GATE 2016 EC – Question 40

Networks, Signals and Systems · Continuous-time Signals · 2 marks · Multiple choice

The Laplace transform of the causal periodic square wave of period $T$ shown in the figure below is

$f(t) = 1$ for $0 \leq t < T/2$ and $f(t) = 0$ for $T/2 \leq t < T$, repeating with period $T$ for all $t \geq 0$.

$F(s) = \frac{1}{1 + e^{-sT/2}}$
$F(s) = \frac{1}{s\left(1 + e^{-\frac{sT}{2}}\right)}$
$F(s) = \frac{1}{s(1 - e^{-sT})}$
$F(s) = \frac{1}{1 - e^{-sT}}$

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Correct answer: (B) $F(s) = \frac{1}{s\left(1 + e^{-\frac{sT}{2}}\right)}$

Explanation

For a causal periodic function, $F(s) = \frac{F_1(s)}{1 - e^{-sT}}$, where $F_1(s)$ is the transform of one period. One period is a pulse of height 1 for time $\frac{T}{2}$, so $F_1(s) = \frac{1 - e^{-sT/2}}{s}$. Using $1 - e^{-sT} = (1 - e^{-sT/2})(1 + e^{-sT/2})$, we get $F(s) = \frac{1}{s(1 + e^{-sT/2})}$.