# Interpreting Hilbert's Filter through Phase spectra

1. Jan 1, 2012

### Bassalisk

So I have been busting my head for quite a while.

I understood what hilbert's transform does to a function.

But I am having problems with getting a feeling for Phase spectra of functions.

Lets consider some basic sine functions.

I know that hilbert's transform adds a $\pi /2$ if $\omega <0$ and $-\pi /2$ if $\omega >0$.

Considering the fact that the Hilbert's filter is a LTI system, arguments of the input function(their FT) and the argument of the hilberts filter add each other up right?

$arg(G(j\omega ))=arg(F(j\omega ))+arg(H(j\omega ))$

Where $arg(H(j\omega ))=-sgn(\omega)\cdot \pi /2$

where G(jω) and F(jω) are respective Fourier transforms of output and input signals.

Let input signal be cos(ω0t).

Its phase spectra is 0 for all terms right?

So if I add that argument of Hilbert's filter, I get a sine, specifically sin(ω0t). Makes sense, that what the transform says.

Let then input signal sin(ω0t).

Its phase spectra is $\pi /2$ if $\omega <0$ and $-\pi /2$ if $\omega >0$. correct?

So if we add the arguments again(one from sine and one from Hilbert's filter) we get

$\pi$ if $\omega <0$ and $-\pi$ if $\omega >0$.

And that would represent a phase spectra of a -cos(ω0t).

Is this correct?

Is phase spectra of -cos(ω0t)
$\pi$ if $\omega <0$ and $-\pi$ if $\omega >0$.

I am bit numb with the math, so thats why I am asking this.