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Bandpass Signaling Principles and Circuits Chap. 4
where A = 0.8 and v 0 = 2p1,000. Taking the Fourier transform by the use of Eq. (2–26) we obtain
the PSD of m(t): †
or
m (f) = A2
4 [d(f-f o) + d(f + f 0 )]
m (f) = 0.16 [d(f-1,000) + d(f + 1,000)]
(4–122)
The autocorrelation for the complex envelope of the AM signal is
R g (t) = 8g * (t)g(t + t)9 = A 2 c 8[1 + m(t)][1 + m(t + t)]9
= A 2 c [819 + 8m(t)9 + 8m(t + t)9 + 8m(t)m(t + t)9]
But 819 = 1, 8m(t)9 = 0, 8m(t + t)9 = 0, and 8m(t) m(t + t)9 = R m (t). Thus,
R g (t) = A c 2 + A c 2 R m (t)
(4–123)
Taking the Fourier transform of both sides of Eq. (4–123), we get
g (f) = A c 2 d(f) + A c 2 m (f)
(4–124)
Substituting Eq. (4–124) into Eq. (4–13), with the aid of Eq. (4–122), we obtain the PSD for the
AM signal:
s (f) = 62,500 d(f-f c ) + 10,000 d(f-f c - 1,000)
+ 10,000 d(f-f c + 1,000) + 62,500 d (f + f c )
+ 10,000 d(f + f c - 1,000) + 10,000 d(f + f c + 1,000)
(4–125)
(Note: We realize that this bandpass PSD for s(t) is found by translating (i.e., moving) the baseband
PSD of g(t) up to f c and down to -f c . Furthermore, for the case of AM, the PSD of g(t) consists of
the PSD for m(t) plus the superposition of a delta function at f = 0).
SA4–3 Average Power for an AM Signal Assume that the AM voltage signal s(t), as
described in SA4–1, appears across a 50-Ω resistive load. Compute the actual average power
dissipated in the load.
Solution.
From Eq. (4–21), the normalized average power is
2
(P s ) norm = (V s ) rms
= 1 2 A c 2 [1 + (V m ) 2 rms ]
= 1 2 (500)2 c1 + a 0.8
12 bd 2
= 165 kW
(4–126a)
† Because m(t) is periodic, Eq. (2–126) can be used as an alternative method of evaluating m (f). That is, by
using Eq. (2–126) with c 1 = c -1 * = A>(2j) = -j0.8>2 = -j0.4 (and the other c n ’s are zero), Eq. (4–122) is obtained.