GATE 2018 IN Q.10: solution and answer

GATE 2018 · INQ.10 · 1 mark · MCQ

Consider signal $x (t) = {1, 0, ∣ t ∣ \leq 2 ∣ t ∣ > 2$ . Let $δ (t)$ denote the unit impulse (Dirac-delta) function. The value of the integral $\int_{0}^{5} 2 x (t - 3) δ (t - 4) d t$ is

(A)
2
(B)
1
(C)
0
(D)
3

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Answer and solution

ANSWER

The integral equals $2 x (4 - 3) = 2 x (1) = 2$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2018 IN (IIT Guwahati)

Sifting property of the unit impulse: for a function $f (t)$ continuous at $t_{0}$ , $\int_{a}^{b} f (t) δ (t - t_{0}) d t = {f (t_{0}), 0, a < t_{0} < b t_{0} < a or t_{0} > b$
Here the impulse $δ (t - 4)$ sits at $t_{0} = 4$ , which lies strictly inside the limits $0 < 4 < 5$ , so the integral picks up the value of the multiplying function at $t = 4$ .
Take $f (t) = 2 x (t - 3)$ and sift: $\int_{0}^{5} 2 x (t - 3) δ (t - 4) d t = 2 x (4 - 3) = 2 x (1)$
$x (t - 3)$ is the pulse shifted right by 3, so it equals 1 for $1 \leq t \leq 5$ ; at $t = 4$ it is inside this range and away from the edges, so $f (t)$ is continuous there. Equivalently, $∣1∣ \leq 2$ , so $x (1) = 1$ .
Result: $2 \times 1 = 2$ , which is option (A).

KEY IDEA

Sifting property: an impulse at $t_{0}$ inside the limits of integration picks out the value of the multiplying function at $t_{0}$ . Here that gives $2 x (t - 3)$ evaluated at $t = 4$ , which is $2 x (1)$ .

WATCH OUT FOR

Substituting the impulse location into the shift the wrong way, for example evaluating $x (4)$ or $x (4 + 3)$ instead of $x (4 - 3) = x (1)$ . Since $∣4∣ > 2$ and $∣7∣ > 2$ , this gives 0 (option C). The other slip is dropping the factor 2 and answering 1 (option B).

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GATE 2018 IN Q.10: solution and answer

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