Line 9: Line 9:
 
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)</math> \rightarrow <math>H(omega)</math> \rightarrow <math>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 19:10, 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>. $

Alumni Liaison

Have a piece of advice for Purdue students? Share it through Rhea!

Alumni Liaison