0
\$\begingroup\$

I couldn't find a proof of the superposition theorem from circuit analysis anywhere online. I thought it might be helpful to ask and provide my proposed proof as an answer to gather feedback and improvements. The superposition theorem I am referring to is the one that says if you only have "linear circuit elements", then the solution to the overall circuit, in the case of DC circuits at least, is the sum of solutions you obtain to variant circuits where only one source is left on (voltage sources become short circuits while current sources become open circuits).

\$\endgroup\$
3
  • \$\begingroup\$ It is a fundamental linear time invariant property. A +B = C= B+A. with scalar properties. dsp.stackexchange.com/questions/19765/… \$\endgroup\$ Jul 14, 2019 at 14:03
  • \$\begingroup\$ I believe it is different from the superposition theorem in circuit analysis where the theorem states that someone can solve a whole bunch of variant circuits by leaving on only one source and add all the solutions to get solution to the overall system. \$\endgroup\$ Jul 14, 2019 at 15:39
  • \$\begingroup\$ No. It is the same proof \$\endgroup\$ Jul 14, 2019 at 16:18

2 Answers 2

2
\$\begingroup\$

A linear system in electronics means a system that can be expressed as a linear differential equation. Hence

$$\dot{x} = \mathbf{A} x + \mathbf{B}u,$$

where \$x\$ is the state vector, \$u\$ is the input vector, and \$\mathbf{A}\$ and \$\mathbf{B}\$ are matrices. Let \$x_1\$ be the state from \$u_1\$ and let \$x_2\$ be the state from \$u_2\$. Thus

$$\frac{\mathrm{d}}{\mathrm{d}t} ( x_1 + x_2) = \dot{x}_1 + \dot{x}_2 = \mathbf{A} x_1 + \mathbf{B}u_1 + \mathbf{A} x_2 + \mathbf{B}u_2.$$

Therefore $$\frac{\mathrm{d}}{\mathrm{d}t} ( x_1 + x_2) = \mathbf{A} (x_1 + x_2) + \mathbf{B}(u_1 + u_2).$$

This means that applying both inputs together is equivalent to adding the states for each input applied individually. For superposition of more than two inputs apply induction.

\$\endgroup\$
7
  • \$\begingroup\$ Would the input vector contain the current and voltage sources in my above example? \$\endgroup\$ Jul 14, 2019 at 15:50
  • \$\begingroup\$ @ShuhengZheng Yes, the input vector holds the current and voltage sources, i.e. the inputs to the system. \$\endgroup\$
    – user110971
    Jul 14, 2019 at 15:51
  • \$\begingroup\$ In general, could x the "state" represent either voltage or current? I can see how Kirchhoff's two laws be casted into the above form although there would be no derivative term. \$\endgroup\$ Jul 14, 2019 at 15:54
  • \$\begingroup\$ @ShuhengZheng There will be derivative terms, if you have capacitors / inductors. Have a look at state-space representations for more information. \$\endgroup\$
    – user110971
    Jul 14, 2019 at 15:56
  • \$\begingroup\$ @ShuhengZheng the state holds enough information to uniquely determine the state of the system, e.g. the state vector for resistors in series is the current, since if you know the current, you can determine the voltages across each resistor. \$\endgroup\$
    – user110971
    Jul 14, 2019 at 16:00
-1
\$\begingroup\$

It is generally said that the superposition theorem is a result that the circuit is linear. I think that is not correct and actually a bit circular. Because one describes the linearity of the input/output relationship of the circuit while the other talks about how to solve the circuit by turning off sources. Here I prove that the superposition theorem in circuit analysis holds if you restrict to certain elements.

Motivating Example

schematic

simulate this circuit – Schematic created using CircuitLab

Let's motivate the general proof with a specific example first. Take a look at the attached circuit.

I'm going to use \$I_C\$ to denote the current of the current source and \$V_V\$ to denote the voltage of voltage source.

Small letter \$v_1, v_2, i_1, i_2\$ denote voltage and current across the resistor elements.

Systems of eqns are: $$ V_V = v_1 + v_2 \\ I_C+i_2=i_1 $$

We switch the current source with open circuit and get the equations $$ V_V = v_1'+v_2' \\ i_2'=i_1' $$ Switch the voltage source with closed circuit and get equations $$ 0=v_1''+v_2'' \\ I_S+i_2''=i_1'' $$ Implicitly, there are also the Ohm's law equations such as $$ v_i=R_ii_i $$

The key thing to notice is that solutions to the "turned-off" circuits, I will now call them the basis circuits, the solutions are all homogenous except for the equations that involve the one element that remains. That means when you add one solution from each basis circuit, most solutions do not affect the "in-homogenousness" of other solutions.

For example, if you add the primed variables to the double primed variables, you will get solution to the original circuit. The primed solutions are homogenous in the second equation, i.e. it won't "touch" \$i_i\$ while the double primed equations won't alter the voltage variables.

General Proof

In general, we can say translate all equations that govern our circuit into several forms, and see that this line of thinking works.

Kirchhoff Current Law

For each node in the circuit not directly attached to a voltage source, The Kirchhoff current law gives us equations of the form $$ i_1+i_2+...i_k=I_{S1}+I_{S2}+...+I_{Sn} $$ where the lowercase i on left hand side denotes unknown currents while right hand side denotes any current source connected to the node. If there are no current source, it would just be zero. The basis circuits obtained by turning off all sources except for one would either leave the right hand side with 0 or with one constant left. The basis circuits obtained by keeping S1, S2, .. etc. would each contribute to each of the term on the right hand side. When you add the solutions of all of them, you can see that the final solution satisfies the above equation. This holds for all nodes.

Kirchhoff Voltage Law

For closed loops that do not have current source, Kirchhoff Voltage law has form $$ v_1+v_2+...+v_k=V_{S1}+...+V_{Sn} $$ The same argument as above holds.

Ohm's law

Ohm's law is homogenous and relates v and i. So adding solutions to the basis circuits would also solve the Ohm's law equations for the full circuit.

This argument also holds for the other linear circuit elements such as inductors, capacitors, and could be extended if newer elements are discovered/created. As long as the equation governing the element is a linear function in voltage, current.

\$\endgroup\$
1
  • \$\begingroup\$ Kirchhoff. One k, one c, two h, two f. There is a built-in schematic editor you can use to add schematics to questions and answers...that is one of the worst schematics I have ever seen on this site. \$\endgroup\$ Jul 14, 2019 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.