ECE 534: Random Pro cesses Spring 2010
Solutions to Problem Set 10
8.1 On filtering a WSS random process
(a) True. Since SY(ω) = |H(ω)|2SX(ω), it follows that SY(ω)≤SX(ω) for all ω. Integrat-
ing over all frequencies yields the result, because the power of Xis R∞
−∞ SX(ω)dω
2πand
similarly for Y.
(c) False. The output power is zero if (and only if) SX(ω)|H(ω)|2≡0. This happens, for
example, if H(ω) = I{|ω|≤2π}and SX(ω) = I{4π≤|ω|≤6π}.
8.5 Slight smoothing
(a) RY X (τ) = h∗RX(τ) = R∞
−∞ h(τ−t)RX(t)dt. Setting τ= 0, and using the symmetry
of hand of RXyields
RY X (0) = h∗RX(0) = Z∞
−∞
h(0 −t)RX(t)dt
=2
aZa/2
0
e−tdt
=2
a(1 −e−a/2) = 2
a1−1 + a
2−1
2a
22+1
3! a
23+· · ·
= 1 −a
4+a2
24 +· · · = 1 −a
4+o(a)
(b) RY(τ) = h∗e
h∗RX(τ) = R∞
−∞(h∗e
h)(τ−t)RX(t)dt. First we find that h∗e
his the
triangle function over the interval [−a, a] with height 1/a. Thus, taking τ= 0 and using
the symmetry of h∗e
hand of RXyields
RY(0) = Z∞
−∞
(h∗e
h)(0 −t)RX(t)dt
=2
a2Za
0
(a−t)e−tdt
=2
a2(a−1 + e−a) = 2
a2a2
2−a3
3! +a4
4! · · ·
= 1 −a
3+a2
12 +· · · = 1 −a
3+o(a)
(c) E[|Xt−Yt|2] = E[(Xt−Yt)(Xt−Yt)∗] = RX(0) −RXY (0) −RY X(0) + RY(0). Since
RX(0) = 1 and RXY (0) = R∗
Y X (−0) = RY X(0), it follows that
E[|Xt−Yt|2] = 1 −2RY X(0) + RY(0) = 1 −2(1 −a
4+o(a)) + 1 −a
3+o(a) = a
6+o(a)
1
