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x[n] = <math>1/{2\pi}</math>
+
x[n] = <math>1/{2\pi} int_{2\pi} X(e^{j\omega}</math>

Revision as of 06:33, 26 July 2009


D.T.F.T.


x[n] = $ 1/{2\pi} int_{2\pi} X(e^{j\omega} $

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EISL lab graduate

Mu Qiao