1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse Fourier transform of -isgn \omega

  1. Aug 8, 2011 #1
    I've been using the Hilbert transform a bit as part of my research work (to analyse time series) and found http://personal.atl.bellsouth.net/p/h/physics/hilberttransforms.pdf document that explains some of the theory in a way that I can understand. I'm just having a problem showing that the inverse Fourier transform of

    \begin{equation}
    H(\omega) = \begin{cases}
    -i, \quad \omega > 0\\
    0, \quad \omega = 0\\
    i, \quad \omega < 0
    \end{cases}\end{equation}

    is

    \begin{equation}h(t) = \frac{1}{\pi t}.\end{equation}

    This is not homework. I'm just doing it for myself to see why the expression for the Hilbert transform looks the way it does. My attempt is as follows:

    \begin{equation}h(t) = \frac{1}{2\pi}\int^{\infty}_{-\infty} H(\omega)\, d\omega = \frac{i}{2\pi}\left[\int^0_{-\infty}e^{i\omega t}\, d\omega - \int^{\infty}_0 e^{i\omega t}\, d\omega\right] = \frac{i}{2\pi}\left\{\left[\frac{e^{i\omega t}}{it}\right]^0_{-\infty} - \left[\frac{e^{i\omega t}}{it}\right]^{\infty}_0\right\}.\end{equation}

    Plugging in the limits gives

    \begin{equation}h(t) = \frac{1}{2\pi t}\left[1 - e^{-i\infty t} - \left(e^{i\infty t} - 1\right)\right] = \frac{1}{\pi t} - \frac{1}{2\pi t}\left(e^{-i\infty t} + e^{i\infty t}\right).\end{equation}

    I don't know what to do about the second term :/. Maybe you can't do the integrals like this, because the exponential is complex. I've not done any complex analysis, so I don't know any other way to do it :/. I did try writing the exponential in terms of cos and sin, but that doesn't help either.

    Any ideas? Is there some identity or other trick that I'm missing? Thanks.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Aug 8, 2011 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You cannot proceed as you did, because you are dealing with integrals that do not converge in the standard sense, but have to be treated as "distributions" (like the Dirac delta function, for example). Anyway, instead of the inverse transform, let's look at the transform of f(t) = 1 for t > 0, and = -1 for t < 0. This is f(t) = H(t) - H(-t), where H(t) =1 for t > 0 and = 0 for t < 0. What is the transform of H? Well, H(t) = integral{g(x) dx, x = -infinity .. t}, where g(x) = Dirac(x). Can you get the transform of Dirac? Can you relate the transform of H to that of g (using some standard results about transforms of derivatives and integrals)? So, you can get the transform of H, etc.

    RGV
     
  4. Aug 15, 2011 #3
    Thanks for that. I've not really had the time to look at this again, but I'll probably get round to it at the weekend.

    Thanks again :).
     
  5. Aug 15, 2011 #4

    hunt_mat

    User Avatar
    Homework Helper

    Or failing that, you're going to have to use the calculus of residues.
     
  6. Aug 15, 2011 #5
    I don't know what that is. I think I can probably do it the way Ray suggested above, since I can revise that stuff.
     
  7. Aug 15, 2011 #6

    hunt_mat

    User Avatar
    Homework Helper

    Have you not taken a course on complex variable theory?
     
  8. Aug 15, 2011 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    One trick you can use to evaluate the integrals is to introduce a convergence factor e-λt. For example, to find the Fourier transform of u(t), you'd have
    \begin{align*}
    \int_{-\infty}^\infty u(t)e^{-i\omega t}\,dt &= \lim_{\lambda \to 0^+} \int_0^\infty e^{-i\omega t}e^{-\lambda t}\,dt \\
    &= \lim_{\lambda \to 0^+} \left[ \frac{ e^{-(\lambda+i\omega)t}}{-(\lambda+i\omega)} \right]_0^\infty \\
    &= \lim_{\lambda \to 0^+} \frac{1}{\lambda+i\omega}
    \end{align*}
    It would be tempting to substitute 0 in for λ at this point, but you have to be a bit more careful.
    \begin{align*}
    \lim_{\lambda \to 0^+} \frac{1}{\lambda+i\omega} &= \lim_{\lambda \to 0^+} \frac{\lambda-i\omega}{\lambda^2+\omega^2} \\
    &= \lim_{\lambda \to 0^+} \frac{\lambda}{\lambda^2+\omega^2} - \lim_{\lambda \to 0^+} \frac{i\omega}{\lambda^2+\omega^2} \\
    &= \pi \delta(\omega) - \frac{i}{\omega}
    \end{align*}
    where [tex]\delta(\omega)=\lim_{\varepsilon \to 0} \frac{1}{\pi}\frac{\varepsilon} {\varepsilon^2+\omega^2}[/tex]is the Dirac delta function.

    In your calculation, what happens is the complex exponentials will vanish thanks to the convergence factor, leaving just the first term, which is the result you were seeking.
     
  9. Aug 21, 2011 #8
    I'll have a look at this again post-viva, as I'm pretty busy with revision and panic. Thanks though! Also, no, I've never done any complex variable theory; there was no course on it during my undergrad and I've not needed it during my PhD.
     
  10. Aug 24, 2011 #9
    Thanks for the limit approach. I'm not too sure where the definition of the delta function there comes from (i.e. I've not seen that before).

    I think I'm making some small amount of progress looking at the way Ray suggested to do it.

    For the delta function, you have

    \begin{equation}\int^b_a f(t)\delta(t - c)\, dt = f(c)\end{equation}

    So, the Fourier transform is

    \begin{equation}\mathcal{F}[\delta(t - c)] = \int^{\infty}_{-\infty} \delta(t - c)e^{-i\omega t}\, dt
    = e^{-i\omega c}
    \end{equation}

    With [itex]c = 0[/itex], [itex]\mathcal{F}[\delta(t)] = 1[/itex].

    Now, [itex]\frac{d u(t)}{dt} = \delta(t)[/itex], where [itex]u(t)[/itex] is the Heaviside step function. Taking the Fourier transform of both sides:

    \begin{equation}\mathcal{F}\left[\frac{d u(t)}{dt}\right] = \mathcal{F}[\delta(t)]
    \Rightarrow -i\omega\mathcal{F}[u(t)] = 1\end{equation}

    This is wrong, because it gives the wrong answer for the Fourier transform of [itex]u(t)[/itex]. I'm stuck figuring out what I've done incorrectly. Maybe you can't use the relation between the delta and step functions, because of the former being a distribution and not a regular function?
     
  11. Aug 24, 2011 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The problem is that the function u(t)+c, where c is a constant, also has as its derivative the delta function. So what you should really have is[tex]-i\omega \mathcal{F}[u(t)+c] = 1[/tex]
     
  12. Aug 24, 2011 #11
    Ah yes, of course it is (since the derivative of a constant is 0). Linearity then means I have

    [tex]-i\omega\mathcal{F}[u(t)] -i\omega\mathcal{F}[c] = 1 \Rightarrow
    -i\omega\mathcal{F}[u(t)] = 1 + i\omega\mathcal{F}[c][/tex]

    So,

    [tex]\mathcal{F}[u(t)] = -\frac{1}{i\omega} -\mathcal{F}[c]
    \Rightarrow \mathcal{F}[u(t)] = \frac{i}{\omega} - \delta(\omega)
    [/tex]

    I hope that's almost correct. It looks similar to the result I see on MathWorld, though missing factors of [itex]2\pi[/itex] (although, obviously I'm using [itex]\omega = 2\pi f[/itex]) and my terms have opposite signs to theirs.

    I suppose the point of working out the Fourier transform of [itex]u(t)[/itex] is that I can rewrite my original integral as

    \begin{equation}\int^{\infty}_{-\infty} H(\omega)e^{i\omega t}\, d\omega = -i\int^{\infty}_{-\infty} u(\omega)e^{i\omega t}\, d\omega\end{equation}


    and then hopefully go on from there, since it's just the inverse transform of the step function. Slightly worried about the delta function, though. Sorry for being thick; I've not touched this stuff since my undergrad really!
     
  13. Aug 24, 2011 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

  14. Aug 29, 2011 #13
    Ah OK, yes. Is my last statement correct, about rewriting my original integral in terms of the step function? If I take the (inverse) Fourier transform of the step function [itex]u(\omega)[/itex] and I end up with two terms (i.e. the delta function and the [itex]\frac{1}{t}[/itex] term that I need, how do I get rid of the delta? *Confused*.
     
  15. Aug 29, 2011 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Go back and read what Ray wrote in post #2 about how to relate the unit step to the function H you're working with.
     
  16. Aug 29, 2011 #15
    Oh, I've got the definition of the step function wrong. Thanks! I almost have the right answer now and it just needs a bit of fiddling :smile:.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse Fourier transform of -isgn \omega
Loading...