So I have been busting my head for quite a while.(adsbygoogle = window.adsbygoogle || []).push({});

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 [itex] \pi /2 [/itex] if [itex] \omega <0 [/itex] and [itex] -\pi /2 [/itex] if [itex] \omega >0 [/itex].

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?

[itex] arg(G(j\omega ))=arg(F(j\omega ))+arg(H(j\omega )) [/itex]

Where [itex] arg(H(j\omega ))=-sgn(\omega)\cdot \pi /2[/itex]

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

Let input signal be cos(ω_{0}t).

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(ω_{0}t). Makes sense, that what the transform says.

Let then input signal sin(ω_{0}t).

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

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

[itex] \pi [/itex] if [itex] \omega <0 [/itex] and [itex] -\pi [/itex] if [itex] \omega >0 [/itex].

And that would represent a phase spectra of a -cos(ω_{0}t).

Is this correct?

Is phase spectra of -cos(ω_{0}t)

[itex] \pi [/itex] if [itex] \omega <0 [/itex] and [itex] -\pi [/itex] if [itex] \omega >0 [/itex].

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

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# Interpreting Hilbert's Filter through Phase spectra

Can you offer guidance or do you also need help?

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