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ECE 459: Solution for Sample Final Exam
Che Lin
December 9, 2004
Problem 1
(a) Averaged power=3. FY (y) = √ 2 π^1 √ 3 e−^
y 62
(b) YˆLM M SE (x) = 23 x Problem 2 (a) Signal power=500 and noise power=10. So SN R = 50. (b) Signal power=250 and noise power=2.5. So SN R = 100. Problem 3 Part a. (a) ψ(t) = kf
∫ (^) t −∞ m(τ^ )dτ (b) The power spectral density is (^) AN 2 rect( (^4) πBω ) and the averaged power is 2 N BA 2
(c) D(ω) = jω and SN R =
k f^2 P (3A^2 ) 8 π^2 N B^3 Part b. (d) D(ω) = ω^2 and Sw′w′^ (ω) = ω^4 ( (^) AN 2 ) from − 2 πB to 2πB.
(e) SN R = k
(^2) f P (5A (^2) ) 32 π^4 N B^5 and the range for^ B^ is 2πB^ ≤
√ 5 / 3 Problem 4 Part a. (a) fY |H 0 (y|H 0 ) ∼ N (0, N/2) and fY |H 1 (y|H 1 ) ∼ N (1/ 2 , N/2) (b) γ = 1/4 and Pe = Q( √^1 /N/^42 )
(c) Yes. Choose h(t) = x(1 − t) from 0 to 1 where x(t) = p(t) ∗ c(t) (because the filter has to be causal). The resulting γ is 1/6 and Pe = Q( √^1 N//^66 ).
Part b. (d)(e) Choose T = 2 and h(t) = x(2 − t). The resulting γ is 1/3 and Pe = Q( √^1 /N/^33 ).
Problem 5 A. TRUE. Passing x(t) through g(t) to get z(t) is equivalent to passing x(t) through h(t) to get y(t) and then passing y(t) through another h(t) to get z(t). Since h(t) is BIBO stable, if x(t) is finite in energy, y(t) is also finite in energy and so is z(t). This implies that g(t) is also BIBO. B. FALSE. YˆLM M SE (x) should be 0. C. FALSE. The counter example is to choose the joint density fX,Y (x, y) = 1 for the region [− 1 / 2 , 1 /2] × [− 1 / 2 , 1 /2]. Then we can find that E[Y |X] is the same as YˆLM M SE (x) and both equal to 0. D. FALSE. σ^2 Y = σ X^21 + 2σX 1 X 2 + σ^2 X 2 not necessarily equals to 2. E. TRUE. Since SXX (ω) is the sinc function and has negative parts, which is not permissible for PSD. F. TRUE. RXX (2) = 0 → uncorrelated → independent. (b/c they are Gaussian)
