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<br>Downsampled signal will only be nonzero for m equal to multiples of D so:<br> | <br>Downsampled signal will only be nonzero for m equal to multiples of D so:<br> | ||
| − | |||
<math>\begin{align} | <math>\begin{align} | ||
\mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty s_d[m]x_1[m]e^{-j\omega \frac{m}{D}} \text{ where}\\ | \mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty s_d[m]x_1[m]e^{-j\omega \frac{m}{D}} \text{ where}\\ | ||
| − | s_d[m] | + | &s_d[m] = \frac{1}{D}\sum_{k=0}^{D-1} e^{jk2\pi \frac{m}{D}} |
\end{align}</math> | \end{align}</math> | ||
==Example== | ==Example== | ||
Revision as of 14:06, 9 October 2014
Downsampling in the Frequency Domain
A slecture by ECE student John S.
Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.
Contents
Introduction
Remember for time domain, Downsampling is defined as:
Image1
Now let's describe this process in the frequency domain.
Derivation
First we'll take the Discrete Time Fourier Transform of the original signal and the downsampled version of it.
$ \begin{align} \mathcal{X}_2(\omega) &= \mathcal{F }\left \{ x_2[n] \right \} = \mathcal{F }\left \{ x_1[Dn] \right \}\\ &= \sum_{n=-\infty}^\infty x_1[Dn]e^{-j\omega n} \end{align} $
make the substitution of $ n=\frac{m}{D} $
$ \begin{align} \mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty x_1[m]e^{-j\omega \frac{m}{D}} \end{align} $
Downsampled signal will only be nonzero for m equal to multiples of D so:
$ \begin{align} \mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty s_d[m]x_1[m]e^{-j\omega \frac{m}{D}} \text{ where}\\ &s_d[m] = \frac{1}{D}\sum_{k=0}^{D-1} e^{jk2\pi \frac{m}{D}} \end{align} $
