GATE 2026 EC Q.43: solution and answer

GATE 2026 · ECQ.43 · 2 marks · MCQ

Consider a real signal $x (t)$ , $- \infty < t < \infty$ , such that

$x (t) = 0$ for $t < 0$ , $x (t) = 2$ for $0 \leq t < 1$ and $x (t) = 0$ for $t \geq 1$ .

Let $E [x (t)] = \int_{- \infty}^{\infty} [x (t)]^{2} d t$ .

Which of the following options correctly represents the ratio,

$E [x (t)] / E [3 x (- 3 t + 5)]$ ?

(A)
$3$
(B)
$1$
(C)
$1/3$
(D)
$1/9$

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Answer and solution

ANSWER

$E [x (t)] = 4$ and $E [3 x (- 3 t + 5)] = 9 \cdot \frac{1}{3} \cdot 4 = 12$ , so the ratio is $\frac{1}{3}$ . The answer is option (C).

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

Energy of the original pulse (height $2$ , width $1$ ): $E [x (t)] = \int_{0}^{1} 2^{2} d t = 4$
Amplitude scaling: $E [A y (t)] = \int_{- \infty}^{\infty} A^{2} [y (t)]^{2} d t = A^{2} E [y (t)]$ , so the factor $3$ multiplies the energy by $3^{2} = 9$ .
Time scaling, reversal and shifting, $y (t) = x (a t + b)$ with $a \neq = 0$ : substitute $τ = a t + b$ , so $d t = d τ / a$ . For $a < 0$ the limits also swap, and the two sign changes cancel, giving $d t \to d τ /∣ a ∣$ in every case: $E [x (a t + b)] = \frac{1}{∣ a ∣} \int_{- \infty}^{\infty} [x (τ)]^{2} d τ = \frac{1}{∣ a ∣} E [x (t)]$ The shift $b$ and the reversal (the sign of $a$ ) do not change the energy.
With $a = - 3$ : $E [x (- 3 t + 5)] = \frac{1}{3} \cdot 4 = \frac{4}{3}$ .
Combine the two rules: $E [3 x (- 3 t + 5)] = 9 \cdot \frac{4}{3} = 12$ .
Direct check: $3 x (- 3 t + 5) = 3 \cdot 2 = 6$ where $0 \leq - 3 t + 5 < 1$ , i.e. $\frac{4}{3} < t \leq \frac{5}{3}$ , an interval of width $\frac{1}{3}$ , and zero elsewhere. Energy $= 6^{2} \cdot \frac{1}{3} = 12$ , which agrees.
Ratio: $\frac{E [ x ( t )]}{E [ 3 x ( - 3 t + 5 )]} = \frac{4}{12} = \frac{1}{3}$ which is option (C).

KEY IDEA

Energy scales with the square of the amplitude and inversely with the magnitude of the time-scale factor: $E [A x (a t + b)] = \frac{A ^{2}}{∣ a ∣} E [x (t)]$ . A time shift or a time reversal leaves the energy unchanged.

WATCH OUT FOR

Three common slips: scaling energy by $A$ instead of $A^{2}$ (this gives $E = 4$ and a ratio of 1, option B); multiplying by $∣ a ∣$ instead of dividing (compression by 3 makes the pulse narrower, so it has less energy, not more; this gives a ratio of 1/27, which is not an option); and thinking the shift $+ 5$ or the minus sign on $t$ changes the energy.

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 18 Signals basics questions, with notes

GATE 2026 EC Q.43: solution and answer

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