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<math>
 
<math>
 
\begin{align}
 
\begin{align}
X_s(f) &= F(\sum_{k = -\infty}^\infty \delta(t-kT))\\
+
X_s(f) &= F(\sum_{k = -\infty}^\infty x(kT)\delta(t-kT))\\
 
&= \sum_{k = -\infty}^\infty x(kT)F(\delta(t-kT))\\
 
&= \sum_{k = -\infty}^\infty x(kT)F(\delta(t-kT))\\
&= X(f)*\frac{1}{T}\sum_{k = -\infty}^\infty \delta(f-\frac{n}{T})\\
+
&= \sum_{k = -\infty}^\infty x(kT)e^{-j2\pi\omega fkT}\\
&= \frac{1}{T}X(f)*P_\frac{1}{T}(f)\\
+
 
&= \frac{1}{T}rep_\frac{1}{T}X(f)\\
+
 
\end{align}
 
\end{align}
 
</math>
 
</math>
 
</div>
 
</div>
 
<font size>
 
<font size>

Revision as of 22:31, 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:

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_{k = -\infty}^\infty x(kT)\delta(t-kT))\\ &= \sum_{k = -\infty}^\infty x(kT)F(\delta(t-kT))\\ &= \sum_{k = -\infty}^\infty x(kT)e^{-j2\pi\omega fkT}\\ \end{align} $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin