Revision as of 15:15, 14 October 2014 by Mdeufel (Talk | contribs)


Upsampling with an emphasis on the frequency domain

By: Michael Deufel


  1. Introduction
  2. Derivation
  3. Examples
  4. Conclusion


1. Introduction

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

  1. Discretize the signal
  2. Pad original signal with zeros
  3. Take the DTFT
  4. Send through a LPF (low pass filter)
  5. Take the inverse DTFT to return to the time domain

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

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

Now LPF with filter that has cutoffs at $ \frac{\pi}{D} $ and $ \frac{- \pi}{D} $ to remove the unwanted repitions


3. Graphical Example in the Frequency Domain

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang