Hilbert transform problem .

1. May 15, 2008

mmzaj

i was trying to come with an easier way of finding the Hilbert transform of a function $$\ x(t)$$ , and here is what i did :

starting with Fourier transform of x :

$$X(f)=\int^{\infty}_{-\infty}\ x(t) \ e^ {\ -i2\pi ft} \ dt$$

now

$$\ e^ {\ -i2\pi ft} = \frac{1}{2\pi i}\oint\frac{e^ {\ -i2\pi fz}}{z-t} \ dz$$

where the contour encloses t .

"this is done by cauchy's integral formula"

=>

$$X(f) = \frac{1}{2\pi i}\int^{\infty}_{-\infty}\ x(t) \oint\frac{e^ {\ -i2\pi fz}}{z-t} \ dz \ dt = \frac{1}{2\pi i} \oint\ e^ {\ -i2\pi fz}\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt \ dz$$

now

$$\pi\hat{x}(z)=\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt$$

where

$$\hat{x}(z)$$ is the Hilbert transform of x .

=>

$$\ X(f)= \frac{1}{2i}\oint\hat{x}(s)\ e^ {\ -i2\pi fs}\ ds$$

"z is replaced with s for conventional reasons"

now the program is to relate the last integral to the inverse laplace (or fourier !!) transform of $$\hat{x}(s)$$ , by choosing the suitable contour(s) , and here is where i'm stuck !! so any help is appreciated .

Last edited: May 15, 2008