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| − | = Practice Question on Nyquist rate = | + | = [[:Category:Problem_solving|Practice Question]] on Nyquist rate = |
What is the Nyquist rate of the signal | What is the Nyquist rate of the signal | ||
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The first step function in the integral is 0 for <math> \omega < \theta - 3\pi </math>. | The first step function in the integral is 0 for <math> \omega < \theta - 3\pi </math>. | ||
The second step function in the integral is 0 for <math> \omega < \theta + 3\pi </math>. | The second step function in the integral is 0 for <math> \omega < \theta + 3\pi </math>. | ||
| − | + | Thus: | |
| − | <math>\mathcal{F}(x(t))= 0 </math> for <math> \omega < -4\pi </math> | + | |
| − | <math>\mathcal{F}(x(t))= 4\pi + \omega </math> for <math> -4\pi < \omega < -2\pi </math> | + | <math>\mathcal{F}(x(t))= 0 </math> for <math> \omega < -4\pi </math> |
| − | <math>\mathcal{F}(x(t))= 2\pi </math> for <math> -2\pi < \omega < 2\pi </math> | + | |
| − | <math>\mathcal{F}(x(t))= 4\pi - \omega </math> for <math> 2\pi < \omega < 4\pi </math> | + | <math>\mathcal{F}(x(t))= 4\pi + \omega </math> for <math> -4\pi < \omega < -2\pi </math> |
| − | <math>\mathcal{F}(x(t))= 0 </math> for <math> \omega > 4\pi </math> | + | |
| + | <math>\mathcal{F}(x(t))= 2\pi </math> for <math> -2\pi < \omega < 2\pi </math> | ||
| + | |||
| + | <math>\mathcal{F}(x(t))= 4\pi - \omega </math> for <math> 2\pi < \omega < 4\pi </math> | ||
| + | |||
| + | <math>\mathcal{F}(x(t))= 0 </math> for <math> \omega > 4\pi </math> | ||
| + | |||
| + | The Nyquist rate is twice the maximum nonzero frequency, or <math> 2(4\pi) = 8\pi </math>. | ||
Latest revision as of 10:31, 11 November 2011
Contents
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} ? $
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Answer 1
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 $.
Answer 2
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Answer 3
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