Fourier Series, Sin(t), but skips every other period Hard

[zenjirou]

Homework Statement

f(t)= sin(t), so as expected the period goes from 0 to 2pi. But 2pi to 4 pi is function is ZERO, and then starts up again from 4pi to 6pi, then zero from 6pi to 8pi,etc

So basically sin(t) that skips every other period.

Homework Equations

The Attempt at a Solution

In my attempt, I did Exponential Fourier series, and got coefficients but it included dividing by 1-(n^2)/4. Thus there is a discontinuity? which mess up everything... since the summation should go from 1 to inf.

I also tried trigonometric Fourier series and got an, bn, that included dividing by 1-n/2, which has the same problem of discontinuity

Question is how to deal with these discontinuities, I'm using MATLAB on this and I get errors becaues its not defined at n=2. But the summation should go from 1 to inf.

any help appreciated.

 

[zenjirou]

So basically for my exponential Fourier coefficient ,
I got Dn= ((-1)*(-1)^k)/{ (8pi*(1-(n^2)/4)}

 

[zenjirou]

Becasue of the 1- (n^2)/4) term, n cannot =2. So when this coefficient is placed in the Summation, It doesn't work. How does one deal with a case like this. Perhaps we change the summation from 3 to inf...

 

[vela]

Take the integral you have for d_n. Before you integrate it, set n=2 and simplify the integrand.

 

[zenjirou]

Thanks,

So Using Exponential Fourier Series, my |Dn| = (-1)^n / 2-(n^2)/2
Is it normal for the complex component, i, to disappear.
I can see how for n=2, we will have to do a separate integral, but for the main summation, it no longer has a complex component which is strange..

 

[HallsofIvy]

Your function does NOT have period [itex]2\pi[/itex], it has period [itex]4\pi[/itex]. Take that into account.