Difference between revisions of "Sampling Rate Conversion - Downsampling" - Rhea

Revision as of 20:40, 22 September 2009

Sample Rate Conversion

Let us take a function $ x(t) $

We wish to covert a signal sampled at $ T_1 $ to 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 \to $ conversion can be accomplished by down-sampling

2. $ T_2 $ is a divider of $ T_1 \to $ conversion can be accomplished by up-sampling followed by LPF

Case 1 - $ T_2 $ is a multiple of $ T_1 $

We are trying to accomplish the following -

We know that

$ x_2[n] = x_1[Dn] \quad \mbox{where } D = \frac{T_2}{T_1} $

as 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}}\quad \mbox{where }m\mbox{ 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,

Let us assume the following plot for $ X_1(\omega) $. We will also assume that the Nyquist condition is satisfied, ie. $ f_{max} < \frac{1}{2T_1} $.

Now, on scaling we have the following plot for $ \frac{1}{D} X_1(\frac{\omega}{D}) $