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#1




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. 
#2




Quote:
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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"The floggings will continue until morale improves!" Last edited by Mark44; 10062008 at 07:49 PM. Reason: Remove extra text at end 
#3




ahhh ok. Thanks Mark.
Won't that result in integrating over TWO periods of the function though? Is that allowed? 
#4




Figured it out...
Apparently i had the right approach, but i had the wrong argument in the exponential. Oops heh. Thanks again, Mark. 
#5




Good point.
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"The floggings will continue until morale improves!" 
#6




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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"The floggings will continue until morale improves!" 
#7




Just the exponential... since ω = 2pi/T = 2, it should have been:
∫[sin(t)exp(j2nt)]dt from 0 to pi 
Tags 
complex, fourier, full, rectifier, series, wave 
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