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If input is <math>(1 + j)cos(\frac{\pi}{2}t + \pi)sin(\frac{\pi}{4}t)</math> | If input is <math>(1 + j)cos(\frac{\pi}{2}t + \pi)sin(\frac{\pi}{4}t)</math> | ||
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| + | <math>= (1 + j)(\frac {e^{j \frac{\pi}{2}t + \pi j} + e^{-j \frac{\pi}{2}t - \pi j}}{2})(\frac{e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}}{2j})</math> | ||
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| + | <math>= \frac{1 + j}{4j}(e^{j \frac{\pi}{2}t}e^{\pi j} + e^{-j \frac{\pi}{2}t}e^{ \pi j})(e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t})</math> | ||
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| + | <math>= \frac{1 + j}{4j}(e^{j \frac{\pi}{2}t}(1) + e^{-j \frac{\pi}{2}t}(-1))(e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t})</math> | ||
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| + | <math>= \frac{1 + j}{4j}(e^{j \frac{3 \pi}{4}t} - e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t} + e^{-j \frac{3 \pi}{4}t})</math> | ||
Latest revision as of 08:55, 26 September 2008
CT Fourier Series
If input is $ (1 + j)cos(\frac{\pi}{2}t + \pi)sin(\frac{\pi}{4}t) $
$ = (1 + j)(\frac {e^{j \frac{\pi}{2}t + \pi j} + e^{-j \frac{\pi}{2}t - \pi j}}{2})(\frac{e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}}{2j}) $
$ = \frac{1 + j}{4j}(e^{j \frac{\pi}{2}t}e^{\pi j} + e^{-j \frac{\pi}{2}t}e^{ \pi j})(e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}) $
$ = \frac{1 + j}{4j}(e^{j \frac{\pi}{2}t}(1) + e^{-j \frac{\pi}{2}t}(-1))(e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}) $
$ = \frac{1 + j}{4j}(e^{j \frac{3 \pi}{4}t} - e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t} + e^{-j \frac{3 \pi}{4}t}) $
