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Old May 2nd 2008, 05:33 PM   #1
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Fourier Series of (x)= abs(sin(x))

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Last edited by nairbdm; May 8th 2008 at 02:26 PM.
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Old May 3rd 2008, 02:18 AM   #2
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Originally Posted by nairbdm View Post
I need to find both the complex and the sine/cosine Fourier series for the function f(x)= abs(sin(x))
Any help would be greatly appreciated,
Thanks.
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Old May 3rd 2008, 10:33 PM   #3
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Originally Posted by nairbdm View Post
since the period of abs(sinx) is pi, should the integral for cn be:
(1/pi) * integral from -pi/2 to pi/2 of abs(sinx) *e^(-2inx/pi)

and if that is correct where should I go from there?
That looks OK. Now you have to integrate. Note |sin x| = -sin x for -pi/2 < x < 0.
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Old May 7th 2008, 06:43 AM   #4
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Originally Posted by nairbdm View Post
okay, now I got:
Cn= [(-2in)(e^(-inpi)-e^(inpi))] / ((-2in)^2 + 1)
can that simplifty any more?
The exponentials will simplify. Note that $\displaystyle e^{i n \pi} = \left( e^{i \pi} \right)^n$ and $\displaystyle e^{-i n \pi} = \left( e^{- i \pi} \right)^n$ ......
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