| Line 20: | Line 20: | ||
Image1<br><br> | Image1<br><br> | ||
Now let's describe this process in the frequency domain. | Now let's describe this process in the frequency domain. | ||
| + | |||
==Derivation== | ==Derivation== | ||
First we'll take the Discrete Time Fourier Transform of the original signal and the downsampled version of it.<br> | First we'll take the Discrete Time Fourier Transform of the original signal and the downsampled version of it.<br> | ||
| + | |||
<math>\begin{align} | <math>\begin{align} | ||
| − | \mathcal{X}(\omega) &= \mathcal{F }\left \{ x_2[n] \right \} = \mathcal{F }\left \{ x_1[Dn] \right \}\\ | + | \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^{- | + | &= \sum_{n=-\infty}^\infty x_1[Dn]e^{-j\omega n} |
| + | \end{align}</math> | ||
| + | <br>make the substitution of <math> n=\frac{m}{D} </math><br> | ||
| + | |||
| + | <math>\begin{align} | ||
| + | \mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty x_1[m]e^{-j\omega \frac{m}{D}} | ||
| + | \end{align}</math> | ||
| + | |||
| + | <br>Downsampled signal will only be nonzero for m equal to multiples of D so:<br> | ||
| + | |||
| + | <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}\\ | ||
| + | s_d[m] | ||
\end{align}</math> | \end{align}</math> | ||
| − | |||
==Example== | ==Example== | ||
==Conclusion== | ==Conclusion== | ||
---- | ---- | ||
Revision as of 14:00, 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] \end{align} $
