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RLC Circuit

Hi everyone. Please i am trying to derive the state space representation of the circuit above. Please i need some help or a textbook for better understanding of this.

RLC Circuit

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  • \$\begingroup\$ Your image is not clear enough to work from, please post a clearer image. \$\endgroup\$ – Reroute Apr 18 '20 at 6:10
  • \$\begingroup\$ A clearer one has been posted \$\endgroup\$ – Boniface Chigozie Apr 18 '20 at 6:21
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Using nodal analysis, along with the obvious relationships between states in the parallel circuits, and also assuming ALL component values are unity (since I just want to show the method, rather than do a complete symbolic solution), I get the following set of equations:

$$\small x_1-x_4=x_3 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)$$ $$\small \dot x_2=x_3-x_2 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(2)$$ $$\small \dot x_1+\dot x_3=V_s-2x_1-x_2-x_3 \:\:\:\:\:(3)$$ $$\small \dot x_4-\dot x_3=x_2+x_3-2x_4 \:\:\:\:\:\:\:\:\:\:\:\:(4)$$

From (1): $$\small \dot x_1=\dot x_3+\dot x_4 $$ Substitute into (3): $$\small 2\dot x_3+\dot x_4=V_s-2x_1-x_2-x_3\:\:\:\:\:\:\:\:\:$$

Re-arranging (4):

$$\small -\dot x_3+\dot x_4=x_2+x_3-2x_4 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$

Solving simultaneously:

\$\hspace{50mm} \small \dot x_3=\left( V_s-2x_1-2x_2-2x_3+2x_4\right)/3\$

\$\hspace{50mm}\small \dot x_4=\left( V_s-2x_1+x_2+x_3-4x_4\right)/3\$

Also:

\$\hspace{50mm}\small \dot x_1=\left(2V_s -4x_1-x_2-x_3-x_4\right)/3\$

and equation (2) is:

\$\hspace{50mm}\small \dot x_2=x_3-x_2\$

These last four expressions give the state-space equation:

\begin{gather} \small \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \\ \dot x_4\end{bmatrix} = \frac{1}{3} \begin{bmatrix} -4 &-1 &-1 &-1 \\ 0&-3&3&0 \\-2&-2&-2&2 \\-2&1&1&-4 \end{bmatrix} \begin{bmatrix} \ x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} + \begin{bmatrix} \ 2/3 \\ 0 \\ 1/3 \\ 1/3\end{bmatrix}V_s \end{gather}

with eigenvalues:

\$\hspace{50mm}\small \lambda_1=-5.42\$

\$\hspace{50mm}\small \lambda_2=-3.61+j2.3\$

\$\hspace{50mm}\small \lambda_3=-3.61-j2.3\$

\$\hspace{50mm}\small \lambda_4=-0.36\$

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  • \$\begingroup\$ Thank you Chu for taking time to respond. Please can you explain to me how linear eqn in (1) becomes differential. \$\endgroup\$ – Boniface Chigozie Apr 19 '20 at 17:31
  • \$\begingroup\$ Just differentiate the entire equation. \$\endgroup\$ – Chu Apr 19 '20 at 20:02
  • \$\begingroup\$ I thought of that too. Thank you Chu \$\endgroup\$ – Boniface Chigozie Apr 21 '20 at 6:23

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