Page 350 - Engineering Electromagnetics, 8th Edition

332 ENGINEERING ELECTROMAGNETICS

receiver now receives only 0.667 W. Because the input impedance of each receiver is
300 , the voltage across the receiver is easily found as
2
1 |V s,L |
0.667 =
2 300
|V s,L |= 20 V
in comparison with the 30 V obtained across the single load.
Before we leave this example, let us ask ourselves several questions about the
voltages on the transmission line. Where is the voltage a maximum and a minimum,
and what are these values? Does the phase of the load voltage still differ from the
input voltage by 288 ? Presumably, if we can answer these questions for the voltage,
◦
we could do the same for the current.
Equation (89) serves to locate the voltage maxima at
1
z max =− (φ + 2mπ)(m = 0, 1, 2,...)
2β
where =| |e . Thus, with β = 0.8π and φ = π,wefind
jφ
z max =−0.625 and −1.875 m
while the minima are λ/4 distant from the maxima;
z min = 0 and −1.25 m
and we find that the load voltage (at z = 0) is a voltage minimum. This, of course,
verifies the general conclusion we reached earlier: a voltage minimum occurs at the
load if Z L < Z 0 , and a voltage maximum occurs if Z L > Z 0 , where both impedances
are pure resistances.
The minimum voltage on the line is thus the load voltage, 20 V; the maximum
voltage must be 40 V, since the standing wave ratio is 2. The voltage at the input end
of the line is
V s,in = I s,in Z in = (0.0756 15.0 )(510 −23.8 ) = 38.5 −8.8 ◦
◦
◦
The input voltage is almost as large as the maximum voltage anywhere on the line
because the line is about three-quarters of a wavelength long, a length which would
place the voltage maximum at the input when Z L < Z 0 .
Finally, it is of interest to determine the load voltage in magnitude and phase.
We begin with the total voltage in the line, using (93).
!
V sT = e − jβz + e jβz " V 0 + (104)
We may use this expression to determine the voltage at any point on the line in terms
of the voltage at any other point. Because we know the voltage at the input to the line,
we let z =−l,
V s,in = e jβl + e − jβl " V 0 + (105)
!
and solve for V ,
+
0
V s,in 38.5 −8.8 ◦
V + = = = 30.0 72.0 V
◦
0
e jβl + e − jβl e j1.6π − e
1 − j1.6π
3