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Dirac delta and fourier transform

  1. Mar 1, 2012 #1
    In my book the dirac delta is described by the equation on the attached picture. This realtion is derived from the fourier transform, but I'm not sure that I understand what it says. If u=t it is clear that one gets f(u) in the fourier inversion theorem. But why wouldn't u=t? In the derivation of the fourier transform from the discrete fourier series t was just changed to u in the expression of the coefficients to avoid confusion.
    Can anyone try to picture what this expression fundamentally says? I should suspect that it is like the analogue of the ortogonality relation of the discrete fourier series, but I can't quite understand it.
    And what would the situation u≠t represent?
     

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  3. Mar 1, 2012 #2

    jbunniii

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    What it is saying is that

    [tex]\delta(t - u)[/tex]
    and
    [tex]\exp(-iu\omega)[/tex]

    are a Fourier transform pair. A complex exponential with "frequency" equal to [itex]u[/itex] has a Fourier transform with all of its energy concentrated at [itex]u[/itex].
     
    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3

    jbunniii

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    P.S. The equation

    [tex]\delta(t - u) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \exp(i \omega(t-u) d\omega[/tex]

    is mathematically nonrigorous. The integral on the right hand side does not actually exist for any values of [itex]u[/itex] and [itex]t[/itex]. What is true is that if [itex]\mathcal{F}[/itex] denotes the Fourier transform operator, then

    [tex]\mathcal{F}(\delta(t-u)) = \exp(-i\omega u)[/tex]
    and
    [tex]\delta(t-u) = \mathcal{F}^{-1}(\exp(-i\omega u))[/tex]

    in the sense of distributions. See here for more details:

    http://en.wikipedia.org/wiki/Distribution_(mathematics)
     
    Last edited: Mar 1, 2012
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