GATE 2023 EE Q.31: solution and answer

GATE 2023 · EEQ.31 · 1 mark · NAT

For the signals $x (t)$ and $y (t)$ shown in the figure, $z (t) = x (t) * y (t)$ is maximum at $t = T_{1}$ . Then $T_{1}$ in seconds is ________ (Round off to the nearest integer).

GATE 2023 EE question 31: Two plots against t. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

$T_{1} = 4 s$ . The peak value is $z (4) = 2$ , reached when the width-2 window $[t - 1, t + 1]$ covers exactly the positive part of $y (t)$ on $[3, 5]$ .

Official answer: 4Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Read the signals from the figure: $x (t) = 1$ for $- 1 < t < 1$ and zero elsewhere. $y (t) = t - 3$ for $1 < t < 5$ and zero elsewhere, so it runs from $- 2$ at $t = 1$ , through $0$ at $t = 3$ , to $+ 2$ at $t = 5$ .
Convolving with a unit rectangular pulse gives a moving integral over a window of width 2. Substitute $λ = t - τ$ : $z (t) = \int_{- \infty}^{\infty} x (τ) y (t - τ) d τ = \int_{- 1}^{1} y (t - τ) d τ = \int_{t - 1}^{t + 1} y (λ) d λ$ So $z (t)$ is the area under $y$ inside the window $[t - 1, t + 1]$ .
Differentiate with the Leibniz rule: $\frac{d z}{d t} = y (t + 1) - y (t - 1)$ . The maximum is where this changes sign from positive to negative.
Full overlap, $2 \leq t \leq 4$ (the window $[t - 1, t + 1]$ lies inside $[1, 5]$ ): $\frac{d z}{d t} = (t - 2) - (t - 4) = 2 > 0$ . Also $z (t) = \int_{t - 1}^{t + 1} (λ - 3) d λ = \frac{( t - 2 ) ^{2} - ( t - 4 ) ^{2}}{2} = 2 (t - 3),$ which rises to $z (4) = 2$ .
Trailing edge, $4 < t < 6$ (the upper end of the window has passed $t = 5$ ): $\frac{d z}{d t} = 0 - (t - 4) < 0$ , and $z (t) = \int_{t - 1}^{5} (λ - 3) d λ = 2 - \frac{( t - 4 ) ^{2}}{2},$ which falls from $2$ at $t = 4$ to $0$ at $t = 6$ .
Leading edge, $0 < t < 2$ : $z (t) = \int_{1}^{t + 1} (λ - 3) d λ = \frac{( t - 2 ) ^{2} - 4}{2} \leq 0,$ so no larger value is possible there. $z (t) = 0$ for $t \leq 0$ and $t \geq 6$ , because the window then misses $(1, 5)$ completely.
So $z (t)$ has its global maximum $z_{m a x} = 2$ at $t = 4$ . Here the window is $[3, 5]$ , which covers exactly the positive part of $y$ . Therefore $T_{1} = 4 s$ .

KEY IDEA

Convolving with a unit rectangular pulse on $(- 1, 1)$ gives a sliding-window integral: $z (t) = \int_{t - 1}^{t + 1} y (λ) d λ$ . $z$ is largest when the window holds the most positive area and none of the negative area. $y$ is positive only on $(3, 5)$ , and that interval is exactly as wide as the window, so the best window is $[3, 5]$ . Its centre gives $t = 4$ . The same result follows from $d z / d t = y (t + 1) - y (t - 1)$ changing from $+ 2$ to negative at $t = 4$ .

WATCH OUT FOR

Answering $5$ because that is where $y (t)$ peaks. The maximum of a convolution is not at the peak of either signal: at $t = 5$ the window $[4, 6]$ loses the area on $(5, 6)$ , so $z (5) = 1.5 < 2$ . Another mistake is answering $3$ , the centre of $y$ 's support. There the positive and negative areas cancel and $z (3) = 0$ . A third error is to use the window $[t, t + 2]$ or $[t - 2, t]$ instead of $[t - 1, t + 1]$ . That forgets the pulse is centred at the origin and moves the answer by one second.

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GATE 2023 EE Q.31: solution and answer

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