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AM, FM, and Digital Modulated Systems Chap. 5
Solution For SSB, g(t) = A c [m(t) ; jmN (t)]. Using Eq. (4–17) yields
P s = 8s 2 (t)9 = 1 2 8|g(t)|2 9 = 1 2 A c8m 2 (t) + [mN (t)] 2 9
or
P s = 8s 2 (t)9 = 1 2 A cE8m 2 (t)9 + 8[mN (t)] 2 9F
(5–136)
But
q
q
8[mN (t)] 2 9 = mN (f) df = |H(f)| 2 m (f) df
L L
-q
where H(f) is the transfer function of the Hilbert transformer. Using Eq. (5–19), we find that
|H(f)| = 1. Consequently,
-q
q
8[mN (t)] 2 9 = mN (f) df = 8m 2 (t)9
L
-q
(5–137)
Substituting Eq. (5–137) into Eq. (5–136), we get
P s = 8s 2 (t)9 = A c 2 8m 2 (t)9
SA5–2 Evaluation of SSB Power An SSB transmitter with A c = 100 is being tested by modulating
it with a triangular waveform that is shown in Fig. 5–14a, where V p = 0.5 V. The transmitter is
connected to a 50-Ω resistive load. Calculate the actual power dissipated into the load.
Solution
Using Eq. (5–25) yields
P actual = (V 2
s) rms
R L
For the waveform shown in Fig. 5–14a,
= 8s2 (t)9
R L
= A c 2
R L
8m 2 (t)9
(5–138)
or
8m 2 (t)9 = 1 T m
m 2 (t) dt = 4 T m >4
a 4V p
2
t - V p b dt
T m L T m L T m
0
0
8m 2 (t)9 = 4V p 2
Substituting Eq. (5–139) into Eq. (5–138), we get
P actual = A c 2 V p
2
T m
a 4 3
t - 1b
T m
3a 4 T m
b
= (100)2 (0.5) 2
3R L 3(50)
∞
T m >4
0
= V p 2
3
= 16.67 W
(5–139)