(New page: Assume a function <math>x(t)</math>. We wish to covert a signal sampled at <math>T_1</math> one sampled at <math>T_2</math> without having to reconstruct the original <math>x(t)</math> an...) |
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2. <math>T_2</math> is a divider of <math>T_1</math> | 2. <math>T_2</math> is a divider of <math>T_1</math> | ||
| − | + | In this discussion, we will look at the first case, ie. where <math>T_2</math> is a multiple of <math>T_1</math>. | |
| + | |||
| + | Conversion can be accomplished by down-sampling <math>x_1[n]</math> | ||
| + | |||
| + | <math>x_1[n] \longrightarrow D = \frac{T_2}{T_1} \longrightarrow x_2[n]</math> | ||
| + | |||
| + | <math>x_2[n] = x_1[Dn]</math> | ||
| + | |||
| + | where <math>D = \frac{T_2}{T_1}</math> | ||
| + | |||
| + | We observe that this is the same as doing <math>x_2[n] = x(T_2n)</math> | ||
| + | |||
| + | This in Fourier Domain becomes | ||
| + | |||
| + | <math>F(x_2[n]) = F(x_1[Dn])</math> | ||
| + | |||
| + | <math>X_2(\omega) = \sum_{n=-\infty}^{\infty} x_1[Dn] e^{-j \omega n}</math> | ||
| + | |||
| + | let <math>m = Dn</math> | ||
| + | |||
| + | <math>X_2(\omega) = \sum_{m=-\infty}^{\infty} x_1[m] e^{-j \omega \frac{m}{D}}</math> | ||
| + | |||
| + | where m is a multiple of D | ||
| + | |||
| + | Now, we can introduce a function <math>s_D[m]</math> such that | ||
| + | |||
| + | <math>s_D[m] = | ||
| + | |||
| + | \begin{cases} | ||
| + | 1, & \mbox{if }m\mbox{ multiple of }D\\ | ||
| + | 0, & \mbox{else } | ||
| + | \end{cases} | ||
| + | |||
| + | </math> | ||
| + | |||
| + | The Fourier series of this function can be represented as | ||
| + | |||
| + | <math>S_D[m] = \frac{1}{D} \sum_{k = 0}^{D-1} (e^{j \frac{2 \pi}{D} m})^k</math> | ||
| + | |||
| + | and therefore we get | ||
| + | |||
| + | <math>X_2(\omega) = \sum_{m = -\infty}^{\infty} S_D[m] e^{-j \omega \frac{m}{D}}</math> | ||
| + | |||
| + | <math>X_2(\omega) = \sum_{m = -\infty}^{\infty} \frac{1}{D}\sum_{k = 0}^{D-1} e^{j k \frac{2 \pi}{D} m} x_1[m] e^{-j \omega \frac{m}{D}}</math> | ||
| + | |||
| + | <math>X_2(\omega) = \frac{1}{D} \sum_{k=0}^{D-1} \sum_{m=-\infty}^{\infty} x_1[m] e^{-j m (\frac{\omega - 2 \pi k}{D})}</math> | ||
| + | |||
| + | And since <math>\sum_{m=-\infty}^{\infty} x_1[m] e^{-j m (\frac{\omega - 2 \pi k}{D})} = X_1(\frac{\omega - 2 \pi k}{D})</math> | ||
| + | |||
| + | Therefore, | ||
| + | <math>X_2(\omega) = \frac{1}{D} \sum_{k=0}^{D-1} X_1(\frac{\omega - 2 \pi k}{D})</math> | ||
Revision as of 18:45, 22 September 2009
Assume a function $ x(t) $.
We wish to covert a signal sampled at $ T_1 $ one sampled at $ T_2 $ without having to reconstruct the original $ x(t) $ and then resampling at a new rate.
There are two cases here.
1. $ T_2 $ is a multiple of $ T_1 $
2. $ T_2 $ is a divider of $ T_1 $
In this discussion, we will look at the first case, ie. where $ T_2 $ is a multiple of $ T_1 $.
Conversion can be accomplished by down-sampling $ x_1[n] $
$ x_1[n] \longrightarrow D = \frac{T_2}{T_1} \longrightarrow x_2[n] $
$ x_2[n] = x_1[Dn] $
where $ D = \frac{T_2}{T_1} $
We observe that this is the same as doing $ x_2[n] = x(T_2n) $
This in Fourier Domain becomes
$ F(x_2[n]) = F(x_1[Dn]) $
$ X_2(\omega) = \sum_{n=-\infty}^{\infty} x_1[Dn] e^{-j \omega n} $
let $ m = Dn $
$ X_2(\omega) = \sum_{m=-\infty}^{\infty} x_1[m] e^{-j \omega \frac{m}{D}} $
where m is a multiple of D
Now, we can introduce a function $ s_D[m] $ such that
$ s_D[m] = \begin{cases} 1, & \mbox{if }m\mbox{ multiple of }D\\ 0, & \mbox{else } \end{cases} $
The Fourier series of this function can be represented as
$ S_D[m] = \frac{1}{D} \sum_{k = 0}^{D-1} (e^{j \frac{2 \pi}{D} m})^k $
and therefore we get
$ X_2(\omega) = \sum_{m = -\infty}^{\infty} S_D[m] e^{-j \omega \frac{m}{D}} $
$ X_2(\omega) = \sum_{m = -\infty}^{\infty} \frac{1}{D}\sum_{k = 0}^{D-1} e^{j k \frac{2 \pi}{D} m} x_1[m] e^{-j \omega \frac{m}{D}} $
$ X_2(\omega) = \frac{1}{D} \sum_{k=0}^{D-1} \sum_{m=-\infty}^{\infty} x_1[m] e^{-j m (\frac{\omega - 2 \pi k}{D})} $
And since $ \sum_{m=-\infty}^{\infty} x_1[m] e^{-j m (\frac{\omega - 2 \pi k}{D})} = X_1(\frac{\omega - 2 \pi k}{D}) $
Therefore, $ X_2(\omega) = \frac{1}{D} \sum_{k=0}^{D-1} X_1(\frac{\omega - 2 \pi k}{D}) $
