(New page: =Homework 1, ECE438, Fall 2011, Prof. Boutin= ---- ==Question 1-5== (under construction) ---- ==Question 6== <math>\text{a)} \;\; \text{General Relation for the d...)
 
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\end{align}</math>
 
\end{align}</math>
  
Replacing D with 3 would be the answer.
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Replacing D with 5 would be the answer.
  
[[Image:HW5Q2sol_3.jpg]]
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[[Image:3.jpg]]
  
 
<math>\text{b)} \;\; \text{General Relation for the upsampling with a factor of } L \,\!</math>.
 
<math>\text{b)} \;\; \text{General Relation for the upsampling with a factor of } L \,\!</math>.
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Since <math>X(w)</math> is periodic with <math>2\pi</math>, <math>Z(w)=X(Lw)</math> is periodic with <math>2\pi/L</math>.
 
Since <math>X(w)</math> is periodic with <math>2\pi</math>, <math>Z(w)=X(Lw)</math> is periodic with <math>2\pi/L</math>.
  
Replaing L with 4 would be the answer.
+
Replaing L with 7 would be the answer.
  
  
[[Image:HW5Q2sol_4.jpg]]
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[[Image:4.jpg]]
  
 
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Revision as of 03:57, 2 October 2011

Homework 1, ECE438, Fall 2011, Prof. Boutin


Question 1-5

(under construction)


Question 6

$ \text{a)} \;\; \text{General Relation for the decimation with a factor of } D \,\! $.

$ \text{Let } X(w) = \mathcal{F}(x[n]) $

$ \begin{align} Y(w) &= \sum_{n=-\infty}^{\infty} y[n]e^{-jwn} = \sum_{n=-\infty}^{\infty} x[3n]e^{-jwn} \\ &= \sum_{m=-\infty, m=Dk}^{\infty} x[m]e^{-j\frac{wm}{D}} = \sum_{m=-\infty}^{\infty} x[m] \left( \sum_{k=-\infty}^{\infty} \delta[m-Dk] \right) e^{-j\frac{wm}{D}} \\ &= \sum_{m=-\infty}^{\infty} x[m] \left( \frac{1}{D} \sum_{k=0}^{D-1} e^{j\frac{2\pi}{D}km} \right) e^{-j\frac{wm}{D}} = \sum_{k=0}^{D-1} \frac{1}{D} \sum_{m=-\infty}^{\infty} x[m]e^{j\left(w-\frac{2\pi}{D}k\right)m} \\ &= \sum_{k=0}^{D-1} \frac{1}{D} X\left(\frac{w-2\pi k}{D}\right) \\ \end{align} $

Replacing D with 5 would be the answer.

File:3.jpg

$ \text{b)} \;\; \text{General Relation for the upsampling with a factor of } L \,\! $.

$ \begin{align} Z(w) &= \sum_{n=-\infty}^{\infty} z[n]e^{-jwn} \\ &= \sum_{n=-\infty}^{\infty} \left( \sum_{k=-\infty}^{\infty} x[k] \delta[n-kL] \right) e^{-jwn} \\ &= \sum_{k=-\infty}^{\infty} x[k] \sum_{n=-\infty}^{\infty} \delta[n-kL] e^{-jwn} = \sum_{k=-\infty}^{\infty} x[k] e^{-jwkL} \\ &= \sum_{k=-\infty}^{\infty} x[k] e^{-jLwk} = X(Lw) \\ &\end{align} $

Since $ X(w) $ is periodic with $ 2\pi $, $ Z(w)=X(Lw) $ is periodic with $ 2\pi/L $.

Replaing L with 7 would be the answer.


File:4.jpg


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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