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===Answer 1=== | ===Answer 1=== | ||
− | + | First off notice that this equation can easily be separated into two functions <math>g[m]=2^{-m}u[m]</math> and <math>h[n]=2^{-n}u[n]</math> where <math>f[m,n]=g[m]h[n]</math>. Then since both equations are the same except for a change of variable where m=n or vice verse we can show the DTFT of either g[m] or h[n] and it should suffice for the other. If they were not the same we would have to evaluate both of them. | |
+ | <math> | ||
+ | \begin{align} | ||
+ | G[u]&=\sum_{m=-\infty}^{\infty}2^{-m}u[m]e^{-j u m} \\ | ||
+ | &=\sum_{m=0}^{\infty}( \frac{1}{2e^{ju}} )^{m} \\ | ||
+ | &=\frac{1}{1-\frac{1}{2e^{ju}}} \\ | ||
+ | &=\frac{-2e^{ju}}{1-2e^{ju}} \\ | ||
+ | \end{align} | ||
+ | </math> | ||
+ | |||
+ | Thus | ||
+ | |||
+ | <math> | ||
+ | F[u,v]= \frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})} | ||
+ | </math> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
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[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] |
Revision as of 09:50, 16 November 2011
Contents
Practice Problem on Discrete-space Fourier transform computation
Compute the discrete-space Fourier transform of the following signal:
$ f[m,n]= 2^{-(n+m)} u[n] u[m] $
(Write enough intermediate steps to fully justify your answer.)
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
First off notice that this equation can easily be separated into two functions $ g[m]=2^{-m}u[m] $ and $ h[n]=2^{-n}u[n] $ where $ f[m,n]=g[m]h[n] $. Then since both equations are the same except for a change of variable where m=n or vice verse we can show the DTFT of either g[m] or h[n] and it should suffice for the other. If they were not the same we would have to evaluate both of them.
$ \begin{align} G[u]&=\sum_{m=-\infty}^{\infty}2^{-m}u[m]e^{-j u m} \\ &=\sum_{m=0}^{\infty}( \frac{1}{2e^{ju}} )^{m} \\ &=\frac{1}{1-\frac{1}{2e^{ju}}} \\ &=\frac{-2e^{ju}}{1-2e^{ju}} \\ \end{align} $
Thus
$ F[u,v]= \frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})} $
Answer 2
Write it here.