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<math>x(t)=4sin(3t)+(1+6j)cos(2t)\!</math> | <math>x(t)=4sin(3t)+(1+6j)cos(2t)\!</math> | ||
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| + | == Solution == | ||
| + | The fundamental period <math>T\!</math> is <math>2\pi\!</math>. Thus we use the equation <math>\omega_0=\frac{2\pi}{T}\!</math> to find <math>\omega_0=1\!</math> | ||
Revision as of 17:15, 24 September 2008
Equations
Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
Signal Coefficients:
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Defined Signal
$ x(t)=4sin(3t)+(1+6j)cos(2t)\! $
Solution
The fundamental period $ T\! $ is $ 2\pi\! $. Thus we use the equation $ \omega_0=\frac{2\pi}{T}\! $ to find $ \omega_0=1\! $
