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\end{align}
 
\end{align}
 
</math>
 
</math>
The result works for <math>\omega</math> between <math>0</math> to <math>2\pi</math>
+
The result works for <math>\omega</math> between <math>0</math> to <math>2\pi</math>.
 +
 
 +
But this is not the final answer. Because Fourier transform must be periodic but the answer we got above is not periodic.

Revision as of 19:17, 29 September 2014


Discrete-time Fourier Transform (DTFT)

A slecture by ECE student Xian Zhang

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.


outline

  • Definition
  • Periodicity property
  • Example of computation of DTFT of a complex exponential


Definition

The discrete-time Fourier transform (DTFT) of a discrete set of real or complex numbers x[n] with n=all integers, is a Fourier series, which produces a periodic function of a frequency variable. With w has units of radians/sample, the Fourier series is:

$ \begin{align} \\ \mathcal{X}_1(\omega) & = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ \end{align} $

Inverse DTFT is :

$ \begin{align} \\ \quad x[n] & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\mathcal{X}(\omega)e^{j\omega n}d\omega \\ \end{align} $



Periodicity property

$ \mathcal{X}(\omega) $ is periodic with period $ 2\pi $. Because,

$ \begin{align} \\ \mathcal{X}(\omega+2\pi) & = \sum_{n=-\infty}^{\infty}x[n]e^{-j(\omega+2\pi)n}\\ & =\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}e^{-j2\pi n}\\ & =\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\\ & =\mathcal{X}(\omega)\\ \end{align} $



Example of computation of DTFT of a complex exponential

Given: $ \begin{align} \\ x[n] =e^{j\omega_o n}\\ \end{align} $

Proof: $ \begin{align} \\ \mathcal{X}(\omega) & = 2\pi rep_{2\pi}(\delta(\omega - \omega_o))\\ \end{align} $


Firstly, let's try:

$ \begin{align} \\ \mathcal{X}(\omega) & = \sum_{n=-\infty}^{\infty}e^{j\omega_o n}e^{-j\omega n}\\ & = \sum_{n=-\infty}^{\infty}e^{j(\omega-\omega_o) n}\\ & = \infty, \omega = \omega_o; 0, else \\ \end{align} $

How to compute something that diverges:


$ \begin{align} \\ e^{j\omega_o n} & =\frac{1}{2\pi}\int_{0}^{2\pi}\mathcal{X}(\omega)e^{j\omega n}d\omega \\ \end{align} $

Assume $ \omega $ is between $ 0 $ to $ 2\pi $.

If $ \begin{align} \\ \mathcal{X}(\omega) & =2\pi \delta(\omega - \omega_o).\\ \end{align} $ The result works for $ \omega $ between $ 0 $ to $ 2\pi $.

But this is not the final answer. Because Fourier transform must be periodic but the answer we got above is not periodic.

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