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I've the following code:

BodePlot[{1/(((5.601443110900379`*^-8)*(5.601443110900379`*^-8)*(1.\
6672176431702952`*^-7)*(56000)*(56000 + 
          56000)*((56000*56000)/(56000 + 56000)))*(2*Pi*f*
         I)^3 + ((5.601443110900379`*^-8)*(5.601443110900379`*^-8)*(\
56000)*(56000 + 
           56000) + \
(5.601443110900379`*^-8)*(1.6672176431702952`*^-7)*(56000)*(56000 + 
           56000))*(2*Pi*f*
         I)^2 + ((5.601443110900379`*^-8)*(56000) + \
(5.601443110900379`*^-8)*(56000 + 56000 + 56000))*(2*Pi*f*I) + 
     1)}, {f, 0, 100}, 
 ScalingFunctions -> {{"Linear", "dB"}, {"Linear", "Degree"}}, 
 ImageSize -> Large]

This code gives two plots, the amplitude and the phase.

How can I find the function that is plotted as the phase? I mean can I aks Mathematica what kind of function (exp, arctan, arccos or arcsin etc.) it has plotted to plot the phase?

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  • $\begingroup$ The phase should be Arg. So try your formula with a simple Plot and wrap Arg around it. Since this will give values from -Pi to Pi, you need to process this. You could use toDegree[arg_] := Mod[arg, 2 Pi]/(2 Pi)*360 or you set ScalingFunctions appropriately. $\endgroup$ – halirutan Jul 2 '18 at 20:27

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