Revision as of 15:00, 26 September 2008 by Choi88 (Talk)

A Periodic DT Signal


$ x[n] = 1 + sin({2\pi \over N})n + 5cos({2\pi \over N})n + 7sin({4\pi \over N}n - {\pi \over 2}) $

The signal above x[n] is periodic with period N.

$ x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {7 \over 2j}[e^{j({4\pi \over N}-{\pi \over 2}n)}-e^{-j({4\pi \over N} - {\pi \over 2})n}] $

The Fourth term on the right hand goes away since $ e^{j{\pi \over 2}n} \text{ and } e^{-j{\pi \over 2}n} $ are zero.

$ x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] $

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