Electronic – DC Black Box and Thenevin Equivalence

Using the superposition theorem you can simply break your circuit down into 3 with one source active at a time - the overall response is the sum of the responses of each.

You might want to evolve your model into a state-space representation instead of the transfer function. State space has multiple advantages:

• Multiple inputs and multiple outputs just as easily done as 1 input/1 output
• Allows initial conditions different than 0
• Quicker to evaluate (takes only matrix arithmetic), really easy to propagate/simulate. You can write your own propagator in a few lines.
• Which means that you can change the coefficients/physical parameters (say, if some depend on time or if some are not linear) in your own propagator without much hassle.

The downside being that it's more complex to obtain the matrices in the first place.

If you wish to do that, then:

1. Calculate using Kirchoff laws, for one voltage/current source at a time, the differential equation :$$f(V_o, \frac{dV_o}{dt}, \frac{d²V_o}{dt²})=0$$As an example, $$V_o=R_3*(\frac{V_i-V_{C1}}{R_1}-C_1\frac{dV_{C1}}{dt}-C_2\frac{dV_o}{dt})$$and $$V_{C1}=R_2*(C_2*\frac{dV_o}{dt}+\frac{V_o}{R_3})+V_o$$ should give you the first differential equation when V1 is the only one ON.
2. Rearrange each differential equation in the following way: $$\ddot{V_o}+a_{1,i}\dot{V_o}+a_{2,i}{V_o}=b_{0,i}U_i$$
3. From there you can build the A, B, C and D matrices that define the state space representation of each differential equation. You have the choice between defining V1 and V2 as inputs to your system, or to define them as state variables which have zero differential over time and set them once and for all in your initial conditions. Here is what the matrices look like if all are inputs: $$A_i=\begin{bmatrix} 0 & 1\\ -a_{2,i} & -a_{1,i} \end{bmatrix}$$ $$B_i=\begin{bmatrix} 0 & 0 & 0\\ b_{0,i} & b_{0,i} & b_{0,i} \end{bmatrix}$$ $$C_i=[1, 0]$$ $$D_i=0$$ For a state vector $$X_i=\begin{bmatrix} V_{o,i}\\ \dot{V_{o,i}} \end{bmatrix}$$ In $$\dot{X_i}=A_iX_i+B_iU_i$$ $$V_{o,i}=C_iX_i+D_iU_i$$ Where U_i is the input vector - a scalar for each of those 3 state space models (in my example, either V1 Ii or V2.
4. Finally, you can either solve those 3 models for each timestep and sum the responses together according to the superposition theorem $$V_o=\Sigma_i V_{o,i}$$ Or concatenate the 3 state space models in a single one and solve this one instead: $$\dot{X}=\begin{bmatrix}\dot{V}_{01} \\ \ddot{V}_{01} \\ \dot{V}_{02} \\ \ddot{V}_{02} \\ \dot{V}_{03} \\ \ddot{V}_{03} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ -\gamma_1/\alpha_1 & -\beta_1/\alpha_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -\gamma_2/\alpha_2 & -\beta_2/\alpha_2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -\gamma_3/\alpha_3 & -\beta_3/\alpha_3 \end{bmatrix} \begin{bmatrix}{V}_{01} \\ \dot{V}_{01} \\ {V}_{02} \\ \dot{V}_{02} \\ {V}_{03} \\ \dot{V}_{03} \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ \delta_1/\alpha_1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \delta_2/\alpha_2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \delta_3/\alpha_3 \end{bmatrix} \begin{bmatrix}I_i \\ V_1 \\ V_2 \end{bmatrix}$$ $$V_o=\begin{bmatrix} 1 && 0 && 1 && 0 && 1 && 0 \end{bmatrix} X+\begin{bmatrix} 0 && 0 && 0 \end{bmatrix}\begin{bmatrix}I_i \\ V_1 \\ V_2 \end{bmatrix}$$

This might help you. I must admit I had the superposition theorem wrong the first time, obareey from the Maths SE helped me put that together.

The circuit you analyse with your formula is essentially this one, with the switch open: two independent braches, each with two resistors in series.

^{simulate this circuit – Schematic created using CircuitLab}

The current in the two braches is independennt, each current is V / ( R + R ).

Note that the potential (voltage) at both sides of the switch is the same, so we can close the switch, without effect on the circuit. Now we have your circuit, except that R1 is represented by TWO parallel resistors, each R, so the equivalent is R/2.

To summarize, you analyzed your circuit as if it were the one I show, which is different from your circuit in the value of R1.

The comment from The Photon gives another view, which amounts to the same thing.