Page 351 - Engineering Electromagnetics, 8th Edition

CHAPTER 10 Transmission Lines 333

We may now let z = 0in (104) to find the load voltage,
V s,L = (1 + )V 0 + = 20 72 = 20 −288 ◦
◦
The amplitude agrees with our previous value. The presence of the reflected wave
causes V s,in and V s,L to differ in phase by about −279 instead of −288 .
◦
◦
EXAMPLE 10.8
In order to provide a slightly more complicated example, let us now place a purely
capacitive impedance of − j300 in parallel with the two 300 receivers. We are
to find the input impedance and the power delivered to each receiver.
Solution. The load impedance is now 150 in parallel with − j300 ,or
150(− j300) − j300
Z L = = = 120 − j60
150 − j300 1 − j2
We first calculate the reflection coefficient and the VSWR:
120 − j60 − 300 −180 − j60
= = = 0.447 −153.4 ◦
120 − j60 + 300 420 − j60
1 + 0.447
s = = 2.62
1 − 0.447
Thus, the VSWR is higher and the mismatch is therefore worse. Let us next calculate
the input impedance. The electrical length of the line is still 288 ,so that
◦
(120 − j60) cos 288 + j300 sin 288 ◦
◦
Z in = 300 = 755 − j138.5
300 cos 288 + j(120 − j60) sin 288 ◦
◦
This leads to a source current of
V Th 60
◦
I s,in = = = 0.0564 7.47 A
Z Th + Z in 300 + 755 − j138.5
Therefore, the average power delivered to the input of the line is P in =
2
1 (0.0564) (755) = 1.200 W. Since the line is lossless, it follows that P L = 1.200 W,
2
and each receiver gets only 0.6 W.

EXAMPLE 10.9

As a final example, let us terminate our line with a purely capacitive impedance, Z L =
− j300 .We seek the reflection coefficient, the VSWR, and the power delivered to
the load.
Solution. Obviously, we cannot deliver any average power to the load since it is a
pure reactance. As a consequence, the reflection coefficient is
− j300 − 300
= =− j1 = 1 −90 ◦
− j300 + 300