Topics on Data and Signal Analysis Coursework I Problem 1. Consider the following three vectors in R3: 1 = [1 1 2]T , 2 = [2 3 2]T , 2 = [3 1 1]T . a) Does the set Φ = { 1, 2, 3} form a basis for R3 b) If Φ = { 1, 2, 3} forms a basis, find its biorthogonal basis Ψ = {ψ1, ψ2, ψ3} . c) For an arbitrary x = [x1 x2 x3] T in R3 describe the procedure and the expression for finding coefficients c1, c2 and c3 such that x = c1 1 + c2 2 + c3 3 . d) Find the largest number A > 0 and the smallest number B <∞ such that A‖x‖2 ≤ 3∑ i=1 |〈 i , x〉|2 ≤ B‖x‖2 for all x ∈ R3. Problem 2. Consider the systems in Figures 1-4. 2 3 2 3 2 4 2 4 1 2 3 4 a) For the signal x[n] which is given in the Fourier domain by the figure below, sketch Y (ejω) for each of the four systems in the above. !" # 2 3 3 1 # 2 !" 2 2 1 b) For any arbitrary signal x[n] express samples of y [n] in terms of samples of x[n] for each of the four systems in the above. Problem 3. Consider a filter h[n]. a) Find the Fourier transform of its autocorrelation sequence a[n] = ∑ k h[k]h [k n] . b)
Show that if 〈h[k n], h[k]〉 = δ[n], then |H(ejω)| = 1, ω. c) Show that if |H(ejω)| = 1, then 〈h[k n], h[k]〉 = δ[n] . d) Show that if |H(ejω)| = 1, ω, then {h[k n], n ∈ Z} is an orthonormal basis for `2(Z). Problem 4. Consider two waveforms 0[n] and 1[n] and two waveform ψ0[n] and ψ1[n] in ` 2(Z). Let h0[n] and h1[n] be two filters such that h0[n] = 0 [ n] and h1[n] = 1 [ n], and g0[n] and g1[n] two filters such that g0[n] = ψ0[n] and g1[n] = ψ1[n]. a) Show that if 〈ψ0[n], 0[n 2k]〉 = δ[k] then H0(z)G0(z) + H0( z)G0( z) = 2 and that if 〈ψ1[n], 1[n 2k]〉 = δ[k] then H1(z)G1(z) + H1( z)G1( z) = 2. b) Show that if 〈ψ0[n], 1[n 2k]〉 = 0 for all k ∈ Z then H1(z)G0(z) + H1( z)G0( z) = 0 and that if 〈ψ1[n], 0[n 2k]〉 = 0 for all k ∈ Z then H0(z)G1(z) + H0( z)G1( z) = 0. c) Using the results of a) and b) show that if 〈ψ0[n], 0[n 2k]〉 = δ[k] , 〈ψ1[n], 1[n 2k]〉 = δ[k], 〈ψ0[n], 1[n 2k]〉 = 0 for all k ∈ Z, and 〈ψ1[n], 0[n 2k]〉 = 0 for all k ∈ Z, then H0(z)G0(z) + H1(z)G1(z) = 2 and H0( z)G0(z) + H1( z)G1(z) = 0 .