GATE 2022 EC Q.44: solution and answer

GATE 2022 · ECQ.44 · 2 marks · MCQ

Consider an even polynomial $p (s)$ given by
$p (s) = s^{4} + 5 s^{2} + 4 + K,$
where $K$ is an unknown real parameter. The complete range of $K$ for which $p (s)$ has all its roots on the imaginary axis is ________.

(A)
$- 4 \leq K \leq \frac{9}{4}$
(B)
$- 3 \leq K \leq \frac{9}{2}$
(C)
$- 6 \leq K \leq \frac{5}{4}$
(D)
$- 5 \leq K \leq 0$

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Answer and solution

ANSWER

All roots of $p (s)$ lie on the imaginary axis for $- 4 \leq K \leq \frac{9}{4}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

$p (s)$ is even, so its roots are symmetric about the origin: every root $s_{0}$ comes with $- s_{0}$ . If a root had a non-zero real part, either it or its mirror $- s_{0}$ would lie in the RHP. So all roots are on the $j ω$ axis exactly when $p (s)$ has no RHP roots.
Routh array: $p (s)$ has no $s^{3}$ or $s^{1}$ terms, so the $s^{3}$ row is entirely zero. $p (s)$ is then its own auxiliary polynomial, $A (s) = s^{4} + 5 s^{2} + (4 + K)$ . Replace the zero row with the coefficients of $\frac{d A}{d s} = 4 s^{3} + 10 s$ .
Complete the array: $s^{4} s^{3} s^{2} s^{1} s^{0} : 154 + K : 410 : \frac{5}{2} 4 + K : \frac{9 - 4 K}{5/2} : 4 + K$ Here $s^{2}$ : $\frac{4 \cdot 5 - 1 \cdot 10}{4} = \frac{5}{2}$ , and $s^{1}$ : $\frac{\frac{5}{2} \cdot 10 - 4 ( 4 + K )}{5/2} = \frac{9 - 4 K}{5/2}$ .
The number of sign changes in the first column from the $s^{3}$ row down equals the number of RHP roots. There are none when $9 - 4 K > 0$ and $4 + K > 0$ , that is, for $- 4 < K < \frac{9}{4}$ . By the mirror symmetry there are then no LHP roots either. For $K > \frac{9}{4}$ the $s^{1}$ entry is negative, giving two sign changes and so two RHP roots. For $K < - 4$ the $s^{0}$ entry is negative, giving one sign change and one RHP root. The two endpoints make a first-column entry zero, so they are checked directly in step 6.
Check with $x = s^{2}$ : $x^{2} + 5 x + (4 + K) = 0$ . For $s = \pm j - x$ to be imaginary (or zero), both $x$ -roots must be real and $\leq 0$ . Real roots: $25 - 4 (4 + K) \geq 0 \Rightarrow K \leq \frac{9}{4}$ . Both non-positive: the sum is $- 5 < 0$ and the product $4 + K \geq 0 \Rightarrow K \geq - 4$ .
Endpoints are included. At $K = - 4$ , $p (s) = s^{2} (s^{2} + 5)$ and the roots are $s = 0, 0, \pm j 5$ . The root $s = 0$ lies on the imaginary axis. At $K = \frac{9}{4}$ , $p (s) = (s^{2} + \frac{5}{2})^{2}$ and the roots are $s = \pm j 5/2$ , each repeated. In both cases every root is on the axis.
So the complete range is $- 4 \leq K \leq \frac{9}{4}$ , which is option (A).

KEY IDEA

The roots of an even polynomial are symmetric about the origin, so 'all roots on the $j ω$ axis' means the same as 'no RHP roots'. In the Routh array the $s^{3}$ row vanishes and is rebuilt from $d A / d s$ . Substituting $x = s^{2}$ turns the question into one about a quadratic: it needs real roots that are both $\leq 0$ . That gives $K \leq 9/4$ from the discriminant and $K \geq - 4$ from the product.

WATCH OUT FOR

Checking only that the coefficients are positive ( $4 + K \geq 0$ ) and missing the discriminant condition $9 - 4 K \geq 0$ . Without it the $s^{2}$ -roots become complex and the roots of $p (s)$ leave the axis in mirror pairs. A second common error is to drop the endpoints. At $K = - 4$ there is a double root at $s = 0$ , which is still on the imaginary axis. At $K = 9/4$ the roots are the repeated pair $\pm j 5/2$ . A third is to give up when the $s^{3}$ row comes out all zero, instead of replacing it with the derivative of the auxiliary polynomial.

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

GATE 2022 EC Q.44: solution and answer

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