Tutorial 8
Problem 1 A sinusoidal voltage source drives the series combination of an impedance,
Z g  50  j50 , and a lossless transmission line of length L, shorted at the load end. The line
characteristic impedance is 50 Ω, and wavelength λ is measured on the line. (a) Determine, in terms of
wavelength, the shortest line length that will result in the voltage source driving a total impedance of
50 Ω. (b) Will other line lengths meet the requirements of part (a)? If so, what are they?
Solution:
Ztot  Z g  Zin  50  50 j  Zin  50  Zin  50 j
Z L  0, Z 0  50 
Zin  Z 0
2
l 
lmin 


Z L cos(  l )  jZ 0 sin(  l )
 j50 tan(  l )  j50  tan(  l )  1
Z 0 cos(  l )  jZ L sin(  l )
l

4
 n , n  0,1,2
l 


 n , n  0,1,2
8
2
8
Zg=50-j50Ω
Zg
Vs
ZL
Z0=50Ω
Vs
Zin
Zin
z=-L
z=0
z
Problem 2 Two lossless transmission lines having different characteristic impedances are to be joined
end to end. The impedances are Z01=100Ω and Z03=25Ω. The operating frequency is 1 GHz.
(a) Find the required characteristic impedance, Z02, of a quarter-wave section to be inserted between
the two, which will impedance-match the joint, thus allowing total power transmission through the
three lines.
(b) The capacitance per unit length of the intermediate line is found to be 100pF/m. Find the shortest
length in meters of this line that is needed to satisfy the impedance-matching condition.
(c) With the three-segment setup as found in parts (a) and (b), the frequency is now doubled to 2 GHz.
Find the input impedance at the line-1-to-line-2 junction, seen by waves incident from line 1.
(d) Under the conditions of part (c) , and with power incident from line 1, evaluate the standing wave
ratio that will be measured in line 1, and power fraction propagates back to the line 1 input.
Solution:
(a)
Z01=100Ω
Z03=25Ω
Z01=100Ω
Zin=Z03=25Ω
Zin
Z01=100Ω Z02
Z03=25Ω
25Ω
Z01=100Ω Z02
Z01=100Ω
Zin
Zin
Zin
Total power transmission means the reflection at line-1-to-line-2 junction is 0, which is
or Zin=Z01. As the length of line 2 is quarter-wave, so
Zin  Z 02
(b)
Z 03 cos(  l )  jZ 02 sin(  l ) Z 022

 Z 01  Z 02  Z 01Z 03  50
Z 02 cos(  l )  jZ 03 sin(  l ) Z 03
   LC ,  


2
1

 0.2m
 LC f LC
25cos(  l )  j50sin(  l )
 m
 100  cos(  l )  0  l    , m  0,1,2
50cos(  l )  j 25sin(  l )
4 2
lmin 

 0.05m
4
f  2GHz   l   LC  l  2 f LCl  
Zin  50
(d)
2
Zin  Z01 
To make
(c)
2  

 4 2
L
 50  L  Z 022 C  0.25μH/m
C
Z 02 
50
l 

25cos(  l )  j50sin(  l )
 25
50cos(  l )  j 25sin(  l )
Z L  Z 0 Zin  Z 01
1 |  |

 0.6  s 
4
Z L  Z 0 Zin  Z 01
1 |  |
Preflected
Pincident
|  |2  0.36
12  0
Problem 3: Slotted line measurements yield a VSWR of 5, a 15-cm spacing between successive
voltage maxima, and the first maximum at a distance of 7.5cm in front of the load. Determine the load
impedance, assuming a 50-Ω impedance for the slotted line.
Solution:
Assume
 |  | e j
Vs ( z )  (V0e j z  V0e j z )  V0e j z (1 |  | e j (2  z  ) )

m
   , m  0,1,2
4
2

 m

    , m  0,1,2
4
4 2
Vmax  1 |  |, 2  z    2m  zmax  
Vmin |1 |  ||, 2  z      2m  zmin
s
Vmax 1 |  |
s 1 2

|  |

Vmin 1 |  |
s 1 3
zmax  zm,max  zm1,max 
z1,max  
2
=15cm   =30cm

  7.5cm    
4
 |  | e j  


2
3
Z L  Z0
 Z L  10
Z L  Z0