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

Xfcbt.png

Xsfcbt.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

Xsfcbt.png

Xdwcbt.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 $


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva