Line 25: Line 25:
 
<math>
 
<math>
 
\begin{align}
 
\begin{align}
x_r(t) &= h_r(t) * X_s(t) \\
+
x_r(t) &= h_r(t) * {\color{OliveGreen} X_s(t)} \\
&= sinc \left (\frac{t}{T}\right) * \sum_k X(kT) \delta(t-kT) \\
+
&= sinc \left (\frac{t}{T}\right) * {\color{Blue} \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}{T}\right) *  \delta(t-kT) \\
 
&= \sum_k X(kT) sinc \left (\frac{t - kT}{T}\right)\\
 
&= \sum_k X(kT) sinc \left (\frac{t - kT}{T}\right)\\
Line 43: Line 43:
  
 
If Nyquist is satisfied, <math>\quad x_r(nT) = X(nT)\quad \forall \quad 't'\,\!</math>
 
If Nyquist is satisfied, <math>\quad x_r(nT) = X(nT)\quad \forall \quad 't'\,\!</math>
 +
 +
Contrast this reconstruction with the zero-order hold,
 +
 +
 +
<math>*** Image \quad goes \quad here***\,\!</math>
 +
 +
 +
<math>\qquad \Rightarrow piecewise\ construct\ approximation\ </math>
 +
 +
 +
<div style="margin-left: 3em;">
 +
<math>
 +
\begin{align}
 +
x_r(t) &= \sum_k x(kT)\ rect \left (\frac{t - \tfrac {T}{2} - kT}{T}\right) \\
 +
&= \sum_k x(kT)\ rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * \delta (t - kT)\\
 +
&= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * \sum_k x(kT)\  \delta (t - kT) \qquad and\ if\ we\ look\ clearly,\ \\
 +
&= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * {\color{Blue}\sum_k x(kT)\  \delta (t - kT)} \\
 +
&= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * {\color{OliveGreen} X_s(t)}\\
 +
\therefore\ x_r(t) &= h_zo (t) * X_s(t)\\
 +
\end{align}
 +
</math>
 +
</div>

Revision as of 05:11, 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) * {\color{OliveGreen} X_s(t)} \\ &= sinc \left (\frac{t}{T}\right) * {\color{Blue} \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'\,\! $

Contrast this reconstruction with the zero-order hold,


$ *** Image \quad goes \quad here***\,\! $


$ \qquad \Rightarrow piecewise\ construct\ approximation\ $


$ \begin{align} x_r(t) &= \sum_k x(kT)\ rect \left (\frac{t - \tfrac {T}{2} - kT}{T}\right) \\ &= \sum_k x(kT)\ rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * \delta (t - kT)\\ &= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * \sum_k x(kT)\ \delta (t - kT) \qquad and\ if\ we\ look\ clearly,\ \\ &= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * {\color{Blue}\sum_k x(kT)\ \delta (t - kT)} \\ &= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * {\color{OliveGreen} X_s(t)}\\ \therefore\ x_r(t) &= h_zo (t) * X_s(t)\\ \end{align} $

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett