| Line 24: | Line 24: | ||
Observe that : <math> X(k.2pi/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 | Observe that : <math> X(k.2pi/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 | ||
| + | The reason behind this is as follow: | ||
| + | <math> X(k.2p/N) = \sum_{n =-\infty}^{\infty} x[n].e^{-J.2pi.kn/N}</math> | ||
| + | |||
| + | <math> = . . . . . + \sum_{n = -N}^{-1} x[n].e^{-J2pi.k.n/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^{-J.2pi.kn/N}</math> | ||
| + | let <math> m = n - lN : </math> | ||
| + | |||
| + | <math> X(k.2pi/N) = \sum_{l=-\infty}^{infty} \sum_{m=0}^{N-1} x[m+lN].e^{-J.2pi.k.(m+lN)/N}</math> | ||
| + | |||
| + | <math> = \sum_{l=-\infty}^{infty} \sum_{m=0}^{N-1} x[m+lN].e^{-J.2pi.km/N}</math> since <math> e^{-J.2pi.k.l}</math> is always = 1 because k.l is always integer | ||
| + | |||
| + | and <math> e^{J*2pi} = \cos (2*pi) + j \sin (2*pi) = 1</math> | ||
| + | |||
| + | <math> = \sum_{m = 0}^{N-1}(\sum_{l=-\infty}^{\infty} x[m+lN]).e^{-J.2pi.km/N}</math> | ||
| + | |||
| + | As a result: <math> x_{p}[n] = \sum_{l=-\infty}^{\infty} x(m + lN)</math> | ||
| + | |||
| + | Also Note that <math> x_{p}[n] </math> is the periodic repetition of <math> x[n] </math> if <math> x[n] </math> | ||
| + | has a finite duration of N | ||
<math> S_{\tau}(t) = P_{T}(t) = \sum_{K=-\infty}^{\infty} \delta (t - KT)</math> [Eq. 2] | <math> S_{\tau}(t) = P_{T}(t) = \sum_{K=-\infty}^{\infty} \delta (t - KT)</math> [Eq. 2] | ||
Revision as of 16:03, 22 September 2009
4) Discrete Fourier Transform
Definition: let x[n] be a DT signal with Period N.
$ X [k] = \sum_{k=0}^{N-1} x[n].e^{-J.2pi.kn/N} $
$ x [n] = (1/N) \sum_{k=0}^{N-1} X[k].e^{J.2pi.kn/N} $
Derivation:
Digital signal are :
- Finite Duration
- Discrete
So the idea is, we need to discretize (ie sample) the Fourier Transform
$ X(w) = \sum_{n=-\infty}^{\infty} x[n].e^{-Jwn} >^{sampling}> X(k.2pi/N) = \sum x[n].e^{-J2pi.n.k/N} $
Note: if X(w) is band-limited and if N is big enough, we can reconstruct X(w)
Observe that : $ X(k.2pi/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
The reason behind this is as follow: $ X(k.2p/N) = \sum_{n =-\infty}^{\infty} x[n].e^{-J.2pi.kn/N} $
$ = . . . . . + \sum_{n = -N}^{-1} x[n].e^{-J2pi.k.n/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^{-J.2pi.kn/N} $
let $ m = n - lN : $
$ X(k.2pi/N) = \sum_{l=-\infty}^{infty} \sum_{m=0}^{N-1} x[m+lN].e^{-J.2pi.k.(m+lN)/N} $
$ = \sum_{l=-\infty}^{infty} \sum_{m=0}^{N-1} x[m+lN].e^{-J.2pi.km/N} $ since $ e^{-J.2pi.k.l} $ is always = 1 because k.l is always integer
and $ e^{J*2pi} = \cos (2*pi) + j \sin (2*pi) = 1 $
$ = \sum_{m = 0}^{N-1}(\sum_{l=-\infty}^{\infty} x[m+lN]).e^{-J.2pi.km/N} $
As a result: $ x_{p}[n] = \sum_{l=-\infty}^{\infty} x(m + lN) $
Also Note that $ x_{p}[n] $ is the periodic repetition of $ x[n] $ if $ x[n] $ has a finite duration of N
$ S_{\tau}(t) = P_{T}(t) = \sum_{K=-\infty}^{\infty} \delta (t - KT) $ [Eq. 2]
