ISRO EC 2025 Q.17: solution and answer

ISRO 2025 · ECQ.17 · 1 mark · MCQ

The $Z$ -transform of a signal $x [n] = δ [n - k]$ , (where $k$ is a positive integer) is :

(A)
$z^{k}$
(B)
$z^{- k}$
(C)
$1$
(D)
$δ [z]$

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Answer and solution

ANSWER

The Z-transform of $δ [n - k]$ is $z^{- k}$ , which is option (b).

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

The Z-transform of a discrete-time signal is $X (z) = \sum_{n = - \infty}^{\infty} x [n] z^{- n}$ .
Here $x [n] = δ [n - k]$ , which equals $1$ only at $n = k$ and $0$ for every other $n$ .
So only one term of the sum survives: $X (z) = 1 \cdot z^{- k} = z^{- k}$ .
Check with the time-shift property: $δ [n] \leftrightarrow 1$ , and delaying a signal by $k$ samples multiplies its transform by $z^{- k}$ , giving $z^{- k}$ again.
The transform is defined for every $z$ except $z = 0$ , because $k > 0$ puts $z^{- k}$ in the denominator.
Option (a), $z^{k}$ , would be the transform of the advanced impulse $δ [n + k]$ . Option (c), $1$ , is the transform of the undelayed impulse $δ [n]$ . Option (d) is not meaningful, since $δ [\cdot]$ is a function of integer time, not of the complex variable $z$ .

KEY IDEA

A delay of $k$ samples in time multiplies the Z-transform by $z^{- k}$ , so $δ [n - k] \leftrightarrow z^{- k}$ . The sum $\sum x [n] z^{- n}$ keeps just the single term at $n = k$ .

WATCH OUT FOR

Mixing up the sign of the exponent: a delay $δ [n - k]$ gives $z^{- k}$ , while an advance $δ [n + k]$ would give $z^{k}$ . Another slip is choosing $1$ , the transform of $δ [n]$ with no shift.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 5 ISRO z-transform questions, with notes

ISRO EC 2025 Q.17: solution and answer

Answer and solution

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