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I found this paper on Hilbert Transform, which is a very nice read. I've studied signal processing, but from a more practical than mathematical perspective. Can someone explain to me how we arrive at equation (2) in this paper?

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up vote 2 down vote accepted

By symmetry if the first equality of (2) holds then the second equality also holds. Expand the integrand using the series expansions:

$$X(v)H(z/v)v^{-1} = \left(\sum_{n=-\infty}^\infty x(n)v^{-n}\right)\left(\sum_{m=-\infty}^\infty y(m)z^{-m}v^m \right)v^{-1} $$

Note that $\oint_\gamma v^{-k}dv=2\pi i$ if $k=1$ and $0$ otherwise - so all powers of $v$ are irrelevant except for when $m-n=0$ (there is already a $v^{-1}$ factor in the integrand). Hence the contour integral reduces as

$$ \frac{1}{2\pi i}\oint_\gamma \left(\sum_{n=-\infty}^\infty x(n)h(n) z^{-m}\right)v^{-1}dv= \frac{1}{2\pi i} \oint Y(z)v^{-1}dv = Y(z). $$

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Thanks, very cool – Phonon Jul 28 '11 at 18:08

We have $$ \oint X(v) H\left(\frac z v\right) v^{-1} \, dv. $$ Let $u = \dfrac z v$, so that $du = \dfrac{-z}{v^2}\,dv$. Then $v$ becomes $\dfrac z u$ and $v^{-1}\,dv$ becomes $\dfrac{-du}{u}$. But as $v$ goes around the unit circle in the counterclockwise direction, $u$ goes around in the clockwise direction. So $$ \oint_{\text{counterclockwise}} X(v) H\left(\frac z v\right) v^{-1} \, dv = \oint_{\text{clockwise}} X\left(\frac{z}{u}\right) H\left(u\right) (-u^{-1}) \, du $$ $$ = \oint_{\text{counterclockwise}} X\left(\frac{z}{u}\right) H\left(u\right) u^{-1} \, du. $$ Then rename the bound variable $u$ so that it's called $v$ again.

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Just a suggestion to make your integrals more readable next time... use a notation such as ccw for counterclockwise and cw for clockwise, it leaves less text in the integral. =) – Patrick Da Silva Jul 29 '11 at 20:42

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