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We can recover <math>x(t)</math> from <math>x_p(t)</math> as follows: | We can recover <math>x(t)</math> from <math>x_p(t)</math> as follows: | ||
| − | <math>x_p(t) \rightarrow H(omega) \rightarrow x_r(t)</math> | + | <math>x_p(t) \rightarrow H(\omega) \rightarrow x_r(t)</math> |
Where <math>H(omega)</math> is a filter with gain equal to the period of the signal and a cutoff frequency of <math>omega_c</math>. | Where <math>H(omega)</math> is a filter with gain equal to the period of the signal and a cutoff frequency of <math>omega_c</math>. | ||
| − | <math>omega_c</math> satisfies <math>omega_m < omega_c < omega_s - omega_m<\math>. | + | <math>\omega_c</math> satisfies <math>\omega_m < \omega_c < \omega_s - \omega_m<\math>. |
Revision as of 19:11, 10 November 2008
-- Proving the Sampling Theorem --
The sampling can be represented by "Impulse-train Sampling."
$ x_p(t) = ? $ $ x_p(t) = x(t)p(t) $ $ x_p(t) = x(t)\sum_{n=-\infty}^{\infty} \delta(t-nT) $
We can recover $ x(t) $ from $ x_p(t) $ as follows:
$ x_p(t) \rightarrow H(\omega) \rightarrow x_r(t) $
Where $ H(omega) $ is a filter with gain equal to the period of the signal and a cutoff frequency of $ omega_c $.
$ \omega_c $ satisfies $ \omega_m < \omega_c < \omega_s - \omega_m<\math>. $
