Smith Chart
5. Locate on a Smith Chart the following load impedances terminating a 50  T-Line.
(a) ZL = 200  , (b) ZL = j25 , (c) ZL = 50 + j50 , and (d) ZL = 25 – j200 .
Fig. P6.21
6. A source with 50  source impedance drives a 50  T-Line 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.
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  L  0.245e j 76 . 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 .
Fig. P6.22a
Fig. P6.22b
7. 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  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+0.1940.400 = 0.294. For the series capacitance we have
j
 j 2.4 
,
CZ o
or C 
2 1x10
Fig. P6.32a
1
9
  50  2.4 
 1.33 pF
Fig. P6.32b
8. You would like to match a 170  load to a 50  T-Line. (a) Determine the
characteristic impedance required for a quarter-wave transformer. (b) What throughline length and stub length are required for a shorted shunt stub matching network?
(a) Z s  Zo RL  92
(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 = 1-jb. This would have resulted
in d = 0.329 and l = 0.396.
Fig. P6.33a
Fig. P6.33b
9. A load impedance ZL = 200 + j160  is to 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 = Fig. P6.34
0.104.)
P6.35: Repeat P6.34 for an open-ended 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=1-jb 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)
Fig. P6.35a
Fig. P6.35b
10. 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)
(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.)
Fig. P6.36
Microstrip Lines
P6.42: A 100  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
w = 0.0666 cm = 0.67 mm.
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  impedance line. (b) Suppose at this frequency the
load impedance is ZL = 150 - j100 . Determine the length of the stubs (dthru and
lstub) required to impedance match the load to the line.
(a) w = 38.6 mils.
Also,
(b) 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 (0.180 - 0.025 = 0.155; or dthru = 0.155 = 9 mm (354 mils))
(4) Move from yopen to 0.336, so lstub = 0.336 = 19.5 mm (768 mils)
Waveguides
11. 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

 fc 
 6.56 
8 m
1     3x108 1  
  2.26 x10
s
 10 
 f 
U
2
3x108 10 x109

 0.0397m,   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.
U
120
Z10TE 

 500
2
2
 fc 
 6.56 
1 
1  

 10 
 f 
12. 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
3x108 m s 1  1in 
fc10 


  403MHz
2 r a
2 2.26 9.75in  0.0254m 
2
 fc 
 403 
F  1    1 
  0.741
 600 
 f 
Now,
c
3x108
m
uU 

 2 x108
s
r
2.26
uP 
2
uU
m
 2.7 x108
F
s
uG  uU F  1.48 x108
m
s
13. 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
  y   j z
H zs  H o cos 
e
 b 
j 
  y   j z
Exs  2
H o sin 
e
2
u   b
 b 
H ys 
j 
  y   j z
H o sin 
e
2
  b
 b 
2
u
14. Find an expression for the magnetic field of the TE11 mode.
H

 j 
x 
 y 
H o sin 
 cos 
 cos(t   z )a x
2
2
u   a
 a 
 b 
j 
x   y 
H o cos 
 sin 
 cos(t   z )a y
2
  b
 a   b 
2
u
x 
 y 
 H o cos 
 cos 
 cos(t   z )a z
 a 
 b 
Resonator
Directional Coupler