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# Fourier Series watch

1. how to find the complex fourier series of

x(t) = |A*sin(pi*t)|

I convert sin function to complex number, mutiply by e^(-jwnt) and integrate over 0 to 1.

I still can't get the correct answer.
May I know is there any wrong with my approach of solving this??

Thank you very much
2. x(t) = c(n) e^(2pi jnt)

where

c(n)
= |A| sin(pi t) e^(-2pi jnt) dt
= (|A|/2j) (e^(j pi t) - e^(-j pi t)) e^(-2pi jnt) dt
= (|A|/2j) e^((1 - 2n)j pi t) - e^(-(1 + 2n)j pi t) dt
= (|A|/2j) [ e^((1 - 2n)j pi t)/((1 - 2n)j pi) + e^(-(1 + 2n)j pi t)/((1 + 2n)j pi) dt ] (from 0 to 1)
= (|A|/2j) [ -2/((1 - 2n)j pi) - 2/((1 + 2n)j pi) ]
= |A| [ 1/((1 - 2n)pi) + 1/((1 + 2n)pi) ]
= (|A|/pi) [ 1/(1 - 2n) + 1/(1 + 2n) ]
= 2|A|/(pi(1 - 4n^2))

So

x(t) = (2|A|/pi) e^(2pi jnt)/(1 - 4n^2)
3. thank you for your great help, Jonny W.

I realise that I make a careless arithmetic mistake.
4. How do I convert Trigo Fourier Series to Complex Fourier Series??

4A/pi 1/(2n-1) sin[(2n-1)wt]

I tried using the formula
C n
where

n > 0 = 0.5(an + jbn)
n = 0 = a0
n < 0 = 0.5(an -jbn)

Thank you

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Updated: April 17, 2006
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