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Suppose I have a linear circuit with 3 voltage sources Vin1, Vin2, and Vin3 in it, and one voltage of interest Vout. It's common practice to solve it using the superposition theorem, so this will lead to a set of 3 equations: $$V_{out,1}=f_1(V_{in1})$$ $$V_{out,2}=f_2(V_{in2})$$ $$V_{out,3}=f_3(V_{in3})$$ Then $$V_{out}=V_{out,1}+V_{out,2}+V_{out,3}$$ However, it seems each of the equations individually must have their own initial conditions, and naturally (in my case anyway) only the initial conditions relating to the total circuit are known - not their breakdown into each "layer".

What I mean is that I know $$V_{out}(t=0)=V_0 \neq 0$$ and $$\frac{dV_{out}}{dt}(t=0)=0$$ But neither $$V_{out,1}(t=0)$$ $$V_{out,2}(t=0)$$ $$V_{out,3}(t=0)$$ nor $$\frac{dV_{out,1}}{dt}(t=0)$$ $$\frac{dV_{out,2}}{dt}(t=0)$$ $$\frac{dV_{out,3}}{dt}(t=0)$$ (example of 2nd order system)

Is there a relationship between what I do not know (the initial conditions of the subcircuits) and what I do know (the initial conditions of the total circuit)?

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  • \$\begingroup\$ Sorry. But I didn't get the question. Are you asking that whether the initial condition for each 'layer' will be [Vo/3,DVDTo/3]? \$\endgroup\$ – nidhin Feb 23 '15 at 15:27
  • \$\begingroup\$ My bad, reformulated the question \$\endgroup\$ – Mister Mystère Feb 23 '15 at 15:47
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Since the system has initial conditions, the equations are differential equations. With your example, there are three 2nd order differential equations.

Since the equations are linear, the voltage sources (inputs) will always appear as the inhomogeneous terms (aka, input functions, forcing functions). When solved, each equation will produce a particular solution and a homogeneous solution with constant(s). With your example, 3 pairs of particular solutions and homogeneous solutions each with 2 integrating constants.

So it appears that there are two many constants to be determined by the initial conditions, but the homogeneous solutions will be linearly dependent, so the number of constants collapse back to just one set. In your example, the 6 constants collapse back to 2. Knowing this, you could just take the 3 particular solutions and any one of the homogeneous solutions, add them together and that is the general solution for the system. Then fix the 2 constants with the initial conditions.

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