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Study the autocorrelation function

  1. Jul 20, 2007 #1
    I would like to know how to find the autocorrelation function.
    If there are any links for free e-books that can I downloaded to study the autocorrelation function and its properties in details.
    Thanks!
     
  2. jcsd
  3. Jul 20, 2007 #2

    EnumaElish

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  4. Jul 20, 2007 #3
    I already know about http://en.wikipedia.org/wiki/Autocorrelation.
    I meant another links that will provide any free download E-books if you know.
    Thanks alot!
     
  5. Jul 21, 2007 #4

    EnumaElish

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  6. Jul 21, 2007 #5
    I am going to find the autocorrelation function for the following equation
    Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n * e^(-t^2)

    So, I didnt find the previous links can help me to solve this problem.
     
    Last edited: Jul 21, 2007
  7. Jul 22, 2007 #6

    EnumaElish

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    You need to solve

    Lims-->0-ss Hn(t)Hn(t+s)ds / (2s).

    Note,
    -ss Hn(t)Hn(t+s)ds = ∫t-zz-t Hn(t)Hn(z)dz
     
    Last edited: Jul 22, 2007
  8. Jul 23, 2007 #7
    You need to solve

    Lims-->0∫-ss Hn(t)Hn(t+s)ds / (2s).

    Note,
    ∫-ss Hn(t)Hn(t+s)ds = ∫t-zz-t Hn(t)Hn(z)dz


    but I want to understand why you suggest this. I want to understad exactly by steps how can I find a solution.
    Thanks alot.
     
  9. Jul 23, 2007 #8

    EnumaElish

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  10. Jul 23, 2007 #9
    Thanks alot EnumaElish!
    I will try to read these and I will reply you.
    Thanks alot.
     
  11. Jul 25, 2007 #10
    I couldnt solve the equation
    t-zz-t Hn(t)Hn(z)dz

    may be I should to follow such courses that explain Autocorrelation function or may be I must make reveiw for the integrals.
    I want to advise me, please.
     
  12. Jul 25, 2007 #11

    EnumaElish

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    For starters,

    ∫Hn(t)Hn(z)dz = Hn(t)∫Hn(z)dz.

    Then, insert Hn(z) = (-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2). Now you have:

    Hn(t)∫(-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2)dz = Hn(t)(-1)^n ∫e^(z^2)* fn(z) dz

    where fn(z) = d^n/dz^n * e^(-z^2).

    You need to work on solving ∫e^(z^2)* fn(z) dz.
    ___________________
    CORRECTION: Lims-->0-ss Hn(t)Hn(t+s)ds should have been LimT-->0-TT Hn(t)Hn(t+s)ds.

    Then, defining z = t+s, we have-TT Hn(t)Hn(t+s)ds = ∫-TT Hn(t)Hn(z)dz.

    This does not change the substance of the advice I have given until now.
     
    Last edited: Jul 25, 2007
  13. Jul 27, 2007 #12


    let g(z)=∫e^(z^2)* fn(z) dz
    for n = 1

    fn(z)= d/dz e^-(z^2)= -2 z e-(z^2)
    So, g(z) = ∫-2 z * e^(z^2)* e-(z^2) dz
    = ∫-2 z *e^[(z^2)-(z^2)] dz
    = ∫-2 z dz
    =-z^2

    is not that true?
     
  14. Jul 27, 2007 #13

    EnumaElish

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    Your calculation looks right.

    Now you need to put in the integration limits (-T, T), multiply with Hn(t)(-1)^n, then take the limit T --> 0.
     
  15. Jul 28, 2007 #14
    Now you need to put in the integration limits (-T, T), multiply with Hn(t)(-1)^n, then take the limit T --> 0.[/QUOTE]

    we have fn(z) = -Z^2 for n =1
    by multiplying it by (-1)^n
    we will have Z^2
    by multiplying it by Hn(t),
    where Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)]
    Then, we will have:
    Z^2 * (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)], where z = t+s
    So, how I will find the integral and I have two variables (t and Z)??
    Can you help me, Please.
     
    Last edited: Jul 28, 2007
  16. Jul 28, 2007 #15

    EnumaElish

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    Remember, the integral is a definite integral. It has limits "from -T to T." You need to evaluate the integral according to the limits.

    Before you multiply the integral with Hn(t), you need to apply the integ. limits to -z^2. This will make z "disappear." The z^2 will be replaced with a quadratic function of T, say q(T).

    Do you understand what q(T) is? Can you derive it explicitly?
     
  17. Jul 29, 2007 #16
    OK, I will have:
    -z^2[tex]\left|^{T}_{-T}[/tex]
    then I will get
    -(T)^2 - [- (-T)^2]= 0
    So, the autocorelation for
    Hn(t)Hn(t+s)ds = 0

    is that true!!!
     
    Last edited: Jul 29, 2007
  18. Jul 29, 2007 #17

    EnumaElish

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    It seems to be true for n=1.

    Can you solve the general case for arbitrary n?
     
  19. Jul 29, 2007 #18
    Actually, this is exactly what I am going to ask you about.
    So, can you just help me a little bit?
    And I have another question:
    What about the autocorrelation of
    Hn(t)Hm(t)dt , where m not equal to n
     
  20. Jul 29, 2007 #19

    EnumaElish

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    I cannot see the LaTex equation for m. There seems to be a typesetting error.

    EDIT: I can see it now.

    For the Hn(t)Hn(t+s) case, you need to work on solving ∫e^(z^2)* fn(z) dz for arbitrary n.

    The Hn(t)Hm(t)dt case is not a regular AC function. Go back to the MathWorld link (http://mathworld.wolfram.com/Autocorrelation.html) and review the definition. The AC function has a single argument (a single function), for example, ac(t) = AC(Hn(t)) or equivalently ac = AC(Hn). What you are proposing has two arguments, Hn and Hm. So it doesn't fit the definition of AC.
     
    Last edited: Jul 29, 2007
  21. Jul 29, 2007 #20
    when I written:the autocorrelation of Hn(t)Hm(t)dt
    I meant that what about the autocorrelation function between the first derivative of H(t) and the second derivative of H(t)
    Such as :
    the autocorrelation function of H1(t)H2(t).
    Thanks alot!
     
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