GATE 2025 EE Q.40: solution and answer

GATE 2025 · EEQ.40 · 2 marks · MCQ

Let continuous-time signals $x_{1} (t)$ and $x_{2} (t)$ be

x_{1} (t) = ⎩ ⎨ ⎧ 1, 2 - t, 0, t \in [0, 1] t \in [1, 2] otherwise and x_{2} (t) = ⎩ ⎨ ⎧ t, 2 - t, 0, t \in [0, 1] t \in [1, 2] otherwise

Consider the convolution $y (t) = x_{1} (t) * x_{2} (t)$ . Then $\int_{- \infty}^{\infty} y (t) d t$ is

(A)
1.5
(B)
2.5
(C)
3.5
(D)
4

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Answer and solution

ANSWER

The area under $y (t)$ equals (area of $x_{1}$ ) times (area of $x_{2}$ ), which is $1.5 \times 1 = 1.5$ , so the answer is option (A).

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

Area property of convolution: if $y (t) = x_{1} (t) * x_{2} (t)$ , then $\int_{- \infty}^{\infty} y (t) d t = (\int_{- \infty}^{\infty} x_{1} (t) d t) (\int_{- \infty}^{\infty} x_{2} (t) d t)$ This is $Y (0) = X_{1} (0) X_{2} (0)$ , because $X (0) = \int_{- \infty}^{\infty} x (t) d t$ . Directly: $\int y (t) d t = \int x_{1} (τ) [\int x_{2} (t - τ) d t] d τ = A_{2} \int x_{1} (τ) d τ = A_{1} A_{2}$ . Swapping the order of integration is valid here because both signals are bounded and last only a finite time.
Area of $x_{1} (t)$ : a unit rectangle on $[0, 1]$ followed by a ramp falling from 1 to 0 on $[1, 2]$ : $A_{1} = \int_{0}^{1} 1 d t + \int_{1}^{2} (2 - t) d t = 1 + \frac{1}{2} = 1.5$
Area of $x_{2} (t)$ : a triangle with base $[0, 2]$ and peak 1 at $t = 1$ : $A_{2} = \int_{0}^{1} t d t + \int_{1}^{2} (2 - t) d t = \frac{1}{2} + \frac{1}{2} = 1$
Multiply the areas: $\int_{- \infty}^{\infty} y (t) d t = A_{1} A_{2} = 1.5 \times 1 = 1.5$ The answer is option (A).

KEY IDEA

The area under a convolution is the product of the areas of the two signals, $\int y d t = (\int x_{1} d t) (\int x_{2} d t)$ , which is the same statement as $Y (0) = X_{1} (0) X_{2} (0)$ . So you never need to work out $y (t)$ itself.

WATCH OUT FOR

Adding the two areas instead of multiplying them: $1.5 + 1 = 2.5$ is exactly option (B). A second trap is counting only the rectangle part of $x_{1} (t)$ and taking its area as 1, which drops the triangle of area $\frac{1}{2}$ on $[1, 2]$ . Working out the full piecewise convolution also reaches 1.5, but it takes much longer and leaves more room for algebra slips.

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GATE 2025 EE Q.40: solution and answer

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