# State variable equations for a RLC circuit

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?

• The purpose is to find the voltage across the capacitor? And it is an AC-circuit? You can use complex analysis. – Jan Oct 28 '19 at 11:26
• She did not specify steady state voltage applied. The voltage source may have a transient component – relayman357 Oct 28 '19 at 11:49
• Carmen, this should help you. – relayman357 Oct 28 '19 at 11:55

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.

Well, the complex voltage across the capacitor is given by:

$$\underline{\text{V}}_\text{C}=\underline{\text{Z}}_\text{C}\cdot\underline{\text{I}}_\text{C}\tag1$$

Now, we know that:

$$\underline{\text{Z}}_\text{C}=\frac{1}{\text{j}\omega\text{C}}\tag2$$

And we can write:

$$\underline{\text{I}}_\text{C}=\frac{\text{R}_2}{\text{R}_2+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\cdot\underline{\text{I}}_\text{in}\tag3$$

Now, we know that:

$$\underline{\text{I}}_\text{in}=\frac{\underline{\text{V}}_\text{in}}{\underline{\text{Z}}_\text{in}}\tag4$$

Now, we can write:

• $$\underline{\text{V}}_\text{in}=\hat{\text{V}}_\text{in}\exp\left(\varphi\text{j}\right)\tag5$$
• $$\underline{\text{Z}}_\text{in}=\text{R}_1+\text{R}_2\text{||}\left(\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}\right)=\text{R}_1+\frac{\text{R}_2\left(\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}\right)}{\text{R}_2+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\tag6$$

So:

$$\underline{\text{V}}_\text{C}=\frac{1}{\text{j}\omega\text{C}}\cdot\frac{\text{R}_2}{\text{R}_2+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}\cdot\frac{\hat{\text{V}}_\text{in}\exp\left(\varphi\text{j}\right)}{\text{R}_1+\frac{\text{R}_2\left(\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}\right)}{\text{R}_2+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}}}=$$ $$\frac{\text{R}_2}{\text{j}\omega\text{C}}\cdot\frac{\hat{\text{V}}_\text{in}\exp\left(\varphi\text{j}\right)}{\text{R}_1\text{R}_2+\left(\text{R}_1+\text{R}_2\right)\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\text{j}}\tag7$$

Which will give an amplitude of:

$$\left|\underline{\text{V}}_\text{C}\right|=\frac{\text{R}_2}{\omega\text{C}}\cdot\frac{\hat{\text{V}}_\text{in}}{\sqrt{\left(\text{R}_1\text{R}_2\right)^2+\left(\left(\text{R}_1+\text{R}_2\right)\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\right)^2}}\tag8$$

• I think OP is asking for the way to the answer using state variables, not for the answer itself found by a different approach. – Huisman Oct 28 '19 at 12:43