Final Practice Problems I

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1. Distributed Parameters Model P6.1: RG223/U coax has an inner conductor radius a = 0.47 mm and inner radius of the outer conductor b = 1.435 mm. The conductor is copper, and polyethylene is the dielectric. Calculate the distributed parameters at 800 MHz.
7
16
for copper: 5.8 10
for polyethylene: 2.26, 10
Cu
r
Sxm
Sm
=
= =
( )( )( )
6 7
3 3 7
1 1 1'2
800 10 4 101 1 1 3.322 0.47 10 1.435 10 5.8 10
c
fRa b
x xx x mx
= + = + =
74 10 1.435' ln ln 2232 2 0.47
b x nHLa m
= = =
( )( )
( )16
182 102' 560 10
ln ln 1.435 0.47 0SG x
b a m = = =
( )( )( )
( )122 2.26 8.854 102' 112
ln ln 1.435 0.47x pFC
b a m = = =
P6.2: MATLAB: Modify MATLAB 6.1 to account for a magnetic conductive material. Apply this program to problem P6.1 if the copper conductor is replaced with nickel.
7for Nickel we have 1.5 10 and 600.Ni rSxm
= = Note that this program has also been modified for P6.04 as well. %Coax distributed parameters % % Modified: P0602 % add rel permeability % also modified for P0604 % clear clc disp('Calc Coax Distributed Parameters') %Some constant values muo=pi*4e7; eo=1e9/(36*pi); %Prompt for input values a=input('inner radius, in mm, = ');

b=input('outer radius, in mm, = '); er=input('relative permittivity, er= '); sigd=input('dielectric conductivity, in S/m, = '); sigc=input('conductor conductivity, in S/m, = '); ur=input('conductor rel. permeability, = '); f=input('input frequency, in Hz, = '); %Perform calulations G=2*pi*sigd/log(b/a); C=2*pi*er*eo/log(b/a); L=muo*log(b/a)/(2*pi); Rs=sqrt(pi*f*ur*muo/sigc); R=(1000*((1/a)+(1/b))*Rs)/(2*pi); omega=2*pi*f; RL=R+i*omega*L; GC=G+i*omega*C; Gamma=sqrt(RL*GC); Zo=sqrt(RL/GC); alpha=real(Gamma); beta=imag(Gamma); loss=exp(2*alpha*1); lossdb=10*log10(loss); %Display results disp(['G/h = ' num2str(G) ' S/m']) disp(['C/h = ' num2str(C) ' F/m']) disp(['L/h = ' num2str(L) ' H/m']) disp(['R/h = ' num2str(R) ' ohm/m']) disp(['Gamma= ' num2str(Gamma) ' /m']) disp(['alpha= ' num2str(alpha) 'Np/m']) disp(['beta= ' num2str(beta) 'rad/m']) disp(['Zo = ' num2str(Zo) ' ohms']) disp(['loss=' num2str(loss) ' /m']) disp(['lossdb=' num2str(lossdb) ' dB/m']) Now run the program for Nickel: Calc Coax Distributed Parameters inner radius, in mm, = 0.47 outer radius, in mm, = 1.435 relative permittivity, er= 2.26 dielectric conductivity, in S/m, = 1e16 conductor conductivity, in S/m, = 1.5e7 conductor rel. permeability, = 600 input frequency, in Hz, = 800e6 G/h = 5.6291e016 S/m C/h = 1.1249e010 F/m

L/h = 2.2324e007 H/m R/h = 159.7792 ohm/m Gamma= 1.78881+25.252i /m alpha= 1.7888Np/m beta= 25.252rad/m Zo = 44.66083.1637i ohms loss=0.027942 /m lossdb=15.5374 dB/m >> Summarizing the distributed parameter data from this routine we have:
18' 160 , ' 223 , ' 560 10 , ' 112 pFnH SR L G x Cm m m m= = = =
P6.3: Modify (6.3) to include internal inductance of the conductors. To simplify the calculation, assume current is evenly distributed across the conductors. Find the new value of L for the coax of Drill 6.1. From Amperes Circuit Law we can find H versus :
2 for 2IH a
a =
for a2
IH b = 2 2
2 2
c for b2
IH cc b
=
0 for H c = Using the energy approach, 2 21
2 2o
mW LI H dv= = , we find
22 2 2 2
2 2 2 2 2 2
1' ln ln2 8 2 4
o o ob c c c c bLa c b b c b c b
+ = + + +
Inserting the given values we find
( ) nH nH' 237 50 41.2 328m m
L = + + = With two significant digits we therefore have L = 330 nH/m. 2. Time Harmonic Waves on Transmission Line P6.4: MATLAB: Modify MATLAB 6.1 to also calculate , , and Zo. Confirm the program using Drill 6.2. See the solution for P6.2. Calc Coax Distributed Parameters inner radius, in mm, = 0.45

outer radius, in mm, = 1.47 relative permittivity, er= 2.26 dielectric conductivity, in S/m, = 1e16 conductor conductivity, in S/m, = 5.8e7 conductor rel. permeability, = 1 input frequency, in Hz, = 1e9 G/h = 5.3078e016 S/m C/h = 1.0606e010 F/m L/h = 2.3675e007 H/m R/h = 3.8112 ohm/m Gamma= 0.0403332+31.4857i /m alpha= 0.040333Np/m beta= 31.4857rad/m Zo = 47.2460.0605221i ohms loss=0.9225 /m lossdb=0.35033 dB/m >> This agrees with the results of Drill 6.2. P6.5: The impedance and propagation constant at 100 MHz for a TLine are determined to be Zo = 18.6 j0.253 and = 0.0638 + j4.68 /m. Calculate the distributed parameters.
( )( )' ' , ' ' ' '' 'o
R j LZ R j L G j CG j C
+= = + ++
' ' 2.37 87.0
' 2.37 , ' 87.0 ' 139
oZ R j L jnHR L so L
m m
= + = + = = =
6' ' 7.63 10 0.252,
' 7.63 , ' 0.252 ' 401
o
G j C x jZ
S pFG and C so Cm m
= + = +
= = =
P6.6: The specifications for RG214 coaxial cable are as follows: 2.21 mm diameter copper inner conductor 7.24 mm inner diameter of outer conductor 9.14 mm outer diameter of outer conductor Teflon dielectric (r = 2.10)
Calculate the characteristic impedance and the propagation velocity for this cable.
60 60 3.62ln ln 49.11.1052.1o r
bZa
= = =

82.07 10pr
c mu xs= =
P6.7: For the RG214 coax of problem P6.6 operating at 1 GHz, how long is this Tline in terms of wavelengths if its physical length is 50 cm?
( )
8
9
2.07 10, 0.2071 10
1( ) 50 2.40.207 100
pp
u xu f mf x
mcmm cm
= = = = = = l
P6.8: If 1 watt of power is inserted into a coaxial cable, and 1 microwatt of power is measured 100 m down the line, what is the lines attenuation in dB/m?
110log 601
60' 0.6100
WA dBW
dB dBAm m
= = + = =
P6.9: Starting with a 1 mm diameter solid copper wire, you are to design a 75 coaxial TLine using mica as the dielectric. Determine (a) the inner diameter of the outer copper conductor, (b) the propagation velocity on the line and (c) the approximate attenuation, in dB/m, at 1 MHz.
( )( ) ( ) ( )( )o60 ln , b=a exp Z 60 0.5 exp 75 5.4 60 9.1o rr
bZ mm mma
= = =
So the inner diameter of the outer conductor is 18 mm. 8
8 82.998 10 1.29 10 , so 1.3 105.4p pr
c x m mu x u xs s= = = =
To calculate , will need . Therefore we calculate R, L, G and C. ( )( )6 7
3 3 7
1 10 4 101 1 1' 87.62 0.5 10 9.1 10 5.8 10
x x mRx x x m
= + =
74 10 9.1' ln 5802 0.5x nHL
m
= = ( )
( )15
152 10
' 2.17 10ln 9.1 0.5
SG xm
= = ( )( )
( )122 5.4 8.854 10
' 103.59.1ln 0.5
x pFCm
= =
Now, with = 2f,

( )( ) 6 1' ' ' ' 585 10 0.049R j L G j C x jm
= + + = +
Finally, 6 38.686585 10 5.1 10Np dB dBx xm Np m
= = This is confirmed using MLP0602. P6.10: MATLAB: A coaxial cable has a solid copper inner conductor of radius a = 1mm and a copper outer conductor of inner radius b. The outer conductor is much thicker than a skin depth. The dielectric has r = 2.26 and eff = 0.0002 at 1 GHz. Letting the ratio b/a vary from 1.5 to 10, generate a plot of the attenuation (in dB/m) versus the line impedance. Use the lossless assumption to calculate impedance. % MLP0610
%
% Plot of alpha vs Zo for a particular coax
clear
clc
%Some constant values
muo=pi*4e7;
eo=8.854e12;
a=1;
er=2.26;
sigd=0.0002;
sigc=5.8e7;
f=1e9;
%Perform calulations
b=1.5:.1:10;
G=2*pi*sigd./log(b./a);
C=2*pi*er*eo./log(b./a);
L=muo*log(b./a)/(2*pi);
Rs=sqrt(pi*f*muo/sigc);

R=(1000*((1./a)+(1./b))*Rs)/(2*pi);
w=2*pi*f;
RL=R+i*w*L;
GC=G+i*w*C;
Gamma=sqrt(RL.*GC);
Zo=abs(sqrt(RL./GC));
alpha=real(Gamma);
loss=exp(2*alpha*1);
lossdb=10*log10(loss);
plot(Zo,lossdb)
xlabel('Characteristic Impedance (ohms)')
ylabel('attenuation (dB/m)')
grid on
3. Terminated TLines P6.11: Start with equation (6.54) and derive (6.55).
Fi P6 10

o o
in oo o
V e V eZ ZV e V e
+ + + +
+= l l
l l
With ,o L oV V += we then have
( )( )Lin oLe e
Z Ze e
+
+ + = l l
l l
We also know that
,L oLL o
Z ZZ Z
= + So now we have
( ) ( )( ) ( )
L o
L o L o L oin o o
L o L oL o
L o
Z Ze eZ Z Z Z e Z Z e
Z Z ZZ Z e Z Z eZ Ze e
Z Z
+ +
+ +
+ + + + = = + +
l ll l
l ll l
and with rearranging, ( ) ( )( ) ( ) .L oin oL o
Z e e Z e eZ Z
Z e e Z e e
+ +
+ + + + = + +l l l l
l l l l
We can convert the exponential terms into hyperbolic functions, given
( ) ( )1 1 sinh(x)sinh( ) , cosh( ) , and tanh(x)= .2 2 cosh(x)x x x xx e e x e e = = + This leads to
( ) ( )( ) ( )
2 cosh 2 sinh,
2 sinh 2 coshL o
in oL o
Z ZZ Z
Z Z
+= +l ll l
or finally ( )( )
tanh.
tanhL o
in oo L
Z ZZ Z
Z Z
+= +ll
P6.12: Derive (6.56) from (6.55) for a lossless line.
( )( )
tanh,
tanhL o
in oo L
Z ZZ Z
Z Z
+= +ll
and ( ) ( ) ( )tanh tanh tanhj j = + =l l l l since = 0 for lossless line. Using the hyperbolic definitions, we have
( ) ( )( )( )( )
sinhtanh .
cosh
j j
j j
e ejj
j e e
+
+ = = +l l
l l
ll
l
Now using Eulers formula,
( ) ( ) ( )( ) ( )( )( )
cos sin( )  cos sin( ) 2sintanh tan( )
cos sin( ) cos sin( ) 2cosj j j
j jj j
+ = = =+ + +l l l l l
l ll l l l l
Plugging this in, we find,

Fig. P6.14
( )( )
tan.
tanL o
in oo L
Z jZZ Z
Z jZ
+= +ll
P6.13: A 2.4 GHz signal is launched on a 1.5 m length of TLine terminated in a matched load. It takes 6.25 ns to reach the load and suffers 1.2 dB of loss. Find the propagation constant.
j = + 1.2 1 0.0921.5 8.686
dB Np Npm dB m
= = 81.5: 2.4 10
6.25pm mu x
t ns s = = = =l
( )98
2 2.4 1062.8
2.4 10p
x radu x m
= = = So
10.092 62.8 jm
= + P6.14: A source with 50 source impedance drives a 50 TLine that is 1/8 of a wavelength long, terminated in a load ZL = 50 j25 . Calculate L, VSWR, and the input impedance seen by the source.
7650 25 50 0.24250 25 50
jL oL
L o
Z Z j eZ Z j
= = =+ +D
11.64
1L
L
VSWR+ = =
2 , tan 18 4 4
= = = l
( )( )
tantan
50 25 505050 50 25
30.8 3.8
L oin o
o L
Z jZZ Z
Z jZj jj
j
+= + += + +
=
ll
P6.15: A 1 m long TLine has the following distributed parameters: R = 0.10 /m, L = 1.0 H/m, G = 10.0 S/m, and C = 1.0 nF/m. If the line is terminated in a 25 resistor in series with a 1 nH inductor, calculate, at 200 MHz, L and Zin. ( )( )6 925 2 200 10 10 25 1.257 LZ j x j = + = + Now, MLP0615 is used to solve the problem. % MLP0615

%
% calculate gamma and char impedance
% given the distributed parameters
% Then, calculate gammaL and Zin
%
% define variables
clc
clear
R=0.1;
L=1.0e6;
G=10e6;
C=1.0e9;
f=200e6;
w=2*pi*f;
length=1;
ZL=25+j*1.257;
% Perform calcuations
A=R+i*w*L;
B=G+i*w*C;
gamma=sqrt(A*B) %Propagation Constant
Zo=sqrt(A/B)
gammaL=(ZLZo)/(ZL+Zo) %Reflection coefficient
TGL=tanh(gamma*length);
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL))
Running the program, Gamma = 0.0017 +39.7384i

Zo = 31.6228  0.0011i gammaL = 0.1164 + 0.0248i Zin = 34.0192  7.4618i >> So the answers are, with the appropriate significant digits,
1680.12 and 34 7.5 jL ine Z j = = D P6.16: The reflection coefficient at the load for a 50 line is measured as L = 0.516ej8.2 at f = 1 GHz. Find the equivalent circuit for ZL.
Rearranging ,L oLL o
Z ZZ Z
= + we find 1 150 30 1
LL o
L
Z Z j+ = = + . This is a resistor in series with an inductor. The inductor is found by considering
( )93030, or 4.8
2 1 10j L j L nH
x = = = ,
So the load is a 150 resistor in series with a 4.8 nH inductor. P6.17: The input impedance for a 30 cm length of lossless 100 impedance Tline operating at 2 GHz is Zin = 92.3 j67.5 . The propagation velocity is 0.7c. Determine the load impedance.
Rearranging ( )( )tan
,tan
L oin o
o L
Z jZZ Z
Z jZ
+= +ll
we find ( )( )tantan
in oL o
o in
Z jZZ Z
Z jZ
= ll
( )( ) ( ) ( )
9
8
2 2 1059.84 ; tan tan 59.84 0.3 1.254
0.7 0.7 3 10
x rad rad mc m mx
= = = = = l Evaluating, we have ( )950 0.016 50 2 2 10 ,LZ j j x L= + = + or L = 1.3 pH. This is a very small inductance, so we have 50 .LZ P6.18: For the lossless TLine circuit shown in Figure 6.51, determine the input impedance Zin and the instantaneous voltage at the load end vL.
25 50 1 2, , tan 025 50 3 2L
= = = = =+ l
0 250
Lin o L
L
ZZ Z ZZ
+= = = + 25 8 2
25 75j z j z
in o oV V V V e V e + += = = ++ ( )2 j jo LV e e + = + l l

cos sin 1, 1,j je j e = + = = ( )1 21 1 2; 3
3 3o o oV V V V+ + + = = =
( ) 11 3 1 23L o L
V V V+ = + = = , so ( )2cos 180Lv t V= + D P6.19: Referring to Figure 6.10, a lossless 75 TLine has up = 0.8c and is 30 cm long. The supply voltage is vs = 6.0 cos(t) V with Zs = 75 . If ZL = 100 + j125 at 600 MHz, find (a) Zin, (b) the voltage at the load end of the TLine, and (c) the voltage at the sending end of the TLine.
, 15.7 , 4.71, tan 418.6pp
raduu m
= = = = =l l ( )
( )( )100 125 75 418.6
7575 100 125 418.6
22 28
in
j jZ
j jj
+ += + +=
Referring to Fig P6.19, 366 2.1
75jin
inin
ZV e VZ
= =+D
( )2.1cos 36inv t V = D 430.593 jL oL
L o
Z Z eZ Z
= =+D
( ) 126 360.70 2.1j j j jin o L oV V e e e V e V + + + = + = =D Dl l 36
90126
2.1 30.70
jj
o j
eV e Ve
+
= =D
DD
( )( )
105.81 4.47
4.5cos 106
jL o L
L
V V e V
v t V+= + =
= +
D
D
P6.20: Suppose the TLine for Figure 6.10 is characterized by the following distributed parameters at 100 MHz: R = 5.0 /m, L = 0.010 H/m, G = 0.010 S/m, and C = 0.020 nF/m. If ZL = 50 j25 ,vs = 10cos(t)V, Zs = 50, and the line length is 1.0 m, find the voltage at each end of the Tline. The following MATLAB routine was used to find the required parameters. % MLP0620
%
% calculate gamma and char impedance
% given the distributed parameters
Fig P6 19

% Then, calculate gammaL and Zin
%
% define variables
clc
clear
R=5;
L=.010e6;
G=.01;
C=.020e9;
f=100e6;
w=2*pi*f;
length=1;
ZL=50j*25;
% Perform calcuations
A=R+i*w*L;
B=G+i*w*C;
gamma=sqrt(A*B)
Zo=sqrt(A/B)
gammaL=(ZLZo)/(ZL+Zo)
TGL=tanh(gamma*length);
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL))
Runningtheprogram,
gamma = 0.2236 + 0.2810i
Zo = 22.3607
gammaL = 0.4479  0.1908i
Zin = 27.2079 15.4134i

>>
( )18.23.97 , 4.0cos 18.2jinin SS inin S
ZV V e V v t VZ Z
= = = +D D
( ) ( ) ( )3.841.504 0.101 1.507 jin o L o oV V e e V j V e + + += + = + = Dl l so
18.222
3.84
3.97 2.631.507
jj
o j
eV ee
+ = =
DD
D
( ) ( )29.61 3.85 , 3.9cos 30jL o L LV V e v t V+ = + = = D D 4. The Smith Chart P6.21: Locate on a Smith Chart the following load impedances terminating a 50 TLine. (a) ZL = 200 , (b) ZL = j25 , (c) ZL = 50 + j50 , and (d) ZL = 25 j200 .
P6.22: Repeat problem P6.14 using the Smith Chart.
Fig P6 21

Fi P6 22
Fig.P6.22b
Fig.P6.23
First we locate the normalized load, zL = 1 j0.5 (point a). By inspection of the Smith Chart, we see that this point corresponds to 760.245 .jL e
= D Also, after drawing the constant circle we can see VSWR = 1.66. Finally, we move from point a, at 0.356 on the WTG scale, clockwise (towards the generator) a distance 0.125 to point b, at 0.481 . At this point we see zin = 0.62 j0.07. Denormalizing we find: Zin = 31 j3.5 .
P6.23: A 0.690 long lossless Zo = 75 TLine is terminated in a load ZL = 15 + j67 . Use the Smith Chart to find (a) L, (b) VSWR, (c) Zin and (d) the distance between the input end of the line and the first voltage maximum from the input end. After normalizing ZL and locating it on the chart (point a), we see 950.80 .jL e = D After drawing the constant circle, we see that VSWR = 9 (point c). We locate the input impedance by moving from the load (point a at WTG = 0.118) clockwise towards the generator to the input point (point b at WTG = 0.118 + 0.690 0.500 = 0.308 ). At this point, zin = 0.8 j2.4, so Zin = 60 j180 . Finally, the distance from the input end of the line (point b) to the first voltage maximum (point c) is simply 0.308 0.250 = 0.058 . Or, using the WTL scale, it is 0.250 0.192 = 0.058 .

P6.24: A 0.269 long lossless Zo = 100 TLine is terminated in a load ZL = 60 + j40 . Use the Smith Chart to find (a) L, (b) VSWR, (c) Zin and (d) the distance from the load to the first voltage maximum. (a) zL = 0.6 + j0.4 located at WTG=0.082. We read off the Smith Chart that this point corresponds to: 1210.34 .jL e = D After drawing the constant circle we notice the VSWR = 2.05 (point c). Moving from this point a distance 0.269 (clockwise, towards generator), we find the input point (point b at WTG = 0.351 ). At this point we have zin = 0.96j0.72, or Zin = 96j72 . Finally, we move from point a towards the generator at point c to reach the voltage maximum, a distance 0.168. P6.25: The input impedance for a 100 lossless TLine of length 1.162 is measured as 12 + j42 . Determine the load impedance. We first locate the normalized input impedance, zin = 0.12 + j0.42, at point a (WTL=0.436). Then we move a distance 1.162 towards the load to point b, at WTL = 0.436 + 1.162 =1.598 ; 1.598 1.500 = 0.098 . At this point, we read zL = 0.15j0.7, or ZL = 15 j70 . P6.26: On a 50 lossless TLine, the VSWR is measured as 3.4. A voltage

Fig P6 27
maximum is located 0.079 away from the load. Determine the load. We can use the given VSWR to draw a constant circle as shown in the figure. Then we move from Vmax at WTG = 0.250 to point a at WTG = 0.250  0.079 = 0.171 . At this point we have zL = 1 +j1.3, or ZL = 50 + j65 . P6.27: Figure 6.52 is generated for a 50 slotted coaxial air line terminated in a short circuit and then in an unknown load. Determine (a) the measurement frequency, (b) the VSWR when the load is attached and (c) the load impedance. From the locations of minima on the shorted line we find :
( )2 7.55 1.25 12.6cm cm cm = = ( ) 2.4ca f GHz= = (b) From the voltage maxima and voltage minimum on the loaded line, we have
4 22
VSWR = = Using VSWR=2 we draw the constant  circle on the Smith Chart. Point a on the circle represents the 1.9 cm minimum. We move from this point towards the load at the 1.25 cm reference location, a move of
1.9 1.25 0.051612.6cm cm
cm
=
At this point (point b on the circle) we have zL = 0.55 j0.25, and upon denormalizing we have (c) ZL = 28 j12 . P6.28: Figure 6.53 is generated for a 50 slotted coaxial air line terminated in a short circuit and then in an unknown load. Determine (a) the measurement frequency, (b) the VSWR when the load is attached and (c) the load impedance. From the location of the maxima on the shorted line, we find :
( )2 9.3 1.7 15cm cm cm = =

( ) 2.0ca f GHz= = (b) From the load line,
10 2.54
VSWR = = Using VSWR=2.5 we draw the constant  circle on the Smith Chart. Point a on the circle represents the minimum at 7.9 cm. We move from this point towards the load at the 5.5 cm reference location, a move of
7.9 5.5 0.1615cm cm
cm
=
At this point (point b on the circle) we have zL = 1 j0.95, and upon denormalizing we have (c) ZL = 50 j48 . P6.29: Referring to Figure 6.20, suppose we measure Zinsc = +j25 and ZinL = 35 + j85 . What is the actual load impedance? Assume Zo = 50 . We normalize the short circuit impedance to zinsc = 0+j0.5 and locate this on the Smith Chart to determine the length of the TLine is 0.074. Then we normalize ZinL to zinL=0.70+j1.70, locate this on the chart at 0.326 (WTL scale) and draw a constant  circle. We then move towards the load, or to 0.336 + 0.074 = 0.400 , and find this point on the Smith Chart (zL = 0.25+j0.7). Denormalizing, we find ZL = 12+j35 . P6.30: MATLAB: Modify MATLAB 6.3 to draw the normalized load point and the constant
L circle, given Zo and ZL. Demonstrate your program with the values from Drill 6.11. Add this to the end of the Matlab 6.3 program: %now add constant gamma circles ZL=50; fudge=0.001+i*0.001; newZL=ZL+fudge; Zo=50; zL=newZL/Zo; gamma=(zL1)/(zL+1); plot(gamma,'o'); constgamma(zL); You must change the value of ZL for each load point. Notice the addition of a fudge factor. This ensures that gamma has both a nonzero and finite real and imaginary part to work with in the plot. Youll also need to add an additional function: function [h]=constgamma(zL) %constgamma(zL) draws the constant gamma circle;

Fig. P6.30
Fig P6 31a
Fig P6 31b
phi=1:1:360; theta=phi*pi/180; a=abs((zL1)/(zL+1)); Re=a*cos(theta); Im=a*sin(theta); z=Re+i*Im; h=plot(z,'k'); axis('equal') axis('off') The program is run for each point of Drill 6.11 by changing the ZL value. Since the MATLAB routine has the hold on, each new point is added to the plot. 5. Impedance Matching P6.31: A matching network, using a reactive element in series with a length d of TLine, is to be used to match a 35 j50 load to a 100 TLine. Find the through line length d and the value of the reactive element if (a) a series capacitor is used, and (b) a series inductor is used. First we normalize the load and locate it on the Smith Chart (point a, at zL = 0.35j0.5, WTG = 0.419). (a) need to move to point b, at z = 1+j1.4, so that a capacitive element of value jx = j1.4 can be added to provide an impedance match. Moving to this point b gives d = 0.500+0.173 0.419 = 0.254 . The capacitance is
( )( )( )91.4,
1 1.142 1 10 100 1.4
o
j jCZ
C pFx
=
= =
(b) Now we need to move to point c, at z = 1j1.4, so that an inductive element of value jx = +j1.4 can be added. Moving to this point c gives d = 0.500 + 0.327 0.419 = 0.408 . The inductance is

Fig P6 32b
Fig.P6.33a
Fig P6 33b
( )( )( )9
1.4 1001.4, 22.3
2 1 10o
j L j L nHZ x
= = = P6.32: A matching network consists of a length of TLine in series with a capacitor. Determine the length (in wavelengths) required of the TLine section and the capacitor value needed (at 1.0 GHz) to match a 10 j35 load impedance to the 50 line. We find the normalized load, zL = 0.2 j0.7, located at point a (WTG = 0.400). Now we move from point a clockwise (towards the generator) until we reach point b, where we have z = 1 + j2.4. Moving from a to b corresponds to d = 0.500+0.1940.400 = 0.294. For the series capacitance we have
( )( )( )92.4 ,
1or 1.332 1 10 50 2.4
o
jjCZ
C pFx
=
= =
P6.33: You would like to match a 170 load to a 50 TLine. (a) Determine the characteristic impedance required for a quarterwave transformer. (b) What throughline length and stub length are required for a shorted shunt stub matching network? (a) 92s o LZ Z R= = (b) (1)Normalize the load (point a, zL = 3.4 + j0). (2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle (d = 0.170) (4) move from the shorted end of the stub (normalized admittance point c) to the point y = 0 jb. (l = 0.354 0.250 = 0.104 .) Note in step 3 we could have gone to the point y = 1jb. This would have resulted in d = 0.329 and l = 0.396 .
P6.34: A load impedance ZL = 200 + j160 is to

Fig P6 36
be matched to a 100 line using a shorted shunt stub tuner. Find the solution that minimizes the length of the shorted stub. Refer to Figure P6.33a for the shunt stub circuit. (1)Normalize the load (point a, zL = 2.0 + j1.6). (2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.170 0.458 = 0.212) (4) move from the shorted end of the stub (normalized admittance point) to the point y = 0 jb. (l = 0.354 0.250 = 0.104 .) P6.35: Repeat P6.34 for an openended shunt stub tuner. (1)Normalize the load (point a, zL = 2.0 + j1.6). (2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1jb circle(0.500 + 0.330 0.458 = 0.372). We choose this point for c so as to minimize the length of the shunt stub. (4) move from the open end of the stub (normalized admittance point) to the point y = 0 + jb. (l = 0.146 )
P6.36: A load impedance ZL = 25 + j90 is to be matched to a 50 line using a shorted shunt stub tuner. Find the solution that minimizes the length of the shorted stub. Refer to Figure P6.33a for the shunt stub circuit. (1)Normalize the load (point a, zL = 0.5 + j1.8). (2) locate the normalized load admittance: yL (point b)

Fig P6 37
(3) move from point b to point c, at the y=1+jb circle(0.500 + 0.198 0.423 = 0.275) (4) move from the shorted end of the stub (normalized admittance point) to the point y = 0 jb. (l = 0.308 0.250 = 0.058 .) P6.37: Repeat P6.36 for an openended shunt stub tuner. Refer to Figure P6.35a. (1)Normalize the load (point a, zL = 0.5 + j1.8). (2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.392 0.423 = 0.379) (4) move from the open end of the stub (normalized admittance point) to the point y = 0 + jb. (l = 0.191 ) P6.38: (a) Design an openended shunt stub matching network to match a load ZL = 70 + j110 to a 50 impedance TLine. Choose the solution that minimizes the length of the through line. (b) Now suppose the load turns out to be ZL = 40 + j100 . Determine the reflection coefficient seen looking into the matching network. (a) Refer to Figure P6.35a. (1)Normalize the load (point a, zL = 1.4 + 2.2). (2) locate the normalized load admittance: yL (point b) (3) move from point b to point c, at the y=1+jb circle(0.500 + 0.185 0.448 = 0.237) (4) move from the open end of the stub (normalized admittance point) to the point y = 0  jb. (l = 0.328 )

Fig.P6.38a
Fig. P6.38b
(b) (1) Normalize the load (point a: zL = 0.8 + j2.0) (2) locate yL (point b) (3) Move a distance 0.237 to point c (0.434 + 0.237 = 0.671 ; or WTG = 0.171 ) (4) Move from yopen to 0.328 (point d) (5) add admittances of point c and d to get ytot = 0.6 j0.2. (6) locate the corresponding ztot (point f) and read the reflection coefficient as:
340.28 je = D

1. Rectangular Waveguide Fundamentals P7.1: Find the cutoff frequency for the first 8 modes of WR430. a = 4.3 in = 0.1092 m, b = 2.15 in = 0.0546 m For airfilled guide we have:
2 2
2mnc m nfc
a b = +
Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find Mode fcmn (GHz) TE10 1.374 TE01 2.747 TE20 2.747 TE11 3.07 TM11 3.07 TE21 3.885 TM21 3.885 TE30 4.121 P7.2: Calculate the cutoff frequency for the first 8 modes of a waveguide that has a = 0.900 inches and b = 0.600 inches. a = 0.900 in = 0.02286 m, b = 0.600in = 0.01524 m For airfilled guide we have:
2 2
2mnc m nfc
a b = +
Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find Mode fcmn (GHz) TE10 6.56 TE01 9.84 TE11 11.83 TM11 11.83 TE20 13.12 TE21 16.40 TM30 19.69 TE02 19.69 P7.3: Calculate the cutoff frequency for the first 8 modes of a waveguide that has a = 0.900 inches and b = 0.300 inches. a = 0.900 in = 0.02286 m, b = 0.300 in = 0.00762 m For airfilled guide we have:

2 2
2mnc m nfc
a b = +
Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find Mode fcmn (GHz) TE10 6.56 TE20 13.12 TE30 19.68 TE01 19.68 TE11 20.75 TM11 20.75 TE21 23.66 TM21 23.66 P7.4: Calculate uG, the wavelength in the guide and the wave impedance at 10 GHz for WR90. From Table 7.1 for WR90 we have fc10 = 6.56 GHz. So
2 28 86.561 3 10 1 2.26 10
10G Ufc mu u x xf s
= = = 8 9
2 2
3 10 10 10 0.0397 , 46.561110
U x x m cmfcf
= = = =
Since fc10 = 6.56 GHz, at 10 GHz only TE10 is present and therefore we only have the Z10TE impedance.
10 2 2
120 5006.561110
TE UZfcf
= = =
P7.5: Consider WR975 is filled with polyethylene. Find (a) uu, (b) up and (c) uG at 600 MHz. From Table 7.1 for WR975 we have a = 9.75 in and b = 4.875 in. Then
8
103 10 1 1 403
9.75 0.02542 2 2.26r
c x m s infc MHzin ma
= = = 2 24031 1 0.741
600fcFf
= = = Now,

Fi P7 6
88
8
8
3 10 2 102.26
2.7 10
1.48 10
Ur
UP
G U
c x mu xs
u mu xF s
mu u F xs
= = =
= =
= =
P7.6: MATLAB: Plot up and wavelength in the guide as a function of frequency over the cited useful frequency range for WR90. % MLP0706 % % Plot propagation velocity and guide wavelength % over the cited useful freq range of WR90. % % 2/2/03 Wentworth % c=3e8; a=0.900;b=0.450; fc=(c/(2*.0254*a)); flo=8.2e9; fhi=12.4e9; N=100; df=(fhiflo)/N; f=flo:df:fhi; A=sqrt(1(fc./f).^2); Lu=c./f; LG=Lu./A; up=c./A; fG=f./1e9; subplot(2,1,1) plot(fG,LG) ylabel('guide wavelength (m)') grid on subplot(2,1,2) plot(fG,up) xlabel('frequency(GHz)') ylabel('propagation velocity (m/s)') grid on P7.7: WR90 waveguide is to be operated at 16 GHz. Tabulate the values of the guide wavelength, phase velocity, group velocity and impedance for each supported mode.

Fig. P7.8
For the TE10 mode we have
10 10, where a = 0.900 in = .02286m, so 6.562 .2cfc fc GHza
= = Then
( )8
92
3 10, where 1.8816 10
1
uu
c x cmf xfc
f
= = = =
( )2.0188 0.0206
6.5621 16
m m = =
( ) ( ) ( )8
8
2 2 2
3 10 3.3 106.56211 1 16
up
u c x mu xsfc fc
f f
= = = =
( ) ( )2 28 86.5621 3 10 1 2.74 1016G u mfcu u x xf s= = = ( ) ( )2 2
120 4136.56211 16
TE umnZ
fcf
= = =
Likewise values are found for the TE20 and TE11 mode. For the TM11 mode, a different expression for impedance is used:
( )21TMmn u fcZ f= Mode fc(GHz) (m) up(m/s) uG(m/s) Z() TE10 6.56 0.0206 3.3x108 2.7x108 413 TE20 13.1 0.0328 5.2x108 1.7x108 659 TE11 14.7 0.0470 7.5x108 1.2x108 945 TM11 14.7 0.0470 7.5x108 1.2x108 150
P7.8: MATLAB: Modify MATLAB 7.1 by plotting uG and up versus frequency for the same guide over the same frequency range. % MFile: MLP0708 % % Waveguide Velocity Plot % Plots uG and uP for TE11 and TM11 % vs freq. for airfilled waveguide % (modifies ML0701) % % Wentworth, 11/26/02

Fi P7 9
% clc %clears command window clear%clears variables % Initialize variables c=2.998e8; %speed of light Zo=120*pi; ainches=0.900; binches=0.450; % convert to metric a=ainches*0.0254; b=binches*0.0254; % calc fc11 fc=c*sqrt((1/a)^2+(1/b)^2)/2; % Perform calculations f=15e9:.1e9:25e9; fghz=f/1e9; Factor=sqrt(1(fc./f).^2); uu=c.*Factor./Factor; %just filling array with c up=c./Factor; ug=c.*Factor; % Display results plot(fghz,up,'.k',fghz,uu,'k',fghz,ug,'k') legend('up','c','uG') xlabel('frequency, (GHz)') ylabel('velocity (m/s)') grid on P7.9: MATLAB: Plot the TE10 wave impedance for WR430 waveguide versus frequency if the guide is filled with Teflon. Choose a suitable frequency range for your plot. % MLP0709 % % Plot TE10 wave impedance for teflon filled % WR430 guide over a suitable frequency range. % % 2/2/03 Wentworth % c=3e8; er=2.1; uu=c/sqrt(er);

a=4.30;b=2.150; fc=(uu/(2*.0254*a)); flo=1.7e9/sqrt(er); fhi=2.6e9/sqrt(er); N=100; df=(fhiflo)/N; f=flo:df:fhi; A=sqrt(1(fc./f).^2); ZTE=(120*pi/sqrt(er))./A; fG=f./1e9; plot(fG,ZTE) xlabel('frequency(GHz)') ylabel('TE10 mode impedance (ohms)') grid on P7.10: Suppose a length of WR137 waveguide operated at 7.0 GHz is terminated in a short circuit. At what distance from this short circuit does the input impedance appear infinite? From our study of TLines, we know that looking into a quarter guidewavelength section of waveguide terminated in a short circuit, the input impedance appears infinite. The cutoff frequency for the TE10 mode is 4.29 GHz. Then,
( ) ( ) ( )8 9
2 2 2
3 10 7 10 0.05424.2911 1 7
U c f x x mfc fc
f f
= = = =
So the quarter wave length is 0.0542m/4 = 0.0136 m. Therefore a distance 1.4 cm away from the short circuit, the input impedance appears infinite. 2. Waveguide Field Equations P7.11: Manipulate (7.41) to get (7.1). Rearranging (7.41), we have
2 22 2 , u
m na b = +
Also from (7.41) we have
( )21u fc f = . So
( ) ( )2 22 2 2 2, 11+ =u u ufc fcf f = and

2 2 22 2 , u
fc m nf a b
= +
2 2 2 222
u
fc m nf a b
= +
2 22
u
fc m nf a b
= + Solving for fc, where we have uu = uf,
2 2 2 21 12 2u
m n m nfc ua b a b
= + = + P7.12: Find expressions for the phasor field components of the TE01 mode. With m = 0 and n = 1, the nonzero field components in equations (7.67)  (7.71)are
2 2
2 2
cos
sin
sin
j zzs o
j zxs o
u
j zys o
u
yH H eb
j yE H eb b
j yH H eb b
= = =