GATE 2025 IN Q.27: solution and answer

GATE 2025 · INQ.27 · 1 mark · MCQ

For the transfer function $G (s) = 1 + \frac{2 s - 1}{s ^{3} + 5 s ^{2} + 3 s + 22}$ , the number of zeros lying in the left half of the $s$ -plane is

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

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Answer and solution

ANSWER

All three zeros of $G (s)$ , the roots of $s^{3} + 5 s^{2} + 5 s + 21$ , lie in the left half of the $s$ -plane. Answer: (D) 3.

Official answer: DSource: Official answer key, GATE 2025 IN (IIT Roorkee)

Put $G (s)$ over one denominator: $G (s) = \frac{( s ^{3} + 5 s ^{2} + 3 s + 22 ) + ( 2 s - 1 )}{s ^{3} + 5 s ^{2} + 3 s + 22} = \frac{s ^{3} + 5 s ^{2} + 5 s + 21}{s ^{3} + 5 s ^{2} + 3 s + 22}$
The zeros of $G (s)$ are the roots of the combined numerator $N (s) = s^{3} + 5 s^{2} + 5 s + 21$ . They are not the root of $2 s - 1$ , which is only the numerator of the second term.
Check for pole-zero cancellation. $N (s) - D (s) = 2 s - 1$ , so a common root would have to be $s = \frac{1}{2}$ . But $N (\frac{1}{2}) = 0.125 + 1.25 + 2.5 + 21 = 24.875 \neq = 0$ . Nothing cancels, so all three zeros remain.
Routh array for $N (s)$ : $s^{3} s^{2} s^{1} s^{0} : 15 : 521 : \frac{5 \cdot 5 - 1 \cdot 21}{5} = \frac{4}{5} : 21$
The first column $1, 5, \frac{4}{5}, 21$ is all positive. There are no sign changes and no zero rows, so $N (s)$ has no roots in the right half-plane and none on the $j ω$ axis.
So all 3 zeros lie in the left half of the $s$ -plane. (Numerical check: one real zero at about $s \approx - 4.86$ and a complex pair at about $- 0.07 \pm j 2.08$ .) Answer: (D) 3.

KEY IDEA

Because of the leading $1 +$ , you first have to rewrite $G (s)$ as a single fraction. Its zeros are then the roots of $D (s) + (2 s - 1) = s^{3} + 5 s^{2} + 5 s + 21$ . Apply the Routh criterion to that combined numerator, not to the denominator. For a cubic $s^{3} + a_{2} s^{2} + a_{1} s + a_{0}$ with all coefficients positive, every root is in the left half-plane exactly when $a_{2} a_{1} > a_{0}$ . Here $5 \cdot 5 = 25 > 21$ , so all three zeros are in the LHP.

WATCH OUT FOR

Reading the zeros off the $2 s - 1$ term gives a single zero at $s = + \frac{1}{2}$ , which leads to the wrong answer (A) 0. A second mistake is running the Routh test on the denominator $s^{3} + 5 s^{2} + 3 s + 22$ . That polynomial gives the poles, not the zeros: its $s^{1}$ entry is $\frac{15 - 22}{5} < 0$ , so it has two right-half-plane roots, and this can mislead you into picking (B) 1.

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GATE 2025 IN Q.27: solution and answer

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