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Autocorrelation Function of a Signal

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm posting this question here at this point. I am having difficulty understanding autocorrelation in terms of solving the problem below. I don't seem to understand the math behind this.

    A white noise process W(t) with unity (N_0/2 = 1) power spectral density is input to a linear system. The output of the linear system is X(t), where

    X(t) = W(t) - W(t - 1)

    Determine the autocorrelation of X(t) and sketch it.

    2. Relevant equations

    We can define the autocorrelation function of a stochastic process X(t) as the expectation of the product of two random variables X(t_1) and X(t_2), obtained by sampling the process X(t) at times t_1 and t_2 respectively. So we can write

    proxy.php?image=http%3A%2F%2Fs12.postimg.org%2Fohb4fg819%2FCapture.png

    f_(X(t_1),X(t_2))(x_1,x_2) is the join probability density function of the process X_(t) sampled at times t_1 and t_2

    M_(XX)(t_1,t_2) is used to emphasize the fact that this is a second order moment. For M_(XX)(t_1,t_2) to dependent on the time difference t_2 - t_1, we have R_(XX)(t_2 - t_1)

    Two different symbols for the autocorrelation function M_(XX)(t_1,t_2) and R_(XX)(t_2 - t_1) to denote that R_(XX)(t_2 - t_1) is the autocorrelation function specifically for a weak stationary process.

    Let τ denote a time shift; that is, t = t_2 and τ = t_1 - t_2

    proxy.php?image=http%3A%2F%2Fs7.postimg.org%2Fihwilnjij%2FCapture.png

    3. The attempt at a solution

    proxy.php?image=http%3A%2F%2Fs28.postimg.org%2Fov8x2e7al%2FCapture.png

    I understand that the first term on the last line is indeed equal to proxy.php?image=http%3A%2F%2Fs13.postimg.org%2Fhh56qatnn%2FCapture.png . I'm however unsure what to do with the other three terms. The solution sets the other three terms to different autocorrelation functions and I'm not sure how these other three terms are autocorrelation functions as well based off of the definition.

    Here's what the solution is. I don't understand how it went from the second line to the third line.

    proxy.php?image=http%3A%2F%2Fs2.postimg.org%2Fdy111sucp%2FCapture.png

    Any help would be greatly appreciated. I also don't understand how how the solution goes from the third line to the fourth line. It seems to just simply replacing the autocorrelation functions with dirac delta functions. I'm not sure how these are equal in any way.
     
  2. jcsd
  3. Nov 29, 2014 #2

    MathematicalPhysicist

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    Gold Member

    From the second line to the third is a simple matter of subtracting the arguments, i.e:

    [tex]E(W(t)W(t+\tau-1) = R_{ww}(t+\tau-1-t) = R_{ww}(\tau-1)[/tex]
    The rest are done in the same way.

    As for the deltas, look at the definition of white noise and its autocorrelation function.
     
  4. Nov 29, 2014 #3
    Hey MathematicalPhysicist,

    Thanks for the reply. I didn't realize this was so simple. My new attempt at a solution

    proxy.php?image=http%3A%2F%2Fs28.postimg.org%2Fov8x2e7al%2FCapture.png
    Capture.png

    I know that Capture.png . But I don't understand how this equation is true Capture.png . It reminds of the simple fact that if

    LHS = RHS
    LHS + 1 = RHS +1

    But that doesn't seem to be the case here. Adding one to both functions inside the expectation operator doesn't change the value of the expectation?

    E[W(t - 1)W(t + tau)] = E[W(t - 1 + 1)W(t + tau + 1)] = E[W(t)W(t + tau + 1)] = R_(WW)(tau + 1)?

    I don't understand this. Is it some property of the expectation operator:

    E[F(t)(F(t)] = E[F(t + 1)F(t + 1)]?
     
  5. Nov 29, 2014 #4

    MathematicalPhysicist

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    For [tex]E(W(t-1)W(t+\tau))=R_{ww}(t+\tau-(t-1))=R_{ww}(\tau+1)[/tex]

    Remember you always take the aboslute value of the difference of times.
     
  6. Nov 29, 2014 #5
    I actually didn't know this

    [tex]E[F(f(t))F(g(t))] = E[F(t)F(g(t) -f(t))] = R_{FF}(g(t) - f(t))[/tex]?

    If I were to try and find the power spectral density of [tex]X(t)[/tex] I would make use of [tex]S_{XX}(f) = F[R_{XX}(\tau)][/tex]. Were [tex]F[][/tex] is the Fourier Transform. [tex]S_{XX}(f) = F[2*delta(\tau) - delta(\tau - 1) - delta(\tau + 1)][/tex] I know that [tex]F[delta(t)] = 1[/tex]. So the Fourier Transform of the first term is simply just two. I however am having difficulty understanding how [tex]F[- delta(\tau - 1) - delta(\tau + 1)] = cos(2\pi*f_{0})[/tex]

    [tex]F[delta(\tau - 1) - delta(\tau - 1)][/tex]
    [tex]F[-e^{i} - e^{-i}][/tex]

    I'm not sure what variable is included in the exponential. But I do know that

    [tex]-e^{ix} - e^{-ix} = -(e^{ix} + e^{-ix}) = -\frac{2}{2}(e^{ix} + e^{-ix}) = -2cos(x)[/tex]

    Thanks for all of your help.
     
  7. Nov 29, 2014 #6

    MathematicalPhysicist

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    You can do what I wrote provided W is WSS (wide sense stationary).

    Read about it here:
    http://en.wikipedia.org/wiki/Stationary_process#Weak_or_wide-sense_stationarity
     
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