| Line 32: | Line 32: | ||
<font size="2"> | <font size="2"> | ||
== 2. Derivation == | == 2. Derivation == | ||
| − | Let there be two discrete signals <math> x_1[n] </math> and <math>x_2[n] </math> | + | Let there be two discrete signals <math> x_1[n] </math> and <math>x_2[n] </math> with <math> x_2</math> being a scalar multiple D of <math> x_1 </math> |
Taking the Discrete Time Fourier Transform of each we get | Taking the Discrete Time Fourier Transform of each we get | ||
| Line 38: | Line 38: | ||
<math>\begin{align} | <math>\begin{align} | ||
\mathcal{X}_2(\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^{-j\omega n} | + | &= \sum_{n=-\infty}^\infty x_1[Dn]e^{-j \omega n} |
\end{align}</math> | \end{align}</math> | ||
| + | |||
| + | substitute using <math> m= D*n </math> resulting with: | ||
| + | |||
| + | <math> \sum_{m = -\infty}^\infty x_1[m] e^{-j \omega {\frac{m}{D}}} </math> | ||
| + | |||
| + | Introduce <math> S_D[m] </math> which is a piece wise function that equals 1 at multiples of D: | ||
| + | |||
| + | <math> \mathcal{F }\left \{ x_2[m] \right \} = \sum_{m = -\infty}^\infty S_D[m]*x_1[m]e^{-j\omega n} | ||
| + | </math> | ||
| + | |||
| + | The Fourier representation | ||
| + | <math> S_D = {\frac{1}{D}} \sum_{k = 0}^{D-1} e^{j\omega \frac{2\pi}{D}mk} </math> | ||
| + | |||
| + | Therefore: <math> \mathcal{F }\left \{ x_2[n] \right \} = {\frac{1}{D}} \sum_{k = 0}^{D-1} \sum_{m=-\infty}^\infty </math> | ||
---- | ---- | ||
Revision as of 02:10, 15 October 2014
Frequency Domain View of Downsampling
A Text slecture by ECE David Klouda
Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.
Contents
Outline
- Introduction
- Derivation
- Example
- Conclusion
1. Introduction
In this slecture, the Frequency Domain view of Downsampling will be discussed. It will begin with the derivation of the formulas and explaining the terms involved. It will then show an example using the DTFT and finish with an explanation as to why filtering is necessary when decimating.
2. Derivation
Let there be two discrete signals $ x_1[n] $ and $ x_2[n] $ with $ x_2 $ being a scalar multiple D of $ x_1 $
Taking the Discrete Time Fourier Transform of each we get
$ \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} $
substitute using $ m= D*n $ resulting with:
$ \sum_{m = -\infty}^\infty x_1[m] e^{-j \omega {\frac{m}{D}}} $
Introduce $ S_D[m] $ which is a piece wise function that equals 1 at multiples of D:
$ \mathcal{F }\left \{ x_2[m] \right \} = \sum_{m = -\infty}^\infty S_D[m]*x_1[m]e^{-j\omega n} $
The Fourier representation $ S_D = {\frac{1}{D}} \sum_{k = 0}^{D-1} e^{j\omega \frac{2\pi}{D}mk} $
Therefore: $ \mathcal{F }\left \{ x_2[n] \right \} = {\frac{1}{D}} \sum_{k = 0}^{D-1} \sum_{m=-\infty}^\infty $
3. Example
4. Conclusion
This slecture demonstrated the use of downsampling as seen from the Fourier domain. It showed that if a signal is below a certain threshold, then it must be filtered before downsampling to eliminate the possibility of aliasing and distorting the reconstructed signal.
NOTE:
Downsampling is always equivalent to resampling with the larger period.
Begin with x(t) as a continuous time signal with $ x_1[n]= x(T_1*n) $ being its discrete time sampling.
Let $ x_2[n]=x(T_2*n)=x_1[T_2/T_1*n] $
with Downsampling factor $ D=T_2/T_1 $
This lets us define $ x_2 $ as $ x_1[D*n] $
In order to prevent Aliasing within this context you need to have $ D*2\pi*T_1*f_{MAX} < \pi $
$ {T_2/T_1}*2\pi*T_1*f_{MAX} < \pi $
$ 2\pi*T_2*f_{MAX} < \pi $
$ f_{MAX} < 1/{2*T_2} $
If $ f_{MAX} > 1/{2*T_2} $ is true, then you must use a low-pass filter before downsampling.
Questions and comments
If you have any questions, comments, etc. please post them on this page.
