(New page: == DT Fourier Series == If input is <math>(2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi)</math>) |
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If input is <math>(2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi)</math> | If input is <math>(2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi)</math> | ||
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| + | <math>= (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})</math> | ||
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| + | <math>= (\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})</math> | ||
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| + | <math>= (\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))</math> | ||
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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> | ||
Revision as of 09:09, 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}) $
