(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 ==
  
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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 {1/(|2T|)}</math>'''
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----
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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:ECE438ssuresh1.jpg]]
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To avoid aliasing,
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'''<math>\frac{1}{T}\ - f_M \ge f_M</math>'''     '''<math>\quad\iff\quad</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>'''
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Let '''<math>x_r(t)\,\!</math>''' be the reconstructed signal. Then,
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'''<math>X_(f) = H_r(f) X_s(f)\,\!</math>'''
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where,
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'''<math>H_r(f) = T rect(f)\,\!</math>'''
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So,
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<div style="margin-left: 3em;">
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<math>
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\begin{align}
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x_r(t) &= h_r(t) * {\color{OliveGreen} X_s(t)} \\
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&= sinc \left (\frac{t}{T}\right) * {\color{Blue} \sum_k X(kT) \delta(t-kT)} \\
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&= \sum_k X(kT) sinc \left (\frac{t}{T}\right) *  \delta(t-kT) \\
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&= \sum_k X(kT) sinc \left (\frac{t - kT}{T}\right)\\
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\end{align}
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</math>
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</div>
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Recall,  <math>\quad sinc(x) = 0 \quad \iff \quad x = \pm 1, \pm 2, \pm 3 ... \,\!</math>
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[[Image:ECE438ssuresh2.jpg]]
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At all integer multiples of T,
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<math>x_r(nT) = X(nT)\,\!</math>
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If Nyquist is satisfied, <math>\quad x_r(nT) = X(nT)\quad \forall \quad 't'\,\!</math>
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----
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Contrast this reconstruction with the zero-order hold,
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[[Image:ECE438ssuresh3.jpg]]
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<math>\qquad \Rightarrow piecewise\ construct\ approximation\ </math>
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<div style="margin-left: 3em;">
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<math>
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\begin{align}
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x_r(t) &= \sum_k x(kT)\ rect \left (\frac{t - \tfrac {T}{2} - kT}{T}\right) \\
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&= \sum_k x(kT)\ rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * \delta (t - kT)\\
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&= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * \sum_k x(kT)\  \delta (t - kT) \qquad and\ if\ we\ look\ clearly,\ \\
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&= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * {\color{Blue}\sum_k x(kT)\  \delta (t - kT)} \\
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&= rect \left (\frac{t - \tfrac {T}{2} }{T}\right) * {\color{OliveGreen} X_s(t)}\\
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\end{align}
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</math>
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</div>
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<div style="font-family: Verdana, sans-serif; font-size: 12px; text-align: center; width: 30%; margin: 0px; border: 1px solid #aaa; padding: 1em;"> <math>\therefore\ x_r(t) = h_{zo} (t) * X_s(t)</math>
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</div>
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[[Image:ECE438ssuresh4.jpg]]
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In the frequency domain,
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<math>X_r(f) = S_s(f) H_{zo}(f) \qquad ; \qquad H_{zo}(f) = T sinc (Tf) e^{ (-j2 \pi fT/T) } </math>
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[[Image:ECE438ssuresh5.jpg]]
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So even though <math>x_r(t)\,\!</math> is not band limited, its higher frequency components are attenuated because the <math>|H_{zo}(f)|\ \,\!</math> decreases as <math>|f| </math> increases.
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----
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--[[User:Ssuresh|Ssuresh]] 13:47, 23 September 2009 (UTC)

Latest revision as of 08:47, 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)\,\! $

ECE438ssuresh1.jpg


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 ... \,\! $


ECE438ssuresh2.jpg

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,


ECE438ssuresh3.jpg


$ \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)}\\ \end{align} $

$ \therefore\ x_r(t) = h_{zo} (t) * X_s(t) $

ECE438ssuresh4.jpg


In the frequency domain,

$ X_r(f) = S_s(f) H_{zo}(f) \qquad ; \qquad H_{zo}(f) = T sinc (Tf) e^{ (-j2 \pi fT/T) } $


ECE438ssuresh5.jpg


So even though $ x_r(t)\,\! $ is not band limited, its higher frequency components are attenuated because the $ |H_{zo}(f)|\ \,\! $ decreases as $ |f| $ increases.

--Ssuresh 13:47, 23 September 2009 (UTC)

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