1. Distributed Parameters Model
P6.1: RG-223/U coax has an inner conductor radius a = 0.47 mm and inner radius of the
outer conductor b = 1.435 mm. The conductor is copper, and polyethylene is the dielectric.
Calculate the distributed parameters at 800 MHz.
for copper: σ Cu = 5.8 x107
S
m
for polyethylene: ε r = 2.26, σ = 10−16
R' =
1
=
2π
L' =
G'=
1
2π
S
m
⎛1 1⎞ π fμ
⎜ + ⎟
⎝ a b ⎠ σc
6
−7
Ω
1
1
⎛
⎞ π ( 800 x10 )( 4π x10 )
+
= 3.32
⎜
−3
−3 ⎟
7
m
1.435 x10 ⎠
⎝ 0.47 x10
( 5.8 x10 )
μ ⎛ b ⎞ 4π x10−7 ⎛ 1.435 ⎞
nH
ln ⎜ ⎟ =
ln ⎜
⎟ = 223
2π ⎝ a ⎠
2π
m
⎝ 0.47 ⎠
2π (10−16 )
2πσ
S
−18
ln ( b a )
=
ln (1.435 0.47 )
= 560 x10
m≈0
2π ( 2.26 ) ( 8.854 x10−12 )
pF
2πε
C'=
=
= 112
m
ln ( b a )
ln (1.435 0.47 )
P6.2: MATLAB: Modify MATLAB 6.1 to account for a magnetic conductive material.
Apply this program to problem P6.1 if the copper conductor is replaced with nickel.
S
and μr = 600.
m
Note that this program has also been modified for P6.04 as well.
for Nickel we have σ Ni = 1.5 x107
%Coax distributed parameters
%
%
Modified: P0602
%
add rel permeability
%
also modified for P0604
%
clear
clc
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'])
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:
R ' = 160 Ω , L ' = 223 nH , G ' = 560 x10−18 S , C ' = 112 pF
m
m
m
m
P6.3: Modify (6.3) to include internal inductance of the conductors. To simplify the
calculation, assume current is evenly distributed across the conductors. Find the new value of
L’ for the coax of Drill 6.1.
From Ampere’s Circuit Law we can find H versus ρ:
Iρ
for ρ ≤ a
2π a 2
I
Hφ =
for a ≤ ρ ≤ b
2πρ
Hφ =
c2 − ρ 2
for b ≤ ρ ≤ c
2πρ c 2 − b 2
H φ = 0 for ρ ≥ c
Hφ =
I
Using the energy approach, Wm =
1 2 μo
LI =
H 2 dv , we find
2
2 ∫
μ
μ ⎡⎛ c 2 ⎞ ⎛ c ⎞ ⎛ c 2 ⎞ 1 ⎛ c 2 + b 2 ⎞ ⎤
b μ
L ' = o ln + o + o ⎢⎜ 2 2 ⎟ ln ⎜ ⎟ − ⎜ 2 2 ⎟ + ⎜ 2 2 ⎟ ⎥
2π a 8π 2π ⎢⎝ c − b ⎠ ⎝ b ⎠ ⎝ c − b ⎠ 4 ⎝ c − b ⎠ ⎥
⎣
⎦
Inserting the given values we find
nH
nH
L ' = ( 237 + 50 + 41.2 )
= 328
m
m
With two significant digits we therefore have L’ = 330 nH/m.
2
2. Time Harmonic Waves on Transmission Line
P6.4: MATLAB: Modify MATLAB 6.1 to also calculate γ, α, β and Zo.
program using Drill 6.2.
See the solution for P6.2.
Calc Coax Distributed Parameters
inner radius, in mm, = 0.45
Confirm the
outer radius, in mm, = 1.47
relative permittivity, er= 2.26
dielectric conductivity, in S/m, = 1e-16
conductor conductivity, in S/m, = 5.8e7
conductor rel. permeability, = 1
input frequency, in Hz, = 1e9
G/h = 5.3078e-016 S/m
C/h = 1.0606e-010 F/m
L/h = 2.3675e-007 H/m
R/h = 3.8112 ohm/m
Gamma= 0.0403332+31.4857i /m
alpha= 0.040333Np/m
beta= 31.4857rad/m
Zo = 47.246-0.0605221i ohms
loss=0.9225 /m
lossdb=0.35033 dB/m
>>
This agrees with the results of Drill 6.2.
P6.5: The impedance and propagation constant at 100 MHz for a T-Line are determined to be
Zo = 18.6 – j0.253 Ω and γ = 0.0638 + j4.68 /m. Calculate the distributed parameters.
R '+ jω L '
, γ = ( R '+ jω L ')( G '+ jωC ')
G '+ jωC '
Z oγ = R '+ jω L ' = 2.37 + j87.0
Zo =
∴ R ' = 2.37
γ
Zo
Ω
nH
, ω L ' = 87.0 so L ' = 139
m
m
= G '+ jωC ' = 7.63 x10−6 + j 0.252,
∴ G ' = 7.63
μS
m
, and ωC ' = 0.252 so C ' = 401
pF
m
P6.6: The specifications for RG-214 coaxial cable are as follows:
• 2.21 mm diameter copper inner conductor
• 7.24 mm inner diameter of outer conductor
• 9.14 mm outer diameter of outer conductor
• Teflon dielectric (εr = 2.10)
Calculate the characteristic impedance and the propagation velocity for this cable.
Zo =
60
⎛b⎞
⎛ 3.62 ⎞
ln ⎜ ⎟ =
ln ⎜
⎟ = 49.1Ω
2.1 ⎝ 1.105 ⎠
εr ⎝ a ⎠
60
up =
c
εr
= 2.07 x108
m
s
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?
up = λ f , λ =
up
f
=
2.07 x108
= 0.207m
1x109
⎛ λ ⎞ ⎛ 1m ⎞
⎟⎜
⎟ = 2.4λ
⎝ 0.207m ⎠ ⎝ 100cm ⎠
l (λ ) = ( 50cm ) ⎜
P6.8: If 1 watt of power is inserted into a coaxial cable, and 1 microwatt of power is
measured 100 m down the line, what is the line’s attenuation in dB/m?
⎛ 1μW ⎞
A = −10 log ⎜
⎟ = +60dB
⎝ 1W ⎠
60dB
dB
= 0.6
A' =
100m
m
P6.9: Starting with a 1 mm diameter solid copper wire, you are to design a 75 Ω 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.
((
) )
((
⎛b⎞
ln ⎜ ⎟ , b=a exp Zo ε r 60 = ( 0.5mm ) exp 75 5.4
εr ⎝ a ⎠
So the inner diameter of the outer conductor is 18 mm.
c
2.998 x108
m
m
up =
=
= 1.29 x108 , so u p = 1.3 x108
s
s
5.4
εr
Zo =
60
) 60) = 9.1mm
To calculate α, will need γ. Therefore we calculate R’, L’, G’ and C’.
6
−7
1
1
mΩ
⎛
⎞ π (1x10 )( 4π x10 )
+
= 87.6
⎜
−3
−3 ⎟
7
m
9.1x10 ⎠
5.8 x10
⎝ 0.5 x10
4π x10−7 ⎛ 9.1 ⎞
nH
L' =
ln ⎜
⎟ = 580
2π
m
⎝ 0.5 ⎠
−15
2π (10 )
S
= 2.17 x10−15
G'=
ln ( 9.1 0.5 )
m
1
R' =
2π
C'=
2π ( 5.4 ) ( 8.854 x10−12 )
(
ln 9.1
0.5
Now, with ω = 2πf,
)
= 103.5
pF
m
1
m
Np
8.686
dB
dB
⎛
⎞
= 5.1x10−3
Finally, α = ⎜ 585 x10−6
⎟
m ⎠ Np
m
⎝
This is confirmed using MLP0602.
γ=
( R '+ jω L ')( G '+ jωC ') = 585 x10−6 + j 0.049
P6.10: MATLAB: A coaxial cable has a solid copper inner conductor of radius a = 1mm and
a copper outer conductor of inner radius b. The outer conductor is much thicker than a skin
depth. The dielectric has 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
Fi P6 10
3. Terminated T-Lines
P6.11: Start with equation (6.54) and derive (6.55).
Z in =
Vo+ e +γ l + Vo− e −γ l
Zo
Vo+ e + γ l − Vo− e−γ l
With Vo− = Γ LVo+ , we then have
Z in
(e
=
(e
+γ l
+γ l
+ Γ L e −γ l )
− Γ L e −γ l )
Zo
We also know that
Z − Zo
ΓL = L
,
Z L + Zo
So now we have
⎛ Z − Z o ⎞ −γ l
e+γ l + ⎜ L
⎟e
Z L + Zo ⎠
Z L + Z o ) e + γ l + ( Z L − Z o ) e −γ l
(
⎝
Z in =
Zo =
Zo
Z L + Z o ) e +γ l − ( Z L − Z o ) e −γ l
⎛ Z L − Z o ⎞ −γ l
(
+γ l
e −⎜
⎟e
⎝ Z L + Zo ⎠
and with rearranging,
Z L ( e + γ l + e −γ l ) + Z o ( e +γ l − e −γ l )
Z in =
Zo .
Z L ( e + γ l − e −γ l ) + Z o ( e +γ l + e −γ l )
We can convert the exponential terms into hyperbolic functions, given
1
1
sinh(x)
sinh( x) = ( e x − e− x ) , cosh( x) = ( e x + e− x ) , and tanh(x)=
.
2
2
cosh(x)
This leads to
2 Z cosh ( γ l ) + 2 Z o sinh ( γ l )
Z in = Z o L
,
2 Z L sinh ( γ l ) + 2Z o cosh ( γ l )
or finally
Z + Z o tanh ( γ l )
Z in = Z o L
.
Z o + Z L tanh ( γ l )
P6.12: Derive (6.56) from (6.55) for a lossless line.
Z L + Z o tanh ( γ l )
, and tanh ( γ l ) = tanh (α l + j β l ) = tanh ( j β l ) since α = 0 for
Z o + Z L tanh ( γ l )
lossless line. Using the hyperbolic definitions, we have
+ jβ l
− e− jβ l )
sinh ( j β l ) ( e
tanh ( j β l ) =
.
=
cosh ( j β l ) ( e+ j β l + e− jβ l )
Z in = Z o
Now using Euler’s formula,
cos ( β l ) + j sin( β l ) - cos ( − β l ) − j sin(− β l ) j 2sin ( β l )
tanh ( j β l ) =
=
= j tan( β l )
cos ( β l ) + j sin( β l ) + cos ( − β l ) + j sin( β l )
2 cos ( β l )
Plugging this in, we find,
Z in = Z o
Z L + jZ o tan ( β l )
.
Z o + jZ L tan ( β l )
P6.13: A 2.4 GHz signal is launched on a 1.5 m length of T-Line terminated in a matched
load. It takes 6.25 ns to reach the load and suffers 1.2 dB of loss. Find the propagation
constant.
γ = α + jβ
1.2dB 1Np
Np
= 0.092
1.5m 8.686dB
m
ω l
1.5m
m
β : up = = =
= 2.4 x108
s
β t 6.25ns
α=
β=
ω
up
=
2π ( 2.4 x109 )
2.4 x10
8
= 62.8
rad
m
So
γ = 0.092 + j 62.8
1
m
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.
D
Z L − Z o 50 − j 25 − 50
=
= 0.242e − j 76
Z L + Z o 50 − j 25 + 50
1 + ΓL
VSWR =
= 1.64
1− ΓL
ΓL =
2π λ π
⎛π ⎞
= , tan ⎜ ⎟ = 1
λ 8 4
⎝4⎠
Z + jZ o tan ( β l )
Z in = Z o L
Z o + jZ L tan ( β l )
βl =
50 − j 25 + j 50
50 + j 50 + 25
= 30.8 − j 3.8 Ω
= 50
Fig. P6.14
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.
Z L = 25 + j 2π ( 200 x106 )(10−9 ) = 25 + j1.257 Ω
Now, MLP0615 is used to solve the problem.
%
MLP0615
%
%
calculate gamma and char impedance
%
given the distributed parameters
%
Then, calculate gammaL and Zin
%
%
define variables
clc
clear
R=0.1;
L=1.0e-6;
G=10e-6;
C=1.0e-9;
f=200e6;
w=2*pi*f;
length=1;
ZL=25+j*1.257;
%
Perform calcuations
A=R+i*w*L;
B=G+i*w*C;
gamma=sqrt(A*B) %Propagation Constant
Zo=sqrt(A/B)
gammaL=(ZL-Zo)/(ZL+Zo)
%Reflection coefficient
TGL=tanh(gamma*length);
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL))
Running the program,
Gamma = 0.0017 +39.7384i
Zo = 31.6228 - 0.0011i
gammaL = -0.1164 + 0.0248i
Zin = 34.0192 - 7.4618i
>>
So the answers are, with the appropriate significant digits,
D
Γ L = 0.12e j168 and Z in = 34 − j 7.5 Ω
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.
Z L − Zo
1 + ΓL
, we find Z L = Z o
= 150 + j 30 Ω .
Z L + Zo
1 − ΓL
This is a resistor in series with an inductor. The inductor is found by considering
30
jω L = j 30, or L =
= 4.8nH ,
2π (1x109 )
Rearranging Γ L =
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.
Z L + jZ o tan ( β l )
Z − jZ o tan ( β l
, we find Z L = Z o in
Z o + jZ L tan ( β l )
Z o − jZ in tan ( β l
Rearranging Z in = Z o
β=
ω
0.7c
=
2π ( 2 x109 )
0.7 ( 3 x10
8
)
= 59.84
rad
;
m
)
)
rad ⎞
⎛⎛
⎞
tan ( β l ) = tan ⎜ ⎜ 59.84
⎟ ( 0.3m ) ⎟ = −1.254
m ⎠
⎝⎝
⎠
Evaluating, we have
Z L = 50 + j 0.016 Ω = 50 + j 2π ( 2 x109 ) L, or L = 1.3 pH.
This is a very small inductance, so we have Z L ≈ 50 Ω.
P6.18: For the lossless T-Line circuit shown in Figure 6.51, determine the input impedance
Zin and the instantaneous voltage at the load end vL.
25 − 50
1
2π λ
= − , βl =
= π , tan π = 0
25 + 50
3
λ 2
Z +0
Z in = Z o L
= Z L = 25Ω
ZL + 0
25
8V = 2V = Vo+ e − j β z + Vo− e + j β z
Vin =
25 + 75
2 = Vo+ ( e jβ l + Γ L e − j β l )
ΓL =
e jπ = cos π + j sin π = −1, e − jπ = −1,
1
⎛
⎞ −2
Vo+ ⎜ −1 − ( −1) ⎟ = Vo+ = 2; Vo+ = −3V
3
⎝
⎠ 3
⎛ 1⎞
VL = Vo+ (1 + Γ L ) = −3 ⎜1 − ⎟ = −2V , so vL = 2 cos (ωt + 180D ) V
⎝ 3⎠
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.
up =
ω
ω
rad
,β=
= 15.7
, β l = 4.71, tan β l = 418.6
β
up
m
Z in = 75
100 + j125 + j 75 ( 418.6 )
75 + j (100 + j125 )( 418.6 )
= 22 − j 28 Ω
Referring to Fig P6.19,
D
Z in
Vin = 6
= 2.1e − j 36 V
Z in + 75
∴ vin = 2.1cos (ωt − 36D )V
Fig P6 19
D
Z − Zo
ΓL = L
= 0.593e j 43
Z L + Zo
Vin = Vo+ ( e + jβ l + Γ L e − jβ l ) = 0.70e − j126 Vo+ = 2.1e − j 36 V
D
D
D
+
o
V =
2.1e − j 36
D
D
= 3e j 90 V
0.70e − j126
D
VL = Vo+ (1 + Γ L ) = 4.47e j105.8 V
vL = 4.5cos (ωt + 106D ) V
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;
B=G+i*w*C;
gamma=sqrt(A*B)
Zo=sqrt(A/B)
gammaL=(ZL-Zo)/(ZL+Zo)
TGL=tanh(gamma*length);
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL))
Running the program, gamma = 0.2236 + 0.2810i
Zo = 22.3607
gammaL = 0.4479 - 0.1908i
Zin = 27.2079 -15.4134i
>>
Vin = VSS
D
Z in
= 3.97e − j18.2 V , ∴ vin = 4.0 cos (ωt − 18.2D ) V
Z in + Z S
(
Vin = Vo+ ( eγ l + Γ L e −γ l ) = Vo+ (1.504 + j 0.101) = Vo+ 1.507e j 3.84
+
o
so V =
3.97e − j18.2
D
)
D
D
D
= 2.63e − j 22
1.507e j 3.84
D
VL = Vo+ (1 + Γ L ) = 3.85e − j 29.6 ,
∴ vL = 3.9 cos (ωt − 30D ) V
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.
Fig P6 21
P6.22: Repeat problem P6.14 using the Smith Chart.
First we locate the normalized load, zL = 1 – j0.5 (point a). By inspection of the Smith Chart,
D
we see that this point corresponds to Γ L = 0.245e j −76 . Also, after drawing the constant Γ
circle we can see VSWR = 1.66. Finally, we move from point a, at 0.356λ on 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.
Fi P6 22
Fig. P6.22b 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.
After normalizing ZL and locating it on the
D
chart (point a), we see Γ L = 0.80e j 95 .
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.
–
Fig. P6.23 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
D
After
corresponds to: Γ L = 0.34e j121 .
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 = 0.96j0.72, or Zin = 96-j72 Ω. Finally, we move
from point a towards the generator at point
to reach the voltage maximum, a distance
0.168λ.
c
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.
P6.27: Figure 6.52 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 locations of minima on the shorted line we find λ:
λ = 2 ( 7.55cm − 1.25cm ) = 12.6cm
(a) f =
c
λ
= 2.4GHz
(b) From the voltage maxima and voltage
minimum on the loaded line, we have
4
VSWR = = 2
2
Using VSWR=2 we draw the constant |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
⎛ 1.9cm − 1.25cm ⎞
⎜
⎟ = 0.0516λ
⎝ 12.6cm λ ⎠
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.
Fig P6 27
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:
λ = 2 ( 9.3cm − 1.7cm ) = 15cm
(a) f =
c
λ
= 2.0GHz
(b) From the load line,
10
VSWR = = 2.5
4
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
⎛ 7.9cm − 5.5cm ⎞
⎜
⎟ = 0.16λ
⎝ 15cm λ
⎠
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
Γ L circle, given Zo and ZL. Demonstrate your program with the values from Drill 6.11.
Add this to the end of the Matlab 6.3 program:
%now add constant gamma circles
ZL=50;
fudge=0.001+i*0.001;
newZL=ZL+fudge;
Zo=50;
zL=newZL/Zo;
gamma=(zL-1)/(zL+1);
plot(gamma,'-o');
constgamma(zL);
You must change the value of ZL for each load point. Notice the addition of a ‘fudge factor’.
This ensures that gamma has both a nonzero and finite real and imaginary part to work with
in the plot.
You’ll also need to add an additional function:
function [h]=constgamma(zL)
%constgamma(zL) draws the constant gamma circle;
phi=1:1:360;
theta=phi*pi/180;
a=abs((zL-1)/(zL+1));
Re=a*cos(theta);
Im=a*sin(theta);
z=Re+i*Im;
h=plot(z,'--k');
axis('equal')
axis('off')
The program is run for each point of
Drill 6.11 by changing the ZL value.
Since the MATLAB routine has the
‘hold on’, each new point is added to
the plot.
Fig. P6.30
5. Impedance Matching
P6.31: A matching network, using a reactive element in series with a length d of T-Line, is to
be used to match a 35 – j50 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 31a
−j
= − j1.4,
ωCZ o
C=
2π (1x10
1
9
) (100 )(1.4 )
= 1.14 pF
Fig P6 31b
(b) Now we need to move to point c, at z = 1-j1.4, so that an inductive element of value jx =
+j1.4 can be added. Moving to this point c gives d = 0.500 λ + 0.327 λ – 0.419 λ = 0.408 λ.
The inductance is
(1.4 )(100 ) = 22.3nH
jω L
= j1.4, L =
Zo
2π (1x109 )
P6.32: A matching network consists of a length of T-Line in series with a capacitor.
Determine the length (in wavelengths) required of the T-Line section and the capacitor value
needed (at 1.0 GHz) to match a 10 – j35 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
−j
− j 2.4 =
,
ωCZ o
or C =
2π (1x10
1
9
) ( 50 )( 2.4 )
= 1.33 pF
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?
(a) Z s = Z o RL = 92Ω
(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
d = 0.329 λ and l = 0.396 λ.
in
Ω
is
to
P6.34: A load impedance ZL = 200 + j160
Fig P6 32b
Fig. P6.33a Fig P6 33b
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
= 0 – jb. (l = 0.354 λ – 0.250 λ = 0.104 λ.)
P6.35: Repeat P6.34 for an open-ended
shunt stub tuner.
y
(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.)
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)
Fig P6 37
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).
(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:
D
Γ = 0.28e j 34
Fig. P6.38a Fig. P6.38b
1. Rectangular Waveguide Fundamentals
P7.1: Find the cutoff frequency for the first 8 modes of WR430.
a = 4.3 in = 0.1092 m, b = 2.15 in = 0.0546 m
For air-filled guide we have:
2
2
c ⎛m⎞ ⎛n⎞
fcmn =
⎜ ⎟ +⎜ ⎟
2 ⎝ a ⎠ ⎝b⎠
Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find
Mode
fcmn (GHz)
TE10
1.374
2.747
TE01
2.747
TE20
3.07
TE11
3.07
TM11
3.885
TE21
3.885
TM21
4.121
TE30
P7.2: Calculate the cutoff frequency for the first 8 modes of a waveguide that has a = 0.900
inches and b = 0.600 inches.
a = 0.900 in = 0.02286 m, b = 0.600in = 0.01524 m
For air-filled guide we have:
2
2
c ⎛m⎞ ⎛n⎞
fcmn =
⎜ ⎟ +⎜ ⎟
2 ⎝ a ⎠ ⎝b⎠
Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find
Mode
fcmn (GHz)
TE10
6.56
9.84
TE01
11.83
TE11
11.83
TM11
13.12
TE20
16.40
TE21
19.69
TM30
19.69
TE02
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 air-filled guide we have:
2
2
c ⎛m⎞ ⎛n⎞
fcmn =
⎜ ⎟ +⎜ ⎟
2 ⎝ a ⎠ ⎝b⎠
Evaluating all the combination of modes for m = 0,1,2,3 and n = 0,1,2,3 we find
Mode
fcmn (GHz)
TE10
6.56
13.12
TE20
19.68
TE30
19.68
TE01
20.75
TE11
20.75
TM11
23.66
TE21
23.66
TM21
P7.4: Calculate uG, the wavelength in the guide and the wave impedance at 10 GHz for
WR90.
From Table 7.1 for WR90 we have fc10 = 6.56 GHz. So
2
uG = uU
λ=
⎛ fc ⎞
⎛ 6.56 ⎞
8 m
1 − ⎜ ⎟ = 3x108 1 − ⎜
⎟ = 2.26 x10
s
⎝ 10 ⎠
⎝ f ⎠
λU
2
3 x108 10 x109
=
= 0.0397m, λ = 4cm
2
2
⎛ fc ⎞
⎛ 6.56 ⎞
1− ⎜
1− ⎜ ⎟
⎟
⎝ 10 ⎠
⎝ f ⎠
Since fc10 = 6.56 GHz, at 10 GHz only TE10 is present and therefore we only have the Z10TE
impedance.
ηU
120πΩ
Z10TE =
=
= 500Ω
2
2
⎛ fc ⎞
⎛ 6.56 ⎞
1− ⎜
1− ⎜ ⎟
⎟
⎝ 10 ⎠
⎝ f ⎠
P7.5: Consider WR975 is filled with polyethylene. Find (a) uu, (b) up and (c) uG at 600 MHz.
From Table 7.1 for WR975 we have a = 9.75 in and b = 4.875 in. Then
c
3 x108 m s 1 ⎛ 1in ⎞
fc10 =
=
⎜
⎟ = 403MHz
2 εr a
2 2.26 9.75in ⎝ 0.0254m ⎠
2
⎛ fc ⎞
⎛ 403 ⎞
F = 1− ⎜ ⎟ = 1− ⎜
⎟ = 0.741
⎝ 600 ⎠
⎝ f ⎠
Now,
2
3 x108
m
uU =
=
= 2 x108
s
2.26
εr
c
uP =
uU
m
= 2.7 x108
F
s
uG = uU F = 1.48 x108
m
s
P7.6: MATLAB: Plot up and wavelength in the guide as a function of frequency over the
cited useful frequency range for WR90.
%
MLP0706
%
%
Plot propagation velocity and guide wavelength
%
over the cited useful freq range of WR90.
%
%
2/2/03 Wentworth
%
c=3e8;
a=0.900;b=0.450;
fc=(c/(2*.0254*a));
flo=8.2e9;
fhi=12.4e9;
N=100;
df=(fhi-flo)/N;
f=flo:df:fhi;
A=sqrt(1-(fc./f).^2);
Lu=c./f;
LG=Lu./A;
up=c./A;
fG=f./1e9;
subplot(2,1,1)
Fi P7 6
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.
For the TE10 mode we have
c
fc10 =
, where a = 0.900 in = .02286m, so fc10 = 6.562GHz.
2a
Then
λu
c
3x108
= 1.88cm
λ=
, where λu = =
2
f 16 x109
fc
1−
f
( )
.0188m
λ=
(
1 − 6.562
16
uu
up =
( f)
1 − ( fc )
f
ηu
TE
Z mn
=
( f)
1 − fc
c
=
2
1 − fc
uG = uu
)
= 0.0206m
2
2
( f)
1 − fc
3x108
=
2
(
1 − 6.562
(
2
= 3x108 1 − 6.562
120πΩ
=
(
1 − 6.562
16
)
2
16 )
2
16
)
2
= 3.3x108
= 2.74 x108
m
s
m
s
= 413Ω
Likewise values are found for the TE20 and TE11 mode. For the TM11 mode, a different
expression for impedance is used:
( f)
TM
= ηu 1 − fc
Z mn
Mode
TE10
TE20
TE11
TM11
2
fc(GHz)
6.56
13.1
14.7
14.7
λ(m)
0.0206
0.0328
0.0470
0.0470
up(m/s)
3.3x108
5.2x108
7.5x108
7.5x108
uG(m/s)
2.7x108
1.7x108
1.2x108
1.2x108
Z(Ω)
413
659
945
150
P7.8: MATLAB: Modify MATLAB 7.1 by plotting uG and up versus frequency for the same
guide over the same frequency range.
% M-File: MLP0708
%
% Waveguide Velocity Plot
% Plots uG and uP for TE11
and TM11
% vs freq. for air-filled
waveguide
%
(modifies ML0701)
%
% Wentworth, 11/26/02
Fig. P7.8
%
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=(fhi-flo)/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 T-Lines, we know that looking into a quarter guide-wavelength section of
waveguide terminated in a short circuit, the input impedance appears infinite. The cutoff
frequency for the TE10 mode is 4.29 GHz. Then,
λU
c f
3 x108 7 x109
λ=
=
=
= 0.0542m
2
2
2
1 − 4.29
1 − fc
1 − fc
7
f
f
( )
( )
(
)
So the quarter wave length is 0.0542m/4 = 0.0136 m. Therefore a distance 1.4 cm away from
the short circuit, the input impedance appears infinite.
2. Waveguide Field Equations
P7.11: Manipulate (7.41) to get (7.1).
Rearranging (7.41), we have
⎛ mπ ⎞ ⎛ nπ ⎞
β −β =⎜
⎟ +⎜
⎟ ,
⎝ a ⎠ ⎝ b ⎠
Also from (7.41) we have
2
2
u
2
2
( ).
β = βu 1 − fc f
So
2
( )
⎛
β − β =, βu ⎜1-1+ fc f
⎝
and
2
u
2
2
2
( )
⎞
2 fc
⎟ =βu
f
⎠
2
2
2
⎛ fc ⎞ ⎡⎛ mπ ⎞ ⎛ nπ ⎞ ⎤ 2
β ⎜ ⎟ = ⎢⎜
⎟ +⎜
⎟ ⎥π ,
f
a
b
⎝
⎠
⎝
⎠ ⎦⎥
⎝ ⎠ ⎣⎢
2
2
u
2
2
⎛ 2π ⎞ ⎛ fc ⎞ ⎡⎛ m ⎞ ⎛ n ⎞ ⎤ 2
⎜
⎟ ⎜ ⎟ = ⎢⎜ ⎟ + ⎜ ⎟ ⎥ π
⎝ λu ⎠ ⎝ f ⎠ ⎣⎢⎝ a ⎠ ⎝ b ⎠ ⎦⎥
2
2
2
2
2 fc
⎛m⎞ ⎛n⎞
= ⎜ ⎟ +⎜ ⎟
λu f
⎝ a ⎠ ⎝b⎠
Solving for fc, where we have uu = λuf,
2
2
1
1
⎛m⎞ ⎛n⎞
fc = uu ⎜ ⎟ + ⎜ ⎟ =
2
2 με
⎝ a ⎠ ⎝b⎠
2
⎛m⎞ ⎛n⎞
⎜ ⎟ +⎜ ⎟
⎝ a ⎠ ⎝b⎠
2
P7.12: Find expressions for the phasor field components of the TE01 mode.
With m = 0 and n = 1, the nonzero field components in equations (7.67) - (7.71)are
⎛ π y ⎞ − jβ z
H zs = H o cos ⎜
⎟e
⎝ b ⎠
Exs =
jωμ π
⎛ π y ⎞ − jβ z
H o sin ⎜
⎟e
2
β −β b
⎝ b ⎠
H ys =
jβ π
⎛ π y ⎞ − jβ z
H o sin ⎜
⎟e
2
β −β b
⎝ b ⎠
2
u
2
u