Revision as of 21:17, 22 September 2009 by Sje (Talk | contribs)

Discrete Fourier Transform

definition

Let X[n] be a DT signal with period N

DFT

$ X [k] = \sum_{k=0}^{N-1} x[n]e^{-j2pkn/N} $

IDFT

$ x [n] = (1/N) \sum_{k=0}^{N-1} X[k]e^{j2pkn/N} $


Derivation

Digital signals are 1) finite duration 2)discrete

want F.T. discrete and finite duration

Idea : discretize (ie. sample) the F.T.

$ X(w) = \sum_{n=-\infty}^{\infty} x[n]e^{-jwn}----sampling---> X(k2p/N) = \sum x[n]e^{-j2pnk/N} $

note : if X(w) band limited can reconstruct X(w) if N big enough.

Oberve :

$ X(k2p/N) = \sum_{n=0}^{N-1} x_{p}[n]e^{-j2pkn/N} $, where $ x_{p}[n] = \sum_{-\infty}^{\infty} x[n-lN] $ is periodic with N.

This is because

$ X(k2p/N) = \sum_{n =-\infty}^{\infty} x[n]e^{-j2pkn/N} $

        $  = . . . + \sum_{n = -N}^{-1} x[n]e^{-j2pkn/N} + \sum_{n = 0}^{N-1} x[n]e^{-j2pkn/N} $
        $  = \sum_{l=-\infty}^{\infty} \sum_{n=lN}^{lN+N-1} x[n]e^{-j2pikn/N} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva