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Question about summation in power spectral density formula

  1. Jul 30, 2015 #1
    1. The problem statement, all variables and given/known data
    This isn't really a problem I've been given, but questions i have about how the author of my textbook, Leon Couch, Digital and Analog communications Systems, found the PSD (power spectral density) of an digital NRZ pulse train.

    2. Relevant equations
    The PSD of a periodic signal is defined as: [itex]
    \mathit{P(s)} = \frac{\left |F(f) \right |^{2}}{T}\sum_{k=- \infty}^{\infty}R(k)e^{j2\pi kfT}
    [/itex]
    The text shows that: [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex]
    In the next step he shows that: [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [1+\sum_{k=- \infty}^{\infty}e^{j2\pi kfT} \right ] [/itex]
    This is where I start being confused. It seems as though he factored the R(0) term out of the sum and then factored the [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex] factors out of the summation leaving [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [1+\sum_{k=- \infty}^{\infty}e^{j2\pi kfT} \right ] [/itex]
    To me this doesn't seem right because when factoring the R(0) term out of the summation, the interval of summation is still from negative infinity to infinity, so zero is still in that interval. Also it seems that when he factors all the R(k) factors out of the summation, he doesn't account for the fact that [itex] R(0) = \frac{A^{2}}{2} [/itex]

    3. The attempt at a solution
    To me it seems that it the step should have looked like this: [itex] \mathit{P(s)} = \frac{A^{2}}{4}\frac{\left |F(f) \right |^{2}}{T}\left [2+\sum_{k=1}^{\infty}e^{j2\pi kfT} +\sum_{k=-\infty}^{-1}e^{j2\pi kfT} \right ] [/itex]
    This way R(0) is not in the summation interval after its been pulled out, and the two separate summations account for the intervals from negative infinity to -1, then from 1 to infinity, and R(0) remains separate.
    Also this way both when [itex] R(0) = \frac{A^{2}}{2}, R(k\neq 0)=\frac{A^{2}}{4} [/itex] are factored out, the number 2 is left where the R(0) factor was, which i think is correct.
    Can someone please point out if my line of thinking is correct, or please explain to me why the text is correct?
    Thanks a lot.
    [itex] [/itex]
     
  2. jcsd
  3. Jul 30, 2015 #2
    As a Fourier series, the first term is ##\frac{A^2/2}{2}=\frac{A^2}{4}## (i.e., divided by ##2##). That, I believe, is what's going on here.

    EDIT (Again D: )
    OK! My bad.

    What's happening here is this. The sum converges, so we rearrange a finite number of terms (i.e., one term) and notice that at ##k=0## we obtain your desired coefficient since ##1+e^{0}=2## and ##2A^2/4=A^2/2##.
     
    Last edited: Jul 30, 2015
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