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What are the state variable equations of this RLC circuit? How do I write the final matrix? I tried to write the equations for the capacitor and inductor:

$$L\frac{di_3}{dt}=v_1-v_C \leftrightarrow \frac{di_3}{dt}=\frac{v_1}{L}-\frac{v_C}{L}$$

$$C\frac{dv_C}{dt}=i_3\leftrightarrow \frac{dv_C}{dt}=\frac{1}{C}i_3$$

I tried to apply the Kirchoff's law:

(1) $$i_1R_1+L\frac{di_3}{dt}+v_C=v(t)\leftrightarrow i_1=\frac{v(t)}{R_1}-\frac{L}{R_1}\frac{di_3}{dt}-\frac{v_C}{R_1}$$

(2)

$$v_1=v(t)-R_1i_1$$

From these equations I am unable to reach the final matrix. How could I go on?

enter image description here

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  • \$\begingroup\$ The purpose is to find the voltage across the capacitor? And it is an AC-circuit? You can use complex analysis. \$\endgroup\$ Oct 28, 2019 at 11:26
  • \$\begingroup\$ She did not specify steady state voltage applied. The voltage source may have a transient component \$\endgroup\$ Oct 28, 2019 at 11:49
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    \$\begingroup\$ Carmen, this should help you. \$\endgroup\$ Oct 28, 2019 at 11:55

1 Answer 1

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The inductor equation: $$L i_3'=v_1-v_C$$

The capacitor equation: $$c v_c'=i_3\ \ \ (1)$$

Kirchoff's current law: $$\frac{u-v_1}{R_1}=i_3+\frac{v_1}{R_2}$$

which can be solved for \$v_1\$ to get $$v_1=\frac{R_2 \left(u-i_3 R_1\right)}{R_1+R_2}$$

Substitute this in the inductor equation $$L i_3'=\frac{R_2 \left(u-i_3 R_1\right)}{R_1+R_2}-v_c \ (2)$$

The equations (2) and (1) can then be put in matrix form as

$$\left( \begin{array}{c} i_3' \\ v_c' \\ \end{array} \right)=\left( \begin{array}{cc} -\frac{R_1 R_2}{L \left(R_1+R_2\right)} & -\frac{1}{L} \\ \frac{1}{c} & 0 \\ \end{array} \right).\left( \begin{array}{c} i_3 \\ v_c \\ \end{array} \right)+ \left( \begin{array}{c} \frac{R_2}{L \left(R_1+R_2\right)} \\ 0 \\ \end{array} \right) u$$

Verifying using Mathematica. enter image description here

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