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Stuck on inverse fourier transform pair

  1. Jul 17, 2011 #1
    I have been trying to solve the inverse fourier transform:

    [itex]\int_{-\infty}^{\infty}\left[e^{-j2\pi ft_0}e^{j\theta}\right]e^{j2\pi ft}df[/itex]

    I know that the fourier transform pair says

    [itex]e^{-j2\pi ft_0}e^{j\theta} \leftrightarrow \delta(t-t_0)[/itex]

    but the extra phase term [itex]e^{j\theta}[/itex] makes it so I can't use this pair. Can I just consider it a constant? If so, then it gives me a weird time based function of an imaginary number.
  2. jcsd
  3. Jul 17, 2011 #2
    Yes, [itex]e^{j\theta}[/itex] is independent of f (unless there's something you're not telling us), so you can just pull it out of the integral. I don't understand why you say it gives you "a weird time based function of an imaginary number." It gives you [itex]e^{j\theta}\delta\left(t-t_0\right)[/itex], which is just the delta function multiplied by a phase factor (assuming theta is real), with maybe a factor of [itex]\frac{1}{\sqrt{2\pi}}[/itex] or something like that.
  4. Jul 17, 2011 #3
    Yes, but this phase factor has an imaginary component for any angle that is not a whole multiple of 0 or pi. I know what this phase means for my application, but I don't understand how a time function can have an imaginary component. In Euler's formula, there are imaginary components, but they always cancel out to a real value, and in this case they don't.
  5. Jul 17, 2011 #4
    Why shouldn't a function of time have an imaginary part? It all depends on the application. Sometimes you're calculating something that, because of the physics it represents, must be a real number, and in those cases the imaginary parts must indeed cancel out. But you say that you know what the phase means for your application, which (correct me if I misunderstand) means that for your purposes, complex numbers do have a physical meaning. This is not uncommon, actually.
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