1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: System Resonse's Autocorrelation Function (using integral)

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    x(t) is wss.
    [tex] R_x(\tau) = e^{-a|\tau|}[/tex]
    The process goes through a system with the impulse response
    [tex] h(t) = u(t)e^{-bt}[/tex]
    What is the output's autocorrelation function?

    3. The attempt at a solution
    For starters, I already found a solution by finding the output power spectral density and finding its inverse Fourier transform. But I cannot make my integration method work. I want someone to tell me where my integration is going wrong:

    [tex]R_y(\tau)=\int_{-\infty}^\infty \int_{-\infty}^\infty h(s)h(r)R_x(\tau +s -r)\,dr \,ds [/tex]
    [tex]R_y(\tau)=\int_{-\infty}^\infty \int_{-\infty}^\infty u(s)e^{-bs}u(r)e^{-br}e^{-a|\tau +s -r|}\,dr \,ds [/tex]

    So I separated the integral. Let us just ignore the second integral for now and focus on my method for the first half. If the first half is working, the second will work. So I write for one integral:
    [tex] \tau +s -r > 0 \rightarrow \tau + s > r[/tex]
    So the first integral becomes, now eliminating the u(r) by incorporating it into the limits:
    [tex]R_y(\tau)=\int_{-\infty}^\infty \int_{0}^{\tau + s} u(s)e^{-bs}e^{-br}e^{-a(\tau +s -r)}\,dr \,ds [/tex]

    But we then realize unless the below condition is true, the integration adds up to zero due to the unit step:
    [tex] \tau + s > 0 \rightarrow s > -\tau[/tex]

    We can use this now as a limit of integration:

    [tex]R_y(\tau)=\int_{-\tau}^\infty \int_{0}^{\tau + s} u(s)e^{-bs}e^{-br}e^{-a(\tau +s -r)}\,dr \,ds [/tex]

    But, due to u(s), we know the following condition must also be true or the answer is zero:
    [tex] -\tau > 0[/tex]
    EDIT: I think the above is not true. If -tau < 0, the integration then goes from 0 to infinity. Is this the root of the error? If so, how do I handle this? uhh, I guess I'm just stuck.

    So I integragted (assuming this solution was for negative tau)
    [tex]R_y(\tau)=\int_{-\tau}^\infty \int_{0}^{\tau + s} e^{-bs}e^{-br}e^{-a(\tau +s -r)}\,dr \,ds [/tex]

    and got (for tau < 0)
    [tex]\frac{e^{b \tau}}{2(a+b)b}[/tex]

    So my total answer will be
    [tex]\frac{e^{-b |\tau|}}{2(a+b)b} \ne \frac{1}{a^2-b^2} \left ( \frac{a}{b} e^{-b|\tau|} - e^{-a |\tau|} \right )[/tex]

    Note, the answer it is not equal to is the one I got using power spectral density.
    Last edited: Mar 29, 2012
  2. jcsd
  3. Mar 29, 2012 #2

    So let's say I have the symbolic solution to that integral (after the second one) in terms of s. I will call it


    Let's say I set up a similar one for the other half and get that solution, calling it xx(r).

    Then the solution for when tau < 0 is actually:
    -x(tau) - xx(0)
    (because x(inf) = 0 and xx(inf) = 0 -- I am just adding the two integrals, one from -tau to inf and the other from 0 to inf)

    And this results in the same answer as from the power spectral density!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook