Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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

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

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.