I'm trying to derive a state space model for this circuit: -


simulate this circuit – Schematic created using CircuitLab

The circuit contains four energy storing elements, but is only of third order because \$L_3\$ is not sitting in a peripheral branch. Therefore, I understand that I need 3 state variables. I define these as

$$x_1 = i_1 \\ x_2 = i_2 \\ x_3 = V_o \\ $$

However, when I try to find the associated differential equations for these state variables \$L_3\$ is causing me trouble. Writing mesh equations and using that \$i_3 = i_2 - C_1\dot{x_3}\$ yields

$$\begin{cases} -V_\text{in} + \dot{x_1}L_1+x_1R_1+\color{red}{L_3(\dot{x_1}-\dot{x_2})} = 0 \\ \\ \dot{x_2}L_2+x_2R_2+x_3+ \color{red}{L_3(\dot{x_2}-\dot{x_1})} = 0 \\ \\ -x_3+i_3R_3 = -x_3+R_3(x_2-C_1\dot{x_3})=0\end{cases} $$

The red terms in the equations are what is causing me trouble, because to find the associated differential equations it requires that only one of the state variables are differentiated. Both as can be seen, two of the state variables are differentiated.

How can I tackle this problem?


1 Answer 1


Rewriting your system of equations in matrix form yields $$ T\cdot \dot{x} = A\cdot x + B\cdot V_{in}$$ where \$x\$ is the state vector and \$T, A, B\$ are matrices of appropiate dimensions. Assuming that \$T\$ is invertable allows us to solve for \$\dot{x}\$ in terms of \$x\$ and \$V_{in}\$ by multiplying both sides by the inverse of \$T\$ resulting in $$ \dot{x} = T^{-1}\cdot A\cdot x + T^{-1}\cdot B\cdot V_{in} .$$ So in order to get the desired form of the state space equation requires you to solve your system of equations for the state derivatives (i.e. solving a system of linear equations for a linear circuit).

  • \$\begingroup\$ how would that equation work when you need to take the derivative of an input? This is something that's long confused me and I can't figure it out. Please see link for an example. \$\endgroup\$ Dec 19, 2023 at 20:37
  • \$\begingroup\$ In that case you need to incorporate the input signal dynamics into the model. Take for example the following system: dx/dt = ax + bu + cdu/dt, with state x and input u. Notice that u depends on du/dt and is fully determined by it (given a suitable initial condition). In that way we can treat du/dt as the 'true' input to our 'whole' system. The input u is then just another state of our 'whole' system. Let du/dt = v and xa = u then: dx/dt = ax + bxa + cv and dxa/dt = v holds. The system can then be expressed in the vector valued standard form dx/dt = Ax + Bv. \$\endgroup\$
    – Yggdrasil
    Dec 21, 2023 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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