A 50 ohm transmission line 0.6 wavelengths long is terminated with a load of 50+j25 ohms. A 30 ohm resistor is connected 0.3 wavelengths from the load. Using the Smith chart, the following input impedance is found: 1) The load admittance and resistor admittance are found and their sum calculated to get the point B admittance on the Smith chart. 2) Point B is used to determine the input admittance Y-IN, which corresponds to the input impedance Z-IN of 1.9-j1.4 ohms. 3) Therefore, the input impedance is calculated to be 95-j70 ohms

1. Problem 2.63 A 50-Ω lossless line 0.6λ long is terminated in a load with
ZL = (50 + j25) Ω. At 0.3λ from the load, a resistor with resistance R = 30 Ω is
connected as shown in Fig. P2.63. Use the Smith chart to find Zin.
Zin ZL
ZL = (50 + j25) Ω
Z0 = 50 Ω Z0 = 50 Ω30 Ω
0.3λ 0.3λ
Figure P2.63: Circuit for Problem 2.63.

2. 0.1
0.1
0.1
0.2
0.2
0.2
0.3
0.3
0.3
0.4
0.4
0.4
0.50.5
0.5
0.60.6
0.6
0.70.7
0.7
0.80.8
0.80.90.9
0.9
1.01.01.0
1.21.2
1.2
1.41.4
1.4
1.61.6
1.6
1.81.8
1.8
2.02.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.01.0
20-20
30-30
40-40
50
-50
60
-60
70
-70
80
-80
90
-90
100
-100
110
-110
120
-120
130
-130
140-140
150-150
160-160
170-170
180±
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.19
0.19
0.2
0.2
0.21
0.21
0.22
0.220.23
0.23
0.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.45
0.45
0.46
0.46
0.47
0.470.48
0.48
0.49
0.49
0.0
0.0
ANGLEOFREFLECTIONCOEFFICIENTINDEGREES
—>WAVELENGTHSTOWARDGENERATOR—>
<—WAVELENGTHSTOWARDLOAD<—
INDUCTIVEREACTANCECOM
PON
EN
T
(+jX/Z
o), O
R
CAPACITIVE SUSCEPTANCE (+jB/Yo)
CAPACITIVEREACTANCECOM
PON
EN
T
(-j
X/Zo),O
R
IN
DUCTIVESUSCEPTANCE(-jB/Yo)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.300 λ
0.300 λ
Z-LOAD
Y-LOAD
A
B
Y-IN
Z-IN
Figure P2.63: (b) Solution of Problem 2.63.
Solution: Refer to Fig. P2.63(b). Since the 30-Ω resistor is in parallel with the input
impedance at that point, it is advantageous to convert all quantities to admittances.
zL =
ZL
Z0
=
(50+ j25) Ω
50 Ω
= 1+ j0.5
and is located at point Z-LOAD. The corresponding normalized load admittance is at
point Y-LOAD, which is at 0.394λ on the WTG scale. The input admittance of the
load only at the shunt conductor is at 0.394λ + 0.300λ − 0.500λ = 0.194λ and is
denoted by point A. It has a value of
yinA = 1.37+ j0.45.

3. The shunt conductance has a normalized conductance
g =
50 Ω
30 Ω
= 1.67.
The normalized admittance of the shunt conductance in parallel with the input
admittance of the load is the sum of their admittances:
yinB = g+yinA = 1.67+1.37+ j0.45 = 3.04+ j0.45
and is located at point B. On the WTG scale, point B is at 0.242λ. The input
admittance of the entire circuit is at 0.242λ + 0.300λ − 0.500λ = 0.042λ and is
denoted by point Y-IN. The corresponding normalized input impedance is at Z-IN
and has a value of
zin = 1.9− j1.4.
Thus,
Zin = zinZ0 = (1.9− j1.4)×50 Ω = (95− j70) Ω.