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| − | A [https://www.projectrhea.org/learning/slectures.php slecture] by [[ECE]] student John | + | A [https://www.projectrhea.org/learning/slectures.php slecture] by [[ECE]] student John Sterrett |
Partly based on the [[2014_Fall_ECE_438_Boutin|ECE438 Fall 2014 lecture]] material of [[user:mboutin|Prof. Mireille Boutin]]. | Partly based on the [[2014_Fall_ECE_438_Boutin|ECE438 Fall 2014 lecture]] material of [[user:mboutin|Prof. Mireille Boutin]]. | ||
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==Introduction== | ==Introduction== | ||
Remember for time domain, Downsampling is defined as:<br><br> | Remember for time domain, Downsampling is defined as:<br><br> | ||
| − | Image1<br><br> | + | [[Image:Image1.png|800px]]<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> | ||
| Line 35: | Line 35: | ||
<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] = \left\{ |
| + | \begin{array}{ll} | ||
| + | 1, & \text{for }m\text{ multiples of } D\\ | ||
| + | 0, & \text{else} | ||
| + | \end{array} | ||
| + | \right.\\ | ||
| + | &\text{ or }=\frac{1}{D}\sum_{k=0}^{D-1} e^{jk2\pi \frac{m}{D}} \text{ so:}\\ | ||
| + | \mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty \frac{1}{D}\sum_{k=0}^{D-1} x_1[m]e^{jk2\pi \frac{m}{D}}e^{-j\omega \frac{m}{D}}\\ | ||
| + | &= \frac{1}{D}\sum_{k=0}^{D-1}\sum_{m=-\infty}^\infty x_1[m]e^{-jm\frac{\omega -k2\pi}{D}} \\ | ||
\end{align}</math> | \end{align}</math> | ||
| + | |||
| + | <br>Comparing to<br> | ||
| + | <math>\begin{align} | ||
| + | \mathcal{X}(\omega) &= \sum_{n=-\infty}^\infty x[n]e^{-jn\omega}\\ | ||
| + | \end{align}</math> | ||
| + | |||
| + | <br> We can rewrite the previous as<br> | ||
| + | <math>\begin{align} | ||
| + | \mathcal{X}_2(\omega) &= \frac{1}{D}\sum_{k=0}^{D-1} \mathcal{X}(\frac{\omega -2\pi k}{D}) \\ | ||
| + | \end{align}</math> | ||
| + | |||
| + | <math>\text{Comared to } \mathcal{X}_1(\omega) \text{, } \mathcal{X}_2(\omega) \text{ is } \frac{1}{D} \text{times the magnitude, has its frequencies stretched by }D \text{ and also repeats every }2\pi \text{ (as every DTFT should)}</math> | ||
| + | ---- | ||
==Example== | ==Example== | ||
| + | [[Image:Image2_1.png|800px]]<br><br> | ||
| + | It is worth noting that for large enough <math>D</math> it is possible for the ends of repetitions to overlap, leading to aliasing! To prevent this:<br> | ||
| + | |||
| + | <math>\begin{align} | ||
| + | D2\pi T_1 f_{max} < \pi\\ | ||
| + | \\ | ||
| + | x_1[Dn] \text{ has a sampling period of } T_2 = DT_1\text{, so:}\\ | ||
| + | \\ | ||
| + | \begin{align} | ||
| + | \frac{T_2}{T_1}2\pi T_1f_{max} < \pi\\ | ||
| + | f_{max} < \frac{1}{2T_2} | ||
| + | \end{align} | ||
| + | \end{align}</math> | ||
| + | |||
| + | To make sure this condition is satisfied, we should first pass the original <math>x_1[n]</math> signal through a low-pass filter with <math>f_c = 1/(2T_2)</math> BEFORE downsampling.<br><br> | ||
| + | Since the possible aliasing is caused by the downsampling, trying to low-pass filter after the downsampling will be too late and won't be able to get rid of the aliasing. Thus, the full process of downsampling should look like this:<br><br> | ||
| + | [[Image:Image3.png|800px]]<br> | ||
| + | ---- | ||
==Conclusion== | ==Conclusion== | ||
| + | There are two important points to take away about downsampling's effects in the frequency domain:<br> | ||
| + | <ol> | ||
| + | <li>The downsampled signal's frequency spectrum will have its magnitude lowered by the downsampling factor <math>D</math>, and will repeat every <math>2\pi</math> | ||
| + | <li>Downsampling can cause aliasing. To prevent this, we need to lowpass filter BEFORE the downsampling causes any aliasing. | ||
| + | </ol> | ||
| + | ---- | ||
| + | ==[[Slecture_Frequency_Downsampling_review|Questions and comments]]== | ||
| + | If you have any questions, comments, etc. please post them on [[Slecture_Frequency_Downsampling_review|this page]]. | ||
---- | ---- | ||
| + | [[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]] | ||
Latest revision as of 09:58, 14 March 2015
Downsampling in the Frequency Domain
A slecture by ECE student John Sterrett
Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.
Introduction
Remember for time domain, Downsampling is defined as:
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] = \left\{ \begin{array}{ll} 1, & \text{for }m\text{ multiples of } D\\ 0, & \text{else} \end{array} \right.\\ &\text{ or }=\frac{1}{D}\sum_{k=0}^{D-1} e^{jk2\pi \frac{m}{D}} \text{ so:}\\ \mathcal{X}_2(\omega) &= \sum_{m=-\infty}^\infty \frac{1}{D}\sum_{k=0}^{D-1} x_1[m]e^{jk2\pi \frac{m}{D}}e^{-j\omega \frac{m}{D}}\\ &= \frac{1}{D}\sum_{k=0}^{D-1}\sum_{m=-\infty}^\infty x_1[m]e^{-jm\frac{\omega -k2\pi}{D}} \\ \end{align} $
Comparing to
$ \begin{align} \mathcal{X}(\omega) &= \sum_{n=-\infty}^\infty x[n]e^{-jn\omega}\\ \end{align} $
We can rewrite the previous as
$ \begin{align} \mathcal{X}_2(\omega) &= \frac{1}{D}\sum_{k=0}^{D-1} \mathcal{X}(\frac{\omega -2\pi k}{D}) \\ \end{align} $
$ \text{Comared to } \mathcal{X}_1(\omega) \text{, } \mathcal{X}_2(\omega) \text{ is } \frac{1}{D} \text{times the magnitude, has its frequencies stretched by }D \text{ and also repeats every }2\pi \text{ (as every DTFT should)} $
Example
It is worth noting that for large enough $ D $ it is possible for the ends of repetitions to overlap, leading to aliasing! To prevent this:
$ \begin{align} D2\pi T_1 f_{max} < \pi\\ \\ x_1[Dn] \text{ has a sampling period of } T_2 = DT_1\text{, so:}\\ \\ \begin{align} \frac{T_2}{T_1}2\pi T_1f_{max} < \pi\\ f_{max} < \frac{1}{2T_2} \end{align} \end{align} $
To make sure this condition is satisfied, we should first pass the original $ x_1[n] $ signal through a low-pass filter with $ f_c = 1/(2T_2) $ BEFORE downsampling.
Since the possible aliasing is caused by the downsampling, trying to low-pass filter after the downsampling will be too late and won't be able to get rid of the aliasing. Thus, the full process of downsampling should look like this:
Conclusion
There are two important points to take away about downsampling's effects in the frequency domain:
- The downsampled signal's frequency spectrum will have its magnitude lowered by the downsampling factor $ D $, and will repeat every $ 2\pi $
- Downsampling can cause aliasing. To prevent this, we need to lowpass filter BEFORE the downsampling causes any aliasing.
Questions and comments
If you have any questions, comments, etc. please post them on this page.
