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Math Help - Convolution of functions

  1. #1
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    Convolution of functions

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  2. #2
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    The convolution of two functions f_1 and f_2 is defined by
    (f_1 \ast f_2)(t) = \int_{-\infty}^\infty f_1(\tau)f_2(t-\tau) \;dx
    The way you have defined f_1 means that it is zero outside the interval [0,1]. So, the product of f_1 and f_2 is also zero outside the same interval. Therefore it should be sufficient to integrate from 0 to 1.

    However, you have not defined the function g(x), and your notation is very inconsistent. Some places you use the variable t, other places x. So, I'm not sure whether this answers your question.
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  3. #3
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    I suggest you start out with drawing a picture of the two signals. If you mirror and timeshift f_1 with x, you end up with these limits ( f_1 has the width 1 which means the right end point is x after you mirror and shift it, while the left one is x-1):

    x < -\pi/2 : (f_1 * f_2)(x) = 0
    -\pi/2 \leq x < 1 - \pi/2: (f_1 * f_2)(x) = \int_{-\pi/2}^{x}f_1(x-\tau)f_2(\tau)d\tau
    1 - \pi/2 \leq x < \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{x}f_1(x-\tau)f_2(\tau)d\tau
    \pi/2 \leq x < 1 + \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{\pi/2}f_1(x-\tau)f_2(\tau)d\tau
    x \geq 1+\pi/2: (f_1 * f_2)(x) = 0
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  4. #4
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    It is strange, but first and third integral divergent. Could it be right?
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  5. #5
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    Probably not... Either way we can't tell unless you post the complete definition of f_2.
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  6. #6
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    I wrote in first topic, how f_2 looks like: for me it's known only:
    f_2(x) is tg(x) when x is in [-pi/2 , pi/2] otherwise it is f_2(x) is 0
    That's all i know about f_2
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  7. #7
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    How exactly did you determine that the first and third integrals are divergent without knowing what tg(x) is? Why is there a t in there when it's a function of x?
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  8. #8
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    Thing with 't' is actually only my attempt to solve it, thus:
    for t <-pi/2, -pi/2 + 1> X=<-pi/2, t>
    for t <-pi/2 + 1, pi/2> X=<t-1, t>
    for t <pi/2, pi/2 + 1> X=<t-1, pi/2>
    it shoud be the same as you wrote, but with 't'. Then I tried to solve integral:
    \int f_1(t-y)f_2(y) dy with appropriate limits. And it gave me really strange conclusion (for quick calculation i used wolfram)

    EDIT: tg(x) is tangens function
    Last edited by token22; May 15th 2011 at 05:30 AM.
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  9. #9
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    I don't know where you got the problem from, but it's poorly written to say the least. I can help you with the last convolution though.

    x\geq 2: (f*f)(x)=\int_1^{x-1}d\tau

    Again, I suggest you draw the two signals. The definition of f here means that it's constantly 1 for [1, \infty). Mirroring it and translating it with x gives the right endpoint the coordinate x-1.
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  10. #10
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    Alright then i understand this second convolution, but i have problem with first.
    By the way, function tg(x) is simply tangens function (i forgot that somebody write it as tan(x))
    I got
    f1*f2(t) = {-infinity for t in <-pi/2, 1 - pi/2>
    ln|cos(t-1)| - ln|cos(t)| for t in <1 - pi/2, pi/2>
    infinity for t in <pi/2, pi/2 + 1>
    0 otherwise

    but i am not sure about infinity
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  11. #11
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    \tan x is not convergent on [-\pi/2, \pi/2], so (f_1*f_2)(x) is also not convergent.
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  12. #12
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    Hm but second integral is like: ln|cos(x-1)| - ln|cos(x)| does it mean that f1*f2(x) is not convergent in every possible case?

    edit: i calculate those integrals over and over again but still do not know what is result
    edit2: could you please write result funcion, becasuse i am getting quite desperate now
    is result only that the function f1*f2(x) is divergent ? (withou any calculation?)
    Last edited by token22; May 15th 2011 at 08:36 AM.
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  13. #13
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    Quote Originally Posted by token22 View Post
    I got
    f1*f2(t) = {-infinity for t in <-pi/2, 1 - pi/2>
    ln|cos(t-1)| - ln|cos(t)| for t in <1 - pi/2, pi/2>
    infinity for t in <pi/2, pi/2 + 1>
    0 otherwise

    but i am not sure about infinity
    That solution is correct. If f_1 is the function defined by f_1(x) = \begin{cases}1&\text{if }0\leqslant x\leqslant 1\\0&\text{otherwise}\end{cases}, and g is any other function, then (f_1*g)(t) is the mean value of g on the interval [t–1,t].

    For the function f_2(x) = \begin{cases}\tan x&\text{if }-\pi/2\leqslant x\leqslant \pi/2\\0&\text{otherwise}\end{cases}, its integral over any interval containing \pm\pi/2 diverges. Thus f_1*f_2 is infinite (or if you prefer, undefined) in those intervals specified in your answer.
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  14. #14
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    Just for sure, so the "undefined state" is caused by divergnet tangent? thus i get some result only in interval [t-1,t], because of f_1 function
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