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.

 

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

 

ahhh ok. Thanks Mark.

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

 

Figured it out...

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

 

ahhh ok. Thanks Mark.

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

Good point.

 

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.

 

Just the exponential... since ω = 2pi/T = 2, it should have been:

∫[sin(t)exp(-j2nt)]dt from 0 to pi