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Thread: 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 $\displaystyle f_1$ and $\displaystyle f_2$ is defined by
    $\displaystyle (f_1 \ast f_2)(t) = \int_{-\infty}^\infty f_1(\tau)f_2(t-\tau) \;dx$
    The way you have defined $\displaystyle f_1$ means that it is zero outside the interval $\displaystyle [0,1]$. So, the product of $\displaystyle f_1$ and $\displaystyle f_2$ 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 $\displaystyle g(x)$, and your notation is very inconsistent. Some places you use the variable $\displaystyle t$, other places $\displaystyle x$. 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 $\displaystyle f_1$ with x, you end up with these limits ($\displaystyle f_1$ 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):

    $\displaystyle x < -\pi/2 : (f_1 * f_2)(x) = 0$
    $\displaystyle -\pi/2 \leq x < 1 - \pi/2: (f_1 * f_2)(x) = \int_{-\pi/2}^{x}f_1(x-\tau)f_2(\tau)d\tau$
    $\displaystyle 1 - \pi/2 \leq x < \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{x}f_1(x-\tau)f_2(\tau)d\tau$
    $\displaystyle \pi/2 \leq x < 1 + \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{\pi/2}f_1(x-\tau)f_2(\tau)d\tau$
    $\displaystyle x \geq 1+\pi/2: (f_1 * f_2)(x) = 0$
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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 $\displaystyle f_2$.
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  6. #6
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    I wrote in first topic, how $\displaystyle f_2$ looks like: for me it's known only:
    $\displaystyle f_2(x)$ is tg(x) when x is in [-pi/2 , pi/2] otherwise it is $\displaystyle f_2(x)$ is 0
    That's all i know about $\displaystyle f_2$
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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 $\displaystyle tg(x)$ 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:
    $\displaystyle \int f_1(t-y)f_2(y) dy$ 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.

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

    Again, I suggest you draw the two signals. The definition of f here means that it's constantly 1 for $\displaystyle [1, \infty)$. 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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    $\displaystyle \tan x$ is not convergent on $\displaystyle [-\pi/2, \pi/2]$, so $\displaystyle (f_1*f_2)(x)$ 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 $\displaystyle f_1$ is the function defined by $\displaystyle f_1(x) = \begin{cases}1&\text{if }0\leqslant x\leqslant 1\\0&\text{otherwise}\end{cases}$, and g is any other function, then $\displaystyle (f_1*g)(t)$ is the mean value of g on the interval [t–1,t].

    For the function $\displaystyle f_2(x) = \begin{cases}\tan x&\text{if }-\pi/2\leqslant x\leqslant \pi/2\\0&\text{otherwise}\end{cases}$, its integral over any interval containing $\displaystyle \pm\pi/2$ diverges. Thus $\displaystyle f_1*f_2$ 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 $\displaystyle f_1$ function
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