Topics on Data and Signal Analysis
Coursework I
Problem 1. Consider the following three vectors in R3:
ϕ1 = [1 − 1 2]T , ϕ2 = [2 3 − 2]T , ϕ2 = [3 1 1]T .
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
A‖x‖2 ≤
3∑
i=1
|〈ϕi , 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 ϕ0[n] and ϕ1[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] = 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]〉 = 0 for all k ∈ Z, and 〈ψ1[n], ϕ0[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 .