Topic: Discrete-space Fourier transform computation

## Question

Compute the discrete-space Fourier transform of the following signal:

$f[m,n]= 2^{-(n+m)} u[n] u[m]$

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

$F[u,v]= \frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})}$

\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)

Instructor's comments: Solutions 1 and solutions 2 are the two main ways to answer the question. The second one is a bit longer to write, in my opinion, but it does not require knowing the separation property of the continuous-space Fourier transform. Note however that, in the exam, if you do not have the separation property in the table, you will need to prove it in order to get full credit. -pm