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Show this relationship in graph below:
 
Show this relationship in graph below:
 +
 +
----
 +
 +
==example==
  
 
[[Image:X1.png]]
 
[[Image:X1.png]]
  
 
[[Image:Xs1.png]]
 
[[Image:Xs1.png]]
 +
 +
----
 +
 +
==Derivation==
  
 
Then we are going to find the relation between <math> X_s(f) </math> and <math> X_d(\omega) </math>
 
Then we are going to find the relation between <math> X_s(f) </math> and <math> X_d(\omega) </math>
Line 109: Line 117:
  
 
from this equation, we can know the relationship between <math> X_s(f) </math> and <math> X_d(\omega) </math> and the relationship is showed in graph as below:
 
from this equation, we can know the relationship between <math> X_s(f) </math> and <math> X_d(\omega) </math> and the relationship is showed in graph as below:
 +
 +
----
 +
 +
==example==
  
 
[[Image:Xs1.png]]
 
[[Image:Xs1.png]]

Revision as of 23:11, 5 October 2014


Frequency domain view of the relationship between a signal and a sampling of that signal

A slecture by ECE student Botao Chen

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.


Outline

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

Introduction

In this slecture I will discuss about the relations between the original signal $ X(f) $ (the CTFT of $ x(t) $ ), sampling continuous time signal $ X_s(f) $ (the CTFT of $ x_s(t) $ ) and sampling discrete time signal $ X_d(\omega) $ (the DTFT of $ x_d[n] $ ) in frequency domain and give a specific example showing the relations.


Derivation

The first thing which need to be clarified is that there two different types of sampling signal: $ x_s(t) $ and $ x_d[n] $. $ x_s(t) $ is created by multiplying a impulse train $ P_T(t) $ with the original signal $ x(t) $ and actually $ x_s(t) $ is $ comb_T(x(t)) $ where T is the sampling period. However the $ x_d[n] $ is $ x(nT) $ where T is the sampling period.

Now we first concentrate on the relationship between $ X(f) $ and $ X_s(f) $.

We know that $ x_s(t) = x(t) \times P_T(t) $, we can derive the relationship between $ x_s(t) $ and $ x(t) $ in the following way:

$ \begin{align} F(comb_T(x(t)) &= F(x(t) \times P_T(t))\\ &= X(f)*F(P_T(t))\\ &= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\ &= \frac{1}{T}X(f)*P_\frac{1}{T}(f)\\ &= \frac{1}{T}rep_\frac{1}{T}X(f)\\ \end{align} $

Show this relationship in graph below:


example

X1.png

Xs1.png


Derivation

Then we are going to find the relation between $ X_s(f) $ and $ X_d(\omega) $

We know another way to express CTFT of $ x_s(t) $:

$ \begin{align} X_s(f) &= F(\sum_{n = -\infty}^\infty x(nT)\delta(t-nT))\\ &= \sum_{n = -\infty}^\infty x(nT)F(\delta(t-nT))\\ &= \sum_{n = -\infty}^\infty x(nT)e^{-j2\pi fnT}\\ \end{align} $

compare it with DTFT of $ x_d[n] $:

$ \begin{align} X_d(\omega) &= \sum_{n = -\infty}^\infty x_d[n]e^{-j\omega n}\\ &= \sum_{n = -\infty}^\infty x(nT)e^{-j\omega n}\\ \end{align} $

we can find that:

$ \begin{align} X_d(2\pi Tf) &= X_s(f)\\ \end{align} $

if $ f = \frac{1}{T} $

we have that:

$ \begin{align} X_d(2\pi ) &= X_s(\frac{1}{T})\\ \end{align} $

from this equation, we can know the relationship between $ X_s(f) $ and $ X_d(\omega) $ and the relationship is showed in graph as below:


example

Xs1.png



conclusion

So the relationship between $ X(f) $ and $ X_s(f) $ is that $ X_s(f) $ is a a rep of $ X(f) $ in frequency domain with period of $ \frac{1}{T} $ and magnitude scaled by $ \frac{1}{T} $. the relationship between $ X(f) $ and $ X_d(\omega) $ is that $ X_d(\omega) $ is also a a rep of $ X(f) $ in frequency domain with period $ 2\pi $ and magnitude is also scaled by $ \frac{1}{T} $, but the frequency is scaled by $ 2\pi T $


Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman