Signals - is this problem supposed to be so long?

[FrogPad]

Ok, I am working through these review problems and I'm coming up short here. I have a TON of problems to do, and I just feel like I am spending way to much time on this one. Could someone tell me if I am on the right track, or is there an easier way to do this.

Thank you.

Question:
Determine the Fourier series of the output of the RC circuit with input shown below:

Image Link (http://farm3.static.flickr.com/2013/2227524302_7129ed8e0c.jpg")

Solution:
So in the previous part of the problem I found the frequency response (given conditions that RC=1) to be: $$H(\omega) = \frac{1}{1+j \omega}$$.

So my thought at the solution is to use the fact that $$Y(\omega) = X(\omega)H(\omega)$$, and then from this I could find $$y(t)$$. Thus, an outline of the solution is:

Outline of possible solution
1: Find $$X(\omega)$$
2: Calculate $$Y(\omega)$$
3: Calculate $$y(t)$$
4: Find coefficients of y(t)

So here are my steps thus far.

1: One period of x(t) is $$T = 2$$. I can therefore write the function as:
x(t) = -2t+1 for t=[0,1]
x(t) = 2(t-2)+1 for 1=[1,2]

The series coefficients are found by:
$$a_k = \frac{1}{T} \int_T x(t) e^{-jk \omega_0 t} dt$$

After a lot of simplifcation I arrive at:
$$a_k = \frac{4}{k^2 \pi^2}, \,\,\, k=odd$$
$$a_k = 0, \,\,\, k=even$$

Finally,
$$X(\omega) = 2 \pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega -k \pi)$$

2: $$Y(\omega) = \left( \frac{1}{1+j\omega} \right) \left( 2 \pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega -k \pi) \right)$$

I stopped here... am I on the right track?

Thanks !

 

[kurt.physics]

Yes, you are on the right track. To finish the problem, you should calculate y(t) using the inverse Fourier transform of Y(ω). After this you can use the Fourier series to find the coefficients of y(t).