Line 45: Line 45:
 
Taking the DTFT of <math>x_2[n]</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>
+
<math>X_2(\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
 
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>
+
<math> X_2(\omega)= \sum_{k}x_1[k] ( \sum_{n}\delta[n-kD]) e^{-j\omega n})</math>
  
 
where,
 
where,
Line 57: Line 57:
 
therefor,  
 
therefor,  
  
<math>X_2(e^{j\omega}) = \sum_{k} x_1[k] e^{-j\omega kD}</math>
+
<math>X_2(\omega) = \sum_{k} x_1[k] e^{-j\omega kD}</math>
  
 
This is just the DTFT of the original signal scaled by D
 
This is just the DTFT of the original signal scaled by D

Revision as of 16:51, 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(\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

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

Fig1 deufel.jpeg

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


3. Conclusion


\url{google.com}

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn