# How to calculate the phase spectrum

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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• Please add to your question the code you have developed so far, displayed in Mathematica format. – bbgodfrey Mar 14 '15 at 15:05
• 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. – bill s Mar 14 '15 at 15:23
• 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. – Nasser Mar 14 '15 at 17:36
• @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. – LLlAMnYP Mar 14 '15 at 17:44

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, 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 