How is the impedance of a circuit calculated when non-linearity is introduced from a switch?
The &quot;transfer&quot; function of the switch consists of a sequence of shortcuts of a virtual infinite resistor R.
The total impedance is R +sL, s is the complex frequency.
R is shortcut or non- shortcut according to the positive and negative step functions. In other words, R is toggled between 0 and infinite.
But here is an important fact:
The shown theoretical circuit can only exist, if and only if the switch is only switched off if the current I is zero.
If the switch is &quot;switched off&quot; when the current I is non-zero, it results in a contradictional circuit. The contradiction is as follows: In this current loop one element makes the current steady, but the other element makes the current non-steady, both by definition. If any inductivity with a non-zero current is &quot;switched off&quot;, the current MUST continue to fulfill the steady condition, i.e. I(t0-) = I(t0+), t0 being the time when the switch is operated. But a switch's function is defined as I(t0-) &lt;&gt; I(t0+) for any non-zero current.
So it is a mathematical contradiction eo ipso if the switch is &quot;switched off&quot; for a non-zero current.
Of course, in reality, there are free-running diodes (which must be fast with low threshold), (stray) capacities, non-perfect switches (e.g. with defined sparking) etc. which are eliminating this contradiction.
EDIT:
It is important to not intermix steadyness and linearity.
This network of a series circuit of a switch and an inductivity is completely linear.
A switch doesn't make a network non- linear. It only makes the current and/or voltage non- steady.