ISRO EC 2025 Q.1: solution and answer

ISRO 2025 · ECQ.1 · 1 mark · MCQ

For the following function, poles and zeros will be : $x (t) = δ (t) - \frac{4}{3} e^{- t} u (t) + \frac{1}{3} e^{2 t} u (t)$

(A)
Poles at -1 and 2, Zero at -1
(B)
Poles at -1 and 2, Zero at 1
(C)
Poles at -1 and -2, Zero at -1
(D)
Poles at 1 and -2, Zero at -1

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Answer and solution

ANSWER

Poles at $- 1$ and $2$ , zero at $1$ (option B).

Official answer: BSource: ISRO's final answer key of 15 May 2026

Take the Laplace transform of each term: $δ (t) \to 1$ , $e^{- t} u (t) \to \frac{1}{s + 1}$ , $e^{2 t} u (t) \to \frac{1}{s - 2}$ .
$X (s) = 1 - \frac{4}{3} \cdot \frac{1}{s + 1} + \frac{1}{3} \cdot \frac{1}{s - 2}$
The poles come from the denominators: $s = - 1$ and $s = 2$ . This rules out options (c) and (d).
Put everything over the common denominator $(s + 1) (s - 2)$ : numerator $= (s + 1) (s - 2) - \frac{4}{3} (s - 2) + \frac{1}{3} (s + 1)$ .
Expand: $s^{2} - s - 2 - \frac{4}{3} s + \frac{8}{3} + \frac{1}{3} s + \frac{1}{3} = s^{2} - 2 s + 1$ .
$X (s) = \frac{( s - 1 ) ^{2}}{( s + 1 ) ( s - 2 )}$
The numerator vanishes at $s = 1$ (a double zero), so the zero is at $s = 1$ , not at $- 1$ . This rules out option (a).
Poles at $- 1$ and $2$ , zero at $1$ : option (b).

KEY IDEA

Transform each term, combine into a single rational function, and read poles from the denominator and zeros from the numerator after simplifying.

WATCH OUT FOR

Reading the zero from the individual terms, or forgetting that the impulse term contributes $1$ to the numerator, which moves the zero from $- 1$ to $1$ .

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO Laplace questions, with notes

ISRO EC 2025 Q.1: solution and answer

Answer and solution

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