Upsampling - Rhea

By: Michael Deufel

1. Introduction

The purpose of Upsampling is to manipulate a signal in order to artificially increase the sampling rate. This is done by...

We will overview the whole process but focus on the effect upsampling has in the frequency domain

2. Derivation

We will start with discrete signal $ x_1[n] $

now we "pad with zeros" to define $ x_2[n] $

$ x_2[n] = \begin{cases}x_1[\frac{n}{D}], & \text{if} \frac{n}{D} \in \mathbb{Z} \\0, &\text{else} \end{cases} f $

note: $ D $ must be an integer greater then one

$ x_2[n] $ can also be defined by

$ x_2[n] = \sum_{k} x_1[k] \delta[n-kD] $

Taking the DTFT of $ x_2[n] $

$ X_2(\omega) = \sum_{n} ( \sum_{k} x_1[k] \delta[n-kD]) e^{-j\omega n}) $

switching the order of the summations you can get

$ X_2(\omega)= \sum_{k}x_1[k] ( \sum_{n}\delta[n-kD]) e^{-j\omega n}) $

where,

$ \sum_{n}\delta[n-kD]) e^{-j\omega n} = e^{-j\omega kD} $

therefor,

$ X_2(\omega) = \sum_{k} x_1[k] e^{-j\omega kD} $

This is just the DTFT of the original signal scaled by D

LPF with filter that has cutoffs at $ \frac{\pi}{D} $ and $ \frac{- \pi}{D} $ to get $ Y_2(\omega) $

3. Graphical Example in the Frequency Domain

Note: The gain in this example is 1

The first graph shows the signal $ X_1(w) $ which is periodic with period $ 2\pi $

The second graph shows the signal $ X_2(W) $ which is also periodic with period $ 2\pi $ but contains unwanted repetitions of the signal

The third graph shows the Final Signal after it has been passed through the LPF to remove the unwanted repetitions

In the second graph it is obvious why the signal must pass through a LPF, because the expansion in the time domain resulted in compression in the frequency domain and that has caused the signal to repeat itself with in the range of $ [-\pi,\pi] $ and this would cause the reconstruction of the signal to be inaccurate

4. Conclusion

Upsampling is an effective way to reduce the sample time of a signal without resampling the original signal.

5. Questions/Comments

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