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480
Random Processes and Spectral Analysis Chap. 6
where h(t) is the impulse response of the filter. Taking the expected value, we get
However, the transfer function of the filter is
so that
q
H(0) = - q
m y = y(t) = h(t) * x(t) = h(t) * x(t) = h(t) * m x
q
q
= h(l)m x dl = m x h(l) dl
L L
-q
q
H(f) = [h(t)] = h(t) e -j2pft dt
L
-q
-q
h(t) dt. Thus, Eq. (6–196) reduces to
m y = m x H(0)
(6–196)
(6–197)
SA6–3 Average Power Out of a Differentiator Let y(t) = dn(t)/dt, where n(t) is a randomnoise
process that has a PSD of n ( f ) = N 0 2 = 10 -6 W/Hz. Evaluate the normalized power of
y(t) over a low-pass frequency band of B = 10 Hz.
Solution
Thus,
or
y ( f ) = |H( f )| 2 n ( f ), where, from Table 2–1, H( f ) = j 2p f for a differentiator.
B
B
P y = y (f) df = (2pf) 2 N 0
L L 2
-B
-B
df =
8p2
3 a N 0
2 bB3
P y = 8p2
3 (10-6 )(10 3 ) = 0.0263 W
(6–198)
p x (f)
2
(a) PSD for x(t)
–2 –1 1 2
p v (f)
f
1
–f c -2 –f c –f c +2 f c -2 f c f c +2
(b) PSD for v(t)
Figure 6–22
PSD for SA6–4.
f