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Interpreting Hilbert's Filter through Phase spectra

  1. Jan 1, 2012 #1
    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 [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(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 [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(ω0t).

    Is this correct?

    Is phase spectra of -cos(ω0t)
    [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.
     
  2. jcsd
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