Line 5: | Line 5: | ||
'''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*** | + | <math>*** Image \quad goes \quad here***\,\!</math> |
To avoid aliasing, | To avoid aliasing, | ||
Line 15: | Line 15: | ||
Let '''<math>x_r(t)\,\!</math>''' be the reconstructed signal. Then, | Let '''<math>x_r(t)\,\!</math>''' be the reconstructed signal. Then, | ||
− | '''<math>X_(f) = H_r(f)X_s(f)\,\!</math>''' | + | '''<math>X_(f) = H_r(f) X_s(f)\,\!</math>''' |
where, | where, | ||
Line 22: | Line 22: | ||
So, | So, | ||
+ | <div style="margin-left: 3em;"> | ||
+ | <math> | ||
+ | \begin{align} | ||
+ | x_r(t) &= h_r(t) * X_s(t) \\ | ||
+ | &= sinc \left (\frac{t}{T}\right) * \sum_k X(kT) \delta(t-kT) \\ | ||
+ | &= \sum_k X(kT) sinc \left (\frac{t}{T}\right) * \delta(t-kT) \\ | ||
+ | &= \sum_k X(kT) sinc \left (\frac{t - kT}{T}\right)\\ | ||
+ | \end{align} | ||
+ | </math> | ||
+ | </div> | ||
− | + | Recall, <math>\quad sinc(x) = 0 \quad \iff \quad x = \pm 1, \pm 2, \pm 3 ... \,\!</math> | |
+ | |||
+ | |||
+ | <math>*** Image \quad goes \quad here***\,\!</math> | ||
+ | |||
+ | At all integer multiples of T, | ||
+ | |||
+ | <math>x_r(nT) = X(nT)\,\!</math> | ||
+ | |||
+ | If Nyquist is satisfied, <math>\quad x_r(nT) = X(nT)\quad \forall \quad 't'\,\!</math> |
Revision as of 04:45, 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 \quad goes \quad 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,
$ \begin{align} x_r(t) &= h_r(t) * X_s(t) \\ &= sinc \left (\frac{t}{T}\right) * \sum_k X(kT) \delta(t-kT) \\ &= \sum_k X(kT) sinc \left (\frac{t}{T}\right) * \delta(t-kT) \\ &= \sum_k X(kT) sinc \left (\frac{t - kT}{T}\right)\\ \end{align} $
Recall, $ \quad sinc(x) = 0 \quad \iff \quad x = \pm 1, \pm 2, \pm 3 ... \,\! $
$ *** Image \quad goes \quad here***\,\! $
At all integer multiples of T,
$ x_r(nT) = X(nT)\,\! $
If Nyquist is satisfied, $ \quad x_r(nT) = X(nT)\quad \forall \quad 't'\,\! $