Difference between revisions of "Fourier transform"

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Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br>
 
Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br>
 
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>
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<br><br>
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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>.
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<br>
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The Fourier Series representation of <math> \gamma(t) </math> is
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<br>
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<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> <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>

Revision as of 09:44, 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 .
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_\alpha}
and  \alpha_k={1\over T_\alpha}\int_{-{T_\alpha\over 2}}^{{T_\alpha\over 2}} \gamma(t) e^{-j2\pi kt}\,dt