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  #1  
Old 10-04-2008, 10:35 PM
guitarguy12387 guitarguy12387 is offline
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Default Complex Fourier series of full wave rectifier

I have a full wave rectified sine wave. Basically, the graph looks like abs(sinx).

When calculating the coefficients, i have tried evaluating:

∫[sin(t)exp(-jnt)]dt from 0 to pi.

and am not getting the correct answer.

I'm wondering if i need to just look for a computation error... or if I am setting the problem up wrong.

I'm having doubts about if i'm allowed to setup the integral as such.

Thanks.
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  #2  
Old 10-06-2008, 02:29 PM
Mark44 Mark44 is offline
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Quote:
Originally Posted by guitarguy12387 View Post
I have a full wave rectified sine wave. Basically, the graph looks like abs(sinx).

When calculating the coefficients, i have tried evaluating:

∫[sin(t)exp(-jnt)]dt from 0 to pi.

and am not getting the correct answer.

I'm wondering if i need to just look for a computation error... or if I am setting the problem up wrong.

I'm having doubts about if i'm allowed to setup the integral as such.

Thanks.
The function you give, f(x) = |sin x|, is an even function, so its Fourier series consists only of cosine terms. An even function is one for which f(-x) = f(x), for all x in the domain of f.

You can also find the Fourier coefficients using the approach you showed, and the integral is c_n = 1/(2pi)∫f(t) exp(-jnt) dt, integrated from -pi to pi.

You need to take into account the fact that your function is equal to -sin(t) on the interval -pi to 0, so the integral above can be rewritten as
c_n = 1/(2pi) ∫ -sin(t) exp(-jnt) dt + 1/(2pi) ∫ -sin(t) exp(-jnt) dt,
with the limits of integration for first integral are -pi to 0, and for the second, 0 to pi.

Mark
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Last edited by Mark44; 10-06-2008 at 07:49 PM. Reason: Remove extra text at end
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  #3  
Old 10-06-2008, 03:00 PM
guitarguy12387 guitarguy12387 is offline
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ahhh ok. Thanks Mark.

Won't that result in integrating over TWO periods of the function though? Is that allowed?
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  #4  
Old 10-06-2008, 07:12 PM
guitarguy12387 guitarguy12387 is offline
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Figured it out...

Apparently i had the right approach, but i had the wrong argument in the exponential. Oops heh. Thanks again, Mark.
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  #5  
Old 10-06-2008, 08:25 PM
Mark44 Mark44 is offline
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Quote:
Originally Posted by guitarguy12387 View Post
ahhh ok. Thanks Mark.

Won't that result in integrating over TWO periods of the function though? Is that allowed?
Good point.
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Old 10-06-2008, 08:27 PM
Mark44 Mark44 is offline
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Quote:
Originally Posted by guitarguy12387 View Post
Figured it out...

Apparently i had the right approach, but i had the wrong argument in the exponential. Oops heh. Thanks again, Mark.
Was the work that you actually did different from what you posted? I didn't notice anything obviously wrong in what you posted.
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  #7  
Old 10-06-2008, 09:29 PM
guitarguy12387 guitarguy12387 is offline
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Just the exponential... since ω = 2pi/T = 2, it should have been:

∫[sin(t)exp(-j2nt)]dt from 0 to pi
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