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Introduction  
 
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 and a sampling of that original signal. We will also take a look at how this relationship translates to the frequency domain.  
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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  
 
Derivation  
  
F = fourier transform  
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F = Fourier transform  
 
<div style="margin-left: 3em;">
 
<div style="margin-left: 3em;">
 
<math>
 
<math>
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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> The comb of a signal is equal to the signal multiplied by an impulse train. <math>
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</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>
 
\begin{align}
 
\begin{align}
 
X_s(f) &= F(x_s(t)) = F(comb_T(x(t))\\
 
X_s(f) &= F(x_s(t)) = F(comb_T(x(t))\\
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\begin{align}
 
\begin{align}
 
X_s(f)&= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\
 
X_s(f)&= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\
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\end{align}
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</math> definition of the Fourier transform of an impulse train.
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<br>
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<div style="margin-left: 3em;">
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<math>
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\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> The result is a repetition of the fourier transformed signal.  
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</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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Xs(f) is a rep of X(f) in the frequency domain with amplitude of 1/T and period of 1/T.
 
Xs(f) is a rep of X(f) in the frequency domain with amplitude of 1/T and period of 1/T.
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Questions
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<br>
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If you have any questions, comments, etc. please post them on [[Slecture_Topic_7_Ryancomments|this page]].
  
 
References
 
References
 
<br>
 
<br>
 
[1] Mireille Boutin, "ECE 438 Digital Signal Processing with Applications," Purdue University. October 6, 2014.
 
[1] Mireille Boutin, "ECE 438 Digital Signal Processing with Applications," Purdue University. October 6, 2014.
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[[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 Johns637 3.jpg

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

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