(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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== DT Fourier Series == | == DT Fourier Series == | ||
| − | If input is <math>(2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi)</math> | + | If input is <math> x[n] = (2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi)</math> |
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <math>= (\frac{1}{4j})(-2e^{j \frac{4 \pi}{3}n} - 2e^{- \frac{2 \pi}{3}n} - 2e^{j \frac{2 \pi}{3}n} - 2e^{-j \frac{4 \pi}{3}n} - e^{j\frac{11 \pi}{6}n} - e^{-j \frac{\pi}{6}n} - e^{j\frac{7 \pi}{6}n} - e^{-j \frac{5 \pi}{6}n})</math> | ||
| + | |||
| + | The fundamental frequency is <math>\frac{\pi}{6}</math> and the fundamental period is 12 | ||
| + | |||
| + | Therefore <math>a_{8} = \frac{-2}{4j}, a_{-4} = \frac{-2}{4j}, a_{4} = \frac{-2}{4j}, a_{-8} = \frac{-2}{4j}, a_{11} = \frac{-1}{4j}, a_{-1} = \frac{-1}{4j}, a_{7} = \frac{-1}{4j}, a_{-5} = \frac{-1}{4j} </math> | ||
| + | |||
| + | The other values of <math>\ a_{k} = 0 </math> within the fundamental period. | ||
Latest revision as of 14:40, 26 September 2008
DT Fourier Series
If input is $ x[n] = (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}) $
$ = (\frac{1}{4j})(-2e^{j \frac{4 \pi}{3}n} - 2e^{- \frac{2 \pi}{3}n} - 2e^{j \frac{2 \pi}{3}n} - 2e^{-j \frac{4 \pi}{3}n} - e^{j\frac{11 \pi}{6}n} - e^{-j \frac{\pi}{6}n} - e^{j\frac{7 \pi}{6}n} - e^{-j \frac{5 \pi}{6}n}) $
The fundamental frequency is $ \frac{\pi}{6} $ and the fundamental period is 12
Therefore $ a_{8} = \frac{-2}{4j}, a_{-4} = \frac{-2}{4j}, a_{4} = \frac{-2}{4j}, a_{-8} = \frac{-2}{4j}, a_{11} = \frac{-1}{4j}, a_{-1} = \frac{-1}{4j}, a_{7} = \frac{-1}{4j}, a_{-5} = \frac{-1}{4j} $
The other values of $ \ a_{k} = 0 $ within the fundamental period.
