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

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood