Difference between revisions of "Obtain DSFT rectangle" - Rhea

Latest revision as of 12:58, 26 November 2013

Question

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} $

--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.

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