Fourier transform: Difference between revisions

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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>
<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>.
<br>
The Fourier Series representation of <math> \gamma(t) </math> is
<br>
<math> \gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt} </math> where <math> f={1\over T_\alpha}
</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)=x(t)ej2πftdt

Suppose that we have some function, say β(t), that is nonperiodic and finite in duration.
This means that β(t)=0 for some Tα<|t|

Now let's make a periodic function γ(t) by repeating β(t) with a fundamental period Tζ.
The Fourier Series representation of γ(t) is
γ(t)=k=αkej2πfkt where f=1Tα
and αk=1TαTα2Tα2γ(t)ej2πktdt