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<math>x(t) = 1 + (1 + \frac{1}{2j})e^{j8\pi t} + (1 - \frac{1}{2j})e^{-j8\pi t} + (\frac{1}{2} e^{j(\frac{\pi}{4})}) e^{j16\pi t} + (\frac{1}{2} e^{-j(\frac{\pi}{4})}) e^{-j16\pi t}</math> | <math>x(t) = 1 + (1 + \frac{1}{2j})e^{j8\pi t} + (1 - \frac{1}{2j})e^{-j8\pi t} + (\frac{1}{2} e^{j(\frac{\pi}{4})}) e^{j16\pi t} + (\frac{1}{2} e^{-j(\frac{\pi}{4})}) e^{-j16\pi t}</math> | ||
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| + | <font size ="3">The Fourier series coefficients are:</font> | ||
Revision as of 16:35, 25 September 2008
CT Fourier Series
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_o t} $
Example
$ x(t) = 1 + sin(8\pi t) + 2cos(8\pi t) + cos(16\pi t + \frac{\pi}{4}) $
The fundamental frequency is $ 8\pi $.
$ x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $
$ x(t) = 1 + (1 + \frac{1}{2j})e^{j8\pi t} + (1 - \frac{1}{2j})e^{-j8\pi t} + (\frac{1}{2} e^{j(\frac{\pi}{4})}) e^{j16\pi t} + (\frac{1}{2} e^{-j(\frac{\pi}{4})}) e^{-j16\pi t} $
The Fourier series coefficients are:
