0
\$\begingroup\$

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

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

enter image description here

I cant figure out how to proceed from here.

\$\endgroup\$
1
  • \$\begingroup\$ So what is your question? \$\endgroup\$
    – Andy aka
    Jun 18 '18 at 9:34
0
\$\begingroup\$

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 $$ B = \frac{1}{L_1L_2+L_1L_3+L_2L_3} \begin{bmatrix} -L_3 & L_1+L_3 & -L_1 \\ -L_2 & -L_1 & L_1+L_2 \end{bmatrix} $$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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