# Complex Fourier series of full wave rectifier

Discussion in 'Homework Help' started by guitarguy12387, Oct 4, 2008.

1. ### guitarguy12387 Thread Starter Active Member

Apr 10, 2008
359
12
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. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
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

Last edited: Oct 6, 2008
3. ### guitarguy12387 Thread Starter Active Member

Apr 10, 2008
359
12
ahhh ok. Thanks Mark.

Won't that result in integrating over TWO periods of the function though? Is that allowed?

4. ### guitarguy12387 Thread Starter Active Member

Apr 10, 2008
359
12
Figured it out...

Apparently i had the right approach, but i had the wrong argument in the exponential. Oops heh. Thanks again, Mark.

Nov 26, 2007
626
1
Good point.

6. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Was the work that you actually did different from what you posted? I didn't notice anything obviously wrong in what you posted.

7. ### guitarguy12387 Thread Starter Active Member

Apr 10, 2008
359
12
Just the exponential... since ω = 2pi/T = 2, it should have been:

∫[sin(t)exp(-j2nt)]dt from 0 to pi