GATE 2024 IN Q.40: solution and answer

GATE 2024 · INQ.40 · 2 marks · MCQ

Laplace transform of a signal $x (t)$ is

X (s) = \frac{1}{s ^{2} + 13 s + 42}

Let $u (t)$ be the unit step function. Choose the signal $x (t)$ from the following options if the region of convergence is $- 7 < Re {s} < - 6$ .

(A)
$- e^{- 6 t} u (t) - e^{- 7 t} u (- t)$
(B)
$- e^{- 6 t} u (- t) - e^{- 7 t} u (t)$
(C)
$e^{- 6 t} u (t) - e^{- 7 t} u (- t)$
(D)
$- e^{- 6 t} u (- t) - e^{- 7 t} u (- t)$

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Answer and solution

ANSWER

$x (t) = - e^{- 6 t} u (- t) - e^{- 7 t} u (t)$ , option (B).

Official answer: BSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

Factor the denominator: $s^{2} + 13 s + 42 = (s + 6) (s + 7)$ , so the poles are at $s = - 6$ and $s = - 7$ .
Partial fractions: $X (s) = \frac{1}{( s + 6 ) ( s + 7 )} = \frac{1}{s + 6} - \frac{1}{s + 7}$ Check: $\frac{( s + 7 ) - ( s + 6 )}{( s + 6 ) ( s + 7 )} = \frac{1}{( s + 6 ) ( s + 7 )}$ .
The ROC $- 7 < Re {s} < - 6$ is a vertical strip between the two poles, so $x (t)$ is two-sided. Each pole's term is right-sided or left-sided depending on which side of the strip the pole lies.
The pole $s = - 6$ lies to the RIGHT of the strip, so its term must have ROC $Re {s} < - 6$ (left-sided). Using the pair $- e^{- a t} u (- t) \leftrightarrow \frac{1}{s + a}$ , $Re {s} < - a$ : $\frac{1}{s + 6} \to - e^{- 6 t} u (- t)$
The pole $s = - 7$ lies to the LEFT of the strip, so its term must have ROC $Re {s} > - 7$ (right-sided). Using the pair $e^{- a t} u (t) \leftrightarrow \frac{1}{s + a}$ , $Re {s} > - a$ : $- \frac{1}{s + 7} \to - e^{- 7 t} u (t)$
Add the two terms: $x (t) = - e^{- 6 t} u (- t) - e^{- 7 t} u (t)$ The intersection of the two ROCs, $Re {s} < - 6$ and $Re {s} > - 7$ , is exactly the given strip, which confirms it. This is option (B).

KEY IDEA

When the ROC is a strip between poles, a pole to the right of the strip gives a left-sided (anti-causal) term, $- e^{- a t} u (- t)$ , and a pole to the left gives a right-sided (causal) term, $e^{- a t} u (t)$ . The ROC of the whole signal is where the individual ROCs overlap.

WATCH OUT FOR

Treating every term as causal (the one-sided Laplace habit), or dropping the minus sign in the left-sided pair $- e^{- a t} u (- t) \leftrightarrow \frac{1}{s + a}$ . Quick check on the other options: in (A) the term $- e^{- 6 t} u (t)$ needs $Re {s} > - 6$ and $- e^{- 7 t} u (- t)$ needs $Re {s} < - 7$ , and no $s$ satisfies both. (C) has the same empty ROC. (D) is purely left-sided with ROC $Re {s} < - 7$ , and its transform is $\frac{1}{s + 6} + \frac{1}{s + 7}$ , which is not $X (s)$ .

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GATE 2024 IN Q.40: solution and answer

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