(New page: ==CT Signal and its Fourier Series Coefficients== Let the signal be <math>\ x(t) = \cos(3t) \sin(9t) </math> Now computing its coefficients: <math>\ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) ...) |
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| − | <math>\ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{ | + | <math>\ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2j}) </math> |
Revision as of 17:58, 25 September 2008
CT Signal and its Fourier Series Coefficients
Let the signal be $ \ x(t) = \cos(3t) \sin(9t) $
Now computing its coefficients:
$ \ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2j}) $
