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Sec. 4–4 Evaluation of Power 245
Realizing that both 8 9 and Re{ } are linear operators, we may exchange the order of the operators
without affecting the result, and the autocorrelation becomes
or
R v (t) = 1 2 Re{8g * (t) g(t + t) e jv ct 9} + 1 2 Re{8g(t) g(t + t) ej2v ct e jv ct 9}
R v (t) = 1 2 Re{8g * (t) g(t + t)9 ejv ct } + 1 2 Re{8g(t) g(t + t)ej2v ct 9 e jv ct }
But 8g *(t)g(t + t)9 = R The second term on the right is negligible because e j2v ct
g (t).
=
cos 2v c t + j sin 2v c t oscillates much faster than variations in g(t)g(t + t). In other words, f c
is much larger than the frequencies in g(t), so the integral is negligible. This is an application
of the Riemann–Lebesque lemma from integral calculus [Olmsted, 1961]. Thus, the autocorrelation
reduces to
R v (t) = 1 2 Re{R g(t)e jv ct }
(4–16)
The PSD is obtained by taking the Fourier transform of Eq. (4–16) (i.e., applying the
Wiener–Khintchine theorem). Note that Eq. (4–16) has the same mathematical form as Eq.
(4–11) when t is replaced by t, so the Fourier transform has the same form as Eq. (4–12). Thus,
v (f) = [R v (t)] = 1 4 [ g1f - f c 2 + g(-f * - f c )]
But * g (f) = g (f), since the PSD is a real function. Hence, the PSD is given by Eq.
(4–13).
4–4 EVALUATION OF POWER
THEOREM.
The total average normalized power of a bandpass waveform v(t) is
q
P v = 8v 2 (t)9 = v (f) df = R v (0) = 1 2 8|g(t)|2 9
L
-q
(4–17)
where “normalized” implies that the load is equivalent to one ohm.
Proof.
Substituting v(t) into Eq. (2–67), we get
q
P v = 8v 2 (t)9 = v (f) df
L
q
But R v (t) = -1 [ v (f)] = v (f)e j2pft df, so
L
-q
q
R v (0) = v (f) df
L
-q
-q