 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  = 1x107S/m. So we have Np . =f=(1000)(5000)(4x10−7)(1x107)=14x103 m In terms of dB, we have  Np d  dB  (dB) = 14  x103  8.686  =122x103m . m B    N  t, then Since atten =  we have  20dB Atten p = 163m. t=  = 122x103 dBm

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 s ur shield thickness (m) uo er shield conductivity (S/m) eo f,w rel permeability c Zo free space permeability A,B,C rel permittivity prop free space permittivity Z1 freq. and ang. freq. (1/s) taud speed of light (m/s) tao0 free space impedance (ohms) ratio calculation variables propagation constant (1/m) impedance (ohms) transmission coeff at z = -d trans coeff at z = 0 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

%clears the command window

clear

%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,'-^') face legend('front 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= 2R + jL vS.

9-13 Therefore, vLf 1 2R 1 = = = jfL . vL 2R + jL 1+ j L 1+ R 2R 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 +   10log1   R    R  =    vLf = vL       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=10log1  .  +   R    2    100x10−9 ) ) (  + (200x106  =16dB. (a) IL =10log 1      10      2   6 9 ) 10 100 x x  ( − ( ) + 200 10 (b) IL=10log1 =   0.17x10−3dB. 3     1010 x     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

(a) IL=10log(1+((200x106)(10)(47x10−1 (b) IL=10log(1+((200x106)(10x103)(47x 2

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 + jL, and Fig. 9.24(b) & (c) − j Z2 = Z1 C Then we have R Z2 R R Z2 vS , vLf = R + jLv2 = v2 ,  vLf = v2 = vS .

Z1 Z1 + Z 2 Z1Z1+Z2 1 2 vL = vS ,

( ).

 v  and we then wish to calculate insertion lossIL as= 20 log L    vLf .  This is calculated using MLP0924. % % MLP0924 % % clc clear A T-Filter Problem

% 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

Z1= R L

−j , C

and Z2 = j L + Z1.

Then, −j Z3=Z 2 . C Using these values we find Z v1 =1+3Rs v S , Z

Z v2 = 1 1+ jv 1 L= vLf , Z Z3 Z1  v Lf = vS Z1+jLZ1+Rs and RL .

RL+ R S Finally, v  IL=20log  L .  Lf  v  Fig. P9.25 This is calculated using MLP0925. vL =

% % MLP0925 % This filter. %

is

a

pi

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 Fig. P6.32b (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 −j2.4= ,

CZo

or C=

1

= 1.33pF

2 (1x109) )2.4) ( ( 50

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 Fig. P10.2(a) (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

10-2 Realizing these components, we have for the series element: (0.1) 50 ()= j L j0.1= , or L = 0.796nH. Zo 2(1x109) So L = 0.80 nH. Then, the shunt element is: j2.0 = jCZo,

or C=

2.0 =6.37pF. 50 2(110 x 9)

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 = We add +j0.22 (shunt 0.44+j0.28. 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 ,

0.22 =0.7pF. or C = 2 1( 10 x 50 9 )(

)

For the series +j1.1 element: j L , j1.1= Zo

or L=

(1.1) 50 ( ) =8.7nH.

2 110 ( x

9 )

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 − 0.9 j = , or  CZo

C=

1

=1.4pF. 2 ( 2.5 x109 ) 0.9) ( ) ( 50

For the shunt +j0.8 we have: j 0.8= jCZ,  or o 0.8 ( ) =1.02pF. C= 2 2.5(x109 50 ) )( The figure shows the final matching Fig. P10.4 An alternate solution replaces network. 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 , or C= −j100=

C

1

=3.18pF.

2(500x106)(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) =0.95pF. j0.3 = jCZo, or C= 2 500 ( x10 6100)) ( For the series j1.2: (1.2 ) (100 ) =38nH. j L j1.2= , or L= Zo 2 (500x 106 ) 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CZo ,

or C =

(0.72)

= 0.20pF.E 2 11.18 ( x10 950 ) ( )

The series inductance is: j L j1.45= , Zo

or L =

) (1.45 ) ( 50 = 1.03nH.

2(11.18x109 )

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 Assume a 50  system capacitor. 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 j0.5= , or Zo

L=

(0.5)(50) =1.33nH.

2(3x109)

and

− jZ − j 1 = o , or L L=

(50) = 2.65nH.

Fig. P10.8

2 x

(3 109 )

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 somewhatsimilar to MATLAB 9.1).

The two fundamental solutions are shown in Figure P10.9a. For the top solution, we have: in1= − j Z (jLm1+ZL)

 Cm1

And for the bottom solution:  −j  +ZL  Zin2=jLm2   Cm2  Once Zin is found, then

Fig. P10.9a

10-8 Zin − Z o, Zin + Zo and we can plot || versus frequency.

From P10.2 we have ZL = 10 + j15 , Lm1 = 0.80 nH, Cm1 = 6.4 pF, L 4.0 nH and m2 = Cm2 = 4.5 pF. The following MATLAB routine generates the plot. =

% %MLP1009 % % % % clc clear ZL=10+j*15; Zo=50; f=0.1:0.01:10; w=2*pi.*f*1e9; Plot magnitude of reflection coefficient Lm1=0.80e-9; XL1=j*w.*Lm1; from 0.1GHz to 10 GHz looking into each Cm1=6.4e-12; XC1=-j./(w.*Cm1); matching network from P10.02.

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 matchinganetwork of P10.3.

Develop 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+ ZL

 Cm1

And for the bottom solution: − j +jLm2ZL Zin2=

 Cm2

Once Zin is found, then Zin−Z  =Z o, in+Zo 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, L 10 nH and Cm2 m2 = = 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 clc clear network from P10.03.

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   j +jLx  ZL Zin=jLm+ Rx− C  p10.11a.  Then, We now−have: Z Z  = in o, Zin + Zo and we can plot|| versus frequency. % % MLP1011 % % % % % clc clear Plot magnitude of reflection coefficient ZL=250-j*250; from 0.1GHz to 10 GHz looking into each Zo=50; matching network of Ex10.1. Compare with f=0.1:0.01:10; plot including parasitic effects. 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   =1  (a)S (b)S =   0 1  0 (a)  −1 0  1 0  S =   (d)S (c)    = 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

Fig. P10.13

The circuit (Figure P10.13(a)) is redrawn in (b) to calculate S11 and S21. Here, 7 −5 1 = =0.2 S1 1 =  = 5 +0 5  V − 7 5  1  1 2  +1 S21 = 2 =   V1+ 5  0  5  3  5    = S21 = 0.40 2 =    5  is used to calculate22S Figure P10.13(c) 7 and S12: = 100 50 = ZL 533.3, 33.3 − 50 S22 =  = = −0.2 33.3 + 50 V−  5 0   1 =0.8  =0.4 S12 = 1 =  V2+  2 100

0.20 0.40  , 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=−20logS21=−20log0.4=8dB S So:  = 

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  4  2 By staring with V1− = S11V1 + + S12V 2+

V2− = S21V1+ + S22V 2+ Fig. P10.14 and considering that 100−50 V2− =1V− , V2+ =  V2− = 3 2 100+50 we can manipulate the scattering parameters to arrive at:

10-14

( ) () ( )

1 1 V− S12 S 2 2 =0.07692 1 = S 11+ 21=0+ V1+ − 3 − S 2 2 3− 1 4 The return loss is then: RL=−20log0.07692=22.3dB. The easier approach: All we need is the reflection coefficient looking into the circuit of Fig P10.14(b). We have ZL=25+50100=58.33,  58.33−5 ZL L= −Zo =0 =0.07693, 58.33+5 L Z+Zo 0 and then RL=−20log0.07693=22.3dB

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 = =−j80 Zc=  C 2( 110 x9 (x − 1 2 ) ) 210 To find S11, 5 − j80−5 S11 =  =0 5 −0 0 − j80 = j80+5 j51 0.625e− = 100− j80 0 And for S21, V− 50 S21= 2 =  V2+ 50− j80

Fig. P10.15

50 ( 0.625e−j51) 50 − j80= 0.78e

= 1+ So

0.625e−j51

S  = 

j39

0.780ej39  , 0.625e−j51  

  0.780ej39 RL = −20 logS11 = 4.1dB IL = −20 logS21 = 2.2dB

10-15 P10.16: A series inductor of value L = 3.5 nH is inserted in a 50T-Line. At 1.0 GHz, determine [S], the return loss and the insertion loss.

The impedance of the inductor at 1 GHz is

110 )x3.5 Z = jL = j2( (9x10 − 922 ) = j  L

To find S11, 5 + j2 − 50 0 S11 =  =5 0 + 2 + 50 77. j2 = 0.215 ej And for S621, 2 V− 50 S21= 2 =  V + 50+ j22 2

( 0.215ej77.6)

= 1+

50 − j 12.4 50+ j22 = 0.977e

Fig. P10.16

So

0.215ej78 0.98e− j12  RL = −20log S11 = 13.4dB , and S= IL=−20logS21 = 0.18dB 0.98e−j12 0.215ej78

P10.17: The scattering matrix for a 3-port network is

0 j 0.80  0.60   S  = 0 1.0ej90 0 .  j 0.80 0.60 0  (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.

Fig. P10.17

(a)

(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 inisa confirmed by evaluation of SS which indeed results unitary

matrix.

V+1 (c) To determine the return loss, we need to determine V−.

1 We start with: (1)V1− = 0.6V1+ + j 0.8V+ 3 = (2)V − =1j90 e V + jV+ 2

2

2

(3)V3− = j 0.8V1++ 0.6V3+ Examining Figure P10.17(b), we have (4)V+=V−e−j45 2 3

(5)V+=V−e−j45 .

3 2 Inserting (5) and (2) into (3), we have V + + 45 (6)V3− = j 0.8 1 0.6 V−e− 0.8V1++ 0.6e− j45 jV+ 2 2

( j

) = j

Now inserting (4) into (6), + + 0.6e− j45 (7)V3− = j 0.8V 3 e j45 = j0.8V1++ 0.6V3− 1

j ( V )− −

and solving (7) forV3−

= j )0.8V1+ , so V3− = j2V+ (8)V3− (1 − 0.6 1 .

Next, inserting (8) into (4) we have

(9)V+=(j2V+)e−j45=j2e−j45V+, 2

1

1

and plugging this into (2),

(

)

− − j + = −2e− j45 V + 2 1 (10)V = j j2e 451V Finally, inserting this value into (5): ( 1 13 ) V + = −1 2 e − j 4 5 1 V. + e − j 4 5 = j 2 V +

(

)

Inserting (11) into (1) leads us to the desired ratio: (12)V− 1=0.6V++j0.8(j2V+ 1)=−1V+1 1

V− or 1 =−1, and RL = 0 dB. V1+ P10.18: (JustAsk): The scattering matrix (assuming a50impedance system) for a 2port network is:  0.5 0.5ej45  .  S= j90 ej45 0.5 0.5 e  



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)V−=SV++SV+=0.5V++0.5ej45 1 111 122 1

V+2

(2)V−=SV++SV+=0.5ej45V++0.5ej90 V+2 2 211 222 1 Finding the return loss we will need the 2 relationship between V− and V+ 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)V+=V−e−j90 .

2 2 Fig. P10.18 Inserting (3) into (2) we can solve for V−2: V (4) 2− =

+ S21V 1 1−S 22e−j90

(4) is now inserted into (3) to find S21e−j90V+ + 1 (5)V2= 1−S e−j90.

22

Finally, inserting (5) into (1) we can find the ratio needed: − j90 S12S21e V − = S11+ − 1− S22ej90 1 j e45 0.5ej45e− j90 V = 0.5+ =1 j90 1− j90 0.5 e + 1 RL=−20log1=0dB ()

(0.5) ( ) ( )e−

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 S11=  L =2 + 1= −3 , 1 2 2 S21 = = 1+ L = .

3 So the scattering matrix is:  −1 2 2  1  S  =2 −1 2  .

3 2 −1 2

 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.  S S 0   31 32 t *  Sthen S =  U If it is also lossless, , which gives the following equations:

S212 + S312 = 1 S212 + S322 = 1 S312 + S322 = 1 and =0 S31S 32 * =0 S21S 32 *

=0 S21S 31 .

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 j j  0.50 0.050e60 0.75e 60     j60 S  =0.75e 0.050ej60  0.50  . 0.050ej60 0.75ej60  0.50   

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=−20logS21=−20log0.75=2.5dB

(d) I=−20logS31=−20log0.05=26dB 1 +  1+ S 1+0.5 11= =3 (e)VSWR= 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

IL=−20log

V2− , V1+

and VSWR=1 +  , 1 −  V− where = 1.

V+1

Fig. P10.22

For a 3 port scattering matrix we have 333, V−=SV++SV++SV+ 3 311 322 and since port 2 is terminated in a matched load we therefore have (1)V−=SV++SV+ 333.

3 311 − With port 3 terminated in a short circuit, V3 += − V 3 , and using (1) we find

(2V ) 3+ =

+ −SV 311 .

1+ S33

Now, for a 3 port scattering matrix matched at port 2 we have 133 (3)V−=SV++SV++SV+=SV++SV+ 1 111 122 133 111 Inserting (2) into (3), we find

(0.75j60 )(0.05ej )=0.513e−j2.2,

e V− 31 1 = S − S13S = ( 0 .

5 ) − V + 111 1+ S33 1+ 0.5 1 +  1 +0.51 = 3.1 so VSWR= 1 − = 1 − 3 0.51 3

60

10-20 Finally, for a 3 port scattering matrix matched at port 2 we also have + + + (4)V−2=S21V1+S 23V3=S21V+1+S23V+3, 22V2+S Inserting (2) into (4), we find

( .05e60)(0.5) 0 =0.733e+j60, =(0.75ej60)− 1+ 0.5

V− SS 2 = S21 − 3133 V + 1 1+ S33

and IL=−20log

j

V2− =−20log0.733=2.7dB V+1

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.

=0.933 S21 =10−0.620 =0.126 Imin =−20logS 31 =18dB, S31 =10−1820 , or S11= 1+ S1 VSWR −1 1.35−1 VSWRma= = =0.15 x 1 1− S11 VSWR +1 1.35+1 ILmax =−20logS 21 =0.6dB,

So, for the symmetrical network we have the following magnitudes for the S-matrix: 0.1 0.1 0.93    5 3 0.13  =    S 0.9 0.1 0.15  3 5 0.1 0.9 P10.24: Verify the scattering matrix (equation (10.27)) for the resistive power divider of 3 3 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 3 Zo 3  Zo   231 1 o  + S21 =  1 ( 1) 4 3= 2   4 Z 4 Z 1 Zo 4 Zo= 3 o 3 3  3      coefficient  is simply 1+ = 1. The first ratio in Here, the transmission  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

1  0 1 2 2  12 . S  =12 0  2  1 1 0 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 S212 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 S212 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  , so P=(P)(10−2010 )=1mW 3 1 3  P 1

10-22

I = 50dB= −10 log S41 (), so S41 = 10−50 10 = 0.00316 2 P4= S41P = 1W IL = 0.25dB= −10 log( S21, so )

S 12 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 = 10−320 = 0.707,

S21= S31= 0.707 and IL = −20log S21 = 3dB. The scattering matrix is:  0 .707 .707 0   0 .707 .707 0  S  =  0 0 .707 .707  .707 .707 0   0  P10.29: (JustAsk): Suppose to port 1 of an ideal ring hybrid coupler we apply the appropriate frequency voltage V1+ =1.0ej0 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    −1 1 0 0

V+3=( V+4= 0 ), then we have Assuming ports 3 and 4 are terminated in matched loads V = S+1 − = 2V2 1 = S+2 V = 1V1 V+2= −V−2so − S+3 Now, at port 2 terminated in a short circuit,

2

1V1

V

S+4

−

2V2

3 V

10-23

V−1= S 1 2 V +122 2 = =− S−S12S21V1 V− +

1ej0)= (

 V −  −j  j − 1=−  2 0.5e j 0 V  2   +  − j V3− = S31V1=  1ej0 = 0.707e− j90 V  2 + V4− = S 4 2 V 22− = −S −42S21V1 S 4 2+ V

()

1( e j0.50e j 0 V) = −

−  + j   − j  V4 = −     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  0 −1 1

With all ports terminated in matched loads, all the incident voltages (V+ k ) will be zero, except, of course, for the injected signals into ports 2 and 3. So we have + V − 1 12V2 = +S +S 13V3 + V−4= 42V2+S S+ 43V3. − j With 2 =0.707e−j90, we have V − =12V S + +S13V3+ 1 2

=

e e ( ) (0.707 ) 5− + e ) 3e− ( ) (0.707 j90

j30

j90

j30

= 3.535 e −j60 + 2.121e−j60 = 5.66e−j60 V and + + + S43V V − 4S=42V 2 3 − j90 e−j90 = − + 3ej30 e

)( ) ( 0.707) 5( j30( 0.707e )

= −1.414e− j60 = −1.414e j120V P10.31: Suppose to port 1 of a quadrature hybrid coupler we apply the appropriate frequency voltage V1+ = 1.0ej 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  j 1 0 0 0 0  1 −1  j  S =21  0 0 j.  1 j  0 0 V3+ = = 0) and port 2 terminated in a With ports 3 and 4 terminated inV4+matched loads (

then = − ,w Ve − 2have ) shortv(V+

V− 1

+ = −S12S21V1

+ V2− = S21V1 + V3− = 31V1 S + V4− S42 S 21V1 = − Evaluating,  j −   j−0 j  V1− = −    e = 0.5e j 0 V  2  2  +  −  V3− = S3 1 V 1 1= ej 0 = −0.707e j 0 V  2 −  − 1   − jj0  V4 = − e = 0.5e− j90 V   2  2 

( 1)

( 1)

( 1)

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 

Zo = 35.4 

w=159 mils  = 0.0893 m = 3520 mils  /4 = 879 mils

w=258 mils  = 3470 mils  /4 = 867 mils

The resulting design is shown in Figure P10.32.

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:

 j 1 0 0 0 0  1 −1  j  S =21 0  0 j.   1 j 0 0 We can expand the matrix to yield the following equations, where we’ve also indicated the equivalent termsA2=V+B1, V− B1, V−3=V+ B4, and V+3=V− B4: AV+2=V− A A

Fig. P10.33

10-26 For the A matrix: + + − − VA−1 = S12VA = S21VB1 + S13VB4 + S13VA 2 3

S VA−2 = S21VA +1 +

+

4 24VA + + VA−3 = S31VA + S34VA 4 1 − − VA−4 = + S43VB4 +3 = S42VB1 + 2 + S43VA S42VA For the B matrix:

VB−1 = S12V B+2 + S13V B+3

− − VB−2 = S21V B+1 + S24V B+4 = S21VA + S24VA 2 3 − − VB−3 = S31V B+1 + S34V B+4 = S31VA + S34VA 2 3

VB−4 =

+SV+ S42V B+2 43 B3 − − − − Note that for the overall matrix, we are only interestedVin A1, VA4 , VB2, andVB3. So we insert reflected values into those equations. For example, + + VA1− =S21V − +S13V − = 21 S (S21V + +S13V + )13 + (S42V SB2+ S43VB3 ), B1 B4 B2 B3 which expanding and rearranging becomes VA1=(S21S21+S13S42)V+B2+(S21S13+S13S43)V+B3 − Inserting the scattering parameter values, we find    −j   − 1   − 1   − j   − j j  −1 −1 B+ 2 +     +     VB+3 = jV+  +    V − B3 VA−1 =      2 − 2   2   2    2   2   2   2    find Likewise, we   jV+   = V A−4 B2 VB−2 = jV+ A4

VB−3 = jV+ A1.

The−overall scattering matrix is then: 0 − j 0  0  0 + j 0 0   Soverall= 0 0 0    −  j 0 + j 0 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 vs. With the filter in place, 2 Ro j L jLR vs,where RojL= vLf= o.

Ro + jL Ro j L + Ro

With no filter present, we haveL= v

10-27 Expanding and rearranging: 1 vLf= . jR 2− o L 1 jR  jR v Now we have L =  2− o  = 1 − o , vLf 2   L  2L

vL = vLf

2   Ro 2   Ro  . =10log 1+  1 +   , so IL   2L     2L  

Now, for a 1 GHz cutoff frequency, Ro Ro 50 =1, or L= = =3.98nH. 9 4 c 4  2 ( 2fc  ) L 110 ( ) x f 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 haveLv = v s .With the filter in place, 2 Ro vs. vLf= j  2 R  o −   C   j  2 −  j vL 1   C  =2 =1− , 2 RoC 1 vLf 2   vL 1   , and IL=10log1 =  +  1  oC vLf  4fR 

2  1   .  +  fRC 4  o   

For a 1 GHz cutoff frequency, 1 1 1 =1, so C= = =1.6pF. 4fcRo 4(1x109 )(50) 4f cRoC

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 ( ) =57.3H ( ) 50

L5 ' = L1 ' g 1.3712 = 27.4mF C2 '= 2 = 50 Ro C4 '= C2 '

L3 ' = g3Ro= (1.975 ( = 98.75H ) 50)

(a)

Frequency Transformation: L1 ' 57.3 = = 4.6nH L1 ''= c 2 ( 2 x109 )

Fig. P10.36 a&b

(b)

(c)

Fig. P10.36 c,d & e

L3''=

L3 ' = c



L5 ''= L1 '' C ' 27.4mF = 2.18pF, C 4''= C2 '' C2 '' = 2 =  c 2(2 x 109 )

(d)

(e)

98.75 =7.86nH. 2 (210 x 9)

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

 C1



R



(R + jL1), vLf = v1  R + jL 

 1   −j Z Z2 = (Z1 + j L2 ,) v1 = v 2  1 1+ j  L 2  C1 Z    Z2 v2 =  vs  Z2 + R +jL 1  So finally we have,    vLf 1  Z1 Z2 R = 2, =    , s v  R + j L 1  Z 1 j +  LZ22 R+ + j L 1   and with  v  vL 1 v vLf vL = Lf = 2 , we divide the two ratios L L. to arrive at IL=20log vs vs v vs vLf   These equations are used in MLP1036 to generate the insertion loss plot of Figure P10.36(f). % MLP1036 % % % clear;clc;clf Design N=5 LPF, 0.1 dB ripple Compare with N=3 LPF of Example 10.14.

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; ZC31=-i./(w*C31);

Fig. P10.36(f) 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: g1 gR =C3'', L2''= 2c .

C1'= '  cR 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,   RXC vLf = Z1v1 , where Z 1 =  R XC+XL ,     Z  andv1 =  2  vs, where Z2 = Z1 XC.  Z2 + R  Then, with vL = vs/2, we arrive at vL R + Z2 = . vLf 2Z1Z2

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=g 3=2.0236, and g2=0.9941, as shown in Figure P10.38(a).

Impedance Transformation:

L1'=g1Ro=101.2, C2'= g

2=0.0199.

Ro

Frequency Transformation: (using Table 10.4) 1 C1 '' =  cL1'

=

1

= 0.655pF 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 and also we let inductor L2’’, Z1=XL(R+XC), where XL = jwL and XC −j = .

C

Then, Fig. P10.38a&b

Z1 R v1, v1 = vs , so R + XC Z1 +R + XC Now, with vL=vs/2, we have v ( R + X C1 ) R( Z +) + X C L = .

Lf v 2RZ1 vLf =

vLf vs

=

Z1 R .

R + XCZ 1 +R+XC

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; 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;

(b)

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 C'= 1=0.06963, L'=gR=38.09, and C'= 3 3=0.09076.

1 2 2 R R The frequency transformation is performed using Table 10.4: 1 1 1 =2.29nH, C2''= =4.18pF, and L3''= cC3'=1.75nH .

L1''=  cC1' cL2' 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 , and XL3=jL3'', XL1=jL1'', XC2=  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 ofin Z1parallel withXL3, andZ3 to be: Z3 = (Z2+ XC2) XL1.

Now we can find the voltage ratio relations:  RXL1  Z2   Z  vLF  v1 , v1 =  v2 , and v 2 =  3 vs. =  Z1   Z3 + R   Z2 +XL2   Combining, and realizing vL = vs/2, we arrive at: ( 3 + R) vL Z1 (Z+XC 2 2) Z = . vLf 2Z2Z 3R(XL1) The insertion loss is then:   IL=20logvL .  vLf    This is plotted by ML1039 in Figure P10.39(c). % % MLP1039 % % fc=1; %corner frequency in GHz wc=2*pi*fc*1e9; g1=3.4817; g2=.7618; g3=4.5381; Design N=5 high pass filter, 3 dB ripple.

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).

transformation, we calculate BW and wo: BW= 2200 ( MHz)

 o = 2 900 ( MHz) ( MHz) 1100

= 2995 MHz ( )

We then have:

(a) Fig. P10.40a&b

(b)

To perform the

10-37 L1 '' =

C1 '' =

L1 ' 101.2 = = 80.5nH BW 2 ( 200 )x10 6

2 (200MHz) BW = = 0.317pF 2  o2L1' ( 2) 995 ( MHz)2 (101.2 )

2 (200MHz) BW  = =1.617nH 2  o2C2 ' 2 )2 995 MHz ) (0.01988) ( ( C 2' 0.01988 = =15.8pF C2''= BW L2''=

2(200x106)

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'' C2''  C1'' We also define Z3 as the impedance seen by the voltage v1: Z3=Z2(R+Z1).

Now we can find the voltage ratio relations:  Z3    R   v s . v1, and v= vLF=  R + Z 1 1  Z 3 + R + Z 1  Combining, and realizing vL = vs/2, we arrive at: v (R + Z1)(R + Z1 + L Lf = 2RZ3 . v

Z3)

The insertion loss is then:   IL=20logvL .  vLf    This is plotted by ML1040 in Figure P10.40(c). % % MLP1040 % clc Design N=3 band pass filter, 1 dB ripple. 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 C' 2= 2 =0.01423 R The frequency transformation is performed using Table 10.4, where L1' is transformed into a series circuit containing L1 ''and C 1 ''. 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200 ( MHz)

 o = 2 900 ( MHz) ( MHz) 1100

= 2995 MHz ( )

We then have: L 1' 167.4 L1'' = BW = = 133.2nH 2 ( 200 )x106 BW 2 (200MHz) = = 0.1921pF C1 '' = 2  o2L1' ( 2) 995 ( MHz)2 (167.4 )

L2''=

2 (200MHz) BW = =2.259nH 2  o2C2 ' (2 ) 995 ( MHz)2 (0.01423)

C 2'' =

C 2' 0.01423 = =11.32pF BW 2(200x106 ) 

To plot the insertion loss using ML1041, we use the impedances shown in Figure P10.41(b), given as − j j , Z2=jL2'' Z1=jL1''−  C1'' C2'' We also define Z3 as the impedance seen by the voltage v1:

10-40

Z3=Z2(R+Z1).

Now we can find the voltage ratio relations:  Z3    R  s. v vLF= v,1 and v 1 =   R + Z 1   Z 3 + R + Z 1 Combining, and realizing vL = vs/2, we arrive at: v Z1 + Z3) ( R +1 Z ) (+ R L Lf = . v 2RZ3 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 BW (1.7−0.57) GHz =5.6 60dB SF= = 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: g2 =0.01988 L1'=L3'=g1R=101.2, and C'= 2 R The frequency transformation is performed using Table 10.4, where L1 ' is transformed into a series circuit containing1'' L and C1 ''. C2' is transformed into a parallel circuit To perform the containing L2 '' and C2 ''. This is shown in Figure P10.42(a).

transformation, we calculate BW and wo: BW= 2400 ( MHz) o

MHz) 2600  = 2 2200 ( MHz) (

(a) Fig. P10.42a&b

= 22392  (MHz)

(b)

10-42 We then have: L' 101.2 = 40.2nH L1 '' = 1 = BW 2 ( 400 )x10 6 BW 2400 ( MHz) = = 0.110pF C1 '' = 2  o2L1' ( 2) 2392 ( MHz)2 (101.2 )

L2''=

2 (400MHz) BW  = =0.560nH 2  o2C2 ' (2 ) 2392 ( MHz)2 (0.01988)

C 2' = C2''= BW

0.01988

=7.93pF

2(400x106)

To plot the insertion loss using ML1043, we use the impedances shown in Figure P10.42(b), given as − j j C2'' , Z2=jL2'' Z1=jL1''−  C1'' We also define Z3 as the impedance seen by the voltage v1: Z3=Z2(R+Z1).

Now we can find the voltage ratio relations:  Z3    R   v s . v1, and v= vLF= 1  R + Z 1   Z 3 + R + Z 1  Combining, and realizing vL = vs/2, we arrive at: v (R + Z1)(R + Z1 + L Lf = 2RZ3 . v The insertion loss is then:   IL=20logvL .  vLf    This is plotted by ML1042 in Figure P10.42(c). % % MLP1042% % clc Design N=3 BPF, 1 dB ripple. clear

R=50; flo=2.2e9;fhi=2.6e9; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi);

Z3)

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 C' 2= R2 =0.01423 The frequency transformation is performed using Table 10.4, where L1' is transformed into a parallel circuit containing L1 ''and C 1 ''. C2 'is transformed into a series circuit containing L2'' and C2''.

This is shown in Figure P10.43(a).

transformation, we calculate BW and wo: BW= 2200 ( MHz)

 o = 2 900 ( MHz) ( MHz) 1100

= 2995 MHz ( )

We then have: 1 1 = = 4.75pF C1''= BWL 1 ' 2 ( 200 )x10 (167.4) 6

L1 '' =

L2 '' = C2 '' =

( ) ( MHz) 167.4 BWL  1 ' 2200 = = 5.39nH 2 2 2 (2  ) (MHz) 995

o

1

1 = = 55.9nH BWC 2 ' 2 (200MHz ) ( 0.01423) BWC'  2

o2

=

2 ( 200 x106 ) ( 0.01423) = 0.458pF 2 2

(2 ) ( 9 9 5 M H z )

To perform the

10-45 To plot the insertion loss using ML1043, we use the impedances shown in Figure P10.43(b), given as − j j Z2=jL2''−  C , Z1=jL1'' C1'' 2'' We also define Z3 as the impedance seen by the voltage v1: Z3=Z2(R+Z1).

Now we can find the voltage ratio relations:  Z3    R  v s . vLF=  v1, and v=1   R + Z 1   Z 3 + R + Z 1  Combining, and realizing vL = vs/2, we arrive at: v ( R + Z 1) ( 3R. + Z + Z ) 1 L = Lf v 2RZ3 The insertion loss is then: v  IL=20log L .  vLf    This is plotted by ML1043 in Figure P10.43(c). % %MLP1043 % % clc;clear Design N=3 band-stop filter, 3 dB ripple.

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));

Fig. P10.43c

L2=1/(BW*C); 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

g2=g4=1.0911 3 = 3.0009 g We first perform transformation: g C1 ' = 1 = 0.0427, Ro L2 ' =g2Ro = 54.56,

the

impedance

(a)

= g3 C3 ' R = 0.0602 o

Figure P10.44(b) shows the filter portion of the circuit after transforming based on Table 10.4. To perform the frequency Fig. P10.44a&b transformation, we first calculate:

BW= 2  2.5 (GHz− 2.3GHz) = 2 200  (MHz)

 o = 2 Then, L1''=

= 2 ( 2.398 GHz) GHz) 2.5 ( GHz) (2.3

1 1 = = 18.6nH BWC 1 ' 2 ( 200 x106 ( 0.0427 ) )

(b)

10-47

C1 '' =

C2 '' = L2 '' = L3 '' = C3 '' =

BW C1'

 1 2 o

=

) ( 0.0427) 2(200MHz = 0.236pF 2 2 (2) (2.398GHz)

1 = 2(200MHz)(54.56) =14.6nH BWL 2'

BW  L2 '



2 o

=

2(200x106)(54.56) =0.302nH (2)(2.398GHz) 2

2

1 1 = = 13.3nH BWC 3 ' 2(200MHz)(0.06002)

BWC  3 ' 2( 200 x106 ) ( 0.06002 ) = =0.332pF 2 2

 o2

(2)(2.398GHz)

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 ' −j Z2 = jL2 '  C2 ' j Z3 = jL3'−  C3 ' Fig. P10.44(c) Some of these impedances are also lumped together: + Z R and Z Za = Z3 ( Z Za ). b=Z1(Z2+ ) 2 1 The voltage ratios are then:  Z1 R      Z1 2=  Za  v ,2 v 2 =  Zb vs , vLf =  1 R +1, v v Z   Za + Z2  R + Zb    so vLf  Z1 R  Za   Zb  .  1 R + Z2 Z a Z2 +R + Zb  vs = Z     Now, since vL=vs/2, then after manipulation we have (Z1 R + Z2 Z a)+ Z(2 R ) Z(b +) vL =  vLf 2ZZ a b (Z1 R) IL v and this is used to find = 20 log L using MLP1044: vLf

10-48 % % MLP1044 % clc;clear R=50; flo=2.3e9;fhi=2.5e9; Design N=5 Band-Stop filter, 1 dB ripple.

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= 0.68ej125 S12=0

S21 =3.6ej40 S22 =0.86e−j74 . (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 2 1− L 2 2 INS2 S21 = S21 =12.96 2 1− 1− S22L

So GT (dB) = 10 log(12.96) = 11.1 dB. (b) Now assuming optimized matching, the transducer gain term reduces to:

Fig. P10.45

10-50 GT=

1 1 2 = 92.6 S21 2 2 1−S 21 1− S22

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 achieve L=S*=0.86ej74 . We move from the open-ended stub in the admittance chart 22 (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=S*=0.68e−j125 . We move from the open-ended stub in the admittance chart (point 11 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 enter the desired impedance: 50 design: enter the substrate thickness: 25 Microstrip Design width & thickness will be in the same units 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)

Fig. P10.46

 G=

c f  eff

=

(d)

3x108 m/s  1mil  4400mils  −6  = 9 m 2x10 1.8015 25.4 x10

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.

We first design the lumped element matching network to achieve  L=S*22=0.86ej74.

Fig P10.47(a)

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: − jZo j = − 2.0 , or L 50 L= = 2nH x 109 )  )(2 2(2 For the series inductance, j L j0.9= , or Zo

L=

(0.9)(50) =3.6nH.

2(2x109)

Now we design the lumped element matching network to achieve  S = S11= * 0.68ej− 125.

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CZo, or

1.8 =2.9pF 2 2 (109x50) ( )

C = Fig. P10.47(b)

For the series capacitance,

10-53

Fig. P10.47(c)

= − 0.07 j C=

− j

CZo

, or

1

= 23pF.

( 0.07 )(2 (50) ) 2 x109

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: S12=0 S11=0.72ej76

S21 =4.4ej125 S22 =0.58e−j30 . (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: G= T

1 − S 2

2 1−  L 2 = S21 2 =19.36 S 21 2 2 1 −  I N  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 GT = 1 2 S21 2 =60.58 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= 0.58j30 e . 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 achieve P10.49(b) we find the shorted shunt stub matching network to S=S*11=0.72e−j76. 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 * 30  L=S22=0.58ej .

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: )j1.6= − j L , or L= 50(1.6 =1.27nH Zo (2 )(10x 109 ) For the shunt capacitance,

j 0.19= jCZo, or

(a)

Fig. P10.50

C=

(0.19)

= .060pF 2 (10x109 )(50)

(b)

10-56 Now we design the lumped element matching network to achieve  S = S11=0.72e−j76. * 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:

j1.25=jCZo, or C=

(1.25)

=0.40pF (2)(10x109)(50)

For the series capacitance, − j , or C= −j0.72= CZo

1

=.44pF 2(10x109 )(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 PRF 1mW = =0.25mW PIF= 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