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What is the impedance of the following circuit if the switch operates at the same frequency with the input voltage.

Since the switch introduces non-linearity to the circuit, how can I calculate the impedance?

My guess is when switch is off impedance is infinity while when the switch is on impedance is Z=2πfinL. However, I believe it is possible to express one impedance for all the period T.

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  • \$\begingroup\$ What, exactly, do you mean by 'operates at the same frequency' i.e. how are the on/off periods defined with respect to the input sinusoidal waveform? Why do you think that the system is non-linear? \$\endgroup\$ – Chu Sep 16 '19 at 23:29
  • \$\begingroup\$ Because the switch is a non-linear element, similarly with the diode. \$\endgroup\$ – Christts Sep 17 '19 at 14:21
  • \$\begingroup\$ The switch is part of the input signal, a diode is a system component. In this case the system is the inductor, and that's a linear component. \$\endgroup\$ – Chu Sep 17 '19 at 15:53
  • \$\begingroup\$ I really cannot believe the answer I just read, they show a complete misunderstanding of the basic mathematical concept of EE.... \$\endgroup\$ – MathieuL Sep 19 '19 at 18:26
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The "transfer" 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 "switched off" 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 "switched off", 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-) <> I(t0+) for any non-zero current.

So it is a mathematical contradiction eo ipso if the switch is "switched off" 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.

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    \$\begingroup\$ Indeed the step function will be 1 and 0 but when 0, that means that impedance is 0 as well? An open switch has infinite impedance. \$\endgroup\$ – Christts Sep 17 '19 at 14:18
  • \$\begingroup\$ The switch bridges/shortcuts a virtual infinite resistor R. The total impedance is R +sL, s is the complex frequency. R is bridged or non- bridged according to the step function. In other words, R is toggled between 0 and infinite. \$\endgroup\$ – xeeka Sep 17 '19 at 15:03
  • \$\begingroup\$ Answer clarified. \$\endgroup\$ – xeeka Sep 20 '19 at 5:57
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You are using the wrong tool for the wrong problem!!!

The impedance doesn't make any sense for lumped elements when they aren't in steady-state! It is a different story for transmission line because delay are introduced.

The problems you show is clearly a Transient problem. The correct tool to analyze this type of problem is the Laplace transform (which is the general case of Fourrier/impedance) or differential equation in the time domain.

Rules of thumbs to analyze a circuit:

Impedance : Steady-state only! Even for short-circuit current computation using the impedance will give 0 insights on the transient nature of the systemm, it will not protect you from a nasty resonance in your system if they are present.

Laplace transfer function: Very to analyze transient behaviour of linear system lumped devices.

As for your specific question when you introduce non-linear component like power electronics or saturation in transformer, the most efficient way to analyze these system is to use a numerical computation of the differential equation with a software like EMTP-RV.

In pratice, impedance & Laplace are nearly useless for details analysis of non-linear system because it is very difficult to convert back from the Laplace/Fourrier domain when the system isn't linear. The best usage of these tools for non-linear system is to develop an "average" model that doesn't capture all the same details but give you enough information at the fundamental frequency.

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