Topics on Data and Signal Analysis Coursework I
Problem 1. Consider the following three vectors in R3:
φ1 =[1 −12]T , φ2 =[23 −2]T , φ2 =[311]T .
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a) Does the set Φ = {φ1, φ2, φ3} form a basis for R3?
b) If Φ = {φ1, φ2, φ3} 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 = c1φ1 + c2φ2 + c3φ3 .
d) Find the largest number A > 0 and the smallest number B < ∞ such that
for all x ∈ R3.
A∥x∥2 ≤ |⟨φi , x⟩|2 ≤ B∥x∥2
Problem 2. Consider the systems in Figures 1-4.
𝑋𝑒 𝑥𝑛23𝑦𝑛1𝑥𝑛32𝑦𝑛
𝐹𝑖𝑔𝑢𝑟𝑒 1 𝐹𝑖𝑔𝑢𝑟𝑒 2
𝑥𝑛24𝑦𝑛𝑥𝑛42𝑦𝑛
𝐹 𝑖 𝑔 𝑢 −𝑟 𝑒𝜋 3 − 𝜔 # 𝜔 # 𝐹 𝑖 𝑔
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.
−𝜋 −𝜋 𝜋 𝜋 22
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] = h[k]h∗[k − n] . k
b) Show that if ⟨h[k − n], h[k]⟩ = δ[n], then
c) Show that if |H(ejω)| = 1, then
|H(ejω)| = 1, ∀ω.
⟨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 l2(Z).
Problem 4. Consider two waveforms φ0[n] and φ1[n] and two waveform ψ0[n] and ψ1[n] in l2(Z). Let h0[n] and h1[n] be two filters such that h0[n] = φ∗0 [−n] and h1[n] = φ∗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], φ0[n − 2k]⟩ = δ[k] then H0(z)G0(z) + H0(−z)G0(−z) = 2 and that if ⟨ψ1[n], φ1[n − 2k]⟩ = δ[k] then H1(z)G1(z) + H1(−z)G1(−z) = 2.
b) Show that if ⟨ψ0[n], φ1[n − 2k]⟩ = 0 for all k ∈ Z then H1(z)G0(z) + H1(−z)G0(−z) = 0 and that if ⟨ψ1[n], φ0[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], φ0[n − 2k]⟩ = δ[k] , ⟨ψ1[n], φ1[n − 2k]⟩ = δ[k], ⟨ψ0[n],φ1[n−2k]⟩=0forallk ∈Z,and⟨ψ1[n],φ0[n−2k]⟩=0forallk ∈Z,then
H0(z)G0(z) + H1(z)G1(z) = 2 and H0(−z)G0(z) + H1(−z)G1(z) = 0 .
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