(New page: == LECTURE on September 11, 2009 == The perfect reconstruction of '''<math>{x(t)}</math>''' from '''<math>x_s(t)</math>''' is possible if '''<math>X(f) = 0</math>''' when '''<math>|f| \ge...)
 
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== LECTURE on September 11, 2009 ==
 
== LECTURE on September 11, 2009 ==
  
The perfect reconstruction of '''<math>{x(t)}</math>''' from '''<math>x_s(t)</math>''' is possible if '''<math>X(f) = 0</math>''' when '''<math>|f| \ge {1/(|2T|)}</math>'''
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The perfect reconstruction of '''<math>{x(t)}</math>''' from '''<math>x_s(t)</math>''' is possible if '''<math>X(f) = 0</math>''' when '''<math>|f| \ge \frac{1}{|2T|}</math>'''
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'''PROOF:''' Look at the graph of '''<math>X_s(f)</math>'''
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*** Image goes here***
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To avoid aliasing,
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'''<math>\frac{1}{T}\ - f_M \ge f_M</math>'''    '''<math>\iff</math>'''    '''<math>\frac{1}{T}\ \ge 2f_M</math>'''
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To recover the signal, we will require a low pass filter with gain '''<math>T</math>''' and cutoff, '''<math>\frac{1}{2T}</math>'''

Revision as of 04:21, 23 September 2009

LECTURE on September 11, 2009

The perfect reconstruction of $ {x(t)} $ from $ x_s(t) $ is possible if $ X(f) = 0 $ when $ |f| \ge \frac{1}{|2T|} $

PROOF: Look at the graph of $ X_s(f) $

      • Image goes here***

To avoid aliasing,

$ \frac{1}{T}\ - f_M \ge f_M $ $ \iff $ $ \frac{1}{T}\ \ge 2f_M $

To recover the signal, we will require a low pass filter with gain $ T $ and cutoff, $ \frac{1}{2T} $

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