Signals - is this problem supposed to be so long?

1. Jan 28, 2008

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:

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 !

Last edited: Jan 28, 2008