I have a problem with this circuit

enter image description here

I can't write the status equation because the system that characterizes the circuit is coupled respect to derivatives of the difference of rings currents.

Let $$i_1:=i_{R_1}=i_{L_1}$$ $$i_2:=i_{R_2}$$ so, applying counterclockwise the KVLs to the two rings we obtain

$$\begin{cases} L_1\frac{\text{d}}{\text{d}t}i_1-L_2\frac{\text{d}}{\text{d}t}[i_2-i_1]+R_1 i_1=0 \\ L_2\frac{\text{d}}{\text{d}t}[i_2-i_1]-e+R_2 i_2=0 \\ \end{cases}$$

or, equivalently

$$\begin{cases} \frac{\text{d}}{\text{d}t}i_1=-\frac{R_1}{L_1+L_2} i_1+\frac{L_2}{L_1+L_2}\frac{\text{d}}{\text{d}t}i_2\\ \frac{\text{d}}{\text{d}t}i_2=-\frac{R_2}{L_2}i_2+\frac{\text{d}}{\text{d}t}i_1+e \\ \end{cases}$$

but now I don't know how to proceed to determinate the vectorial status equation $$\dot{x}=Ax+Bu$$

  • \$\begingroup\$ Just use substitution (i.e. replace \$di_2/dt\$ from the first equation with the second equation). You will be able to represent the currents derivatives in terms of currents and \$e\$. \$\endgroup\$ May 2, 2016 at 12:23
  • \$\begingroup\$ @Vicente Cunha, that will still leave \$i_2\$ in the equation. \$\endgroup\$
    – Chu
    May 2, 2016 at 12:27
  • \$\begingroup\$ @Chu \$x\$ is a vertical vector with both \$i_1\$ and \$i_2\$; should I expand these comments to an answer? \$\endgroup\$ May 2, 2016 at 12:45
  • \$\begingroup\$ @Vicente Cunha, state space form would be a very useful answer \$\endgroup\$
    – Chu
    May 2, 2016 at 13:14

1 Answer 1


Starting from your equations:

$$\begin{cases} \frac{\text{d}}{\text{d}t}i_1=-\frac{R_1}{L_1+L_2} i_1+\frac{L_2}{L_1+L_2}\frac{\text{d}}{\text{d}t}i_2\\ \frac{\text{d}}{\text{d}t}i_2=-\frac{R_2}{L_2}i_2+\frac{\text{d}}{\text{d}t}i_1+e \\ \end{cases}$$

Substitution of second equation into first equation, and of first equation into second:

$$\begin{cases} \frac{\text{d}}{\text{d}t}i_1=-\frac{R_1}{L_1+L_2} i_1+\frac{L_2}{L_1+L_2}\left(-\frac{R_2}{L_2}i_2+\frac{\text{d}}{\text{d}t}i_1+e\right) \\ \frac{\text{d}}{\text{d}t}i_2=-\frac{R_2}{L_2}i_2+\left(-\frac{R_1}{L_1+L_2} i_1+\frac{L_2}{L_1+L_2}\frac{\text{d}}{\text{d}t}i_2\right)+e \\ \end{cases}$$

Rewriting equations:

$$\begin{cases} \frac{\text{d}}{\text{d}t}i_1\times\left(1-\frac{L_2}{L_1+L_2}\right)= -\frac{R_1}{L_1+L_2}i_1-\frac{R_2}{L_1+L_2}i_2+\frac{L_2}{L_1+L_2}e\\ \frac{\text{d}}{\text{d}t}i_2\times\left(1-\frac{L_2}{L_1+L_2}\right)= -\frac{R_1}{L_1+L_2}i_1 -\frac{R_2}{L_2}i_2 + e\\ \end{cases}$$

$$\begin{cases} \frac{\text{d}}{\text{d}t}i_1\times L_1= -R_1i_1-R_2i_2+L_2e\\ \frac{\text{d}}{\text{d}t}i_2\times L_1= -R_1i_1 -\frac{(L_1+L_2)R_2}{L_2}i_2 + (L_1+L_2)e\\ \end{cases}$$

Matrix representation:

$$ A = \frac{1}{L_1}\left( \begin{array}{cc} -R_1 & -R_2 \\ -R_1 & -\frac{(L_1+L_2)R_2}{L_2} \end{array} \right) $$

$$ B = \frac{1}{L_1}\left( \begin{array}{c} L_2 \\ L_1+L_2 \end{array} \right) $$


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