# Practice Question on Nyquist rate

What is the Nyquist rate of the signal

$x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( 3 \pi t )}{\pi t} ?$

From property 9 of the table: $\mathcal{F}(\frac{\sin( W t)}{\pi t }) = \left\{\begin{array}{ll}1, & \text{ if }|\omega| <W,\\ 0, & \text{else.}\end{array} \right. \$

By property 16 of the table: The Fourier transform of the product of two functions is (1/2pi) times the convolution of the Fourier transforms of the individual functions.

So in this case $\mathcal{F}(x(t))=(u(\omega + \pi) - u(\omega - \pi))*(u(\omega + 3\pi) - u(\omega - 3\pi))$

$\mathcal{F}(x(t))=\int_{-\infty}^{\infty} (u(\theta + \pi)-u(\theta - \pi))(u(\omega -\theta + 3\pi)-u(\omega - \theta - 3\pi)) d\theta$

$\mathcal{F}(x(t))=\int_{-\pi}^{\pi} (u(\omega -\theta + 3\pi)-u(\omega - \theta - 3\pi)) d\theta$

The first step function in the integral is 0 for $\omega < \theta - 3\pi$. The second step function in the integral is 0 for $\omega < \theta + 3\pi$. Thus:

$\mathcal{F}(x(t))= 0$ for $\omega < -4\pi$

$\mathcal{F}(x(t))= 4\pi + \omega$ for $-4\pi < \omega < -2\pi$

$\mathcal{F}(x(t))= 2\pi$ for $-2\pi < \omega < 2\pi$

$\mathcal{F}(x(t))= 4\pi - \omega$ for $2\pi < \omega < 4\pi$

$\mathcal{F}(x(t))= 0$ for $\omega > 4\pi$

The Nyquist rate is twice the maximum nonzero frequency, or $2(4\pi) = 8\pi$.

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