I'm studyng electrochemical impedance spectroscopy. On my book there is the sentence: "Impedance response of a linear system is independent of the perturbation amplitude.".

Why? If I have a non-linear system, when I zoom the voltage-current curve, this will be linear and, if I chose a bias voltage \$V_B\$ and if I superimpose a perturbation with small amplitude \$v=v^* sin(\omega t)\$ to \$V_B\$, for the frequency response theorem, I'll get:

$$i(t)=I_B+i^*sin(\omega t+\phi)$$

The impedance is:

$$Z=\frac{\mid V \mid}{\mid i \mid}e^{j(0-\phi)}=\frac{\mid V \mid}{\mid i \mid}e^{-j\phi}$$

Maybe the impedance is independent of the perturbation amplitude because, in the region in which the V-i curve is linear, the ratio \$\frac{\mid V \mid}{\mid i \mid}\$ is constant?

Thank you for your time.

  • \$\begingroup\$ Yes, exactly that. You must assume that or else all linear theory is busted and you have to solve a nonlinear system. \$\endgroup\$
    – Janka
    Jun 12, 2018 at 17:36
  • 1
    \$\begingroup\$ Your question goes in circles. The definition of impedance absolutely requires linearity, and the passage you quote is about a linear system. You then try to apply it to a non-linear one. You cannot do that, you can only use a linear model to approximate your system to the degree to which your system's behavior in a limited regime is sufficiently linear for the result to have meaning. \$\endgroup\$ Jun 12, 2018 at 17:37
  • \$\begingroup\$ Well that is pretty much the definition of a linear system. \$\endgroup\$
    – user16324
    Jun 12, 2018 at 21:54

1 Answer 1


The response of a linear system does not depend on the perturbation amplitude.

That's the definition of what it means to be a linear system.

If I have a non-linear system,

... then you can't expect a result specified to apply to a linear system to apply to your system.

In the real world, just about all systems are non-linear, but often it is useful to approximate their behavior with a linear model.

  • \$\begingroup\$ Hello @ThePhoton, thank you so much for your very clear reply. \$\endgroup\$ Jun 12, 2018 at 18:02

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