# Thread: Power of a signal, rather urgent

1. ## Power of a signal, rather urgent

Hello,

I've been working on this problem for a while now (hopefully the image worked), specifically part 4b i. I thought I had a solution, but then I realized that the square of a summation is not the summation of a square and that little mistake destroyed my proof.

I know that the power (P) is the limit as t --> infinity of (1/t) multiplied by the integral from -T/2 to T/2 of [x(t)]^2, but taking the square of the summation is where I get stuck. Does anybody know how to integrate the square of the summation, or how to simplify the square of the summation so I can integrate it?

Thank you very much

2. Originally Posted by bnay

Hello,

I've been working on this problem for a while now (hopefully the image worked), specifically part 4b i. I thought I had a solution, but then I realized that the square of a summation is not the summation of a square and that little mistake destroyed my proof.

I know that the power (P) is the limit as t --> infinity of (1/t) multiplied by the integral from -T/2 to T/2 of [x(t)]^2, but taking the square of the summation is where I get stuck. Does anybody know how to integrate the square of the summation, or how to simplify the square of the summation so I can integrate it?

Thank you very much
What did you find for the average power in 4a i?

CB

3. I found the average power to be [(C1)^2 + (C2)^2]/2

4. Originally Posted by bnay
I found the average power to be [(C1)^2 + (C2)^2]/2
So you have shown that if $\displaystyle \omega_i, \omega_j \ne 0$ and $\displaystyle \omega_i \ne \omega_j$ when $\displaystyle i \ne j$:

$\displaystyle \lim_{T\to \infty}\frac{1}{T}\int_{-T/2}^{T/2} \cos(\omega_it+\theta_i) \cos(\omega_jt+\theta_j)\;dt=\begin{cases}0,&i \ne j\\1,&i=j\end{cases}$

which is all you need for 4b i

CB