GATE 2026 EC Q.16: solution and answer

GATE 2026 · ECQ.16 · 1 mark · MCQ

The response of a discrete time system $y [n]$ obeys the following relation:

y [n] = \frac{5}{6} y [n - 1] - \frac{1}{6} y [n - 2] + x [n] .

The input to the system is $x [n] = δ [n] - \frac{1}{3} δ [n - 1]$ .

Which of the following options is TRUE for $y [n]$ ?

(A)
Stable and causal response
(B)
Stable and non-causal response
(C)
Unstable and causal response
(D)
Unstable and non-causal response

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Answer and solution

ANSWER

$y [n] = (\frac{1}{2})^{n} u [n]$ . It is zero for $n < 0$ and absolutely summable (sum $= 2$ ), so the response is stable and causal: option (A).

Official answer: ASource: Official answer key, GATE 2026 EC (IIT Guwahati)

Take the z-transform of the difference equation, assuming initial rest (the system is run forward from $y [- 1] = y [- 2] = 0$ ): $Y (z) (1 - \frac{5}{6} z^{- 1} + \frac{1}{6} z^{- 2}) = X (z)$
Factor the denominator. The roots of $z^{2} - \frac{5}{6} z + \frac{1}{6} = 0$ are $z = \frac{1}{2}$ and $z = \frac{1}{3}$ , since $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ and $\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ . So $H (z) = \frac{1}{( 1 - \frac{1}{2} z ^{- 1} ) ( 1 - \frac{1}{3} z ^{- 1} )}$
Transform the input using $δ [n - k] \leftrightarrow z^{- k}$ : $X (z) = 1 - \frac{1}{3} z^{- 1}$ .
Multiply. The input's zero at $z = \frac{1}{3}$ cancels the pole at $z = \frac{1}{3}$ : $Y (z) = H (z) X (z) = \frac{1}{1 - \frac{1}{2} z ^{- 1}}$
A recursive difference equation run forward from rest gives a right-sided output, so the ROC is $∣ z ∣ > \frac{1}{2}$ . Inverting, $y [n] = (\frac{1}{2})^{n} u [n]$ .
Check by direct recursion: $y [0] = x [0] = 1$ , $y [1] = \frac{5}{6} \cdot 1 - \frac{1}{3} = \frac{1}{2}$ , $y [2] = \frac{5}{6} \cdot \frac{1}{2} - \frac{1}{6} \cdot 1 + 0 = \frac{1}{4}$ . This matches $(\frac{1}{2})^{n}$ .
$y [n] = 0$ for $n < 0$ , so the response is causal. Also $\sum_{n} ∣ y [n] ∣ = \sum_{n \geq 0} (\frac{1}{2})^{n} = 2 < \infty$ , and the ROC $∣ z ∣ > \frac{1}{2}$ contains the unit circle $∣ z ∣ = 1$ , so the response is stable (absolutely summable). The answer is (A).

KEY IDEA

Find $Y (z) = H (z) X (z)$ . The input's zero at $z = \frac{1}{3}$ cancels one system pole, which leaves one pole at $z = \frac{1}{2}$ . A recursion run forward from rest gives a right-sided output, so the ROC is $∣ z ∣ > \frac{1}{2}$ . That ROC contains the unit circle, so $y [n] = (\frac{1}{2})^{n} u [n]$ is both causal and absolutely summable.

WATCH OUT FOR

Taking the left-sided ROC $∣ z ∣ < \frac{1}{2}$ gives $y [n] = - (\frac{1}{2})^{n} u [- n - 1]$ , which grows like $2^{∣ n ∣}$ as $n \to - \infty$ . That points to option (D), "unstable and non-causal". But a difference equation computed forward in time from rest always gives a right-sided response. A second slip is to judge the response by the system's poles alone and miss the pole-zero cancellation. Here both poles ( $\frac{1}{2}$ and $\frac{1}{3}$ ) lie inside the unit circle anyway, so the conclusion is the same, but the cancellation is what makes $y [n]$ a single geometric sequence.

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 12 z-transform questions, with notes

GATE 2026 EC Q.16: solution and answer

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