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= Discrete Fourier Transform =
 
= Discrete Fourier Transform =
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Revision as of 05:51, 21 April 2013

Collective Table of Formulas

Discrete Fourier transforms (DFT)

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Discrete Fourier Transform

Discrete Fourier Transform Pairs and Properties (info)
Definition Discrete Fourier Transform and its Inverse
Let x[n] be a periodic DT signal, with period N.
N-point Discrete Fourier Transform $ X [k] = \sum_{n=0}^{N-1} x[n]e^{-j 2\pi \frac{k n}{N}} \, $
Inverse Discrete Fourier Transform $ \,x [n] = (1/N) \sum_{k=0}^{N-1} X[k] e^{j 2\pi\frac{kn}{N}} \, $
Discrete Fourier Transform Pairs (info)
$ x[n] \ \text{ (period } N) $ $ \longrightarrow $ $ X_N[k] \ \ (N \text{ point DFT)} $
$ \ \sum_{k=-\infty}^\infty \delta[n+Nk] = \left\{ \begin{array}{ll} 1, & \text{ if } n=0, \pm N, \pm 2N , \ldots\\ 0, & \text{ else.} \end{array}\right. $ $ \ 1 \text{ (period } N) $
$ \ 1 \text{ (period } N) $ $ \ N\sum_{m=-\infty}^\infty \delta[k+Nm] = \left\{ \begin{array}{ll} N, & \text{ if } n=0, \pm N, \pm 2N , \ldots\\ 0, & \text{ else.} \end{array}\right. $
$ \ e^{j2\pi k_0 n} $ $ \ N\delta[((k - k_0))_N] $
$ \ \cos(\frac{2\pi}{N}k_0n) $ $ \ \frac{N}{2}(\delta[((k - k_0))_N] + \delta[((k+k_0))_N]) $
Discrete Fourier Transform Properties
$ x[n] \ $ $ \longrightarrow $ $ X[k] \ $
Linearity $ ax[n]+by[n] \ $ $ aX[k]+bY[k] \ $
Circular Shift $ x[((n-m))_N] \ $ $ X[k]e^{(-j\frac{2 \pi}{N}km)} \ $
Duality $ X[n] \ $ $ NX[((-k))_N] \ $
Multiplication $ x[n]y[n] \ $ $ \frac{1}{N} X[k]\circledast Y[k], \ \circledast \text{ denotes the circular convolution} $
Convolution $ x(t) \circledast y(t) \ $ $ X[k]Y[k] \ $
$ \ x^*[n] $ $ \ X^*[((-k))_N] $
$ \ x^*[((-n))_N] $ $ \ X^*[k] $
$ \ \Re\{x[n]\} $ $ \ X_{ep}[k] = \frac{1}{2}\{X[((k))_N] + X^*[((-k))_N]\} $
$ \ j\Im\{x[n]\} $ $ \ X_{op}[k] = \frac{1}{2}\{X[((k))_N] - X^*[((-k))_N]\} $
$ \ x_{ep}[n] = \frac{1}{2}\{x[n] + x^*[((-n))_N]\} $ $ \ \Re\{X[k]\} $
$ \ x_{op}[n] = \frac{1}{2}\{x[n] - x^*[((-n))_N]\} $ $ \ j\Im\{X[k]\} $
Other Discrete Fourier Transform Properties
Parseval's Theorem $ \sum_{n=0}^{N-1}|x[n]|^2 = \frac{1}{N} \sum_{k=0}^{N-1}|X[k]|^2 $

Go to Relevant Course Page: ECE 438

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