GATE 2026 IN Q.41: solution and answer

GATE 2026 · INQ.41 · 2 marks · MCQ

A signal $x (t)$ is shown below.

[Figure: $x (t)$ ]

The plot of $y (t) = 1 - x (4 - t)$ is shown in ______.

GATE 2026 IN question 41: Plot of x(t) against t. — Figure from the GATE 2026 IN paper. It is part of the question.

(A)
Figure — y(t) = 0 for t < -1; jumps to 1 at t = -1, then falls linearly through 0 at t = 0 to -1 at t = 1; drops to -2 at t = 1 and stays at -2 until t = 5; jumps to 1 at t = 5 and stays at 1 until t = 6; falls linearly from 1 at t = 6 to -1 at t = 7; returns to 0 for t > 7. (t labels -1, 0, 1, 5, 7; y ticks 1, -1, -2.)
(B)
Figure — y(t) = 0 for t < -1; jumps to 3 at t = -1, then falls linearly through 2 at t = 0 to 1 at t = 1; drops to 0 at t = 1 and stays at 0 until t = 5; jumps to 3 at t = 5 and stays at 3 until t = 6; falls linearly from 3 at t = 6 to 1 at t = 7; drops to 0 for t > 7. (t labels -1, 0, 1, 5, 6, 7; y ticks 1, 2, 3.)
(C)
Figure — y(t) = 0 for t < -4; falls linearly from 0 at t = -4 to -2 at t = -3; stays at -2 until t = -2; jumps to 1 at t = -2 and stays at 1 until t = 2; drops to 0 at t = 2, then falls linearly to -2 at t = 4; jumps back to 0 at t = 4 and stays 0. (t labels -4, -3, -2, 0, 2, 4; y ticks 1, -1, -2.)
(D)
Figure — y(t) = 0 for t < -4; jumps to 1 at t = -4, then falls linearly to -1 at t = -3; stays at -1 until t = -2; jumps to 2 at t = -2 and stays at 2 until t = 2; drops to 1 at t = 2, then falls linearly through 0 at t = 3 to -1 at t = 4; jumps back to 0 at t = 4 and stays 0. (t labels -4, -3, -2, 0, 2, 3, 4; y ticks 2, 1, -1.)

Answer and solution

ANSWER

The answer is (B). $y (t) = 1 - x (4 - t)$ falls from 3 to 1 on $(- 1, 1)$ , is 0 on $(1, 5)$ , is 3 on $(5, 6)$ and falls from 3 to 1 on $(6, 7)$ . Like every option, (B) draws 0 outside $(- 1, 7)$ , where strictly $y (t) = 1$ .

Official answer: BSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Read $x (t)$ from the figure piece by piece: $x (t) = 0$ for $t < - 3$ ; a ramp from $0$ at $t = - 3$ down to $- 2$ at $t = - 2$ ; $x (t) = - 2$ on $(- 2, - 1)$ ; $x (t) = 1$ on $(- 1, 3)$ ; a ramp from $0$ at $t = 3$ down to $- 2$ at $t = 5$ ; $x (t) = 0$ for $t > 5$ .
Time operation first. $x (4 - t) = x (- (t - 4))$ , which is $x (- t)$ delayed by 4. Equivalently, it is $x (t)$ reflected about $t = 2$ . The quickest method is to map each point directly: the value $x (τ)$ appears at $t = 4 - τ$ . So the breakpoints $τ = - 3, - 2, - 1, 3, 5$ move to $t = 7, 6, 5, 1, - 1$ , in reverse order.
Segment by segment, $x (4 - t)$ is: a ramp from $- 2$ at $t = - 1$ up to $0$ at $t = 1$ (the image of the ramp on $(3, 5)$ ); $1$ on $(1, 5)$ ; $- 2$ on $(5, 6)$ ; a ramp from $- 2$ at $t = 6$ up to $0$ at $t = 7$ (the image of the ramp on $(- 3, - 2)$ ); and $0$ everywhere else.
Amplitude operations: negate, then add 1. So $y (t) = 1 - x (4 - t)$ falls linearly from $3$ to $1$ on $(- 1, 1)$ , is $1 - 1 = 0$ on $(1, 5)$ , is $1 - (- 2) = 3$ on $(5, 6)$ , and falls linearly from $3$ to $1$ on $(6, 7)$ .
Check two points. First, $y (0) = 1 - x (4)$ . $x (4)$ is halfway down the ramp from $0$ to $- 2$ , so $x (4) = - 1$ and $y (0) = 2$ , which matches the label 2 at $t = 0$ in (B). Second, $y (5.5) = 1 - x (- 1.5) = 1 - (- 2) = 3$ , which matches the level-3 plateau of (B) on $(5, 6)$ .
Eliminate the other options. (C) is $x (t + 1)$ , a shift with no reversal: its breakpoints are at $- 4, - 3, - 2, 2, 4$ . (D) is $1 + x (t + 1)$ on $(- 4, 4)$ . (A) is $- 1 - x (4 - t)$ on $(- 1, 7)$ . It has the right shape but the constant has the wrong sign, so it sits at $- 2$ instead of $0$ on $(1, 5)$ . Only (B) agrees with $1 - x (4 - t)$ on every segment where the signal varies.
Caveat on the options: strictly, for $t < - 1$ and $t > 7$ we have $x (4 - t) = 0$ , so $y (t) = 1$ there. Every option draws $0$ on those intervals, so no option is exact everywhere. (B) is the only option that matches $y (t)$ on $- 1 < t < 7$ , where the signal actually varies, so (B) is the intended answer.

KEY IDEA

Do the time operation on the argument first and the amplitude operations last. For $x (4 - t)$ , the value $x (τ)$ lands at $t = 4 - τ$ , which reflects $x (t)$ about $t = 2$ and reverses the order of the breakpoints. Then negate and add 1. One point check, such as $y (0) = 1 - x (4) = 2$ , separates the options quickly.

WATCH OUT FOR

Shifting in the wrong direction is the usual error. Writing $x (4 - t)$ as $x (- t)$ advanced by 4, i.e. $x (- (t + 4))$ , puts the breakpoints near $t = - 7 \dots - 1$ instead of $- 1 \dots 7$ . Other slips: shifting without reversing, which gives (C), $x (t + 1)$ . Letting the minus sign act on the constant as well, which gives (A), $- 1 - x (4 - t)$ . Adding 1 without negating first.

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GATE 2026 IN Q.41: solution and answer

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