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I have the following frequency characteristics in the Fourier domain:

$$H(\omega)=\frac{-\omega^2}{63170s^{-2}-\omega^2+355.1s^{-1}i\omega}$$

How do I find the phase spectrum from this? I should plot both the negative and positive frequencies.

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  • $\begingroup$ Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. $\endgroup$ – bbgodfrey Mar 14 '15 at 15:04
  • $\begingroup$ Please add to your question the code you have developed so far, displayed in Mathematica format. $\endgroup$ – bbgodfrey Mar 14 '15 at 15:05
  • $\begingroup$ Have you looked at the command FourierTransform? Or maybe (because you have the symbol s), you want the LaplaceTransform? Plot is useful for showing your results. $\endgroup$ – bill s Mar 14 '15 at 15:23
  • $\begingroup$ Could you clarify what the terms are. What is s there? s is normally used for Laplace transform variable. But you say the above is Fourier transform. $\endgroup$ – Nasser Mar 14 '15 at 17:36
  • $\begingroup$ @Nasser - I would (and already have) put my money on s standing for seconds. The function given is not really a Fourier transform of anything physically meaningful in the time domain, unless of course the force driving the oscillator had the form of a Dirac delta function w.r.t. time, then it's the Fourier transform of the acceleration of the oscillator vs. time. $\endgroup$ – LLlAMnYP Mar 14 '15 at 17:44
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Given the standard definitions of amplitude and phase spectra, I believe that s^-1 should be interpreted as units of inverse seconds. With this supposition,

h = -ω^2/(63170 + 355.1 I ω - ω^2)

and the amplitude and phase spectra are Abs[h] and Arg[h], respectively, with ω measured in inverse seconds. The curves have the typical shapes,

enter image description here

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H[\omega] looks a lot like a function for a driven damped oscillator. I'll use x instead of \omega for brevity. Its not the response function, though, this looks more like the second derivative w.r.t time. So I would define another function, which would be proportional to the complex amplitude response function: a[x_]:=-H[x]/x^2. As it's the complex amplitude, it already contains information about both phase and amplitude, there's really no need to look further for the phase spectrum. The magnitude of it Abs[a[x]] will give you the amplitude response, while Arg[a[x]] gives the phase response. Here's some code:

h[x_]:=-x^2/(63170-x^2+355.1 I x)
a[x_]:=-h[x]*63170/x^2
(* The multiplier 63170 serves to normalize the function
to unity at zero frequency, it hasn't got any other special significance *)
Plot[{Abs[a[x]],Arg[a[x]]},{x,-1000,1000}]

It returns

spectra

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