Difference between revisions of "Fourier transform"

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<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> \lim_{T_\zeta \to \infty}\gamma(t)=\beta(t) </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> \gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt} </math> where <math> f={1\over T_\alpha}
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<math> \gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt} </math> where <math> f={1\over T_\zeta}
</math> <br>and <math> \alpha_k={1\over T_\alpha}\int_{-{T_\alpha\over 2}}^{{T_\alpha\over 2}} \gamma(t) e^{-j2\pi kt}\,dt</math>
+
</math> <br>and <math> \alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt</math>
  +
<br>
  +
<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
  +
<math>
  +
\alpha_k={1\over T_\zeta}\Beta(kf)
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</math>
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<br> and
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<math>
  +
\gamma(t)
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</math>
  +
becomes
  +
<math>
  +
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}
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</math>

Revision as of 09:02, 8 December 2004

An initially identity that is useful: 

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

Suppose that we have some function, say  \beta (t) , that is nonperiodic and finite in duration.
This means that  \beta(t)=0 for some  T_\alpha < \left | t \right |

Now let's make a periodic function  \gamma(t) by repeating  \beta(t) with a fundamental period  T_\zeta . Note that  \lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)
The Fourier Series representation of  \gamma(t) is
 \gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt} where  f={1\over T_\zeta}
and  \alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt
 \alpha_k can now be rewritten as  \alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt
From our initial identity then, we can write  \alpha_k as 
\alpha_k={1\over T_\zeta}\Beta(kf)
and 
\gamma(t)
becomes 
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}