# How do I write the equations for three inductors connected together in state space form?

The system is basically three inductors connected to each other in one end and the other end connected to three independent voltage sources.

I want to get the equations in state space form. For this I need to eliminate the intermediary variable VT.

I cant figure out how to proceed from here.

• So what is your question? Jun 18 '18 at 9:34

You need to use Kirchhoff's current law on the node with voltage $$\V_T\$$: $$i_1 = i_2 + i_3$$ Taking the time derivative at both sides results in $$\frac{\mathrm{d}i_1}{\mathrm{d}t} = \frac{\mathrm{d}i_2}{\mathrm{d}t} + \frac{\mathrm{d}i_3}{\mathrm{d}t}$$ You now have a set of 4 equations in 4 unknowns ($$\\frac{\mathrm{d}i_1}{\mathrm{d}t}\$$, $$\\frac{\mathrm{d}i_2}{\mathrm{d}t}\$$, $$\\frac{\mathrm{d}i_3}{\mathrm{d}t}\$$, and $$\V_T\$$): $$\begin{cases} L_1 \frac{\mathrm{d}i_1}{\mathrm{d}t} & = V_T - V_1 \\ L_2 \frac{\mathrm{d}i_2}{\mathrm{d}t} & = V_2 - V_T \\ L_3 \frac{\mathrm{d}i_3}{\mathrm{d}t} & = V_3 - V_T \\ \frac{\mathrm{d}i_1}{\mathrm{d}t} & = \frac{\mathrm{d}i_2}{\mathrm{d}t} + \frac{\mathrm{d}i_3}{\mathrm{d}t} \end{cases}$$
After eliminating $$\\frac{\mathrm{d}i_1}{\mathrm{d}t}\$$ and $$\V_T\$$, you get $$\begin{bmatrix} \frac{\mathrm{d}i_2}{\mathrm{d}t} \\ \frac{\mathrm{d}i_3}{\mathrm{d}t} \end{bmatrix} = \frac{1}{L_1L_2+L_1L_3+L_2L_3} \begin{bmatrix} -L_3 & L_1+L_3 & -L_1 \\ -L_2 & -L_1 & L_1+L_2 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$$ which is a state equation of the form $$\\frac{\mathrm{d}x}{\mathrm{d}t}=Ax+Bu\$$. The state variables are $$\x_1=i_2\$$ and $$\x_2=i_3\$$. The inputs are $$\u_1=V_1\$$, $$\u_2=V_2\$$, and $$\u_3=V_3\$$. The matrix $$\A\$$ is a zero matrix and the matrix $$\B\$$ is $$B = \frac{1}{L_1L_2+L_1L_3+L_2L_3} \begin{bmatrix} -L_3 & L_1+L_3 & -L_1 \\ -L_2 & -L_1 & L_1+L_2 \end{bmatrix}$$