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<math>= (\frac{2 + j}{4j})(-e^{j \frac{4 \pi}{3}n} - e^{- \frac{2 \pi}{3}n} - e^{j \frac{2 \pi}{3}n} - e^{-j \frac{4 \pi}{3}n})</math> | <math>= (\frac{2 + j}{4j})(-e^{j \frac{4 \pi}{3}n} - e^{- \frac{2 \pi}{3}n} - e^{j \frac{2 \pi}{3}n} - e^{-j \frac{4 \pi}{3}n})</math> | ||
| + | |||
| + | <math>= (\frac{2 + e^{\frac{\pi}{2}}}{4j})(-e^{j \frac{4 \pi}{3}n} - e^{- \frac{2 \pi}{3}n} - e^{j \frac{2 \pi}{3}n} - e^{-j \frac{4 \pi}{3}n})</math> | ||
Revision as of 13:17, 26 September 2008
DT Fourier Series
If input is $ (2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi) $
$ = (2 + j)(\frac{e^{j \frac{\pi}{3}n} + e^{-j \frac{\pi}{3}n}}{2})(\frac{e^{j \pi n - j \pi} - e^{-j \pi n + j \pi}}{2j}) $
$ = (\frac{2 + j}{4j})(e^{j \frac{\pi}{3}n} + e^{-j \frac{\pi}{3}n})(e^{j \pi n}e^{ - j \pi} - e^{-j \pi n}e^{j \pi}) $
$ = (\frac{2 + j}{4j})(e^{j \frac{\pi}{3}n} + e^{-j \frac{\pi}{3}n})(e^{j \pi n}(-1) - e^{-j \pi n}(1)) $
$ = (\frac{2 + j}{4j})(-e^{j \frac{4 \pi}{3}n} - e^{- \frac{2 \pi}{3}n} - e^{j \frac{2 \pi}{3}n} - e^{-j \frac{4 \pi}{3}n}) $
$ = (\frac{2 + e^{\frac{\pi}{2}}}{4j})(-e^{j \frac{4 \pi}{3}n} - e^{- \frac{2 \pi}{3}n} - e^{j \frac{2 \pi}{3}n} - e^{-j \frac{4 \pi}{3}n}) $
