| Line 41: | Line 41: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | + | F(comb_T(x(t)) &= F(x(t)*P_T(t))\\ | |
&= X(f)*\mathcal{F}(\sum_{k = -\infty}^\infty \delta(t-\frac{k}{f_s}))\\ | &= X(f)*\mathcal{F}(\sum_{k = -\infty}^\infty \delta(t-\frac{k}{f_s}))\\ | ||
&= X(f)*f_s\sum_{k = -\infty}^\infty \delta(f-kf_s)\\ | &= X(f)*f_s\sum_{k = -\infty}^\infty \delta(f-kf_s)\\ | ||
Revision as of 21:50, 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
- Introduction
- Derivation
- Example
- 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)*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)*P_T(t))\\ &= X(f)*\mathcal{F}(\sum_{k = -\infty}^\infty \delta(t-\frac{k}{f_s}))\\ &= X(f)*f_s\sum_{k = -\infty}^\infty \delta(f-kf_s)\\ &= f_s\sum_{k = -\infty}^\infty X(f)*\delta(t-\frac{k}{f_s})\\ &= f_s\sum_{k = -\infty}^\infty X(f-kf_s)\\ \end{align} $
