GATE 2024 EE Q.16: solution and answer

GATE 2024 · EEQ.16 · 1 mark · MCQ

If $u (t)$ is the unit step function, then the region of convergence (ROC) of the Laplace transform of the signal

x (t) = e^{t^{2}} [u (t - 1) - u (t - 10)]

(A)
$- \infty < Re (s) < \infty$
(B)
$Re (s) \geq 10$
(C)
$Re (s) \leq 1$
(D)
$1 \leq Re (s) \leq 10$

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Answer and solution

ANSWER

The signal is non-zero only for $1 \leq t < 10$ and is bounded there, so its Laplace integral converges for every $s$ . The ROC is the entire $s$ -plane, $- \infty < Re (s) < \infty$ : option (A).

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Window: $u (t - 1) - u (t - 10)$ is $1$ for $1 \leq t < 10$ and $0$ everywhere else. So $x (t) = e^{t^{2}}$ only on $[1, 10)$ and $x (t) = 0$ outside it. The signal has finite duration.
It is also bounded on that interval. $e^{t^{2}}$ increases for $t > 0$ , so $∣ x (t) ∣ \leq e^{100}$ : a very large number, but finite.
Write out the transform: $X (s) = \int_{1}^{10} e^{t^{2}} e^{- s t} d t .$
Bound the integrand for $s = σ + j ω$ : $∣ e^{t^{2}} e^{- s t} ∣ = e^{t^{2}} e^{- σ t}$ . For a fixed real $σ$ , $e^{- σ t}$ is monotonic in $t$ , so on $[1, 10]$ it is largest at one of the endpoints. The integrand therefore stays below $e^{100} max (e^{- σ}, e^{- 10 σ})$ , which is finite.
The integrand is bounded and the interval has finite length $9$ , so $\int_{1}^{10} e^{t^{2}} e^{- s t} d t \leq 9 e^{100} max (e^{- σ}, e^{- 10 σ}) < \infty$ for every real $σ$ . The integral converges absolutely whatever $Re (s)$ is.
Rule: a finite-duration signal that is absolutely integrable has an ROC covering the whole $s$ -plane. So the ROC is $- \infty < Re (s) < \infty$ , which is option (A).

KEY IDEA

How fast a signal grows only matters when the signal lasts forever. A bounded signal of finite duration has a Laplace transform that converges for every $s$ , so its ROC is the whole $s$ -plane. This holds even for something like $e^{t^{2}}$ , which on its own grows faster than any exponential and has no Laplace transform at all.

WATCH OUT FOR

There are two traps. The first is taking the window edges $t = 1$ and $t = 10$ as limits on $Re (s)$ and picking (D) $1 \leq Re (s) \leq 10$ . Time limits never become ROC boundaries: those come from exponential growth rates or pole locations. The second is seeing $e^{t^{2}}$ and deciding the transform cannot exist, or that it needs a large $Re (s)$ as in (B). Because of the window, the integral runs over a finite interval of a bounded function, so it converges everywhere.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 8 Laplace questions, with notes

GATE 2024 EE Q.16: solution and answer

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