GATE 2024 EC Q.38: solution and answer

GATE 2024 · ECQ.38 · 2 marks · MCQ

A satellite attitude control system, as shown below, has a plant with transfer function $G (s) = \frac{1}{s ^{2}}$ cascaded with a compensator $C (s) = \frac{K ( s + α )}{s + 4}$ , where $K$ and $α$ are positive real constants.

[Figure: unity negative feedback loop with $C (s)$ and $G (s)$ in cascade]

In order for the closed-loop system to have poles at $- 1 \pm j 3$ , the value of $α$ must be ________.

GATE 2024 EC question 38: Block diagram: input R(s) enters a summing junction with sign + on the reference input and sign - on the feedback input. — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
0
(B)
1
(C)
2
(D)
3

Answer and solution

ANSWER

$α = 1$ (with $K = 8$ , and the third closed-loop pole at $s = - 2$ ). Option (B).

Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Unity negative feedback, so the characteristic equation is $1 + C (s) G (s) = 0$ : $1 + \frac{K ( s + α )}{s ^{2} ( s + 4 )} = 0 \Rightarrow s^{3} + 4 s^{2} + K s + K α = 0$
The closed loop is third order, but the question fixes only two poles. The desired pair $- 1 \pm j 3$ gives the quadratic factor $(s + 1)^{2} + 3 = s^{2} + 2 s + 4$ .
Let the third closed-loop pole be at $s = - p$ . The desired characteristic polynomial is then $(s^{2} + 2 s + 4) (s + p) = s^{3} + (2 + p) s^{2} + (4 + 2 p) s + 4 p$
Match the $s^{2}$ coefficients: $2 + p = 4 \Rightarrow p = 2$ , so the third pole is at $s = - 2$ . (Equivalently, the three poles must sum to $- 4$ , and $- 1 - 1 + (- p) = - 4$ gives $p = 2$ .)
Match the $s^{1}$ coefficients: $K = 4 + 2 p = 8$ .
Match the constant terms: $K α = 4 p = 8 \Rightarrow α = \frac{8}{8} = 1$ .
Check: $(s^{2} + 2 s + 4) (s + 2) = s^{3} + 4 s^{2} + 8 s + 8$ , which equals $s^{3} + 4 s^{2} + 8 s + 8 \cdot 1$ . Both $K = 8$ and $α = 1$ are positive, as the question requires, and all three closed-loop poles ( $- 1 \pm j 3$ and $- 2$ ) lie in the left half-plane. So $α = 1$ .

KEY IDEA

The loop $\frac{K ( s + α )}{s ^{2} ( s + 4 )}$ gives a third-order characteristic equation $s^{3} + 4 s^{2} + K s + K α = 0$ , so there are three closed-loop poles. Two of them are specified. Add an unknown third pole $(s + p)$ and match coefficients. The $s^{2}$ coefficient contains no $K$ or $α$ , so it fixes the third pole first: the poles must sum to $- 4$ , which puts it at $s = - 2$ . The other two coefficients then give $K = 8$ and $K α = 8$ .

WATCH OUT FOR

Treating the closed loop as second order, so that $s^{2} + 2 s + 4$ is matched straight against a truncated characteristic equation and the third pole is dropped. A second trap is assuming the third closed-loop pole stays at the compensator pole $s = - 4$ . Closing the loop moves it: the pole sum forces it to $s = - 2$ . Assuming $- 4$ gives $K α = 16$ and an $s^{2}$ coefficient of $6 \neq = 4$ , which is inconsistent. Some also try pole-zero cancellation ( $α = 4$ ), which is not an option and does not place the poles at $- 1 \pm j 3$ .

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 10 Compensators & PID questions, with notes

GATE 2024 EC Q.38: solution and answer

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