The value of -+ e-t (2t - 2) \, dt, where (t) is the Dirac delta function, is

GATE 2016 EE – Question 19

Signals and Systems · Representation of continuous and discrete time signals, shifting and scaling properties · 1 mark · Multiple choice

The value of $\int_{-\infty}^{+\infty} e^{-t} \delta(2t - 2) \, dt$, where $\delta(t)$ is the Dirac delta function, is

$\frac{1}{2e}$
$\frac{2}{e}$
$\frac{1}{e^2}$
$\frac{1}{2e^2}$

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Correct answer: (A) $\frac{1}{2e}$

Explanation

Since $\delta(2t - 2) = \frac{1}{2}\delta(t - 1)$, the integral picks out the value at $t = 1$ and scales it by $\frac{1}{2}$. The result is $\frac{1}{2}e^{-1} = \frac{1}{2e}$.