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

Revision as of 09:29, 8 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