Line 9: Line 9:
 
DFT
 
DFT
  
<math> X [k] = \sum_{k=0}^{N-1} x[n]e^{-j2pkn/N}</math>
+
<math> X [k] = \sum_{k=0}^{N-1} x[n]e^{-j2\pi kn/N}</math>
  
 
IDFT
 
IDFT
  
<math> x [n] = (1/N) \sum_{k=0}^{N-1} X[k]e^{j2[pi]kn/N}</math>
+
<math> x [n] = (1/N) \sum_{k=0}^{N-1} X[k]e^{j2\pi kn/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^{-jwn}----sampling---> X(k2p/N) = \sum x[n]e^{-j2pnk/N} </math>
+
<math> X(w) = \sum_{n=-\infty}^{\infty} x[n]e^{-jwn}----sampling---> X(k2\pi /N) = \sum x[n]e^{-j2\pi nk/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(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>  
+
<math> X(k2\pi /N) = \sum_{n=0}^{N-1} x_{p}[n]e^{-j\2pi kn/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^{-j2pkn/N}</math>
+
<math> X(k2p/N) = \sum_{n =-\infty}^{\infty} x[n]e^{-j2\pi kn/N}</math>
 
            
 
            
         <math> = . . . + \sum_{n = -N}^{-1} x[n]e^{-j2pkn/N} + \sum_{n = 0}^{N-1} x[n]e^{-j2pkn/N}</math>
+
         <math> = . . . + \sum_{n = -N}^{-1} x[n]e^{-j2\pi kn/N} + \sum_{n = 0}^{N-1} x[n]e^{-j\2pi kn/N}</math>
  
         <math> = \sum_{l=-\infty}^{\infty} \sum_{n=lN}^{lN+N-1} x[n]e^{-j2pkn/N}</math>
+
         <math> = \sum_{l=-\infty}^{\infty} \sum_{n=lN}^{lN+N-1} x[n]e^{-j2\pi kn/N}</math>
 
Let m=n-lN
 
Let m=n-lN
 
<math>X(k2
 
<math>X(k2

Revision as of 22:21, 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^{-j2\pi kn/N} $

IDFT

$ x [n] = (1/N) \sum_{k=0}^{N-1} X[k]e^{j2\pi kn/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(k2\pi /N) = \sum x[n]e^{-j2\pi nk/N} $

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

Oberve :

$ X(k2\pi /N) = \sum_{n=0}^{N-1} x_{p}[n]e^{-j\2pi kn/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^{-j2\pi kn/N} $

        $  = . . . + \sum_{n = -N}^{-1} x[n]e^{-j2\pi kn/N} + \sum_{n = 0}^{N-1} x[n]e^{-j\2pi kn/N} $
        $  = \sum_{l=-\infty}^{\infty} \sum_{n=lN}^{lN+N-1} x[n]e^{-j2\pi kn/N} $

Let m=n-lN $ X(k2 $

Alumni Liaison

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

Francisco Blanco-Silva