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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> | ||
| − | We know another way to express CTFT of <math> | + | We know another way to express CTFT of <math> x_s(t) </math>: |
<div style="margin-left: 3em;"> | <div style="margin-left: 3em;"> | ||
| Line 63: | Line 63: | ||
&= \sum_{k = -\infty}^\infty x(kT)F(\delta(t-kT))\\ | &= \sum_{k = -\infty}^\infty x(kT)F(\delta(t-kT))\\ | ||
&= \sum_{k = -\infty}^\infty x(kT)e^{-j2\pi fkT}\\ | &= \sum_{k = -\infty}^\infty x(kT)e^{-j2\pi fkT}\\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | </div> | ||
| + | <font size> | ||
| + | compare it with DTFT of <math> x_d[n] </math>: | ||
| + | |||
| + | <div style="margin-left: 3em;"> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | X_d(\omega) &= \sum_{k = -\infty}^\infty x_d(k)e^{-j\omegak}\\ | ||
| + | &= \sum_{k = -\infty}^\infty x(kT)e^{-j\omegak}\\ | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
</div> | </div> | ||
<font size> | <font size> | ||
Revision as of 21:35, 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) \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 fkT}\\ \end{align} $
compare it with DTFT of $ x_d[n] $:
$ \begin{align} X_d(\omega) &= \sum_{k = -\infty}^\infty x_d(k)e^{-j\omegak}\\ &= \sum_{k = -\infty}^\infty x(kT)e^{-j\omegak}\\ \end{align} $
