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:<span style="color:green">Instructor's comment: This approach is correct, but it may not be obvious to other students reading the solution how you obtain the last expression from the previous line. Can somebody else clarify? -pm </span> | :<span style="color:green">Instructor's comment: This approach is correct, but it may not be obvious to other students reading the solution how you obtain the last expression from the previous line. Can somebody else clarify? -pm </span> | ||
| + | |||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | F [u,v] &= \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)}\\ | ||
| + | &= \sum_{m=-\infty}^{\infty} \left( \delta[n+1] + delta[n] + delta[n-1] \right) e^{-j(mu)} \sum_{n=-\infty}^{\infty} \left( \delta[n] + delta[n-1] + delta[n-2] \right)e^{-j(nv)}\\ | ||
| + | &_{by. looking. up. in. table. or. compute. shifted. delta. function's. DFT, on. can. get}\\ | ||
| + | &= (e^{jmu} + 1 + e^{-jmu})\cdot(1 + e^{-jnv} + e^{-2jnv})\\ | ||
| + | \end{align}</math> | ||
| + | |||
| + | --[[User:Xiao1|Xiao1]] 13:12, 25 November 2011 (UTC) | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
---- | ---- | ||
[[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 08:12, 25 November 2011
Contents
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.)
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
$ \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} $
--Xiao1 23:03, 19 November 2011 (UTC)
- Instructor's comment: This approach is correct, but it may not be obvious to other students reading the solution how you obtain the last expression from the previous line. Can somebody else clarify? -pm
$ \begin{align} F [u,v] &= \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)}\\ &= \sum_{m=-\infty}^{\infty} \left( \delta[n+1] + delta[n] + delta[n-1] \right) e^{-j(mu)} \sum_{n=-\infty}^{\infty} \left( \delta[n] + delta[n-1] + delta[n-2] \right)e^{-j(nv)}\\ &_{by. looking. up. in. table. or. compute. shifted. delta. function's. DFT, on. can. get}\\ &= (e^{jmu} + 1 + e^{-jmu})\cdot(1 + e^{-jnv} + e^{-2jnv})\\ \end{align} $
--Xiao1 13:12, 25 November 2011 (UTC)
Answer 2
Write it here.
