solution for a problem

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
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0.80.90.9
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1.01.01.0
1.21.2
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1.41.4
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1.61.6
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1.81.8
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2.02.0
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0.470.48
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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.