Wentworth solutions chapter 6 - LT

Prévia do material em texto

6- !1
Solutions for Chapter 6 Problems 
1. Distributed Parameters Model 
P6.1: RG-223/U coax has an inner conductor radius a = 0.47 mm and inner radius of the 
outer conductor b = 1.435 mm. The conductor is copper, and polyethylene is the 
dielectric. Calculate the distributed parameters at 800 MHz. 
! 
! 
! 
! 
! 
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. 
! 
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 
7
16
for copper: 5.8 10
for polyethylene: 2.26, 10
Cu
r
S
x
m
S
m
σ
ε σ −
=
= =
( )( )
( )
6 7
3 3 7
1 1 1'
2
800 10 4 101 1 1 3.32
2 0.47 10 1.435 10 5.8 10
c
f
R
a b
x x
x x mx
π µ
π σ
π π
π
−
− −
⎛ ⎞= +⎜ ⎟
⎝ ⎠
Ω⎛ ⎞= + =⎜ ⎟
⎝ ⎠
74 10 1.435' ln ln 223
2 2 0.47
b x nHL
a m
µ π
π π
−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )
( )
( )
16
18
2 102' 560 10
ln ln 1.435 0.47 0
S
G x
b a m
ππσ
−
−= = =
≈
( )
( )( )
( )
122 2.26 8.854 102' 112
ln ln 1.435 0.47
x pF
C
b a m
ππε
−
= = =
7for Nickel we have 1.5 10 and 600.Ni r
S
x
m
σ µ= =
6- !2
disp('Calc Coax Distributed Parameters') 
%Some constant values 
muo=pi*4e-7; 
eo=1e-9/(36*pi); 
%Prompt for input values 
a=input('inner radius, in mm, = '); 
b=input('outer radius, in mm, = '); 
er=input('relative permittivity, er= '); 
sigd=input('dielectric conductivity, in S/m, = '); 
sigc=input('conductor conductivity, in S/m, = '); 
ur=input('conductor rel. permeability, = '); 
f=input('input frequency, in Hz, = '); 
%Perform calulations 
G=2*pi*sigd/log(b/a); 
C=2*pi*er*eo/log(b/a); 
L=muo*log(b/a)/(2*pi); 
Rs=sqrt(pi*f*ur*muo/sigc); 
R=(1000*((1/a)+(1/b))*Rs)/(2*pi); 
omega=2*pi*f; 
RL=R+i*omega*L; 
GC=G+i*omega*C; 
Gamma=sqrt(RL*GC); 
Zo=sqrt(RL/GC); 
alpha=real(Gamma); 
beta=imag(Gamma); 
loss=exp(-2*alpha*1); 
lossdb=-10*log10(loss); 
%Display results 
disp(['G/h = ' num2str(G) ' S/m']) 
disp(['C/h = ' num2str(C) ' F/m']) 
disp(['L/h = ' num2str(L) ' H/m']) 
disp(['R/h = ' num2str(R) ' ohm/m']) 
disp(['Gamma= ' num2str(Gamma) ' /m']) 
disp(['alpha= ' num2str(alpha) 'Np/m']) 
disp(['beta= ' num2str(beta) 'rad/m']) 
disp(['Zo = ' num2str(Zo) ' ohms']) 
disp(['loss=' num2str(loss) ' /m']) 
disp(['lossdb=' num2str(lossdb) ' dB/m']) 
6- !3
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, = 1e-16 
conductor conductivity, in S/m, = 1.5e7 
conductor rel. permeability, = 600 
input frequency, in Hz, = 800e6 
G/h = 5.6291e-016 S/m 
C/h = 1.1249e-010 F/m 
L/h = 2.2324e-007 H/m 
R/h = 159.7792 ohm/m 
Gamma= 1.78881+25.252i /m 
alpha= 1.7888Np/m 
beta= 25.252rad/m 
Zo = 44.6608-3.1637i ohms 
loss=0.027942 /m 
lossdb=15.5374 dB/m 
>> 
Summarizing the distributed parameter data from this routine we have: 
! 
P6.3: Modify (6.3) to include internal inductance of the conductors. To simplify the 
calculation, assume current is evenly distributed across the conductors. Find the new 
value of L’ for the coax of Drill 6.1. 
From Ampere’s Circuit Law we can find H versus ρ: 
! 
! 
! 
! 
18' 160 , ' 223 , ' 560 10 , ' 112 pFnH SR L G x Cm m m m
−Ω= = = =
2 for 2
IH a
aφ
ρ
ρ
π
= ≤
 for a
2
IH bφ ρπρ
= ≤ ≤
2 2
2 2
c
 for b
2
IH c
c bφ
ρ
ρ
πρ
−
= ≤ ≤
−
0 for H cφ ρ= ≥
6- !4
Using the energy approach, ! , we find 
! 
Inserting the given values we find 
! 
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 γ, α, β and Zo. Confirm the 
program using Drill 6.2. 
See the solution for P6.2. 
Calc Coax Distributed Parameters 
inner radius, in mm, = 0.45 
outer radius, in mm, = 1.47 
relative permittivity, er= 2.26 
dielectric conductivity, in S/m, = 1e-16 
conductor conductivity, in S/m, = 5.8e7 
conductor rel. permeability, = 1 
input frequency, in Hz, = 1e9 
G/h = 5.3078e-016 S/m 
C/h = 1.0606e-010 F/m 
L/h = 2.3675e-007 H/m 
R/h = 3.8112 ohm/m 
Gamma= 0.0403332+31.4857i /m 
alpha= 0.040333Np/m 
beta= 31.4857rad/m 
Zo = 47.246-0.0605221i ohms 
loss=0.9225 /m 
lossdb=0.35033 dB/m 
>> 
This agrees with the results of Drill 6.2. 
2 21
2 2
o
mW LI H dv
µ
= = ∫
22 2 2 2
2 2 2 2 2 2
1' ln ln
2 8 2 4
o o ob c c c c bL
a c b b c b c b
µ µ µ
π π π
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+⎛ ⎞= + + − +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟− − −⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
( ) nH nH' 237 50 41.2 328
m m
L = + + =
6- !5
P6.5: The impedance and propagation constant at 100 MHz for a T-Line are determined 
to be Zo = 18.6 – j0.253 Ω and γ = 0.0638 + j4.68 /m. Calculate the distributed 
parameters. 
! 
! 
! 
P6.6: The specifications for RG-214 coaxial cable are as follows: 
• 2.21 mm diameter copper inner conductor 
• 7.24 mm inner diameter of outer conductor 
• 9.14 mm outer diameter of outer conductor 
• Teflon dielectric (εr = 2.10) 
Calculate the characteristic impedance and the propagation velocity for this cable. 
! 
! 
P6.7: For the RG-214 coax of problem P6.6 operating at 1 GHz, how long is this T-line in 
terms of wavelengths if its physical length is 50 cm? 
! 
P6.8: If 1 watt of power is inserted into a coaxial cable, and 1 microwatt of power is 
measured 100 m down the line, what is the line’s attenuation in dB/m? 
( )( )' ' , ' ' ' '
' 'o
R j L
Z R j L G j C
G j C
ω
γ ω ω
ω
+
= = + +
+
' ' 2.37 87.0
' 2.37 , ' 87.0 ' 139
oZ R j L j
nHR L so L
m m
γ ω
ω
= + = +
Ω
∴ = = =
6' ' 7.63 10 0.252,
' 7.63 , ' 0.252 ' 401
o
G j C x j
Z
S pF
G and C so C
m m
γ
ω
µ
ω
−= + = +
∴ = = =
60 60 3.62ln ln 49.1
1.1052.1o r
b
Z
aε
⎛ ⎞ ⎛ ⎞= = = Ω⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
82.07 10p
r
c m
u x
sε
= =
( )
8
9
2.07 10, 0.207
1 10
1( ) 50 2.4
0.207 100
p
p
u x
u f m
f x
m
cm
m cm
λ λ
λ
λ λ
= = = =
⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
l
6- !6
! 
P6.9: Starting with a 1 mm diameter solid copper wire, you are to design a 75 Ω coaxial 
T-Line using mica as the dielectric. Determine (a) the inner diameter of the outer copper 
conductor, (b) the propagation velocity on the line and (c) the approximate attenuation, in 
dB/m, at 1 MHz. 
! 
So the inner diameter of the outer conductor is 18 mm. 
! 
To calculate α, will need γ. Therefore we calculate R’, L’, G’ and C’. 
! 
! 
! 
! 
Now, with ω = 2πf, 
! 
Finally, ! 
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 
1
10log 60
1
60
' 0.6
100
WA dB
W
dB dBA
m m
µ⎛ ⎞= − = +⎜ ⎟
⎝ ⎠
= =
( )( ) ( ) ( )( )o60 ln , b=a exp Z 60 0.5 exp 75 5.4 60 9.1o r
r
b
Z mm mm
a
ε
ε
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
8
8 82.998 10 1.29 10 , so 1.3 10
5.4p pr
c x m m
u x u x
s sε
= = = =
( )( )6 7
3 3 7
1 10 4 101 1 1' 87.6
2 0.5 10 9.1 10 5.8 10
x x m
R
x x x m
π π
π
−
− −
Ω⎛ ⎞= + =⎜ ⎟
⎝ ⎠
74 10 9.1' ln 580
2 0.5
x nHL
m
π
π
− ⎛ ⎞= =⎜ ⎟
⎝ ⎠
( )
( )
15
15
2 10
' 2.17 10
ln 9.1 0.5
S
G x
m
π −
−= =
( )( )
( )
122 5.4 8.854 10
' 103.5
9.1ln 0.5
x pF
C
m
π −
= =
( )( ) 6 1' ' ' ' 585 10 0.049R j L G j C x j
m
γ ω ω −= + + = +
6 38.686585 10 5.1 10Np dB dBx x
m Np m
α − −⎛ ⎞= =⎜ ⎟
⎝ ⎠
6- !7
a skin depth. The dielectric has εr = 2.26 and σeff = 0.0002 at 1 GHz. Letting the ratio b/
a vary from 1.5 to 10, generate a plot of the attenuation (in dB/m) versus the line 
impedance. Use the lossless assumption to calculate impedance. 
% MLP0610 
% 
% Plot of alpha vs Zo for a particular coax 
clear 
clc 
%Some constant values 
muo=pi*4e-7; 
eo=8.854e-12; 
a=1; 
er=2.26; 
sigd=0.0002; 
sigc=5.8e7; 
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 T-Lines 
P6.11: Start with equation (6.54) and derive (6.55). 
6- !8
! 
With ! we then have 
! 
We also know that 
! 
So now we have 
! 
and with rearranging, 
! 
We can convert the exponential terms into hyperbolic functions, given 
! 
This leads to 
o o
in o
o o
V e V e
Z Z
V e V e
γ γ
γ γ
+ + − −
+ + − −
+
=
−
l l
l l
,o L oV V
− += Γ
( )
( )
L
in o
L
e e
Z Z
e e
γ γ
γ γ
+ −
+ −
+Γ
=
−Γ
l l
l l
,L oL
L o
Z Z
Z Z
−
Γ =
+
( ) ( )
( ) ( )
L o
L o L o L o
in o o
L o L oL o
L o
Z Ze e
Z Z Z Z e Z Z e
Z Z Z
Z Z e Z Z eZ Ze e
Z Z
γ γ
γ γ
γ γ
γ γ
+ −
+ −
+ −
+ −
⎛ ⎞−
+ ⎜ ⎟+ + + −⎝ ⎠= =
+ − −⎛ ⎞−
− ⎜ ⎟+⎝ ⎠
l l
l l
l l
l l
( ) ( )
( ) ( )
.L oin o
L o
Z e e Z e e
Z Z
Z e e Z e e
γ γ γ γ
γ γ γ γ
+ − + −
+ − + −
+ + −
=
− + +
l l l l
l l l l
( ) ( )1 1 sinh(x)sinh( ) , cosh( ) , and tanh(x)= .
2 2 cosh(x)
x x x xx e e x e e− −= − = +
Fig. P6.10
6- !9
! 
or finally 
! 
P6.12: Derive (6.56) from (6.55) for a lossless line. 
! and ! since α = 0 for 
lossless line. Using the hyperbolic definitions, we have 
! 
Now using Euler’s formula, 
! 
Plugging this in, we find, 
! 
P6.13: A 2.4 GHz signal is launched on a 1.5 m length of T-Line terminated in a matched 
load. It takes 6.25 ns to reach the load and suffers 1.2 dB of loss. Find the propagation 
constant. 
! 
! 
! 
! 
So 
! 
( ) ( )
( ) ( )
2 cosh 2 sinh
,
2 sinh 2 cosh
L o
in o
L o
Z Z
Z Z
Z Z
γ γ
γ γ
+
=
+
l l
l l
( )
( )
tanh
.
tanh
L o
in o
o L
Z Z
Z Z
Z Z
γ
γ
+
=
+
l
l
( )
( )
tanh
,
tanh
L o
in o
o L
Z Z
Z Z
Z Z
γ
γ
+
=
+
l
l ( ) ( ) ( )tanh tanh tanhj jγ α β β= + =l l l l
( ) ( )
( )
( )
( )
sinh
tanh .
cosh
j j
j j
e ej
j
j e e
β β
β β
β
β
β
+ −
+ −
−
= =
+
l l
l l
ll l
( ) ( ) ( )
( ) ( )
( )
( )
cos sin( ) - cos sin( ) 2sin
tanh tan( )
cos sin( ) cos sin( ) 2cos
j j j
j j
j j
β β β β β
β β
β β β β β
+ − − −
= = =
+ + − +
l l l l ll ll l l l l
( )
( )
tan
.
tan
L o
in o
o L
Z jZ
Z Z
Z jZ
β
β
+
=
+
l
l
jγ α β= +
1.2 1 0.092
1.5 8.686
dB Np Np
m dB m
α = =
81.5: 2.4 10
6.25p
m m
u x
t ns s
ω
β
β
= = = =
l
( )9
8
2 2.4 10
62.8
2.4 10p
x rad
u x m
πω
β = = =
10.092 62.8 j
m
γ = +
6- !10
P6.14: A source with 50 Ω source impedance drives a 50 Ω T-Line that is 1/8 of a 
wavelength long, terminated in a load ZL = 50 – j25 Ω. Calculate ΓL, VSWR, and the 
input impedance seen by the source. 
! 
! 
! 
! 
P6.15: A 1 m long T-Line has the following distributed parameters: R’ = 0.10 Ω/m, L’ = 
1.0 µH/m, G’ = 10.0 µS/m, and C’ = 1.0 nF/m. If the line is terminated in a 25 Ω resistor 
in series with a 1 nH inductor, calculate, at 200 MHz, ΓL and Zin. 
! 
Now, MLP0615 is used to solve the problem. 
% MLP0615 
% 
% calculate gamma and char impedance 
% given the distributed parameters 
% Then, calculate gammaL and Zin 
% 
% define variables 
clc 
clear 
R=0.1; 
L=1.0e-6; 
G=10e-6; 
C=1.0e-9; 
f=200e6; 
7650 25 50 0.242
50 25 50
jL o
L
L o
Z Z j
e
Z Z j
−− − −Γ = = =
+ − +
�
1
1.64
1
L
L
VSWR
+ Γ
= =
− Γ
2 , tan 1
8 4 4
π λ π π
β
λ
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
l
( )
( )
tan
tan
50 25 5050
50 50 25
30.8 3.8 
L o
in o
o L
Z jZ
Z Z
Z jZ
j j
j
j
β
β
+
=
+
− +
=
+ +
= − Ω
l
l
( )( )6 925 2 200 10 10 25 1.257 LZ j x jπ −= + = + Ω
Fig. P6.14
6- !11
w=2*pi*f; 
length=1; 
ZL=25+j*1.257; 
% Perform calcuations 
A=R+i*w*L; 
B=G+i*w*C; 
gamma=sqrt(A*B) %Propagation Constant 
Zo=sqrt(A/B) 
gammaL=(ZL-Zo)/(ZL+Zo) %Reflection coefficient 
TGL=tanh(gamma*length); 
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL)) 
Running the program, 
Gamma = 0.0017 +39.7384i 
Zo = 31.6228 - 0.0011i 
gammaL = -0.1164 + 0.0248i 
Zin = 34.0192 - 7.4618i 
>> 
So the answers are, with the appropriate significant digits, 
! 
P6.16: The reflection coefficient at the load for a 50 Ω line is measured as ΓL = 0.516ej8.2° 
at f = 1 GHz. Find the equivalent circuit for ZL. 
Rearranging ! we find ! . 
This is a resistor in series with an inductor. The inductor is found by considering 
! , 
So the load is a 150 Ω resistor in series with a 4.8 nH inductor. 
P6.17: The input impedance for a 30 cm length of lossless 100 Ω impedance T-line 
operating at 2 GHz is Zin = 92.3 – j67.5 Ω. The propagation velocity is 0.7c. Determine 
the load impedance. 
Rearranging ! we find ! 
1680.12 and 34 7.5 jL ine Z jΓ = = − Ω
�
,L oL
L o
Z Z
Z Z
−
Γ =
+
1 150 30 
1
L
L o
L
Z Z j
+Γ
= = + Ω
−Γ
( )9
30
30, or 4.8
2 1 10
j L j L nH
x
ω
π
= = =
( )
( )
tan
,
tan
L o
in o
o L
Z jZ
Z Z
Z jZ
β
β
+
=
+
l
l
( )
( )
tan
tan
in o
L o
o in
Z jZ
Z Z
Z jZ
β
β
−
=
−
l
l
6- !12
! 
Evaluating, we have 
! or L = 1.3 pH. 
This is a very small inductance, so we have ! 
P6.18: For the lossless T-Line circuit shown in Figure 6.51, determine the input 
impedance Zin and the instantaneous voltage at the load end vL. 
! 
! 
! 
! 
! 
! 
! , so ! 
P6.19: Referring to Figure 6.10, a lossless 75 Ω T-Line has up = 0.8c and is 30 cm long. 
The supply voltage is vs = 6.0 cos(ωt) V with Zs = 75 Ω. If ZL = 100 + j125 Ω at 600 
MHz, find (a) Zin, (b) the voltage at the load end of the T-Line, and (c) the voltage at the 
sending end of the T-Line. 
! 
! 
Referring to Fig P6.19, 
( )
( )
( ) ( )
9
8
2 2 10
59.84 ; tan tan 59.84 0.3 1.254
0.7 0.7 3 10
x rad rad
m
c m mx
πω
β β
⎛ ⎞⎛ ⎞= = = = = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
l
( )950 0.016 50 2 2 10 ,LZ j j x Lπ= + Ω = +
50 .LZ ≈ Ω
25 50 1 2, , tan 0
25 50 3 2L
π λ
β π π
λ
−
Γ = = − = = =
+
l
0 25
0
L
in o L
L
Z
Z Z Z
Z
+
= = = Ω
+
25 8 2
25 75
j z j z
in o oV V V V e V e
β β+ − − += = = +
+
( )2 j jo LV e eβ β+ −= +Γl l
cos sin 1, 1,j je j eπ ππ π −= + = − = −
( )1 21 1 2; 3
3 3o o o
V V V V+ + +
−⎛ ⎞− − − = = = −⎜ ⎟
⎝ ⎠
( ) 11 3 1 2
3L o L
V V V+ ⎛ ⎞= +Γ = − − = −⎜ ⎟
⎝ ⎠ ( )2cos 180Lv t Vω= + �
, 15.7 , 4.71, tan 418.6p
p
radu
u m
ω ω
β β β
β
= = = = =l l
( )
( )( )
100 125 75 418.6
75
75 100 125 418.6
22 28 
in
j j
Z
j j
j
+ +
=
+ +
= − Ω
Fig. P6.19
6- !13
! 
! 
! 
! 
! 
! 
P6.20: Suppose the T-Line for Figure 6.10 is characterized by the following distributed 
parameters at 100 MHz: R’ = 5.0 Ω/m, L’ = 0.010 µH/m, G’ = 0.010 S/m, and C’ = 0.020 
nF/m. If ZL = 50 – j25 Ω,vs = 10cos(ωt)V, Zs = 50Ω, and the line length is 1.0 m, find the 
voltage at each end of the T-line. 
The following MATLAB routine was used to find the required parameters. 
% MLP0620 
% 
% calculate gamma and char impedance 
% given the distributed parameters 
% Then, calculate gammaL and Zin 
% 
% define variables 
clc 
clear 
R=5; 
L=.010e-6; 
G=.01; 
C=.020e-9; 
f=100e6; 
w=2*pi*f; 
length=1; 
ZL=50-j*25; 
% Perform calcuations 
A=R+i*w*L; 
366 2.1
75
jin
in
in
ZV e V
Z
−= =
+
�
( )2.1cos 36inv t Vω∴ = − �
430.593 jL oL
L o
Z Z
e
Z Z
−
Γ = =
+
�
( ) 126 360.70 2.1j j j jin o L oV V e e e V e Vβ β+ + − − + −= +Γ = =
� �l l
36
90
126
2.1 3
0.70
j
j
o j
e
V e V
e
−
+
−
= =
�
�
�
( )
( )
105.81 4.47
4.5cos 106
j
L o L
L
V V e V
v t Vω
+= +Γ =
= +
�
�
6- !14
B=G+i*w*C; 
gamma=sqrt(A*B) 
Zo=sqrt(A/B) 
gammaL=(ZL-Zo)/(ZL+Zo) 
TGL=tanh(gamma*length); 
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL)) 
Running the program, 
gamma = 0.2236 + 0.2810i 
Zo = 22.3607 
gammaL = 0.4479 - 0.1908i 
Zin = 27.2079 -15.4134i 
>> 
! 
! 
so ! 
! 
4. The Smith Chart 
P6.21: Locate on a Smith Chart the following load impedances terminating a 50 Ω T-
Line. (a) ZL = 200 Ω , (b) ZL = j25 Ω, (c) ZL = 50 + j50 Ω, and (d) ZL = 25 – j200 Ω. 
 
P6.22: Repeat problem P6.14 using the Smith Chart. 
First we locate the normalizedload, zL = 1 – j0.5 (point a). By inspection of the Smith 
Chart, we see that this point corresponds to ! Also, after drawing the 
constant Γ circle we can see VSWR = 1.66. Finally, we move from point a, at 0.356λ on 
( )18.23.97 , 4.0cos 18.2jinin SS in
in S
ZV V e V v t V
Z Z
ω−= = ∴ = −
+
� �
( ) ( ) ( )3.841.504 0.101 1.507 jin o L o oV V e e V j V eγ γ+ − + += +Γ = + = �l l
18.2
22
3.84
3.97 2.63
1.507
j
j
o j
e
V e
e
−
+ −= =
�
�
�
( ) ( )29.61 3.85 , 3.9cos 30jL o L LV V e v t Vω+ −= +Γ = ∴ = −
� �
760.245 .jL e
−Γ =
�
6- !15
the WTG scale, clockwise (towards the generator) a distance 0.125 λ to point b, at 0.481 
λ. At this point we see zin = 0.62 – j0.07. Denormalizing we find: 
Zin = 31 – j3.5 Ω. 
 
P6.23: A 0.690λ long lossless Zo = 75 Ω T-Line is terminated in a load ZL = 15 + j67 Ω. 
Use the Smith Chart to find (a) ΓL, (b) VSWR, (c) Zin and (d) the distance between the 
input end of the line and the first voltage maximum from the input end. 
Fig. P6.22a
Fig. P6.21
6- !16
After normalizing ZL and locating it on the chart (point a), we see After 
drawing the constant Γ circle, we see that VSWR = 9 (point c). We locate the input 
impedance by moving from the load (point a at WTG = 0.118λ) clockwise towards the 
generator to the input point (point b at WTG = 0.118 λ + 0.690 λ – 0.500 λ = 0.308 λ). 
At this point, zin = 0.8 – j2.4, so Zin = 60 – j180 Ω. Finally, the distance from the input 
end of the line (point b) to the first voltage maximum (point c) is simply 0.308 λ – 0.250 
λ = 0.058 λ. Or, using the WTL scale, it is 0.250 λ – 0.192 λ = 0.058 λ. 
P6.24: A 0.269λ long lossless Zo = 100 Ω T-Line is terminated in a load ZL = 60 + j40 Ω. 
Use the Smith Chart to find (a) ΓL, (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=0.082λ. 
We read off the Smith Chart that this point corresponds to: ! After drawing 
the constant Γ circle we notice the VSWR = 2.05 (point c). 
Moving from this point a distance 0.269 λ (clockwise, towards generator), we find the 
input point (point b at WTG = 0.351 λ). At this point we have zin = 0.96-j0.72, or Zin = 
96-j72 Ω. Finally, we move from point a towards the generator at point c to reach the 
voltage maximum, a distance 0.168λ. 
950.80 .jL eΓ =
�
1210.34 .jL eΓ =
�
Fig. P6.22b Fig. P6.23
6- !17
P6.25: The input impedance for a 100 Ω 
lossless T-Line of length 1.162λ is 
measured as 12 + j42 Ω. Determine the 
load impedance. 
We first locate the normalized input 
impedance, zin = 0.12 + j0.42, at point a 
(WTL=0.436λ). Then we move a distance 
1.162 λ towards the load to point b, at WTL 
= 0.436 λ + 1.162 λ =1.598 λ; 1.598 λ – 
1.500 λ = 0.098 λ. At this point, we read zL 
= 0.15-j0.7, or ZL = 15 – j70 Ω. 
P6.26: On a 50 Ω lossless T-Line, the VSWR is measured as 3.4. A voltage maximum is 
located 0.079λ away from the load. Determine the load. 
We can use the given VSWR to draw a constant Γ circle as shown in the figure. Then we 
move from Vmax at WTG = 0.250λ to point a at WTG = 0.250 λ - 0.079 λ = 0.171 λ. At 
this point we have zL = 1 +j1.3, or ZL = 50 + j65 Ω. 
Fig. P6.25
Fig. P6.26
6- !18
VSWR when the load is attached and (c) the 
load impedance. 
From the locations of minima on the 
shorted line we find λ: 
! 
! 
(b) From the voltage maxima and voltage 
minimum on the loaded line, we have 
! 
Using VSWR=2 we draw the constant |Γ| 
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 
! 
At this point (point b on the circle) we have 
zL = 0.55 – j0.25, and upon denormalizing 
we have (c) ZL = 28 – j12 Ω. 
P6.28: Figure 6.53 is generated for a 50 Ω 
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 λ: 
! 
! 
(b) From the load line, 
! 
( )2 7.55 1.25 12.6cm cm cmλ = − =
( ) 2.4ca f GHz
λ
= =
4 2
2
VSWR = =
1.9 1.25 0.0516
12.6
cm cm
cm
λ
λ
⎛ ⎞−
=⎜ ⎟
⎝ ⎠
( )2 9.3 1.7 15cm cm cmλ = − =
( ) 2.0ca f GHz
λ
= =
10 2.5
4
VSWR = =
Fig. P6.27
Fig. P6.28
6- !19
Using VSWR=2.5 we draw the constant |Γ| 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 
! 
At this point (point b on the circle) we have zL = 1 – j0.95, and upon denormalizing we 
have (c) ZL = 50 – j48 Ω. 
P6.29: Referring to Figure 6.20, suppose we measure Zinsc = +j25 Ω and ZinL = 35 + j85 
Ω. What is the actual load impedance? Assume Zo = 50 Ω. 
We normalize the short circuit impedance to zinsc = 0+j0.5 and locate this on the Smith 
Chart to determine the length of the T-Line is 0.074λ. Then we normalize ZinL to 
zinL=0.70+j1.70, locate this on the chart at 0.326λ (WTL scale) and draw a constant |Γ| 
circle. We then move towards the load, or to 0.336λ + 0.074 λ = 0.400 λ, and find this 
point on the Smith Chart (zL = 0.25+j0.7). Denormalizing, we find ZL = 12+j35 Ω. 
P6.30: MATLAB: Modify MATLAB 6.3 to draw the normalized load point and the 
constant circle, given Zo and ZL. Demonstrate your program with the values from 
Drill 6.11. 
Add this to the end of the Matlab 6.3 program: 
%now add constant gamma circles 
ZL=50; 
fudge=0.001+i*0.001; 
newZL=ZL+fudge; 
Zo=50; 
zL=newZL/Zo; 
gamma=(zL-1)/(zL+1); 
plot(gamma,'-o'); 
constgamma(zL); 
You must change the value of ZL for each load point. Notice the addition of a ‘fudge 
factor’. This ensures that gamma has both a nonzero and finite real and imaginary part to 
work with in the plot. 
You’ll also need to add an additional function: 
function [h]=constgamma(zL) 
%constgamma(zL) draws the constant gamma circle; 
7.9 5.5 0.16
15
cm cm
cm
λ
λ
⎛ ⎞−
=⎜ ⎟
⎝ ⎠
LΓ
6- !20
phi=1:1:360; 
theta=phi*pi/180; 
a=abs((zL-1)/(zL+1)); 
Re=a*cos(theta); 
Im=a*sin(theta); 
z=Re+i*Im; 
h=plot(z,'--k'); 
axis('equal') 
axis('off') 
The program is run for each point of 
Drill 6.11 by changing the ZL value. 
Since the MATLAB routine has the 
‘hold on’, each new point is added to 
the plot. 
5. Impedance Matching 
P6.31: A matching network, using a reactive element in series with a length d of T-Line, 
is to be used to match a 35 – j50 Ω load to a 100 Ω T-Line. Find the through line length 
d and the value of the reactive element if (a) a series capacitor is used, and (b) a series 
inductor is used. 
First we normalize the load and locate it on the Smith Chart (point a, at zL = 0.35-j0.5, 
WTG = 0.419λ). 
(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 = 0.500λ
+0.173 λ -0.419 λ = 0.254 λ. The capacitance is 
Fig. P6.30
Fig. P6.31a Fig. P6.31b
6- !21
! 
(b) Now we need to move to point c, at z = 1-j1.4, so that an inductive element of value jx 
= +j1.4 can be added. Moving to this point c gives d = 0.500 λ + 0.327 λ – 0.419 λ = 
0.408 λ. The inductance is 
! 
P6.32: A matching network consists of a length of T-Line in series with a capacitor. 
Determine the length (in wavelengths) required of the T-Line section and the capacitor 
value needed (at 1.0 GHz) to match a 10 – j35 Ω load impedance to the 50 Ω line. 
We find the normalized load, zL = 0.2 – j0.7, located at point a (WTG = 0.400λ). 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 = 0.500λ+0.194λ-0.400λ = 
0.294λ. For the series capacitance we have 
! 
P6.33: You would like to match a 170 Ω load to a 50 Ω T-Line. (a) Determine the 
characteristic impedance required for a 
quarter-wave transformer.(b) What 
through-line length and stub length are 
required for a shorted shunt stub matching 
network? 
( )( )( )9
1.4,
1 1.14
2 1 10 100 1.4
o
j
j
CZ
C pF
x
ω
π
−
= −
= =
( )( )
( )9
1.4 100
1.4, 22.3
2 1 10o
j L j L nH
Z x
ω
π
= = =
( )( )( )9
2.4 ,
1
or 1.33
2 1 10 50 2.4
o
j
j
CZ
C pF
x
ω
π
−
− =
= =
Fig. P6.32a
Fig. P6.32b
6- !22
(a) ! 
(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 = 0.170λ) 
(4) move from the shorted end of the stub (normalized admittance point c) to the point y 
= 0 – jb. (l = 0.354 λ – 0.250 λ = 0.104 λ.) 
 
Note in step 3 we could have gone to the 
point y = 1-jb. This would have resulted in 
d = 0.329 λ and l = 0.396 λ. 
 
P6.34: A load impedance ZL = 200 + j160 Ω is to be matched to a 100 Ω 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(0.500λ + 0.170λ -0.458λ = 
0.212λ) 
(4) move from the shorted end of the stub 
(normalized admittance point) to the point 
y = 0 – jb. (l = 0.354 λ – 0.250 λ = 0.104 
λ.) 
P6.35: Repeat P6.34 for an open-ended shunt stub tuner. 
92s o LZ Z R= = Ω
Fig. P6.33a
Fig. P6.33b
Fig. P6.34
6- !23
(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(0.500λ + 0.330λ -0.458λ = 0.372λ). 
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 λ) 
P6.36: A load impedance ZL = 25 + j90 Ω is to be matched to a 50 Ω line using a shorted 
shunt stub tuner. Find the solution that minimizes the length of the shorted stub. 
Refer to Figure P6.33a for the shunt stub 
circuit. 
(1)Normalize the load (point a, zL = 0.5 + 
j1.8). 
(2) locate the normalized load admittance: yL 
(point b) 
(3) move from point b to point c, at the 
y=1+jb circle(0.500λ + 0.198λ -0.423λ = 
0.275λ) 
(4) move from the shorted end of the stub 
(normalized admittance point) to the point y 
= 0 – jb. (l = 0.308 λ – 0.250 λ = 0.058 λ.) 
Fig. P6.35a
Fig. P6.35b
Fig. P6.36
6- !24
P6.37: Repeat P6.36 for an open-ended shunt stub tuner. 
Refer to Figure P6.35a. 
(1)Normalize the load (point a, zL = 0.5 + 
j1.8). 
(2) locate the normalized load admittance: yL 
(point b) 
(3) move from point b to point c, at the 
y=1+jb circle(0.500λ + 0.392λ -0.423λ = 
0.379λ) 
(4) move from the open end of the stub 
(normalized admittance point) to the point y 
= 0 + jb. (l = 0.191 λ) 
P6.38: (a) Design an open-ended shunt stub matching network to match a load ZL = 70 + 
j110 Ω to a 50 Ω impedance T-Line. Choose the solution that minimizes the length of the 
through line. (b) Now suppose the load turns out to be ZL = 40 + j100 Ω. 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). 
Fig. P6.37
Fig. P6.38a Fig. P6.38b
6- !25
(2) locate the normalized load admittance: yL (point b) 
(3) move from point b to point c, at the y=1+jb circle(0.500λ + 0.185λ -0.448λ = 0.237λ) 
(4) move from the open end of the stub (normalized admittance point) to the point y = 0 - 
jb. (l = 0.328 λ) 
(b) 
(1) Normalize the load (point a: zL = 0.8 + j2.0) 
(2) locate yL (point b) 
(3) Move a distance 0.237λ to point c (0.434 λ + 0.237 λ = 0.671 λ; or WTG = 0.171 λ) 
(4) Move from yopen to 0.328 λ (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: 
! 
6. Microstrip 
P6.39: A 6.00 cm long microstrip transmission line is terminated in a 100. Ω resistive 
load. The signal line is 0.692 mm wide atop a 0.500 mm thick polyethylene substrate. 
What is the input impedance of this line at 1.0 GHz? What is the maximum frequency at 
which this microstrip can operate? 
This can be solved using either the Smith Chart or ML0604 in conjunction with the 
Zinput function from Matlab 6.2. Using the latter approach we have: 
Microstrip Analysis 
enter width & thickness in the same units 
 
enter the line width: 0.692e-3 
enter the substrate thickness: 0.500e-3 
enter substrate rel permittivity: 2.26 
eeff = 1.8326 
up = 221461941.7986m/s 
Zo = 80.2454ohms 
To run the Zinput routine, we also need the propagation constant. Assuming lossless line, 
we have ! 
>> Zinput(80.2,100,j*28.4,0.06) 
ans = 64.7278 + 3.7906i 
>> 
340.28 jeΓ =
�
( )9
8
2 1 10 128.4 .
2.215 10p
x
j j j j
u x m
πω
γ β= = = =
6- !26
So we have Zin = 65 + j3.8 Ω 
To find fmax, we have 
! 
P6.40: A 75 Ω impedance microstrip line is to be designed on a 2.0 mm thick Teflon 
substrate using copper metallization. What is the maximum operating frequency for this 
microstrip? Now determine w, and the physical length of a quarter wave section of line 
at 800. MHz. 
! 
Using ML0605: 
Microstrip Design 
width & thickness will be in the same units 
 
enter the desired impedance: 75 
enter the substrate thickness: 2 
enter substrate rel permittivity: 2.1 
w = 3.2929 
eeff = 1.741 
up = 227209857.0703m/s 
>> 
so w = 3.29mm and εeff = 1.741. 
The guide wavelength is: 
! 
The quarter wave section length is then: ! 
P6.41: Analysis of a 2.56 cm long microstrip line reveals that it has a 50 Ω characteristic 
impedance and an effective relative permittivity of 5.49. It is terminated in a 60 Ω 
resistor in series with a 1.42 pF capacitor. Determine the input impedance looking into 
this terminated line at 1.60 GHz. 
( )
8
max 3
3 10
100
4 4 0.5 10 2.26r
c x m sf GHz
h x mε −
= = =
( )
8
max 3
3 10
26
4 4 2 10 2.1r
c x m sf GHz
h x mε −
= = =
8 63 10 800 10 0.284
1.741
o
G
eff eff
c f x x
m
λ
λ
ε ε
= = = =
0.071 7.1
4
G m cmλ= = =l
6- !27
This problem may be solved analytically or with the Smith Chart. For the analytical 
solution we have: 
! ! 
! 
The load capacitance has an impedance: 
! 
so the total load impedance is ZL = 60-j70 Ω. 
Then, the input impedance is 
! 
With the Smith Chart, the answer is ! 
P6.42: A 100 Ω impedance microstrip line is to be designed using copper metallization on 
a 0.127 cm thick dielectric of relative permittivity 3.8. Determine (a) w, (b) fmax, and at 2 
GHz find (c) up and λG. 
From ML0605: 
Microstrip Design 
width & thickness will be in the same units 
 
enter the desired impedance: 100 
enter the substrate thickness: 0.127 
enter substrate rel permittivity: 3.8 
w = 0.066625 
eeff = 2.6865 
up = 182909468.597m/s 
>> 
(a) So we have w = 0.0666 cm = 0.67 mm. 
(b) 
! 
(c) We know up, from the program (up = 1.83x108 m/s), so at 2 GHz 
( )8
9
3 10 5.49
0.080 ,
1.6 10
effp xcu m
f f x
ε
λ = = = =
.0256 0.320 .
0.080
m
m
λ
λ
= =l
( )2 0.320 , tan 2.125πβ λ β
λ
= = −l l
70c
jZ j
Cω
−
= = − Ω
( )
( )( )
60 70 50 2.125
50 31.8 48.2 
50 60 70 2.125in
j j
Z j
j j
− + −
= = + Ω
+ − −
32 48 inZ j≈ + Ω
( )
8
max 3
3 10
30
4 4 1.27 10 3.8r
c x m sf GHz
h x mε −
= = =
6- !28
! 
P6.43: MATLAB: Modify MATLAB 6.4 to calculate attenuation. Try out your program 
using the parameters of Drill 6.21 and Drill 6.22. 
% M-File: MLP0640 
% 
% Microstrip Analysis 
% 
% Given the physical dimensions and er, this 
% will calculate eeff, Zo and up for microstrip. 
% 
% Wentworth, 8/3/02 
% modified ML0604 on 9/5 to calculate attenuation 
% 
% Variables: 
% w line width (m) 
% h substrate thickness (m) 
% t conductor thickness (m) 
% sigc conductor conductivity (S/m) 
% ur conductor rel permeability% er substrate relative permittivity 
% eeff effective relative permittivity 
% up propagation velocity (m/s) 
% Zo characteristic impedance (ohms) 
% ad dielectric attenuation(dB/m) 
% ac conductor attenuation dB/m) 
% atot total attenuation (dB/m) 
% ds skin depth (m) 
% Rs skin effect resistance (ohms/square) 
% tand dielectric loss tangent 
clc %clears the command window 
clear %clears variables 
disp('Microstrip Analysis') 
disp(' ') 
% Prompt for input values 
0.0915 9.15pG
u
m cm
f
λ = = =
6- !29
w=input('enter the line width (m): '); 
h=input('enter the substrate thickness(m): '); 
er=input('enter substrate rel permittivity: '); 
t=input('enter conductor thickness (m): '); 
sigc=input('enter conductor conductivity (S/m): '); 
ur=input('enter conductor relative permeability: '); 
tand=input('enter dielectric loss tangent: '); 
f=input('enter frequency (Hz): '); 
uo=pi*4e-7; 
eo=8.854e-12; 
c=2.998e8; 
u=ur*uo; 
e=er*eo; 
% Perform Calculations 
eeff=((er+1)/2)+(er-1)/(2*sqrt(1+12*h/w)); 
up=2.998e8/sqrt(eeff); 
if w/h<=1 
 Zo=(60/sqrt(eeff))*log((8*h/w)+(w/(4*h))); 
else if w/h>1 
 Zo=120*pi/(sqrt(eeff)*((w/h)+1.393+0.667*log((w/h)
+1.444))); 
 end 
end 
ds=1/sqrt(pi*f*u*sigc); 
Rs=1/(sigc*ds*(1-exp(-t/ds))); 
ac=8.686*Rs/(Zo*w); 
ad=8.686*2*pi*f*er*(eeff-1)*tand/(c*2*sqrt(eeff)*(er-1)); 
atot=ac+ad; 
% Display results 
disp(['eeff = ' num2str(eeff) ]) 
disp(['up = ' num2str(up) 'm/s']) 
disp(['Zo = ' num2str(Zo) 'ohms']) 
disp(['ac = ' num2str(ac) 'dB/m']) 
disp(['ad = ' num2str(ad) 'dB/m']) 
 
Now we run the program using the information from Drill 6.21 and Drill 6.22. Note that 
we’ve changed the dimensions to metric units. 
6- !30
Microstrip Analysis 
 
enter the line width (m): 9.8e-4 
enter the substrate thickness(m): 1.016e-3 
enter substrate rel permittivity: 9.9 
enter conductor thickness (m): 6e-6 
enter conductor conductivity (S/m): 5.8e7 
enter conductor relative permeability: 1 
enter dielectric loss tangent: 0.0001 
enter frequency (Hz): 1e9 
eeff = 6.6638 
up = 116137011.5308m/s 
Zo = 49.8369ohms 
ac = 1.5554dB/m 
ad = 0.022214dB/m 
>> 
P6.44: A 50 Ω impedance microstrip line is desired for operation at 2.4 GHz. It is to be 
built on a 20 mil thick mica substrate using a 10 µm thick copper conductor. Calculate 
(a) w, (b) αc, (c) αd, and (d) αtot at this frequency. 
Mica has εr = 5.4 and tanδ = 0.0003. Using ML0605: 
Microstrip Design 
width & thickness will be in the same units 
 
enter the desired impedance: 50 
enter the substrate thickness: 20 
enter substrate rel permittivity: 5.4 
w = 32.6859 
eeff = 3.9617 
up = 150623255.8531m/s 
>> 
To use this information in ML0640, we need to convert h and w to metric units. We find 
h = 5.08x10-4m and w = 8.3x10-4m. 
Microstrip Analysis 
 
enter the line width (m): 8.3e-4 
enter the substrate thickness(m): 5.08e-4 
6- !31
enter substrate rel permittivity: 5.4 
enter conductor thickness (m): 10e-6 
enter conductor conductivity (S/m): 5.8e7 
enter conductor relative permeability: 1 
enter dielectric loss tangent: 0.0003 
enter frequency (Hz): 2.4e9 
eeff = 3.9616 
up = 150624957.4337m/s 
Zo = 50.1512ohms 
ac = 2.6687dB/m 
ad = 0.11967dB/m 
>> 
With more appropriate significant digits we find: 
w = 33 mils 
αc = 2.67 dB/m 
αd = 0.12 dB/m 
αtot = 2.79 dB/m 
P6.45: One type of board routinely used to build microwave circuits is 50 mils thick 
Rogers Corporation RT/Duroid©, with εr = 10.8 and tanδ = 0.0028. It is coated on both 
sides by “1/4 oz copper”. This translates to a 0.35 mil thickness of copper. Find w and up 
for a 50 Ω line. Then determine the αc, αd and αtot at three frequencies: 1, 10 and 20 
GHz. What is the maximum frequency of operation for this microstrip? 
Using ML0605: 
Microstrip Design 
width & thickness will be in the same units 
 
enter the desired impedance: 50 
enter the substrate thickness: 50 
enter substrate rel permittivity: 10.8 
w = 44.3241 
eeff = 7.1852 
up = 111844030.4942m/s 
>> 
Now convert w and h and t to metric (1.125x10-3m, 1.27x10-3m, and 8.89x10-6m, 
respectively), and run MLP0640 at each frequency. For instance, at 1 GHz we have: 
Microstrip Analysis 
 
6- !32
enter the line width (m): 1.125e-3 
enter the substrate thickness(m): 1.27e-3 
enter substrate rel permittivity: 10.8 
enter conductor thickness (m): 8.89e-6 
enter conductor conductivity (S/m): 5.8e7 
enter conductor relative permeability: 1 
enter dielectric loss tangent: 0.0028 
enter frequency (Hz): 1e9 
eeff = 7.1847 
up = 111847474.32m/s 
Zo = 49.8031ohms 
ac = 1.2975dB/m 
ad = 0.64805dB/m 
>> 
Tabulating the results for each frequency: 
The maximum frequency is 
! 
P6.46: A 1.5 inch length of microstrip line of width 48.86 mils sits atop a 50 mil thick 
substrate with dielectric constant 4. Determine the impedance looking into this circuit at 
2 GHz if it is terminated in a 300 Ω resistor. Assume ideal conductors and lossless 
dielectric. 
From ML0604 we find: 
Microstrip Analysis 
enter width & thickness in the same units 
 
enter the line width: 48.86 
enter the substrate thickness: 50 
enter substrate rel permittivity: 4 
α in dB/m 1 GHz 10 GHz 20 GHz
αc= 1.3 4.0 5.7
αd= 0.65 6.5 13.0
αtot= 2.0 10.5 18.7
( )
8
max 6
3 10 1
18
25.2 104 4 50 10.8r
c x m s milf GHz
x mh milsε −
= = =
6- !33
eeff = 2.9116 
up = 175697087.6994m/s 
Zo = 74.9641ohms 
>> 
So we have Zo = 75 Ω. Also, we find 
! 
We can also calculate β: ! 
Now either a Smith Chart or the Zinput equation may be used to evaluate Zin. From the 
Zinput function we defined earlier in MATLAB, 
>> Zo=75; 
>> ZL=300; 
>> G=j*71.5; 
>> L=0.0381; 
>> Zinput(Zo,ZL,G,L) 
ans = 8.6567e+001 +1.2031e+002i 
>> 
So Zin = 87 + j120 Ω. 
A Smith Chart approach is more approximate, yielding Zin 90 + j120 Ω. 
P6.47: The top-down view of a microstrip circuit is shown in Figure 6.54. If the 
microstrip is supported by a 40 mil thick alumina substrate, (a) determine the line width 
required to achieve a 50 Ω impedance line. (b) What is the guide wavelength on this 
microstrip line at 2 GHz? (c) Suppose at this frequency the load impedance is ZL = 150 - 
j100 Ω. Determine the length of the stubs (dthru and lstub) required to impedance match the 
load to the line. 
We first apply ML0605: 
Microstrip Design 
width & thickness will be in the same units 
 
enter the desired impedance: 50 
enter the substrate thickness: 40 
( ) 0.02541.5 0.0381 , 0.08785 ,
0.0381
so 0.434
0.08785
p
G
G
G
umin m m
in f
m
m
λ
λ
λ
⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
= =
l
l
171.5 .
pu m
ω
β = =
≈
6- !34
enter substrate rel permittivity: 9.9 
w = 38.6273 
eeff = 6.6644 
up = 116131354.2486m/s 
>> 
From this we have (a) w = 38.6 mils. 
Also, 
(b) ! 
(c) Now we use a Smith Chart to determine 
the open-ended shunt stub matching network. 
(1) Normalize the load (point a: zL = 3.0 - 
j2.0) 
(2) locate yL (point b) 
(3) Move to point c (0.180λ - 0.025λ = 0.155λ; or dthru = 0.155λ = 9 mm (354 mils)) 
(4) Move from yopen to 0.336λ, so lstub = 0.336λ = 19.5 mm (768 mils) 
P6.48: Suppose the microstrip circuit shown in Figure 6.54 is realized atop the RT/
Duroid© board of problem P6.45. Assuming the board material is lossless, (a) determine 
the line width required to achieve a 75 Ω impedance line. (b) Now suppose at 1.0 GHz 
the load impedance is ZL = 150 + j150 Ω. Find the length of the stubs (dthru and lstub) 
required to impedance match the load to the line. 
Applying ML0605: 
Microstrip Design 
width & thickness will be in the same units 
 
enter the desired impedance: 75 
enter the substrate thickness: 50 
enter substrate rel permittivity: 10.8 
w = 15.7817 
eeff = 6.6844 
up = 115957584.2884m/s 
>> 
So we have (a) w = 15.8 mils. 
We also have: ! 
(b) Now we employ a Smith Chart to 
determine the open-ended shunt stub 
8
9
1.161 10 0.058 2.29
2 10
p
G
u x
m in
f x
λ = = = =0.116pG
u
m
f
λ = =
Fig. P6.47
Fig. P6.48
6- !35
matching network. 
(1) Normalize the load (point a: zL = 2.0 + j2.0) 
(2) locate yL (point b) 
(3) Move to point c (0.178λ + 0.500λ - 0.459λ = 0.219λ; or dthru = 0.219λ = 2.54cm (1 
in)) 
(4) Move from yopen to 0.339λ, so lstub = 0.339λ = 3.93cm (1.55 in) 
7. Transients 
P6.49: Consider Figure 6.33 with the following values: Vs = 10 V, Zs = 30 Ω, Zo = 50 Ω, 
up = 0.666c, ZL = 150 Ω, l = 10 cm. Plot, out to 2 ns, (a) the voltage at the source end, (b) 
the voltage at the middle, and (c) the voltage at the load end of the T-Line. 
! 
! 
! 
 
P6.50: Repeat problem P6.49 for a 10 V pulse of duration 0.4 ns. 
 
P6.51: MATLAB: Consider a 12 cm long 50 Ω transmission line terminated in a 25 Ω 
load and having a matched source impedance (Zs = 50 Ω). Propagation velocity on the T-
8
0.1 0.5
0.666 3 10
m
t ns
mx
s
= =
⎛ ⎞
⎜ ⎟
⎝ ⎠
l
150 50 1 30 50 1; 
150 50 2 30 50 4L S
− −
Γ = = Γ = = −
+ +
5010 6.25
50 30o
V V= =
+
Fig. P6.49b
6- !36
Line is 0.67c. The source is a 0.4 ns square pulse of amplitude 6 V. Modify MATLAB 
Fig. P6.50a
Fig. P6.50b
6- !37
6.6 to plot v(z,t) at two points: z = 2 cm and z = 10 cm. 
% M-File: ML0651 
% modifies ML0606 
% replaces triangular pulse with rectangular pulse 
% 
% ML0606: 
% Analysis of a triangular pulse 
% (matched source impedance) traveling 
% down a T-Line and reflecting off a 
% resistive load. We want to be able to 
% trace the voltage at an arbitrary 
% point along the line. 
% 
% Wentworth, 4/25/03 
% Variables 
% Vo pulse height (V) 
% t1 pulse start (ns) 
% t2 pulse end (ns) 
% L line length (cm) 
% T transit time (ns) 
% z location to find pulse (cm) 
% tau time "location" to find pulse (ns) 
% up propagation velocity (m/s) 
% Zo,ZL line,load impedance (ohms) 
% N number of points 
% GL load reflection coefficient 
clc 
clear 
%enter variables 
Vo=6; 
t1=0; 
t2=.4; 
L=12; 
z=2; 
up=2e8; 
Zo=50; 
ZL=25; 
T=1e9*(L/up)/100; 
tau=1e9*(z/up)/100; 
N=500; 
GL=(ZL-Zo)/(ZL+Zo); 
%initialize array 
for i=1:N+1 
 v(i)=0; 
end 
dt=2*T/N; 
6- !38
%enter rectangular pulse function 
t=0:dt:2*T; 
vo=0.5*Vo*(step(t,t1)-step(t,t2)); 
%Note that “step” is a function defined 
%in ML0606. 
%Generate + wave data 
for i=1:N+1 
 ta=i*dt; 
 if ta>tau 
 j=ceil((ta-tau)/dt); 
 vplus(i)=vo(j); 
 end 
end 
%Generate - wave data 
for i=1:N+1 
 ta=i*dt; 
 tb=2*T-tau; 
 if ta>tb 
 j=ceil((ta-tb)/dt); 
 vmin(i)=GL*vo(j); 
 end 
end 
%Sum the data 
for i=1:N+1 
 v(i)=vplus(i)+vmin(i); 
end 
plot(t,v) 
xlabel('time (ns)') 
ylabel('voltage') 
AXIS([0 2*T -Vo Vo]) 
grid on 
P6.52: MATLAB: Modify MATLAB 6.6 to plot v(z,t) at z = 4.5 cm if the source pulse is 
as indicated in Figure 6.55. 
% M-File: MLP0652 
% Analysis of a triangular pulse 
% (matched source impedance) traveling 
% down a T-Line and reflecting off a 
% resistive load. We want to be able to 
% trace the voltage at an arbitrary 
% point along the line. 
% 
% Wentworth, 4/25/03 
% Variables 
Fig. P6.51 (plot at z = 2 cm)
6- !39
% Vo pulse height (V) 
% t1 pulse start (ns) 
% t2 pulse middle (ns) 
% t3 pulse end (ns) 
% L line length (cm) 
% T transit time (ns) 
% z location to find pulse (cm) 
% tau time "location" to find pulse (ns) 
% up propagation velocity (m/s) 
% Zo,ZL line,load impedance (ohms) 
% N number of points 
% GL load reflection coefficient 
clc 
clear 
%enter variables 
Vo=10; 
t1=0; 
t2=1; 
t3=2; 
L=6; 
z=4.5; 
up=3e7; 
Zo=50; 
ZL=0; 
T=1e9*(L/up)/100; 
tau=1e9*(z/up)/100; 
N=500; 
GL=(ZL-Zo)/(ZL+Zo); 
%initialize array 
for i=1:N+1 
 v(i)=0; 
end 
dt=2*T/N; 
%enter triangular pulse function 
m1=0.5*Vo/(t2-t1); 
b1=0.5*Vo-m1*t2; 
m2=0.5*Vo/(t2-t3); 
b2=0.5*Vo-m2*t2; 
for i=1:N+1 
 t(i)=i*dt; 
 if t(i)<t1 
 vo(i)=0; 
 end 
 if and(t(i)>t1,t(i)<=t2) 
 vo(i)=m1*t(i)+b1; 
 end 
 if and(t(i)>t2,t(i)<=t3) 
 vo(i)=m2*t(i)+b2; 
 end 
 if t(i)>t3 
6- !40
 vo(i)=0; 
 end 
end 
%Generate + wave data 
for i=1:N+1 
 ta=i*dt; 
 if ta>tau 
 j=ceil((ta-tau)/dt); 
 vplus(i)=vo(j); 
 end 
end 
%Generate - wave data 
for i=1:N+1 
 ta=i*dt; 
 tb=2*T-tau; 
 if ta>tb 
 j=ceil((ta-tb)/dt); 
 vmin(i)=GL*vo(j); 
 end 
end 
%Sum the data 
for i=1:N+1 
 v(i)=vplus(i)+vmin(i); 
end 
plot(t,v) 
xlabel('time (ns)') 
ylabel('voltage') 
AXIS([0 2*T -Vo Vo]) 
grid on 
P6.53: The expression for iL(t) and vL(t) of equations (6.106) and (6.107) were derived for 
a T-Line terminated in an inductor. Find similar expressions for a T-Line terminated in a 
capacitor. 
! 
Suppressing U(τ), we have 
! 
! 
This is in the form: 
( ) ( )o
( )
( ) , ( ) ( ), Z ( ) ( )i r i rLL L o o L o o
dv t
i t C v t V V U i t V V U
dt
τ τ= = + = −
( )2 ( ) ( ) ( )i Lo L o L L o
dv t
V v t Z i t v t Z C
dt
= + = +
2 2( ) ( )1 1
( ) , or ( ) .
i i
o oL L
L L
o o o o
V Vdv t dv t
v t v t
dt Z C Z C dt Z C Z C
−
+ = = +
Fig P6.52
6- !41
! 
solving the integral 
! 
we have 
! 
Now since at t = 0, v = 0, we have ! 
! 
Solving for v: ! 
! 
Now for iL(t), 
! 
! 
P6.54: For Figure 6.42, Zo = 100 Ω and up = 0.1c. Estimate L if the VL vs t is given in 
Figure 6.56. 
Using (6.107), ! here we have τ = t – 2 ns. Choosing the voltage 
at 2 ns we have ! Then, at approximately 2.2 ns we have 0.2 
V, so 
! 
Solving for L: 
! 
21, where , and 
i
o
o o
Vdv
Av B A B
dt Z C Z C
−
= + = =
dv dt
Av B
=
+∫ ∫
( )1 ln .Av B t C
A
+ = +
1 ln( ), and C B
A
=
1 ln .Av Bt
A B
+⎛ ⎞= ⎜ ⎟
⎝ ⎠
( )1 2 1o
t
Z CAt i
o
B
v e V e
A
−⎛ ⎞= − = − −⎜ ⎟
⎝ ⎠
( ) 2 1 ( ).o
t
Z Ci
L ov t V e U τ
−⎛ ⎞∴ = −⎜ ⎟
⎝ ⎠
2( ) 1( ) 2 o o
it t
Z C Z Ci oL
L o
o o
Vdv t
i t C C V e e
dt Z C Z
− −⎛ ⎞
= = =⎜ ⎟
⎝ ⎠
2( ) ( ).o
i t
Z Co
L
o
V
i t e U
Z
τ
−
∴ =
( ) 2 ( ),
oZ
Li
L ov t V e U
τ
τ
−⎛ ⎞
⎜ ⎟
⎝ ⎠=
0.9 2 , or 0.45 .i io oV V V V= =
( )100 0.2
200.20.2 2 , or .
0.45
ns
Li ns L
oV V e e
−⎛ ⎞
⎜ ⎟ −⎝ ⎠= =
( )
20 25 .
ln 0.2 0.45
nsL nH−= =
6- !42
P6.55: A 50 Ω T-Line with up = 0.5c is terminated in some load such that the TDR is 
given by Figure 6.57. Determine the location and the value of the load. 
From the figure we see the two-way travel time is 7 ns, so therefore one-way is 3.5 ns and 
we have 
! 
Also from the figure we can calculate Γ: 
! 
and then 
! 
So we have a 12.5 Ω terminating resistor located 0.525 m along the line. 
P6.56: The TDR plot for a 75 Ω T-Line with up = 0.2c is given in Figure 6.58. What type 
components terminate the line? Estimate the component values. 
By inspection we see it is a series combination of resistance and inductance. The location 
is: 
! 
The resistance is found by determining the reflection coefficient: 
! 
and then the resistance is: ! 
Only a very rough estimate can be found for the inductance. At τ = 10 ns – 10 ns = 0,. we 
have Then, at (very roughly) τ = 11 ns – 10 ns = 1ns,, vL(t) has dropped to 
1.75 V, so 
! 
So our rough estimate is an inductance between 500 and 600 nH. 
( )( ) 83.5 0.5 3 10 0.525 .ml ns x m
s
⎛ ⎞= =⎜ ⎟
⎝ ⎠
0.4 1 0.6
1
totV V
V
+
+
− −
Γ = = = −
1 0.650 12.5 .
1 0.6
R −= = Ω
+
( )( )( )9 81 10 10 sec 0.2 3 10 sec 0.302l x x m m
−= =
1.75 1 0.75,
1
−
Γ = =
1.7550 350 .
0.25
R = Ω = Ω
2 2 .ioV V=
( )
( )
75 1 751.75 2 , or 560 .
ln 1.75 2
ns L nse L nH− −= = =
6- !43
8. Dispersion 
P6.57: MATLAB: Use Fourier Series to construct a 5 V pulse of duration 5 ns that 
repeats every 10 ns. 
% M-File: MLP0657 
% 
% This program assembles a pulse using 
% Fourier series. It modifies ML0607. 
% 
% Wentworth, 8/3/02 
% 
% Variables: 
% N number of Fourier coefficients 
% aO avg value of the waveform (volts) 
% T period (s) 
% fo fundamental frequency (Hz) 
% wo fund angular freq (rad/s) 
% ttime (sec) 
% ftot fourier sum at a particular time(volts) 
clc %clears the command window 
clear %clears variables%MstripDesign 
% Initialize variables 
clear 
N=1000; 
a0=2.5; 
T=10e-9; 
fo=1/T; 
wo=2*pi*fo; 
% Evaluate Fourier Series Coefficients 
for n=1:N 
 a(n)=(10/(pi*n))*sin(n*pi/2); 
end 
% Generate data and plot 
for i=1:180 
 t(i)=i*T/90; 
 for n=1:N 
 f(n)=a(n)*cos(n*wo*t(i)); 
 end 
 ftot(i)=a0+sum(f); 
end 
plot(t,ftot) 
6- !44
xlabel('time(s)') 
ylabel('volts') 
grid on 
 
P6.58: MATLAB: Actual pulses have some slope to the leading and trailing edge. 
Suppose a symmetrical pulse is 5 V from –2 ns to +2ns, and has a linear slope to 0 V on 
each edge of duration 0.2 ns. The pulse repeats every 20 ns. Construct this pulse using 
Fourier Series for N = 10, 100 and 1000. Comment on how this pulse compares to the 
one of Figure 6.48. 
We must first find the Fourier coefficients: 
ao = 1.05 
! 
% M-File: MLP0658 
% 
% This program assembles a pulse using 
% Fourier series. It modifies ML0607. 
% 
% Wentworth, 8/3/02 
% 
( ) ( )( )
( )
( ) ( )( ) ( ) ( )( )2
10 50
sin 2.2sin 0.22 2.0sin 0.20
5
500 110
cos 0.22 cos 0.20 sin 0.22 sin 0.20
n
na n n
n n
n n n n
nn
π
π π
π π
π π π π
ππ
−⎛ ⎞= + −⎜ ⎟
⎝ ⎠
− − + −
Fig. P6.57
6- !45
% Variables: 
% N number of Fourier coefficients 
% aO avg value of the waveform (volts) 
% T period (ns) 
% fo fundamental frequency (Hz) 
% wo fund angular freq (rad/s) 
% t time (ns) 
% ftot fourier sum at a particular time(volts) 
clc %clears the command window 
clear %clears variables%MstripDesign 
% Initialize variables 
clear 
N=1000; 
a0=1.05; 
T=20; 
fo=1/T; 
wo=2*pi*fo; 
% Evaluate Fourier Series Coefficients 
for n=1:N 
 a1(n)=(10/(pi*n))*sin(n*pi/5); 
 a2a(n)=(-50/(pi*n))*(2.2*sin(.22*pi*n)-2.0*sin(.
20*pi*n)); 
 a2b(n)=(-500/((pi*n)^2))*(cos(.22*pi*n)-cos(.20*pi*n)); 
 a3(n)=(110/(pi*n))*(sin(.22*pi*n)-sin(.2*pi*n)); 
 a(n)=a1(n)+a2a(n)+a2b(n)+a3(n); 
end 
% Generate data and plot 
for i=1:180 
 t(i)=i*T/90; 
 for n=1:N 
 f(n)=a(n)*cos(n*wo*t(i)); 
 end 
 ftot(i)=a0+sum(f); 
end 
plot(t,ftot) 
xlabel('time(ns)') 
ylabel('volts') 
grid on 
save 'Fourier1000' t ftot -ascii 
6- !46
 
To see the difference between the N = 100 and N = 1000 cases requires looking at the 
plot over a reduced portion of time, as shown in Figure P6.59b. 
P6.59: MATLAB: A material has a constant εr = 4 from DC up to 20 GHz. Then 
Fig. P6.58a (the N=100 and N=100 cases are indistinguishable) 
Fig. P6.58b
6- !47
 
! 
for 20 GHz < f < 50 GHz. Show the pulse from problem P6.58 after it has traveled along 
a coaxial T-Line with this dielectric. 
% M-File: MLP0659 
% 
% This program modifies ML0607. 
% Wentworth, 2/2/03 
clc %clears the command window 
clear %clears variables%MstripDesign 
% Initialize variables 
clear 
N=1000; 
a0=1.2; 
T=20e-9; 
fo=1/T; 
wo=2*pi*fo; 
z=10; 
%evaluate Fourier Series Coefficients 
for n=1:N 
 a(n)=(12/(pi*n))*sin(n*pi/5); 
end 
%Generate data 
for i=1:180 
 t(i)=i*T/90; 
 for n=1:N 
 f(n)=n*50e6; 
 er(n)=4; 
 if f(n)>20e9 
 er(n)=4*cos((f(n)-20e9)/60e9); 
 end 
 beta(n)=2*pi*f(n)*sqrt(er(n))/3e8; 
 V(n)=a(n)*cos(n*wo*t(i)-beta(n)*z); 
 end 
 Vtot(i)=a0+sum(V); 
end 
plot(t,Vtot) 
xlabel('time(ns)') 
9
9
20 10
4cos ,
60 10r
f x
x
ε
⎛ ⎞−
= ⎜ ⎟
⎝ ⎠
6- !48
ylabel('volts') 
grid on 
save 'dispoff' t Vtot -ascii 
 
Fig. P6.59