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Outline | Outline | ||
| − | + | <br> | |
| − | Derivation | + | Introduction<br>Derivation<br>Example<br>Conclusion<br>References |
| − | + | <br> | |
| − | + | Introduction | |
| − | + | Hello! My name is Ryan Johnson! You might be wondering what a slecture is! A slecture is a student lecture that gives a brief overview about a particular topic! In this slecture, I will discuss the relationship between an original signal, x(t), and a sampling of that original signal, x_s(t). We will also take a look at how this relationship translates to the frequency domain, (X(f) & X_s(f)). | |
| + | |||
| + | Derivation | ||
| − | + | F = Fourier transform | |
<div style="margin-left: 3em;"> | <div style="margin-left: 3em;"> | ||
<math> | <math> | ||
| Line 17: | Line 19: | ||
comb_T(x(t)) &= x(t) \times p_T(t) &= x_s(t)\\ | comb_T(x(t)) &= x(t) \times p_T(t) &= x_s(t)\\ | ||
\end{align} | \end{align} | ||
| − | </math> | + | </math> The comb of a signal is equal to the signal multiplied by an impulse train which is equal to the sampled signal. Essentially, the comb is grabbing points on the graph x(t) at a set interval, T. <math> |
| − | + | ||
| − | <math> | + | |
\begin{align} | \begin{align} | ||
| − | X_s(f) &= F(x_s(t)) = F(comb_T)\\ | + | X_s(f) &= F(x_s(t)) = F(comb_T(x(t))\\ |
&= F(x(t)p_T(t))\\ | &= F(x(t)p_T(t))\\ | ||
| − | &= X(f) * F(p_T( | + | &= X(f) * F(p_T(f)\\ |
| − | &= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\ | + | \end{align} |
| + | </math> Multiplication in time is equal to convolution in frequency. <br> <math> | ||
| + | \begin{align} | ||
| + | X_s(f)&= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\ | ||
| + | \end{align} | ||
| + | </math> definition of the Fourier transform of an impulse train. | ||
| + | <br> | ||
| + | <div style="margin-left: 3em;"> | ||
| + | <math> | ||
| + | \begin{align} | ||
&= \frac{1}{T}X(f)*p_\frac{1}{T}(f)\\ | &= \frac{1}{T}X(f)*p_\frac{1}{T}(f)\\ | ||
&= \frac{1}{T}rep_\frac{1}{T}X(f)\\ | &= \frac{1}{T}rep_\frac{1}{T}X(f)\\ | ||
\end{align} | \end{align} | ||
| − | </math> | + | </math> The result is a repetition of the Fourier transformed signal. |
</div> | </div> | ||
<font size="size"></font> | <font size="size"></font> | ||
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Example | Example | ||
| + | [[Image:Johns637_3.jpg]] | ||
| + | <br> | ||
| + | |||
| + | Conclusion | ||
| + | |||
| + | Xs(f) is a rep of X(f) in the frequency domain with amplitude of 1/T and period of 1/T. | ||
| + | |||
| + | Questions | ||
| + | <br> | ||
| − | [[ | + | If you have any questions, comments, etc. please post them on [[Slecture_Topic_7_Ryancomments|this page]]. |
| − | + | References | |
| + | <br> | ||
| + | [1] Mireille Boutin, "ECE 438 Digital Signal Processing with Applications," Purdue University. October 6, 2014. | ||
| + | ---- | ||
| + | [[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]] | ||
Latest revision as of 08:57, 14 March 2015
Outline
Introduction
Derivation
Example
Conclusion
References
Introduction
Hello! My name is Ryan Johnson! You might be wondering what a slecture is! A slecture is a student lecture that gives a brief overview about a particular topic! In this slecture, I will discuss the relationship between an original signal, x(t), and a sampling of that original signal, x_s(t). We will also take a look at how this relationship translates to the frequency domain, (X(f) & X_s(f)).
Derivation
F = Fourier transform
$ \begin{align} comb_T(x(t)) &= x(t) \times p_T(t) &= x_s(t)\\ \end{align} $ The comb of a signal is equal to the signal multiplied by an impulse train which is equal to the sampled signal. Essentially, the comb is grabbing points on the graph x(t) at a set interval, T. $ \begin{align} X_s(f) &= F(x_s(t)) = F(comb_T(x(t))\\ &= F(x(t)p_T(t))\\ &= X(f) * F(p_T(f)\\ \end{align} $ Multiplication in time is equal to convolution in frequency.
$ \begin{align} X_s(f)&= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\ \end{align} $ definition of the Fourier transform of an impulse train.
$ \begin{align} &= \frac{1}{T}X(f)*p_\frac{1}{T}(f)\\ &= \frac{1}{T}rep_\frac{1}{T}X(f)\\ \end{align} $ The result is a repetition of the Fourier transformed signal.
Example
Conclusion
Xs(f) is a rep of X(f) in the frequency domain with amplitude of 1/T and period of 1/T.
Questions
If you have any questions, comments, etc. please post them on this page.
References
[1] Mireille Boutin, "ECE 438 Digital Signal Processing with Applications," Purdue University. October 6, 2014.
Back to ECE438 slectures, Fall 2014
