- #1
Nylex
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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.
\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.
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