GATE 2020 EE Q.3: solution and answer

GATE 2020 · EEQ.3 · 1 mark · MCQ

Which of the following statements is true about the two sided Laplace transform?

(A)
It exists for every signal that may or may not have a Fourier transform.
(B)
It has no poles for any bounded signal that is non-zero only inside a finite time interval.
(C)
The number of finite poles and finite zeroes must be equal.
(D)
If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its Laplace Transform will have no poles.

See the question in the paper

Answer and solution

ANSWER

(B) is true. A bounded signal that is non-zero only over a finite time interval has the entire s-plane as its ROC, so its two-sided Laplace transform has no poles.

Official answer: BSource: Official answer key, GATE 2020 EE (IIT Delhi)

(A) is false. The two-sided transform $X (s) = \int_{- \infty}^{\infty} x (t) e^{- s t} d t$ exists only if some vertical strip of the $s$ -plane is a region of convergence (ROC). Take $x (t) = 1$ for all $t$ . The part for $t > 0$ converges only for $σ > 0$ , and the part for $t < 0$ converges only for $σ < 0$ . No $σ$ satisfies both, so the ROC is empty and the transform does not exist.
ROC of a finite-duration signal. Let $x (t)$ be bounded, $∣ x (t) ∣ \leq M$ , and zero outside $[T_{1}, T_{2}]$ . For every $σ$ , $∣ X (s) ∣ \leq \int_{T_{1}}^{T_{2}} M e^{- σ t} d t < \infty,$ because a continuous function is integrated over a finite interval. So the ROC is the entire $s$ -plane.
(B) is true. A pole is a point where $X (s)$ becomes unbounded, and a pole can never lie inside the ROC. Here the ROC is the whole $s$ -plane, so $X (s)$ is finite everywhere and has no finite poles.
Check with a rectangular pulse: $x (t) = u (t) - u (t - T)$ gives $X (s) = \frac{1 - e ^{- s T}}{s}$ . The apparent pole at $s = 0$ is not a pole, because $lim_{s \to 0} \frac{1 - e ^{- s T}}{s} = T$ is finite. The singularity is removable.
(C) is false. For $u (t)$ the transform is $\frac{1}{s}$ with ROC $Re (s) > 0$ . It has one finite pole and no finite zero. In general, a strictly proper rational $X (s)$ has fewer finite zeros than finite poles.
(D) is false. Take one shifted one-sided exponential: $e^{- a (t - t_{0})} u (t - t_{0}) \leftrightarrow \frac{e ^{- s t_{0}}}{s + a}$ , with ROC $Re (s) > - a$ . It has a pole at $s = - a$ . A weighted sum of such terms has poles at $s = - a_{k}$ unless the terms are specially arranged to cancel, as in the rectangular pulse of step 4.
Only statement (B) holds. The answer is (B).

KEY IDEA

A bounded, finite-duration signal gives a Laplace integral over a finite interval of a bounded integrand. That integral converges for every s, so the ROC is the whole s-plane. Poles cannot lie inside the ROC, so there are none. Any apparent pole, such as s = 0 in (1 - e^{-sT})/s, is a removable singularity.

WATCH OUT FOR

Seeing the factor 1/s in the rectangular-pulse transform (1 - e^{-sT})/s and concluding that finite-duration signals have a pole at s = 0. The numerator also vanishes there, so the limit is T and there is no pole. Two other errors: picking (A) because the Laplace transform 'generalises' the Fourier transform (a constant or two-sided signal can have an empty ROC), and picking (D) because only the delay factor e^{-st_0} is remembered and the 1/(s+a) factor is forgotten.

From the GATE 2020 EE paper. Every question of this paper · Download the full paper · All 8 Laplace questions, with notes

GATE 2020 EE Q.3: solution and answer

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