(→CT Signal and its Fourier Series Coefficients) |
(→CT Signal and its Fourier Series Coefficients) |
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Let the signal be | Let the signal be | ||
<math>\ x(t) = \cos(3t) \sin(9t) </math> | <math>\ x(t) = \cos(3t) \sin(9t) </math> | ||
| − | + | which has a fundamental frequency of | |
Now computing its coefficients: | Now computing its coefficients: | ||
<math>\ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2j}) </math> | <math>\ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2j}) </math> | ||
| + | |||
| + | |||
| + | |||
| + | <math>\ x(t)= \frac{e^{j12t} + e^{-j6t} + e^{j6t} - e^{-j12t}}{4j} </math> | ||
| + | |||
| + | Now rearranging the terms: | ||
| + | |||
| + | <math>\ x(t)= \frac{e^{j12t} + e^{-j12t} + e^{j6t} - e^{-j6t}}{4j} </math> | ||
Revision as of 18:36, 25 September 2008
CT Signal and its Fourier Series Coefficients
Let the signal be $ \ x(t) = \cos(3t) \sin(9t) $ which has a fundamental frequency of Now computing its coefficients:
$ \ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2j}) $
$ \ x(t)= \frac{e^{j12t} + e^{-j6t} + e^{j6t} - e^{-j12t}}{4j} $
Now rearranging the terms:
$ \ x(t)= \frac{e^{j12t} + e^{-j12t} + e^{j6t} - e^{-j6t}}{4j} $
