| Line 9: | Line 9: | ||
DFT | DFT | ||
| − | <math> X [k] = \sum_{k=0}^{N-1} x[n]e^{- | + | <math> X [k] = \sum_{k=0}^{N-1} x[n]e^{-j2pkn/N}</math> |
IDFT | IDFT | ||
| − | <math> x [n] = (1/N) \sum_{k=0}^{N-1} X[k]e^{ | + | <math> x [n] = (1/N) \sum_{k=0}^{N-1} X[k]e^{j2pkn/N}</math> |
| Line 24: | Line 24: | ||
Idea : discretize (ie. sample) the F.T. | Idea : discretize (ie. sample) the F.T. | ||
| − | <math> X(w) = \sum_{n=-\infty}^{\infty} x[n]e^{- | + | <math> X(w) = \sum_{n=-\infty}^{\infty} x[n]e^{-jwn}----sampling---> X(k2p/N) = \sum x[n]e^{-j2pnk/N} </math> |
note : if X(w) band limited can reconstruct X(w) if N big enough. | note : if X(w) band limited can reconstruct X(w) if N big enough. | ||
| Line 30: | Line 30: | ||
Oberve : | Oberve : | ||
| − | <math> X( | + | <math> X(k2p/N) = \sum_{n=0}^{N-1} x_{p}[n]e^{-j2pkn/N}</math>, where <math> x_{p}[n] = \sum_{-\infty}^{\infty} x[n-lN]</math> |
is periodic with N. | is periodic with N. | ||
This is because | This is because | ||
| − | <math> X(k2p/N) = \sum_{n =-\infty}^{\infty} x[n]e^{- | + | <math> X(k2p/N) = \sum_{n =-\infty}^{\infty} x[n]e^{-j2pkn/N}</math> |
| − | <math> = . . . + \sum_{n = -N}^{-1} x[n]e^{- | + | <math> = . . . + \sum_{n = -N}^{-1} x[n]e^{-j2pkn/N} + \sum_{n = 0}^{N-1} x[n]e^{-j2pkn/N}</math> |
<math> = \sum_{l=-\infty}^{\infty} \sum_{n=lN}^{lN+N-1} x[n]e^{-j2pikn/N}</math> | <math> = \sum_{l=-\infty}^{\infty} \sum_{n=lN}^{lN+N-1} x[n]e^{-j2pikn/N}</math> | ||
Revision as of 21:17, 22 September 2009
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} $
