Difference between revisions of "Fourier transform"
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Line 2:  Line 2:  
<math> 
<math> 

−  +  X(f)=\int_{\infty}^{\infty} x(t) e^{j2\pi ft}\, dt 

</math> 
</math> 

Line 10:  Line 10:  
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left  t \right  </math> 
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left  t \right  </math> 

<br><br> 
<br><br> 

−  Now let's make a periodic function <math> \gamma(t) </math> by repeating <math> \beta(t) </math> with a fundamental period <math> T_\zeta </math>. 

+  Now let's make a periodic function 

⚫  
+  <math> 

+  \gamma(t) 

+  </math> 

+  by repeating 

+  <math> 

+  \beta(t) 

+  </math> 

+  with a fundamental period 

+  <math> 

+  T_\zeta 

+  </math>. 

+  Note that 

+  <math> 

⚫  
+  </math> 

<br> 
<br> 

The Fourier Series representation of <math> \gamma(t) </math> is 
The Fourier Series representation of <math> \gamma(t) </math> is 

<br> 
<br> 

⚫  
+  <math> 

⚫  
⚫  
+  </math> 

+  where 

+  <math> 

+  f={1\over T_\zeta} 

+  </math> 

+  <br>and 

+  <math> 

⚫  
+  </math> 

<br> 
<br> 

−  <math> \alpha_k </math> can now be rewritten as 
+  <math> \alpha_k </math> can now be rewritten as 
+  <math> 

+  \alpha_k={1\over T_\zeta}\int_{\infty}^{\infty} \beta(t) e^{j2\pi kt}\,dt 

+  </math> 

<br>From our initial identity then, we can write <math> \alpha_k </math> as 
<br>From our initial identity then, we can write <math> \alpha_k </math> as 

<math> 
<math> 

−  \alpha_k={1\over T_\zeta}\Beta(kf) 
+  \alpha_k={1\over T_\zeta}\Beta(kf) 
</math> 
</math> 

<br> and 
<br> and 

<math> 
<math> 

−  \gamma(t) 
+  \gamma(t) 
</math> 
</math> 

becomes 
becomes 

<math> 
<math> 

−  \gamma(t)=\sum_{k=\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt} 
+  \gamma(t)=\sum_{k=\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt} 
+  </math> 

+  <br> 

+  Now remember that 

+  <math> 

+  \beta(t)=\lim_{T_\zeta \to \infty}\gamma(t) 

+  </math> 

+  and 

+  <math> 

+  {1\over {T_\zeta}} = f. 

+  </math> 

+  <br> 

+  Which means that 

+  <math> 

+  \beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=\infty}^\infty f \Beta(kf) e^{j2\pi fkt} 

+  </math> 

+  <br> 

+  Which is just to say that 

+  <math> 

+  \beta(t)=\int_{\infty}^\infty f \Beta(f) e^{j2\pi fkt}\,df 

+  </math> 

+  <br> 

+  <br> 

+  So we have that the Fourier Transform of 

+  <math> 

+  \beta(t) 

+  </math> 

+  is 

+  <math> 

+  X(f)=\int_{\infty}^{\infty} x(t) e^{j2\pi ft}\, dt 

</math> 
</math> 
Revision as of 10:21, 9 December 2004
An initially identity that is useful:
Suppose that we have some function, say , that is nonperiodic and finite in duration.
This means that for some
Now let's make a periodic function
by repeating
with a fundamental period
.
Note that
The Fourier Series representation of is
where
and
can now be rewritten as
From our initial identity then, we can write as
and
becomes
Now remember that
and
Which means that
Which is just to say that
So we have that the Fourier Transform of
is