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<math>=-\frac{1}{4}e^{j3\pi n}-\frac{1}{4}e^{-j3\pi n}+\frac{1}{j2}e^{3\pi n}-\frac{1}{j2}e^{-j3\pi n}</math>
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<math>   =-\frac{1}{4}e^{j3\pi n}-\frac{1}{4}e^{-j3\pi n}+\frac{1}{j2}e^{3\pi n}-\frac{1}{j2}e^{-j3\pi n}</math>

Revision as of 15:32, 26 September 2008

$ x[n]=-\frac{1}{2}cos(3\pi n)+sin(3\pi n)\! $.


$ x[n]=-\frac{1}{2}[\frac{e^{j3\pi n}+e^{-j3\pi n}}{2}]+\frac{e^{j3\pi n}-e^{-3\pi n}}{j2} $


$ =-\frac{1}{4}e^{j3\pi n}-\frac{1}{4}e^{-j3\pi n}+\frac{1}{j2}e^{3\pi n}-\frac{1}{j2}e^{-j3\pi n} $

Alumni Liaison

EISL lab graduate

Mu Qiao