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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 \frac{1}{|2T|}</math>''' | + | 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>''' |
| − | '''PROOF:''' Look at the graph of '''<math>X_s(f)</math>''' | + | '''PROOF:''' Look at the graph of '''<math>X_s(f)\,\!</math>''' |
*** Image goes here*** | *** Image goes here*** | ||
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To avoid aliasing, | To avoid aliasing, | ||
| − | '''<math>\frac{1}{T}\ - f_M \ge f_M</math>''' '''<math>\iff</math>''' '''<math>\frac{1}{T}\ \ge 2f_M</math>''' | + | '''<math>\frac{1}{T}\ - f_M \ge f_M</math>''' '''<math>\quad\iff\quad</math>''' '''<math>\frac{1}{T}\ \ge 2f_M</math>''' |
| − | To recover the signal, we will require a low pass filter with gain '''<math>T</math>''' and cutoff, '''<math>\frac{1}{2T}</math>''' | + | To recover the signal, we will require a low pass filter with gain '''<math>T\,\!</math>''' and cutoff, '''<math>\frac{1}{2T}</math>''' |
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| + | Let '''<math>x_r(t)\,\!</math>''' be the reconstructed signal. Then, | ||
| + | |||
| + | '''<math>X_(f) = H_r(f)X_s(f)\,\!</math>''' | ||
| + | |||
| + | where, | ||
| + | |||
| + | '''<math>H_r(f) = T rect(f)\,\!</math>''' | ||
| + | |||
| + | So, | ||
| + | |||
| + | '''<math>x_r(t) = h_r(t) * X_s(t)\,\!</math>''' | ||
Revision as of 05:20, 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 $ $ \quad\iff\quad $ $ \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} $
Let $ x_r(t)\,\! $ be the reconstructed signal. Then,
$ X_(f) = H_r(f)X_s(f)\,\! $
where,
$ H_r(f) = T rect(f)\,\! $
So,
$ x_r(t) = h_r(t) * X_s(t)\,\! $
