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1.INTRODUCTION:  
 
1.INTRODUCTION:  
 
</font size>
 
</font size>
<font size= 3>
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<font size= 4>
  
 
The topic 3 is the Fourier Transform of the Comb and Rep function. In my selecture, I am going to introduce the definition, the Fourier Transformation and the relationship of Comb function and Rep function.</font size>
 
The topic 3 is the Fourier Transform of the Comb and Rep function. In my selecture, I am going to introduce the definition, the Fourier Transformation and the relationship of Comb function and Rep function.</font size>
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<font size= 3>(1)According to the definition of the comb function: </font size>
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<font size= 4>(1)According to the definition of the comb function: </font size>
  
  
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<font size= 3> where</font size> <math>P_T(t)= \sum_{n=-\infty}^\infty \delta(t-nT)</math>
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<font size= 4> where</font size> <math>P_T(t)= \sum_{n=-\infty}^\infty \delta(t-nT)</math>
  
  
<font size= 3>Do the Fourier Transform to the function:</font size>
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<font size= 4>Do the Fourier Transform to the function:</font size>
  
  
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<font size= 3>
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<font size= 4>
 
According to the property of Fourier Transformation, the multiplication in the time domain is equal to the convolution in the frequency domain.</font size>
 
According to the property of Fourier Transformation, the multiplication in the time domain is equal to the convolution in the frequency domain.</font size>
  
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                   <math>=x(f)*F\big(P_T(t)\big)</math>
 
                   <math>=x(f)*F\big(P_T(t)\big)</math>
  
<font size= 3>Because </font size>  <math>P_T(t)= \sum_{n=-\infty}^\infty \delta(t-nT)</math>  <font size= 3> is a periodic function , so we can expand it to Fourier series. </font size>
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<font size= 4>Because </font size>  <math>P_T(t)= \sum_{n=-\infty}^\infty \delta(t-nT)</math>  <font size= 4> is a periodic function , so we can expand it to Fourier series. </font size>
  
  
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       <math>=\frac{1}{T}</math>
 
       <math>=\frac{1}{T}</math>
  
<font size= 3>So, </font size>  <math>P_T(t) = \frac{1}{T}\sum_{n=-\infty}^\infty F_n e^{jn\cdot 2\pi t/T} </math>
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<font size= 4>So, </font size>  <math>P_T(t) = \frac{1}{T}\sum_{n=-\infty}^\infty F_n e^{jn\cdot 2\pi t/T} </math>
  
 
           <math>=\sum_{n=-\infty}^\infty \frac{1}{T} F(e^{jn\cdot 2\pi t/T}) </math>
 
           <math>=\sum_{n=-\infty}^\infty \frac{1}{T} F(e^{jn\cdot 2\pi t/T}) </math>
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           <math>= \frac{1}{T}P_{1/T}(f)</math>
 
           <math>= \frac{1}{T}P_{1/T}(f)</math>
  
<font size= 3>So, </font size>  <math>F\bigg(comb_T\big(x(t)\big)\bigg)=X(f)*\frac{1}{T}P_{1/T}(f)</math>
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<font size= 4>So, </font size>  <math>F\bigg(comb_T\big(x(t)\big)\bigg)=X(f)*\frac{1}{T}P_{1/T}(f)</math>
  
 
                     <math>=\frac{1}{T}X(f)*P_{1/T}(f)</math>
 
                     <math>=\frac{1}{T}X(f)*P_{1/T}(f)</math>
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Reference:https://engineering.purdue.edu/~bouman/ece637/notes/pdf/RepComb.pdf
 
Reference:https://engineering.purdue.edu/~bouman/ece637/notes/pdf/RepComb.pdf
  
<font size= 3>(2)According to the definition of Rep function:</font size>
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<font size= 4>(2)According to the definition of Rep function:</font size>
  
  
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<font size= 3>Use the impluse-train we get previously, according to the conclusion we get from Fourier Transformation of it, we know:</font size>
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<font size= 4>Use the impluse-train we get previously, according to the conclusion we get from Fourier Transformation of it, we know:</font size>
  
 
           <math>F\big(P_T(t)\big)=\frac{1}{T}P_{1/T}(f)</math>
 
           <math>F\big(P_T(t)\big)=\frac{1}{T}P_{1/T}(f)</math>
  
<font size= 3>So, </font size> <math>F\bigg(rep_T\big(x(t)\big)\bigg)=x(f)\cdot\frac{1}{T}P_{1/T}(f)</math>
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<font size= 4>So, </font size> <math>F\bigg(rep_T\big(x(t)\big)\bigg)=x(f)\cdot\frac{1}{T}P_{1/T}(f)</math>
  
 
                     <math>=\frac{1}{T}x(f)\cdot P_{1/T}(f)</math>  
 
                     <math>=\frac{1}{T}x(f)\cdot P_{1/T}(f)</math>  

Revision as of 18:35, 1 October 2014


Topic 3:Fourier transform of "rep" and "comb"

A slecture by ECE student Youqin Liu

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.



ECE438 SELECTURE


1.INTRODUCTION:

The topic 3 is the Fourier Transform of the Comb and Rep function. In my selecture, I am going to introduce the definition, the Fourier Transformation and the relationship of Comb function and Rep function.

2.THEORY:

Combfunction3.jpg

Reference:https://engineering.purdue.edu/~bouman/ece637/notes/pdf/RepComb.pdf


(1)According to the definition of the comb function:


$ comb_T\big(X(t)\big)= x(t)\cdot\ P_T(t) $


where $ P_T(t)= \sum_{n=-\infty}^\infty \delta(t-nT) $


Do the Fourier Transform to the function:


$ F\bigg(comb_T\big(x(t)\big)\bigg) = F\big(x(t)\cdot P_T(t)\big) $


According to the property of Fourier Transformation, the multiplication in the time domain is equal to the convolution in the frequency domain.

$ F\bigg(comb_T\big(x(t)\big)\bigg) = F\big(x(t)\big)* F\big(P_T(t)\big) $

                 $ =x(f)*F\big(P_T(t)\big) $

Because $ P_T(t)= \sum_{n=-\infty}^\infty \delta(t-nT) $ is a periodic function , so we can expand it to Fourier series.


$ P_T(t)=\sum_{n=-\infty}^\infty F_n e^{jn\cdot 2\pi t/T}  $


$ \Rightarrow F_n = \frac{1}{T}\int\limits_{-T/2}^{T/2}P_T(t)e^{jn\cdot 2\pi t/T}dt $

      $ =\frac{1}{T} $

So, $ P_T(t) = \frac{1}{T}\sum_{n=-\infty}^\infty F_n e^{jn\cdot 2\pi t/T} $

         $ =\sum_{n=-\infty}^\infty \frac{1}{T} F(e^{jn\cdot 2\pi t/T})  $
         $ =\sum_{n=-\infty}^\infty \frac{1}{T} \delta(f-\frac{n}{T}) $
         $ = \frac{1}{T}P_{1/T}(f) $

So, $ F\bigg(comb_T\big(x(t)\big)\bigg)=X(f)*\frac{1}{T}P_{1/T}(f) $

                    $ =\frac{1}{T}X(f)*P_{1/T}(f) $
  
                    $ =\frac{1}{T}rep_{1/T}X(f) $


Repfunction3.jpg

Reference:https://engineering.purdue.edu/~bouman/ece637/notes/pdf/RepComb.pdf

(2)According to the definition of Rep function:


        $ rep_T\big(x(t)\big):= x(t)*P_T(t) $
                   $ =x(t)*\sum_{n=-\infty}^\infty \delta(t-nT) $


So, $ F\bigg(rep_T\big(x(t)\big)\bigg)=F\bigg(x(t)*\sum_{n=-\infty}^\infty \delta(t-nT)\bigg) $


Use the impluse-train we get previously, according to the conclusion we get from Fourier Transformation of it, we know:

         $ F\big(P_T(t)\big)=\frac{1}{T}P_{1/T}(f) $

So, $ F\bigg(rep_T\big(x(t)\big)\bigg)=x(f)\cdot\frac{1}{T}P_{1/T}(f) $

                   $ =\frac{1}{T}x(f)\cdot P_{1/T}(f) $ 


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