ISRO EC 2025 Q.75: solution and answer

ISRO 2025 · ECQ.75 · 1 mark · MCQ

Which of the following systems is "not" causal?

(A)
$h [n] = δ [n]$
(B)
$h [n] = u [n]$
(C)
$h [n] = δ [n] + δ [n + 1]$
(D)
$h [n] = (0.5)^{n} u [n]$

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Answer and solution

ANSWER

Option (C): $h [n] = δ [n] + δ [n + 1]$ is not causal, because $h [- 1] = 1 \neq = 0$ .

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

A discrete-time LTI system is causal if and only if its impulse response satisfies $h [n] = 0$ for all $n < 0$ , because the output at time $n$ is $y [n] = \sum_{k} h [k] x [n - k]$ and a term with $k < 0$ would use the future input $x [n - k]$ .
(a) $h [n] = δ [n]$ is nonzero only at $n = 0$ , so $h [n] = 0$ for $n < 0$ . Causal (the identity system).
(b) $h [n] = u [n]$ is $0$ for $n < 0$ . Causal (an accumulator).
(c) $h [n] = δ [n] + δ [n + 1]$ . Since $δ [n + 1] = 1$ exactly when $n + 1 = 0$ , that is $n = - 1$ , we get $h [- 1] = 1 \neq = 0$ . The impulse response is nonzero for a negative $n$ , so the system is not causal.
Equivalently, $y [n] = x [n] + x [n + 1]$ , so the output at time $n$ needs the future input $x [n + 1]$ .
(d) $h [n] = (0.5)^{n} u [n]$ contains the factor $u [n]$ , so it is $0$ for $n < 0$ . Causal (and also stable, since $0.5 < 1$ ).
Only option (c) has a nonzero response for $n < 0$ , so it is the system that is not causal.

KEY IDEA

An LTI system is causal exactly when $h [n] = 0$ for all $n < 0$ . Find where each shifted impulse or step is nonzero. A term like $δ [n + 1]$ sits at $n = - 1$ , which is in the past, so it uses a future input.

WATCH OUT FOR

Reading $δ [n + 1]$ as a delay. The shift $n + 1$ moves the impulse to $n = - 1$ , which is an advance, and it makes the system depend on the future input $x [n + 1]$ .

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 3 ISRO LTI & convolution questions, with notes