Download ECE 534 Random Processes: Solutions to Problem Set 6 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE 534 RANDOM PROCESSES SPRING 2008
SOLUTIONS TO PROBLEM SET 6
6.1. On filtering a WSS random process (a) True. Since SY (ω) = |H(ω)|^2 SX (ω), it follows that SY (ω) ≤ SX (ω) for all ω. Integrating over all frequencies yields the result, because the power of X is
∫ (^) ∞ −∞ SX^ (ω)^
dω 2 π and similarly for^ Y^. (b) True. The process Y is WSS because X is WSS and the system is time-invariant. Since X is WSS and periodic, RX is periodic with some period T. Thus, RY (τ +T ) =
∫ (^) ∞ −∞ RX^ (τ^ +T^ −t)(h∗ ˜h)(t)dt = ∫ (^) ∞ −∞ RX^ (τ^ −^ t)(h^ ∗^ ˜h)(t)dt^ =^ RY^ (τ^ ), so that^ RY^ is also periodic with period^ T^.^ Therefore,^ Y^ is periodic. (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|≤ 2 π} and SX (ω) = I{ 4 π≤|ω|≤ 6 π}.
6.3. Modulating and filtering a stationary process (Note: Since cos(80πt + Θ) = cos(Θ) for any integer time t, parts (a) and (b) of the problem as stated are degenerate. It would make more sense either in continuous time, or if the frequency of the cosine term were non-integer, such as Yt = Xt cos(2πt/40 + Θ). The form of the answer given here would be correct for non-integer frequencies, thinking of SX as a period 2π function.) (a) Since Θ and ((Θ + c) mod 2π) have the same distribution, the process cos(80πt + Θ) is station- ary. Two independent stationary processes are jointly stationary, and the product of two jointly stationary processes is also stationary. (b) Using the fact cos(a) cos(b) = (cos(a − b) + cos(a + b))/2, we find RY (s, t) = E[XsXt]E[cos(80π(s − t)) + cos(80π(s + t) + 2Θ)]/2 = RX (s, t) cos(80π(s − t))/2. So RY (τ ) = cos(80 2 πτ^ )RX (τ ). In the transform domain this is SY (ω) = 14 [SX (ω − 80 π) + SX (ω + 80π)]. (c) The impulse response function is given by h(τ ) =
∑∞ ∑^ k=0^ a(1^ −^ a)kδ(τ^ −^ k).^ Thus,^ H(ω) = ∞ k=0 a[(1^ −^ a)e −jω]k (^) = a 1 −(1−a)e−jω^. (d) Note that | 1 − (1 − a)e−jω| ≥ 1 − (1 − a) = a, so that |H(ω)| ≤ 1 for all ω. Thus, SZ (ω) ≤ SY (ω) for all ω. Therefore the power of Z is less than or equal to the power of Y , which is given by RY (0) = 12 RX (0) = 1/ 2. (e) Examining H further shows that |H(ω)| = 1 if and only if ω is a multiple of 2π, and in that case H(ω) = 1. Now, if X is a periodic process with power one and period one (or a smaller period that divides one), then Y will have power 0.5 and also have period one (or a smaller period that divides one) because the frequencies in Y are the frequencies in X plus or minus 40 Hz. Then, Y will pass through the filter H unchanged, and the power of Z will therefore be 0.5.
6.5. Slight smoothing (a) RY X (τ ) = h ∗ RX (τ ) =
∫ (^) ∞ −∞ h(τ^ −^ t)RX^ (t)dt. Setting^ τ^ = 0, and using the symmetry of^ h^ and of RX yields
RY X (0) = h ∗ RX (0) =
∫ (^) ∞
−∞
h(0 − t)RX (t)dt
a
∫ (^) a/ 2
0
e−tdt
a (1 − e−a/^2 ) =
a
( 1 − 1 +
a 2
( a 2
) 2
( a 2
) 3
)
a 4
a^2 24
a 4
(b) RY (τ ) = h ∗ ˜h ∗ RX (τ ) =
∫ (^) ∞ −∞(h^ ∗^ ˜h)(τ^ −^ t)RX^ (t)dt.^ First we find that^ h^ ∗^ h˜^ is the triangle function over the interval [−a, a] with height 1/a. Thus, taking τ = 0 and using the symmetry of h ∗ ˜h and of RX yields
RY (0) =
∫ (^) ∞
−∞
(h ∗ ˜h)(0 − t)RX (t)dt
=
a^2
∫ (^) a
0
(a − t)e−tdt
a^2 (a − 1 + e−a) =
a^2
( a^2 2
a^3 3!
a^4 4!
)
a 3
a^2 12
a 3
(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 − 2 RY X (0) + RY (0) = 1 − 2(1 − a 4 + o(a)) + 1 − a 3 + o(a) = a 6 + o(a)
6.11. Sampling a cubed Gaussian process (a) Yes, Y is WSS, since it is obtained from the stationary process Y by sliding a time-invariant function along the stationary process X. By the hint, RY (τ ) = Cov(X τ^3 , X^30 ) = 6RX (τ )^3 + 9 RX (0)^2 RX (τ ). Therefore, SY (ω) = 6(SX ∗ SX ∗ SX )(ω) + 9RX (0)^2 SX (ω). That is, a triple convolution is involved. (b) Due to the triple convolution, we can see that Y is a baseband random process with one-sided band limit 3fo. Therefore, Y can be linearly reconstructed from its samples taken at sampling rate 1 T ′^ = 6fo. (c) If nonlinear processing is allowed, then Y can be recovered from its samples taken at sampling frequency 2fo, because XnT = Y (^) nT^1 / 3. That is:
Yt =
( (^) ∞ ∑
n=−∞
Y (^) nT^1 /^3 sinc
( t − nT T
)) 3
7.7. Predicting the future of a simple WSS process (a) First finding that the denominator of SX vanishes for ω^2 = −4 and for ω^2 = −9, we find the factorization: SX (ω) =
(jω + 2)(jω + 3) ︸ ︷︷ ︸ S+(ω)
(−jω + 2)(−jω + 3) ︸ ︷︷ ︸ S−(ω)
(b) The optimal filter is given by H(ω) = (^) S+^1 X (ω)
[ S X+ (ω)ejωT^
]
. The technique of partial frac-
tion expansions yields that S X+ (ω) = (^) jω^1 +2 − (^) jω^1 +3 ↔ I{t≥ 0 }(e−^2 t^ − e−^3 t). Thus S X+ (ω)ejωT^ ↔
I{t≥−T }(e−2(t+T^ )−e−3(t+T^ )) so
[ S X+ (ω)ejωT^
]
↔ I{t≥ 0 }(e−2(t+T^ )−e−3(t+T^ )), so that
[ S X+ (ω)ejωT^
]
e−^2 T jω+2 +^
e−^3 T jω+3. Therefore,^ H(ω) =^ e − 2 T (^) (jω + 3) − e− 3 T (^) (jω + 2). Therefore, X̂ t+T |t = Xt(3e− 2 T (^) − 2 e−^3 T^ ) + X t′(e−^2 T^ − e−^3 T^ ). (c) The MSE for the optimal predictor is RX (0)−
∫ (^) ∞ −∞ |
[ S+ X (ω)ejωT^
]
|^2 dω 2 π = RX (0)−
∫ (^) ∞ 0 ((e −2(t+T )−
e−3(t+T^ ))^2 dt = RX (0) − e−^4 T^ / 4 − 2 e−^5 T^ /5 + e−^6 T^ /6. The constant RX (0) can be found by noting
(1 + jω)^2 x̂(ω) = 2 n̂ (ω), or ((jω)^2 + 2jω + 1)̂ x(ω) = 2 n̂ (ω) or x′′^ + 2x′^ + x = 2n, so that X satisfies the stochastic differential equation X′′^ + 2X′^ + X = 2N.
7.15. A continuous-time Wiener filtering problem (a) The optimal filter without the causality constraint is given by
H(ω) = SXY (ω) SY (ω)
SX (ω) SX (ω) + SN (ω)
4 4+ω^2 4 4+ω^2 +^
No 2
=
8 No ω^2 + α^2
where α =
√ 4 + (^) N^8 o , or in the time domain, h(t) = 4 e
−α|t| αNo.^ The corresponding MMSE is given by
MMSE (noncausal) =
∫ (^) ∞
−∞
( SX (ω) − SXY (ω)^2 SY (ω)
) dω 2 π
∫ (^) ∞
−∞
SX (ω) N 2 o SX (ω) + N 2 o
dω 2 π
=
∫ (^) ∞
−∞
ω^2 + α^2
dω 2 π
√^1
1 + (^) N^2 o
As expected, the MMSE increases monotonically from 0 to 1 as N 0 ranges from 0 to ∞. (b) We shall use the formula for the optimal filter H(ω). The factorization of SY is given by
SY (ω) =
4 + ω^2
No 2
No 2
(4 + (^) N^8 o ) + ω^2 4 + ω^2
=
√ No 2
(jω + α) (jω + 2) ︸ ︷︷ ︸ S Y+ (ω)
√ No 2
(−jω + α) (−jω + 2) ︸ ︷︷ ︸ S Y− (ω)
Thus
SXY (ω) S− Y (ω)
4 (jω+2)(−jω+2) S− Y (ω)
√ 2 No
(jω + 2)(−jω + α)
= γ
{ 1 jω + 2
−jω + α
}
where γ =
4
√ 2 No 2+α.^ Therefore SXY (ω) S− Y (ω)
↔ γe−^2 tI{t≥ 0 } + γeαtI{t< 0 },
so that [ ejωT^ SXY S Y−
]
{ γe−2(t+T^ )I{t ≥ 0 } if T ≥ 0 γe−2(t+T^ )I{t ≥ −T } + γeα(t+T^ )I{ 0 ≤ t ≤ T } if T < 0
Converting back to the transform domain yields [ ejωT^ SXY S Y−
]
γe−^2 T jω+2 if^ T^ ≥^0 γejωT jω+2 +^
γ[eiωT^ −eαT^ ] −jω+α if^ T <^0
Dividing by S Y+ yields the optimal transfer function. For the remainder of this problem solution assume T ≥ 0. The optimal transfer function is thus
H(ω) =
γe−^2 T^
√ 2 No jω + α so that the optimal impulse response function is given by
h(t) =
4 e−^2 T No(1+
√ 1+ (^) No^2 )
e−αt^ t ≥ 0
0 t < 0
The MMSE is given by
MMSE = RX (0) −
∫ (^) ∞
−∞
h(s)RXY (s + T )ds = 1 −
∫ (^) ∞
0
h(s)e−2(s+T^ )ds
e−^4 T 1 + No +
√ N (^) o^2 + 2No
As expected, the MMSE converges to one as either T → ∞ or No → ∞, and it converges to 1−e−^4 T as No → 0.
