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| − | It will be helpful -- necessary even -- to find the fundamental period of the signal. In our case, the period of the overall signal is <math>2\pi</math>, so <math>\ | + | It will be helpful -- necessary even -- to find the fundamental period of the signal. In our case, the period of the overall signal is <math>2\pi</math>, so <math>\omega_0</math> will be <math>\frac{2\pi}{2\pi}=1</math>. |
A good place to start is the calculation of <math>a_0</math>, which is the average of the signal. Plotting the signal makes it look like the average is 0, but we can integrate to check. | A good place to start is the calculation of <math>a_0</math>, which is the average of the signal. Plotting the signal makes it look like the average is 0, but we can integrate to check. | ||
Revision as of 13:03, 23 September 2008
The Signal
Consider the signal $ 7\sin(2t)+(1+j)\cos(3t) $.
The Formulae
Recall the Fourier Series formulae for the continuous time signal case:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
and
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Finding the Series
It will be helpful -- necessary even -- to find the fundamental period of the signal. In our case, the period of the overall signal is $ 2\pi $, so $ \omega_0 $ will be $ \frac{2\pi}{2\pi}=1 $.
A good place to start is the calculation of $ a_0 $, which is the average of the signal. Plotting the signal makes it look like the average is 0, but we can integrate to check.
$ a_0=\frac{1}{T}\int_0^T[7\sin(2t)+(1+j)\cos(3t)]e^{-jk\omega_0t}dt $
$ =\frac{7}{2\pi}\int_0^{2\pi}\sin(2t)dt + \frac{1+j}{2\pi}\int_0^{2\pi}\cos(3t)dt $
$ =\frac{-7}{4\pi}\cos(2t)|_0^{2\pi}+\frac{1+j}{6\pi}\sin(3t)|_0^{2\pi} $
$ =\frac{-7}{4\pi}(\cos(4\pi)-\cos(0))+\frac{1+j}{6\pi}(\sin(6\pi)-\sin(0))=0 $
