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<math>x(t)p(t) = x_p(t)</math> | <math>x(t)p(t) = x_p(t)</math> | ||
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+ | <math>x_p(t) = x(t) \sum_{n = -\infty}^\infty \delta(t - nT)</math> | ||
+ | |||
+ | <math>x_p(t) = \sum_{n = -\infty}^\infty x(t) \delta(t - nT)</math> | ||
+ | |||
+ | <math>x_p(t) = \sum_{n = -\infty}^\infty x(xT) \delta(t - nT)</math> |
Revision as of 11:33, 10 November 2008
Sampling Theorem
Let $ \omega_m $ be a non-negative number.
Let $ x(t) $ be a signal with $ X(\omega) = 0 $ when $ |\omega| > \omega_m $.
Consider the samples $ x(nT) $ for $ n = 0, +-1, +-2, ... $
If $ T < \frac{1}{2}(\frac{2\pi}{\omega_m}) $ then $ x(t) $ can be uniquely recovered from its samples.
Variable Definitions
$ T $ Sampling Period
$ \frac{2\pi}{T} = \omega_s $ Sampling Frequency
$ T < \frac{1}{2}(\frac{2\pi}{\omega_m}) <==> \omega_s>2\omega_m $
$ \omega_m $ Maximum frequencye for a band limited signal
$ NQ = 2\omega_m $ Nyquist Rate - The frequencye the sampling frequency should be, or greater.
$ \omega_c $ Cut off frequency for a filter
Impulse-Train Sampling
Let $ x(t) $ be a continuous signal
Let $ p(t) = \sum_{n = -\infty}^\infty \delta(t - nT) $
$ x(t)p(t) = x_p(t) $
$ x_p(t) = x(t) \sum_{n = -\infty}^\infty \delta(t - nT) $
$ x_p(t) = \sum_{n = -\infty}^\infty x(t) \delta(t - nT) $
$ x_p(t) = \sum_{n = -\infty}^\infty x(xT) \delta(t - nT) $