(a) A signal is given by

x[t] = cos(28πt) + 2cos(40πt) + 3cos(70πt)

This signal is sampled at 90 samples/s to get discrete-time signal x(n).

(i) Find the periodicity of the individual components in the signal and hence find the periodicity N₀ of t

Question

(a) A signal is given by

x[t] = cos(28πt) + 2cos(40πt) + 3cos(70πt)

This signal is sampled at 90 samples/s to get discrete-time signal x(n).

(i) Find the periodicity of the individual components in the signal and hence find the periodicity N₀ of the signal x(n).

(ii) Find the harmonic indices m (0 ≤ m < N₀) of the complex DTFS coefficient Dₘ, where Dₘ is non-zero.

(iii) By inspection, write the magnitude of the coefficients |Dₘ| for the indices found above. 20 marks

(b) For a unity feedback time delay system with open-loop transfer function

G(s) = Ke⁻ᵀˢ/s(s+2)

calculate—

(i) the maximum tolerable value of delay T, when K = 1;

(ii) phase margin when K = √5 and delay T = 0·5 second. 20 marks

(c) Given a system in state space representation as

[ẋ₁]   [0  1][x₁]   [0]
[ẋ₂] = [0 -3][x₂] + [1] u

y = [1 0][x₁]
        [x₂]

(i) Check whether the system is observable or not.

(ii) Find the state transition matrix.

(iii) Design a state feedback controller to place closed-loop poles at −1±2j. 20 marks

UPSC Answer Check · Accepted Answer

(a)(i) 
The component frequencies are 14 Hz, 20 Hz and 35 Hz, so their continuous periods are 1/14 s, 1/20 s and 1/35 s. With sampling frequency fₛ = 90 samples/s, t = n/90. Hence

x(n) = cos(28πn/90) + 2cos(40πn/90) + 3cos(70πn/90)
= cos(14πn/45) + 2cos(4πn/9) + 3cos(7πn/9).

For a discrete component cos(Ωn), period N satisfies ΩN = 2πm, m integer.
For Ω₁ = 14π/45: (14π/45)N₁ = 2πm ⇒ 7N₁ = 45m ⇒ N₁ = 45.
For Ω₂ = 4π/9: (4π/9)N₂ = 2πm ⇒ 2N₂ = 9m ⇒ N₂ = 9.
For Ω₃ = 7π/9: (7π/9)N₃ = 2πm ⇒ 7N₃ = 18m ⇒ N₃ = 18.

Thus N₀ = lcm(45, 9, 18) = 90 samples.

(a)(ii) 
With N₀ = 90, the DTFS basis is e^(j2πmn/90) = e^(jπmn/45). A cosine A cos(Ωn) contributes at m and at N₀ − m.
14πn/45 = π(14)n/45 ⇒ m = 14 and m = 90 − 14 = 76.
4πn/9 = 20πn/45 ⇒ m = 20 and m = 70.
7πn/9 = 35πn/45 ⇒ m = 35 and m = 55.

Non-zero harmonic indices are m = 14, 20, 35, 55, 70, 76.

(a)(iii) 
Using A cos(Ωn) = (A/2)e^(jΩn) + (A/2)e^(−jΩn),
|D₁₄| = 1/2, |D₇₆| = 1/2
|D₂₀| = 2/2 = 1, |D₇₀| = 1
|D₃₅| = 3/2, |D₅₅| = 3/2

So the non-zero magnitudes are
|D₁₄| = 1/2, |D₂₀| = 1, |D₃₅| = 3/2, |D₅₅| = 3/2, |D₇₀| = 1, |D₇₆| = 1/2.

(b)(i) 
For G(s) = Ke^(−Ts)/[s(s+2)],

|G(jω)| = K/[ω√(ω² + 4)]

and phase

∠G(jω) = −π/2 − arctan(ω/2) − ωT.

For K = 1, gain crossover is given by

1 = 1/[ω√(ω² + 4)] ⇒ ω²(ω² + 4) = 1.

Let u = ω². Then u² + 4u − 1 = 0, so

u = −2 + √5 = √5 − 2.

Therefore

ω_gc = √(√5 − 2) ≈ 0.485868 rad/s.

Phase margin without delay is

PM = π/2 − arctan(ω_gc/2)
= 1.332479 rad.

At marginal stability, PM = ω_gc T_max. Hence

T_max = 1.332479/0.485868 ≈ 2.742 s.

So the maximum tolerable delay is T_max ≈ 2.742 s.

(b)(ii) 
For K = √5, gain crossover satisfies

√5/[ω√(ω² + 4)] = 1 ⇒ ω²(ω² + 4) = 5.

Let u = ω². Then u² + 4u − 5 = 0 ⇒ u = 1. Thus ω_gc = 1 rad/s.

Phase margin is

PM = π + ∠G(jω_gc)

= π − π/2 − arctan(1/2) − (1)(0.5)

= π/2 − arctan(0.5) − 0.5

= arctan(2) − 0.5 rad

≈ 0.607149 rad

= 34.787°.

(c)(i) 
A = [0 1; 0 −3], C = [1 0].

Observability matrix:

O = [C; CA] = [[1, 0], [0, 1]].

Since det(O) = 1 ≠ 0 and rank(O) = 2, the system is completely observable.

(c)(ii) 
The state transition matrix is Φ(t) = e^(At) = L⁻¹[(sI − A)⁻¹].

sI − A = [[s, −1], [0, s + 3]].

Thus

(sI − A)⁻¹ = [[1/s, 1/(s(s+3))], [0, 1/(s+3)]].

Taking inverse Laplace,

Φ(t) = [[1, (1 − e^(−3t))/3], [0, e^(−3t)]], for t ≥ 0.

So

Φ(t) = [[1, (1 − e^(−3t))/3], [0, e^(−3t)]], t ≥ 0.

(c)(iii) 
Controllability matrix:

[B AB] = [[0, 1], [1, −3]], det = −1 ≠ 0.

So pole placement is possible. Let u = −Kx, where K = [k₁ k₂]. Then

A − BK = [[0, 1], [−k₁, −3 − k₂]].

Characteristic polynomial:

det(sI − (A − BK)) = s² + (3 + k₂)s + k₁.

Desired poles are −1 ± 2j, so desired polynomial is

(s + 1 − 2j)(s + 1 + 2j) = s² + 2s + 5.

Compare:

3 + k₂ = 2 ⇒ k₂ = −1,

k₁ = 5.

Therefore

K = [5, −1], and the control law is u = −5x₁ + x₂.

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