GATE 2023 IN Q.37: solution and answer

GATE 2023 · INQ.37 · 2 marks · MCQ

The Laplace transform of the continuous-time signal $x (t) = e^{- 3 t} u (t - 5)$ is ______, where $u (t)$ denotes the continuous-time unit step signal.

(A)
$\frac{e ^{- 5 s}}{s + 3}, R e a l {s} > - 3$
(B)
$\frac{e ^{- 5 (s - 3)}}{s - 3}, R e a l {s} > 3$
(C)
$\frac{e ^{- 5 (s + 3)}}{s + 3}, R e a l {s} > - 3$
(D)
$\frac{e ^{- 5 (s - 3)}}{s + 3}, R e a l {s} > - 3$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The answer is (C): $X (s) = \frac{e ^{- 5 (s + 3)}}{s + 3}$ with $Re {s} > - 3$ .

Official answer: CSource: Official answer key, GATE 2023 IN (IIT Kanpur)

The step switches on at $t = 5$ , but the exponential is written in terms of $t$ , not $(t - 5)$ . To use the time-shift property, rewrite it in terms of $(t - 5)$ : $x (t) = e^{- 3 t} u (t - 5) = e^{- 3 (t - 5) - 15} u (t - 5) = e^{- 15} e^{- 3 (t - 5)} u (t - 5)$
Base pair: $e^{- 3 t} u (t) ⟷ \frac{1}{s + 3}$ , with $ROC: Re {s} > - 3$ .
Time-shift property: $g (t - t_{0}) u (t - t_{0}) ⟷ e^{- s t_{0}} G (s)$ . It leaves the ROC unchanged because $e^{- s t_{0}}$ is finite for every $s$ . With $t_{0} = 5$ : $e^{- 3 (t - 5)} u (t - 5) ⟷ \frac{e ^{- 5 s}}{s + 3}$ , $Re {s} > - 3$ .
Multiply by the constant $e^{- 15}$ : $X (s) = e^{- 15} \cdot \frac{e ^{- 5 s}}{s + 3} = \frac{e ^{- 5 s - 15}}{s + 3} = \frac{e ^{- 5 (s + 3)}}{s + 3}, Re {s} > - 3$
Check by direct integration. $x (t)$ is zero for $t < 5$ , so $X (s) = \int_{5}^{\infty} e^{- 3 t} e^{- s t} d t = \int_{5}^{\infty} e^{- (s + 3) t} d t = [\frac{e ^{- (s + 3) t}}{- ( s + 3 )}]_{5}^{\infty} = \frac{e ^{- 5 (s + 3)}}{s + 3}$ The upper limit goes to zero only when $Re {s + 3} > 0$ , which gives $Re {s} > - 3$ . The signal is right-sided, so this right half-plane ROC is what we expect.
Rule out the others. (A) leaves out the factor $e^{- 15}$ , so it treats the signal as $e^{- 3 (t - 5)} u (t - 5)$ . (B) is the transform of $e^{+ 3 t} u (t - 5)$ . (D) has the wrong sign inside the exponent of the numerator. The answer is option (C).

KEY IDEA

The time-shift property $g (t - t_{0}) u (t - t_{0}) ⟷ e^{- s t_{0}} G (s)$ only works when the whole function is shifted, not just the step. So first write $e^{- 3 t} = e^{- 15} e^{- 3 (t - 5)}$ . The constant $e^{- 15}$ multiplies the transform, and the ROC stays $Re {s} > - 3$ . The same result follows from shifting in $s$ : $e^{- a t} f (t) ⟷ F (s + a)$ , where $u (t - 5) ⟷ \frac{e ^{- 5 s}}{s}$ gives $\frac{e ^{- 5 (s + 3)}}{s + 3}$ .

WATCH OUT FOR

The usual mistake is to apply the time shift to the step alone and write $\frac{e ^{- 5 s}}{s + 3}$ (option A). That drops the factor $e^{- 15}$ , because $e^{- 3 t} u (t - 5)$ is not the same as $e^{- 3 (t - 5)} u (t - 5)$ . Getting the sign of the exponential wrong leads to (B) or (D).

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 8 Laplace questions, with notes

GATE 2023 IN Q.37: solution and answer

Answer and solution

More Laplace questions