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Fourier transform/Convolution theorem

  1. Nov 23, 2011 #1
    2v3lzlx.png

    Ok, so first we need to find h(u).

    By letting

    h(u) = Integral -1 to 1 of (1/2)*g(u-x) dx

    Then we can change the limits about by setting u = 2x

    so now we have:


    h(u) = Integral -2 to 2 of (1/4) du

    so h(u) = 1

    and I find the fourier transform of this between -2 and 2 and I don't get sin^2(k) / k^2

    Can anyone help me here?

    Thanks
     
  2. jcsd
  3. Nov 23, 2011 #2
    That's confussing to read. k is being used as the transform variable which is unusual. I'll use "z" below. How about I re-word it:

    If I have the function f(x), then the Fourier transform of f(x), which will be a function of the Fourier integral variable "z" say and not k is:

    [tex]F\left\{f(x)\right\}=\int_{-\infty}^{\infty} f(x)e^{izx}dx=\hat{f}(z)[/tex]

    and if I have another function h(x) defined by:

    [tex]h(x)=\int_{-\infty}^{\infty} g(x-y)f(y)dy[/tex]

    then by the convolution theorem, the Fourier transform of h(x) which will be a function of z will be:

    [tex]F\left\{h\right\}=\hat{g} \hat{f}[/tex]

    So for your g,

    [tex]F\left\{g\right\}=\int_{-\infty}^{\infty}g(x)e^{izx}dx=1/2\int_{-1}^{1}e^{izx}dx[/tex]

    Ok, now you have the transform of g which is a function of z but if you have to, call it a function of k then. Now use the convolution theorem to find the transform of g*g.
     
  4. Nov 23, 2011 #3
    Hi Jackmell, thanks for the reply

    I believe you're doing the 2nd part? Verifying the solution via the convolution theorem.

    What I need first is to show the result of the transform of h(u), thus I need to find what h(u) is.
     
  5. Nov 23, 2011 #4
    I don't think that's gonna' happen dude. Don't see how anyway. Maybe I'm wrong but I think the way I said is the way to go. Maybe others can confirm the matter for us though.

    Edit:

    Ok, I might be wrong about that. How about we consider the double integral:

    [tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x)g(u-x)e^{iwu}dudx=\int_{-\infty}^{\infty}g(x)\int_{-\infty}^{\infty} g(u-x)e^{iwu}dudx=\int_{-\infty}^{\infty}g(x)F\left\{g(u-x)\right\}dx[/tex]

    Then maybe use some shifting theorem to finish it off.
     
    Last edited: Nov 23, 2011
  6. Nov 23, 2011 #5

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    Hi Firepanda! :smile:

    Let's pick u=10.
    Since we have x between -1 and +1 in your integral (before your substitution).
    What would f(u-x)=f(10-x) be?

    In other words, I think something went wrong with your substitution.
    h(u)≠1.
     
  7. Nov 23, 2011 #6
    hmm, do you mean what the range of f(10-x) would be?
     
  8. Nov 23, 2011 #7

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    Yes...
     
  9. Nov 23, 2011 #8
    Ok, sorry I'm not too confident with all of this

    would it be 9<x<11

    so the range is still 2, just shifted. So am I right in thinking my subsitution should be.. something like u = x+1 or u = x-1?
     
  10. Nov 23, 2011 #9

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    Sorry. I said f(u-x), but I meant g(u-x), since that is the one you need for h(u).

    What is the value of g(10-x) if 10-x is between 9 and 11?
     
  11. Nov 23, 2011 #10
    0 I assume by the definition of g
     
  12. Nov 23, 2011 #11

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    Yes. So what will h(10) be?

    As for your substitution, I recommend y=u-x.
     
  13. Nov 23, 2011 #12
    h(10) = 0.5

    so my integral is

    h(u) = Integral -(y+1) to (y-1) of (1/4) du ?
     
  14. Nov 23, 2011 #13

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    Let's revisit your formula for h(u).
    [tex]h(u)=\int_{-\infty}^\infty g(u-x)g(x) dx[/tex]
    You already reduced this to:
    [tex]h(u)=\int_{-1}^1 g(u-x) {1 \over 2} dx[/tex]

    For u is 10, you said that g(u-x) would be 0 (for all x).
    I'm afraid the integral won't be 0.5 then...

    Hmm, in the integral for h(u) there is no integration with respect to u.
    It couldn't, because then you'd never get a function of u.

    After substitution the integration should be with respect to y.
     
  15. Nov 23, 2011 #14
    Ah shoot, so my reduction is wrong too?

    So I need the integral to be 0.5 when u = 10, perhaps I multiply it all by 0.25, or change my limits?
     
  16. Nov 23, 2011 #15

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    You have:
    [tex]h(u)=\int_{-1}^1 g(u-x) {1 \over 2} dx[/tex]

    For u is 10, you said that g(u-x) would be 0 (for all x).
    This means:
    [tex]h(10)=\int_{-1}^1 0 \cdot {1 \over 2} dx = 0[/tex]
     
  17. Nov 23, 2011 #16
    and that integral is equal to 2, which is why I said we may need to multiply it by 0.25, or change the limits?

    Sorry it's late, and I'm trying my best! :)

    I just need to get the integral before I go to bed then I'm sure I can do it!
     
  18. Nov 23, 2011 #17

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    Huh? :confused:
    Which integral is equal to 2?
    What would you need to multiply by 0.25 and why?

    When you make a substitution, say y=u-x, then yes, you need to change the limits.
    But you need to make a proper substitution wrt y.


    I'll be off to bed soon too.
     
  19. Nov 23, 2011 #18
    NO no no noooo

    wait please, we can get this!

    I was just trying to make it easier to get to 0.5, the integral of h(u) in your earlier post.... oh wait sorry, I though the integral of 0 was x, I realise now its a constant!

    Forget what I was saying before.

    y = u - x and so i am doing the integral of

    h(u) = int from u+1 to u-1 0.5*g(y) dy

    correct?
     
  20. Nov 23, 2011 #19

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    The anti-derivative of 0 is a constant, but the integral between any 2 boundaries is 0.

    Yes, you have the right integral for h(u).
    Now you need to consider on which part of your range g(y) is zero, and on which part it is (1/2).
     
  21. Nov 23, 2011 #20
    g(y) is 0.5 between u-1< 0 < u+1 no?
     
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