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I Fourier transform of Dirac delta

  1. May 1, 2016 #1
    In lectures, I have learnt that [tex]F(k)= \int_{-\infty}^{\infty} e^{-ikx}f(x)dx[/tex] where F(k) is the fourier transform of f(x) and the inverse fourier transform is [tex]f(x)= \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx}f(k)dk[/tex] .
    But on the same chapter in the lecture notes, there is an example solving for the fourier transform of [itex]f(x)=(\delta(x+d))+(\delta(x-d))[/itex].
    I set the transform as [tex]F(k)= \int_{-\infty}^{\infty} e^{-ikx}(\delta(x+d))+(\delta(x-d))dx[/tex]Splitting the integral into two and using the sifting property, I got [tex]F(k)=e^{ikd}+e^{-ikd}[/tex] But the solution has [itex]\frac{1}{\sqrt{2\pi}}[/itex] in front, hence from there they used the trig identity to get it in terms of cosine.
    I searched on google and there seems to be different conventions on the transform. But I am only familiar with the one I mentioned. How can I get to the solution using the integral form I used?
     
  2. jcsd
  3. May 1, 2016 #2
    Hmm, it could be that he used a different convention just for this problem. Is my solution still a valid answer?
     
  4. May 1, 2016 #3
    Actually from what I see the answer is [tex] 2\cos(kd) [/tex], but depending on the convention used you can have [tex] \frac{1}{\sqrt{2\pi}}[/tex] in front of both the forward and backward transforms.
     
  5. May 1, 2016 #4
    Yes, the 1/√2π convention that Septim mentions is used by the more mathematically oriented scientists, since in that case both the Fourier transform and its inverse are isometries on the space L2 of square-integrable functions, which is often a very convenient thing to have.*

    The other most common convention, with no coefficient (okay, it's actually 1) in the forward transform tends to be preferred by engineers, who then don't have to worry about any coefficient every time they do a (forward) transform, and never have to worry about a square root even with the inverse transform.

    ___________________
    * Technically, any two square-integrable functions are considered to be the same in L2 if they differ only on a set of measure 0. Thus L2 is not precisely a set of functions, but rather a set of equivalence classes.
     
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