# status equations of 2° order circuit

I have a problem with this circuit

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 $$\dot{x}=Ax+Bu$$

• 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 $e$. May 2, 2016 at 12:23
• @Vicente Cunha, that will still leave $i_2$ in the equation.
– Chu
May 2, 2016 at 12:27
• @Chu $x$ is a vertical vector with both $i_1$ and $i_2$; should I expand these comments to an answer? May 2, 2016 at 12:45
• @Vicente Cunha, state space form would be a very useful answer
– Chu
May 2, 2016 at 13:14

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

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:

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

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