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Math Help - Power of a signal, rather urgent

  1. #1
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    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
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bnay View Post


    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
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  3. #3
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    I found the average power to be [(C1)^2 + (C2)^2]/2
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by bnay View Post
    I found the average power to be [(C1)^2 + (C2)^2]/2
    So you have shown that if \omega_i, \omega_j \ne 0 and \omega_i \ne \omega_j when i \ne j:

    \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
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