GATE 2022 EE Q.19: solution and answer

GATE 2022 · EEQ.19 · 1 mark · MCQ

The open loop transfer function of a unity gain negative feedback system is given by $G (s) = \frac{k}{s ^{2} + 4 s - 5}$ . The range of $k$ for which the system is stable, is

(A)
$k > 3$
(B)
$k < 3$
(C)
$k > 5$
(D)
$k < 5$

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Answer and solution

ANSWER

The closed-loop characteristic equation is $s^{2} + 4 s + (k - 5) = 0$ , so the system is stable for $k > 5$ . The answer is option (C).

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

For unity negative feedback the closed-loop characteristic equation is $1 + G (s) = 0$ . Multiplying through by $s^{2} + 4 s - 5$ gives $s^{2} + 4 s - 5 + k = 0 \Rightarrow s^{2} + 4 s + (k - 5) = 0$
Routh array: row $s^{2}$ : $1, k - 5$ ; row $s^{1}$ : $4, 0$ ; row $s^{0}$ : $\frac{4 ( k - 5 ) - 1 \cdot 0}{4} = k - 5$ .
By the Routh criterion the system is stable only if the first column $1, 4, k - 5$ has no sign change. The first two entries are positive, so we need $k - 5 > 0$ .
Cross-check: a second-order polynomial is Hurwitz if and only if all its coefficients have the same sign. Here that means $1 > 0$ , $4 > 0$ and $k - 5 > 0$ , which gives the same condition.
Boundary check: at $k = 5$ the equation becomes $s (s + 4) = 0$ . It has a pole at the origin, so the system is only marginally stable there, and the inequality must be strict.
So the system is stable for $k > 5$ , which is option (C).

KEY IDEA

Stability depends on the closed-loop poles, the roots of $1 + G (s) = 0$ , not on the open-loop poles. The open-loop plant has a right-half-plane pole at $s = 1$ , since $s^{2} + 4 s - 5 = (s + 5) (s - 1)$ . Feedback gain $k$ moves the constant term from $- 5$ to $k - 5$ . For a quadratic, all coefficients positive is both necessary and sufficient for stability.

WATCH OUT FOR

One mistake is to apply the Routh test to the open-loop denominator $s^{2} + 4 s - 5$ and conclude the system is unstable for every $k$ . Another is dropping the sign of the $- 5$ and writing $k + 5 > 0$ . A third is treating $k = 5$ as stable, which gives $k \geq 5$ : the characteristic equation is then $s (s + 4) = 0$ , a pole at the origin, which is only marginal stability. Options (A) and (B) use 3, a number that has no role in this characteristic equation.

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

GATE 2022 EE Q.19: solution and answer

Answer and solution

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