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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
$ \begin{align} F [u,v] &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f[m,n]e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} 2^{-(n+m)} u[n] u[m] e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} 2^{-m} u[m] e^{-j(mu)} \sum_{n=-\infty}^{\infty} 2^{-n} u[n] e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} 2^{-m} e^{-j(mu)} \sum_{n=0}^{\infty} 2^{-n} e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} (2e^{ju})^{-m} \sum_{n=0}^{\infty} (2e^{jv})^{-n}\\ &= \frac{1}{1-\frac{1}{2e^{ju}}}\cdot\frac{1}{1-\frac{1}{2e^{jv}}}\\ &= \frac{1}{1-\frac{1}{2}e^{-ju}}\cdot\frac{1}{1-\frac{1}{2}e^{-jv}}\\ &= \frac{1}{(1-\frac{1}{2}e^{-ju})(1-\frac{1}{2}e^{-jv})} \end{align} $
--Xiao1 23:03, 19 November 2011 (UTC)
