The 3-dB bandwidth of the low-pass signal $e^{-t}u(t)$, where $u(t)$ is the unit step function, is given by

$\frac{1}{2\pi}$ Hz
$\frac{1}{2\pi}\sqrt{2-1}$ Hz
$\infty$
1 Hz

Correct answer: $\frac{1}{2\pi}$ Hz

Solution

The Fourier transform of $e^{-t}u(t)$ is $X(j\omega)=\frac{1}{1+j\omega}$, so $|X(j\omega)|=\frac{1}{\sqrt{1+\omega^2}}$. The 3-dB point occurs when the magnitude becomes $1/\sqrt{2}$ of its zero-frequency value, giving $\omega=1$ rad/s, i.e. bandwidth $=\frac{1}{2\pi}$ Hz.

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