Page 349 - Engineering Electromagnetics, 8th Edition

CHAPTER 10 Transmission Lines 331

to the line. Because there is no reflection and no attenuation, the voltage at the load
is 30 V, but it is delayed in phase by 1.6π rad. Thus,
8
V in = 30 cos(2π10 t)V
whereas
8
V L = 30 cos(2π10 t − 1.6π)V
The input current is
V in 8
I in = = 0.1 cos(2π10 t)A
300
while the load current is
8
I L = 0.1 cos(2π10 t − 1.6π)A
The average power delivered to the input of the line by the source must all be delivered
to the load by the line,
1
P in = P L = × 30 × 0.1 = 1.5W
2
Now let us connect a second receiver, also having an input resistance of 300 ,
across the line in parallel with the first receiver. The load impedance is now 150 ,
the reflection coefficient is
150 − 300 1
= =−
150 + 300 3
and the standing wave ratio on the line is
1
1 + 3
s = 1 = 2
1 −
3
The input impedance is no longer 300 ,but is now
Z L cos βl + jZ 0 sin βl 150 cos 288 + j300 sin 288 ◦
◦
Z in = Z 0 = 300
Z 0 cos βl + jZ L sin βl 300 cos 288 + j150 sin 288 ◦
◦
= 510 −23.8 = 466 − j206
◦
which is a capacitive impedance. Physically, this means that this length of line stores
more energy in its electric field than in its magnetic field. The input current phasor is
thus
60
◦
I s,in = = 0.0756 15.0 A
300 + 466 − j206
and the power supplied to the line by the source is
1
2
P in = × (0.0756) × 466 = 1.333 W
2
Since there are no losses in the line, 1.333 W must also be delivered to the load.
Note that this is less than the 1.50 W which we were able to deliver to a matched
load; moreover, this power must divide equally between two receivers, and thus each