(New page: = Practice Question on Nyquist rate = What is the Nyquist rate of the signal <math>x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( 3 \pi t )}{\pi t} ?</math> ---- == Share your answ...) |
|||
| Line 9: | Line 9: | ||
---- | ---- | ||
=== Answer 1 === | === Answer 1 === | ||
| − | + | ||
| + | From property 9 of the table: | ||
| + | <math>\mathcal{F}(\frac{\sin( W t)}{\pi t }) = \left\{\begin{array}{ll}1, & \text{ if }|\omega| <W,\\ 0, & \text{else.}\end{array} \right. \ </math> | ||
| + | |||
| + | 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 <math>\mathcal{F}(x(t))=(u(\omega + \pi) - u(\omega - \pi))*(u(\omega + 3\pi) - u(\omega - 3\pi))</math> | ||
| + | |||
| + | <math>\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</math> | ||
| + | |||
| + | |||
| + | <math>\mathcal{F}(x(t))=\int_{-\pi}^{\pi} (u(\omega -\theta + 3\pi)-u(\omega - \theta - 3\pi)) d\theta</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>. | ||
| + | So for | ||
| + | <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))= 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> | ||
| + | |||
| + | |||
=== Answer 2 === | === Answer 2 === | ||
Write it here | Write it here | ||
Revision as of 18:41, 21 April 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} ? $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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 $. So for $ \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 $
Answer 2
Write it here
Answer 3
Write it here.
