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Fourier transform of a function

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data

    a) Find the Fourier transform of the function f(x) defined as:

    f(x) = 1-3|x| , |x|<2 and 0 for |x|>2

    b) Find the values of the inverse Fourier transform of the function F(k) obtained in a)

    2. Relevant equations

    F(k) = [itex]\frac{1}{\sqrt{2π}}[/itex][itex]\int[/itex] f(t) eikx dx
    f(x) = [itex]\frac{1}{\sqrt{2π}}[/itex][itex]\int[/itex] F(k) e-ikx dk

    (Both integrals are from -inf to +inf)
    3. The attempt at a solution

    I solved the a) and got the result:

    F(k) = [itex]\frac{1}{\sqrt{2π}}[/itex](-[itex]\frac{10}{k}[/itex]sin(2k) - [itex]\frac{6}{k^2}[/itex]cos(2k) + [itex]\frac{6}{k^2}[/itex])

    The problem arrises at question b)
    If I insert the function F(k) into the formula
    f(x) = [itex]\frac{1}{\sqrt{2π}}[/itex][itex]\int[/itex] F(k) e-ikx dk
    I'll get an integral that I can't solve.. I tried inserting the definition of F(k) into the formula above, and I got:

    f(x) = [itex]\frac{1}{2π}[/itex][itex]\int[/itex] dk e-ikx [itex]\int[/itex] dx f(x) eikx

    I get stuck here tho... I suspect it has something to do with delta dirac's function, but it is not making much sense in my head right now...

    If someone could point me in the right direction I'd appreciate!

    Thanks.
     
  2. jcsd
  3. Dec 15, 2012 #2

    BruceW

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    yep, you've got part a) correct. And you've almost finished part b). You just need to slightly rearrange your final equation, and use the definition of the delta dirac, then you will show that the inverse Fourier transform of F is simply f, the equation you started with. Nice work, you're so close now.

    Edit: also, your final equation isn't quite right. (Or at least, it is bad notation). You should have two different x variables (the one that gets integrated over, and the one which doesn't). So you should use two different variables, like x and y for example.
     
  4. Dec 15, 2012 #3
    Ok, rewriting the equation which I had I get:

    f(x) = [itex]\frac{1}{2π}[/itex][itex]\int[/itex] dk e-iky [itex]\int[/itex] dx f(x) eikx =
    = [itex]\int[/itex]dx f(x) eikx [itex]\frac{1}{2π}[/itex][itex]\int[/itex] dk e-iky =
    = [itex]\int[/itex]dx f(x) [itex]\frac{1}{2π}[/itex][itex]\int[/itex]dk e-iky eikx =
    = [itex]\int[/itex]dx f(x) [itex]\frac{1}{2π}[/itex][itex]\int[/itex]dk eik(x-y)

    Since [itex]\frac{1}{2π}[/itex][itex]\int[/itex]dk eik(x-y) is equal to the function f(x) I can conclude that:
    [itex]\frac{1}{2π}[/itex][itex]\int[/itex]dk eik(x-y) = δ(x-y)

    And therefore:

    f(x) = [itex]\int[/itex]dx f(x) δ(x-y) = f(y)

    Is this correct or have I done some illegal step along the way?

    If this is correct I still have a little problem. I'm asked the values of the inverse Fourier transform at x=0 and x=2. At x=0 I just need to substitute the value of x, however at x=2 I have a descontinuity point, and the function is not defined, laying between the values of -5 and 0. Should I take the average between those two, as is done with fourier series descontinuity points?
     
  5. Dec 15, 2012 #4

    BruceW

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    Almost right. The first line should have had f(y) on the left-hand side, since that is the variable which is not being integrated over on the right-hand side. So that means in your final equation you get f(y)=f(y) :)
     
  6. Dec 15, 2012 #5

    BruceW

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    hmm. The function should have been defined for x=2. Are you sure it didn't have |x|=>2 in the definition ? Well, I think your answer of average of the two nearest points is right, but I am not sure. It looks like they intended for you to explicitly calculate the inverse Fourier transform of F(k). I don't think it should be too tricky. Have you used the sinc function before?
     
  7. Dec 15, 2012 #6
    Sadly I'm sure it is not defined at x=2 ...
    I know that if I integrate the delta dirac function I'll get something involving the sinc function, I don't know what to do with it tho.. I'll still have an integral that I won't be able to solve.
     
  8. Dec 15, 2012 #7

    BruceW

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    hint: wikipedia the sinc function. Its integral is not too difficult when the limits are +- infinity

    Edit: another hint because it is quite tricky. I think you need to make use of the fact that the sinc function times by a complex exponential, then integrated over infinity gives a form of the rectangular function. This is the explicit way to find the inverse Fourier transform of the Fourier transform. But the way you have done it already is just as correct. So which way you want to do it depends on which way you would guess the teacher wants you to do it.

    another edit: but there is the problem (as you said) of when x=2. So if you want to find this from the inverse Fourier transform, then I think you will need to do it explicitly.
     
    Last edited: Dec 15, 2012
  9. Dec 15, 2012 #8
    My professor said the solution was quite simple, so I think it involves delta dirac's function. My only remaining doubt is at x=2. Since the sinc function is delta dirac's function won't the integral of it be equal to 1? I'm not sure how that would help me.
     
  10. Dec 15, 2012 #9

    BruceW

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    The sinc function is not the same as the dirac delta. I am not sure what you mean. We have two different ways to calculate the inverse fourier transform of the fourier transform of the given function:
    1)simply calculate it for a general function (which is what you have done already), which involves using the definition of the dirac delta function.
    2)actually calculating it for the specific function given. And for the specific function, its fourier transform is a sum of sinc functions. So then you would also need to calculate the inverse fourier transform of the sum of sinc functions. (Which is not too bad, as long as you know that the transform of a sinc function gives a rectangular function).

    Now, by definition, method 1) will give you back the same function that you started with. But method 2) will give a function that is the same except at a few special points. (since the function is discontinuous).

    You already guessed that method 2) would give a value at x=2 which is an average of the points around it. I am not sure if this is true, but it sounds plausible. Maybe this is the simple solution which your professor mentioned. Otherwise, I can only think that you are supposed to use method 2), but if you haven't learned about the sinc and rectangular functions, then it seems unlikely that your professor would call this a simple solution.
     
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