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To find the value of <math>a_0\!</math> we simply plug and chug: | To find the value of <math>a_0\!</math> we simply plug and chug: | ||
<br> | <br> | ||
| − | <math>a_0=\frac{1}{ | + | <math>a_0=\frac{1}{2\pi}\int_0^T[4sin(3t)+(1+6j)cos(2t)]e^{0}dt</math> |
| + | <br> | ||
| + | <br> | ||
| + | <math> =\frac{2}{\pi}\int_0^T4sin(3t)dt+\frac{1+6j}{2\pi}\int_0^Tcos(2t)dt</math> | ||
| + | <br> | ||
| + | <br> | ||
Revision as of 16:24, 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\! $
To find the value of $ a_0\! $ we simply plug and chug:
$ a_0=\frac{1}{2\pi}\int_0^T[4sin(3t)+(1+6j)cos(2t)]e^{0}dt $
$ =\frac{2}{\pi}\int_0^T4sin(3t)dt+\frac{1+6j}{2\pi}\int_0^Tcos(2t)dt $
