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===Answer 1===
 
===Answer 1===
Write it here.
+
<math>
 +
\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}  \left( u[n]-u[n-3] \right) \left( u[m+1]-u[m-2] \right)e^{-j(mu + nv)}\\
 +
&= \sum_{m=-\infty}^{\infty} \left( u[m+1]-u[m-2] \right) e^{-j(mu)}  \sum_{n=-\infty}^{\infty} \left( u[n]-u[n-3] \right)e^{-j(nv)}\\
 +
&= (e^{jmu} + 1 + e^{-jmu})\cdot(1 + e^{-jnv} + e^{-2jnv})\\
 +
\end{align}</math>
  
 
===Answer 2===
 
===Answer 2===

Revision as of 19:03, 19 November 2011

Practice Problem on Discrete-space Fourier transform computation

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

$ f[m,n]= \left( u[n]-u[n-3] \right) \left( u[m+1]-u[m-2] \right) $

(Write enough intermediate steps to fully justify your answer.)


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Answer 1

$ \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} \left( u[n]-u[n-3] \right) \left( u[m+1]-u[m-2] \right)e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} \left( u[m+1]-u[m-2] \right) e^{-j(mu)} \sum_{n=-\infty}^{\infty} \left( u[n]-u[n-3] \right)e^{-j(nv)}\\ &= (e^{jmu} + 1 + e^{-jmu})\cdot(1 + e^{-jnv} + e^{-2jnv})\\ \end{align} $

Answer 2

Write it here.


Back to ECE438 Fall 2011 Prof. Boutin

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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