I have a complex impedance which I want to back transform into the time domain. My attempt in MM is:

InverseFourierTransform[1/((I \[Omega] Cs)/(1 - Cs Ls \[Omega]^2) + 1/Rp + I \[Omega] Cp + 1/(I \[Omega] Lp)), \[Omega], t, Assumptions->Cp > 0 && Lp > 0 && Cs > 0 && Ls > 0 && Rp > 0]

However this spits out some nasty expressions expressions with Root[...] arguments, but given how nasty the current output is, or, I'm not sure how to deal with the output.

The objective is I want to weight the 'time-transient impedance' and then then Fourier transform it back into the frequency domain to see what my weighting does.

  • 1
    $\begingroup$ You can use ToRadicals to get more familiar notation, but the result is far uglier. Symbolic expressions for temporal behavior of coupled resonators are generally extremely complicated. That's just mathematics: there is no cure. $\endgroup$ – John Doty Aug 6 '19 at 14:42


There something wrong below (...~p~... should be replace simply by ...+... ?). I will see that later



I'm entering difficulties : once my error corrected, I have the root objects. (Nothing glorious in fact.)


Apparently you are calculating the transient response of this :


Here is how I would do it :

x_ ~p~ y_ := x y /(x + y) (* definition of a "parallel operator on impedances *)  

ti00 = (1/(Lp s)) ~p~ (Cp s) ~p~ (Rp)~p~ (1/(1/(Ls s) + (Cs s))) // 
   Expand // Together (* this is the circuit above *)  

tf = TransferFunctionModel[{{ti00}}, s]  

OutputResponse[tf, UnitStep[t], t] (* litteral output response *)  

 (* example of numeric output response to a unit step (not a dirac) : *)
outR00 = OutputResponse[
   tf /. {Lp -> 1, Cp -> 1, Rp -> 1, Ls -> 1, Cs -> 1}, 
   UnitStep[t], {t, 0, 10}]  

Plot[outR00 , {t, 0, 10}] 

enter image description here

Note : The stimulus is a voltage step. The response is a the current.

| improve this answer | |
  • $\begingroup$ The schematic was created with a tool used on ElectricalEngineering Stack Exchange. (Very convenient) $\endgroup$ – andre314 Aug 6 '19 at 13:56
  • $\begingroup$ This is absolutely glorious. $\endgroup$ – Q.P. Aug 6 '19 at 15:13
  • $\begingroup$ One final question, once I weight my data, I actually want to apply a Hanning window, how would one transform back into the frequency domain? $\endgroup$ – Q.P. Aug 6 '19 at 15:35
  • $\begingroup$ I don't know exactly and I don't want to run the risk to mislead the readers. $\endgroup$ – andre314 Aug 6 '19 at 15:44
  • $\begingroup$ Okay, maybe I will ask on the EESE -- Thanks!! $\endgroup$ – Q.P. Aug 6 '19 at 17:41

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