(New page: 4) Discrete Fourier Transform Definition: let x[n] be a DT signal with Period N. <math>DFT : X[k] = sum</math>) |
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4) Discrete Fourier Transform | 4) Discrete Fourier Transform | ||
| − | Definition: let x[n] be a DT signal with Period N. | + | Definition: let x[n] be a DT signal with Period N. |
| − | <math> | + | |
| + | <math> X [k] = \sum_{k=0}^{N-1} x[n].e^{-J.2pi.kn/N}</math> | ||
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| + | <math> x [n] = (1/N) \sum_{k=0}^{N-1} X[k].e^{J.2pi.kn/N}</math> | ||
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| + | Derivation: | ||
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| + | Digital signal are : | ||
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| + | * Finite Duration | ||
| + | * Discrete | ||
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| + | So the idea is, we need to discretize (ie sample) the Fourier Transform | ||
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| + | <math> X(w) = \sum_{n=-\infty}^{\infty} x[n].e^{-Jwn} >^{sampling}> X(k.2pi/N) = \sum x[n].e^{-J2pi.n.k/N} </math> | ||
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| + | Note: if X(w) is band-limited and if N is big enough, we can reconstruct X(w) | ||
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| + | ---------------------------------------------------------------------------- | ||
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| + | 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 | ||
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| + | <math> S_{\tau}(t) = P_{T}(t) = \sum_{K=-\infty}^{\infty} \delta (t - KT)</math> [Eq. 2] | ||
Revision as of 15:18, 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
$ S_{\tau}(t) = P_{T}(t) = \sum_{K=-\infty}^{\infty} \delta (t - KT) $ [Eq. 2]
