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now we "pad with zeros" to define <math>x_2[n]</math>
 
now we "pad with zeros" to define <math>x_2[n]</math>
  
<math>x_2[n] = \begin{cases}x[\frac{n}{D}], & \text{if} \frac{n}{D} \in \mathbb{Z} \\0, &\text{else} \end{cases} f</math>
+
<math>x_2[n] = \begin{cases}x_1[\frac{n}{D}], & \text{if} \frac{n}{D} \in \mathbb{Z} \\0, &\text{else} \end{cases} f</math>
 +
 
 +
note: <math>D</math> must be an integer greater then one
  
 
<math>x_2[n]</math> can also be defined by
 
<math>x_2[n]</math> can also be defined by
  
<math>x_2[n] = \sum_{n = -\inf}^{\inf} </math>
+
<math>x_2[n] = \sum_{k} x_1[k] \delta[n-kD] </math>
 +
 
 +
Taking the DTFT of <math>x_2[n]</math>
 +
 
 +
<math>X_2(e^{j\omega}) = \sum_{n} ( \sum_{k} x_1[k] \delta[n-kD]) e^{-j\omega n})</math>
 +
 
 +
switching the order of the summations you can get
 +
 
 +
<math> X_2(e^{j\omega})= \sum_{k}x_1[k] ( \sum_{n}\delta[n-kD]) e^{-j\omega n})</math>
 +
 
 +
where,
 +
 
 +
<math>\sum_{n}\delta[n-kD]) e^{-j\omega n} = e^{-j\omega kD}</math>
 +
 
 +
therefor,
 +
 
 +
<math>X_2(e^{j\omega}) = \sum_{k} x_1[k] e^{-j\omega kD}</math>
 +
 
 +
This is just the DTFT of the original signal scaled by D!
 +
 
 +
</font>
 +
----
 +
<font size = 4> 3. Example </font>
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 +
 
 +
 
 +
 
 +
 
 +
 
  
  

Revision as of 14:28, 14 October 2014


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!


3. Example

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn