2007 Final Exam for Random Processes. x(t)

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1 2007 Final Exam for Random Processes x(t ϕ n (τ y n (t. (a (8 pt. Let {ϕ n (τ} n and {λ n } n satisfy that and ϕ n (τϕ m(τdτ δ[n m] R xx (t sϕ n (sds λ n ϕ n (t, where R xx (τ is the autocorrelation function of WSS process x(t, and δ[ ] is the Kronecker delta function. Let y n (t be the output process through the linear time-invariant filter ϕ n (τ due to input process x(t as shown in the above figure. Prove that E[y n (ty m(t] λ n δ[n m]. (b (6 pt. Prove that R xx (t sϕ(sds C ϕ(t is valid for any continuous function ϕ(τ, provided that x(t is white and WSS with autocorrelation function R xx (τ C δ(τ. (c (6 pt. By following (b, what is the eigenvalue for a white noise with two-sided power spectrum density N 0 /2? (a [( E[y n (ty m(t] E ( ] ϕ n (ux(t udu ϕ m (vx(t vdv ϕ n (uϕ m(ve [x(t ux (t v] dudv ϕ n (uϕ m(vr xx (v ududv ( R xx (v uϕ n (udu ϕ m(vdv λ n δ[n m]. λ n ϕ n (vϕ m(vdv

2 (b By the replication property, R xx (t sϕ(sds Cϕ(t. C δ(t sϕ(sds (c N 0 /2. 2. (8 pt. Prove that if x(t is zero-mean, white and WSS, then X(ω x(te jωt dt is also zero-mean, white and WSS. It is obvious that X(ω is zero-mean; hence, we omit its proof. Continuing from the proof in slide -73, we derive that the autocorrelation function of X(ω is R XX (u, v 2πS xx (uδ(u v 2πCδ(u v, where S xx (u C for some C, by the assumption of whiteness. Hence, X(ω is also a white process. By the way, the autocorrelation of X(ω only depends on the frequencydifference; hence, it is WSS since it is zero-mean. 3. (a (8 pt. Is there a continuous-in-time process that is simultaneously regular and predictable? Justify your answer with a formal proof. (Hint: A process is predictable if, and only, if it has non-degenerate line spectra of countably many lines. (b (8 pt. Let ˆx[t] k a kx[t k] be the best linear MS estimator of discretein-time non-predictable WSS process x[t] based upon its past. Find a whitening filter of x[t]. (Hint: The answer is inside the proof of the Wold s decomposition. (c (8 pt. Does (b imply that all WSS processes can be whitened by a linear filter? Justify your answer. (Hint: A whitening filter of process x[t], by definition, shall make the output becoming a white process with power spectrum of non-zero height. In other words, nullification (zero-output cannot be considered as a whitening procedure. (a Suppose x(t is both regular and predictable. Then, S xx (ω L(ω 2 for some causal filter L(ω. Since x(t is also predictable, L(ω 2 only consists of countably many impulses, i.e., L(ω n c nδ(ω ω n for complex numbers {c n }. The impulse response of L(ω is therefore equal to 2π n c ne jωnτ, which is not zero for τ < 0 a violation to causality. Hence, such a process does not exist. (b By the proof of the Wold s decomposition, e[t] x[t] ˆx[t] is the output due to input x[t] through filter A[z] k a kz k, and {e[t]} is white. Hence, A[z] is a whitening filter of x[t]. 2

3 (c By Wold s decomposition, x[t] x p [t]+x r [t], where x p [t] is predictable, and x r [t] is regular. From the proof, we learn that x p [t] k a kx p [t k] y[t] 0 with probability one. Hence, ( ( x[t] a k x[t k] x p [t] a k x p [t k] + x r [t] a k x r [t k] k x r [t] k a k x r [t k]. k Hence, it seems that all processes can be whitened by its corresponding A[z]. However, if x[t] is predictable itself, then the output x[t] ˆx[t] will be zero. In such case, A[z] cannot be claimed to be the whitening filter of x[t]. 4. (a (8 pt. Prove that a discrete-in-time zero-mean white process i[t] of finite power is mean-ergodic in the sense that for some η, where E [ η T η 2] 0 as T η T 2T + (b (8 pt. By following (a, prove that any process of the form x[t] n k0 h ki[t k] is mean-ergodic, provided that n k0 h k <. (c (6 pt. Answer without proof that whether a regular process is mean-ergodic. t T i[t]. k (a For η 0, (b E [ η T 2] 2T + E[ i[t] 2 ] 0. η T 2T + 2T + t T k0 u T n n h k i[t k] w u i[u], where w u n k0 h k w max. Hence, E [ η T 2] w max 2 2T + E[ i[t] 2 ] 0. 3

4 (c It follows from (b that a regular process should be mean-ergodic. 5. (a (8 pt. Prove that estimation variance is equal to the sum of variance of estimation and the square of the bias. In other words, E [ g(x x 2] E [ g(x E[g(x] 2] + E[g(x] x 2, where g(x is an estimator of x based on the random variable x. (Note that x is in general a complex number, and x is a complex random variable. (b (8 pt. By following (a, the linear estimator of x based on the observation x is of the form g(x c x for some complex constant c. Find the best c that minimizes the estimation variance. (Hint: You may denote µ x E[x] and σ 2 x E[ x µ x 2 ] for convenience. (a For convenience, abbreviate g g(x and µ E[g(x] in my solution. E [ g(x E[g(x] 2] + E[g(x] x 2 E [ g µ 2] + µ x 2 E[gg ] µe[g ] µ E[g] + µµ + µµ xµ x µ + xx E[gg ] µµ µ µ + µµ + µµ xµ x µ + xx E[ g x 2 ]. (b Denote the mean and variance of x by E[x] µ x and E[ x µ x 2 ] σ 2 x. Then, and bias 2 cµ x x 2 c 2 µ x 2 c µ xx cµ x x + x 2 variance of estimate c 2 σ 2 x. Hence, the estimation variance is given by: c 2 (σx 2 + µ x 2 cµ x x xc µ x + x 2 ( (σx 2 + µ x 2 c 2 µ x x c σx 2 + µ x xµ 2 c x + x 2 σx 2 + µ x 2 (σx 2 + µ x 2 c xµ 2 x σx 2 σx 2 + µ x 2 + σx 2 + µ x 2 x 2. Hence, the best linear estimator is g(x 4 xµ x σx+ µ 2 x x. 2

5 6. (0 pt. Can we find a real data window c(t to exactly emulate the effect of the real spectral window W (y for all input processes? Justify your answer. (Hint: S T,w (ω; c 4πT X T (ω C(ω 2 W (ω, where C(ω is the Fourier transform of c(t, and denotes the convolution operation. The problem is equivalent to finding C(ω (for a given W (ω such that X T (ω C(ω 2 X T (ω 2 W (ω. The right-hand-side is a system with input X T (ω and output X T (ω 2 W (ω, and the left-hand-side is a system with input X T (ω and output X T (ω C(ω 2. If the two systems are equivalent, then the outputs should be the same for any input X T (ω. Yet, the right-hand-side system is irrelevant to the phase of the complex X(ω, while the left-hand-side is a function of the phase of the complex X(ω. So, no matter how we design C(ω, there always exists a specific input process to fail the equivalence. So, the answer of the problem is NO. 5