GATE 2022 IN Q.15: solution and answer

GATE 2022 · INQ.15 · 1 mark · MCQ

A unity-gain negative-feedback control system has a loop-gain $L (s)$ given by
$L (s) = \frac{6}{s ( s - 5 )}$
The closed-loop system is ________

(A)
Causal and stable
(B)
Causal and unstable
(C)
Non-causal and stable
(D)
Non-causal and unstable

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Answer and solution

ANSWER

The closed-loop characteristic equation is $s^{2} - 5 s + 6 = (s - 2) (s - 3) = 0$ . Both poles are in the right half-plane and the system is physically realised, so it is causal and unstable. The answer is (B).

Official answer: BSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Closed-loop transfer function (unity negative feedback): $T (s) = \frac{L ( s )}{1 + L ( s )} = \frac{6}{s ( s - 5 ) + 6} = \frac{6}{s ^{2} - 5 s + 6}$
Characteristic equation: $s^{2} - 5 s + 6 = 0$ .
Routh array: row $s^{2}$ : $1, 6$ ; row $s^{1}$ : $- 5, 0$ ; row $s^{0}$ : $6$ . The first column is $1, - 5, 6$ . It changes sign twice ( $1 \to - 5$ and $- 5 \to 6$ ), so there are two right-half-plane poles.
Check by factoring: $s^{2} - 5 s + 6 = (s - 2) (s - 3)$ , so the closed-loop poles are $s = 2$ and $s = 3$ . Both lie in the right half-plane, so the closed loop is unstable.
(Quick check without Routh: a second-order polynomial is Hurwitz only when all its coefficients have the same sign. Here they are $+ 1, - 5, + 6$ , so instability is clear straight away.)
Causality: $T (s)$ is a strictly proper rational function (numerator degree $0$ < denominator degree $2$ ). It comes from a physical feedback loop built around a causal plant, so its region of convergence is the right-sided one, $Re (s) > 3$ , and its impulse response is zero for $t < 0$ . The system is therefore causal. Instability does not make it non-causal. The growing modes $e^{2 t}$ and $e^{3 t}$ exist only for $t \geq 0$ . It is unstable because the ROC $Re (s) > 3$ does not contain the $j ω$ axis.
So the closed-loop system is causal and unstable, which is option (B).

KEY IDEA

Stability depends on the closed-loop poles, which are the roots of $1 + L (s) = 0$ , not the open-loop poles. Here they are $s = 2$ and $s = 3$ , so the system is unstable. Causality is a separate property. A feedback system built from causal physical parts is always causal, and its right-sided ROC ( $Re (s) > 3$ ) leaves out the $j ω$ axis, which is why it is unstable.

WATCH OUT FOR

Treating instability as if it meant non-causality and choosing (D). Another mistake is judging stability from the open-loop pole at $s = 5$ alone, or thinking that feedback with positive gain $6$ must stabilise it. The closed-loop polynomial $s^{2} - 5 s + 6$ has mixed-sign coefficients, so the closed loop is still unstable.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 15 Routh–Hurwitz questions, with notes

GATE 2022 IN Q.15: solution and answer

Answer and solution

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