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I want to find the state space equations for the following circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The equations I find are:

  1. Ir = I1 + I2 = (U-Vn)/2
  2. I2 = I3
  3. I1 = 2*X1'
  4. I2 = 1*X2'
  5. I3 = 3*X3'
  6. -X1 = X2+X3
  7. Y = X3

Plugging in (3) and (4) to (1) I get: (U+X1)/2 = 2*X1'+ X2'

I need to decouple the derivatives, but I don't see a relation that can do that.

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  • \$\begingroup\$ What are you considering your state? What is it dimension? \$\endgroup\$ – jDAQ Nov 14 '19 at 0:32
  • \$\begingroup\$ I consider the voltages across my capacitors my states. Respectively they're X1,X2,X3. \$\endgroup\$ – Kevin Silken Nov 14 '19 at 0:42
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The current going through \$C_2\$ and \$C_3\$ is the same, so, by, $$ Q = Cv \leftrightarrow i=C\dot{v}$$ For that branch,

$$\frac{\dot{v_3} - \dot{v_2} }{C_2} = \frac{-\dot{v_3}}{C_3}.$$

Also, the current that goes through the resistor either goes through \$C_1\$ or the branch with \$C_2, C_3\$. By associating them in series and then parallel, we find that,

$$\frac{v_1 - v_2 }{R} = \frac{ \dot{v_2}}{C_1+(C_2^{-1}+C_3^{-1})^{-1}}.$$

I did not go through the whole process of figuring out the state equations, but these should show that \$\dot{v_3} = k \dot{v_2}\$, those derivatives are related by a constant.

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  • \$\begingroup\$ Thanks, I forgot that in the current equation for C2 you can write it as i = C2*(V2') = C2*(-V1-V3)'= -C2*(V1'+V3'). \$\endgroup\$ – Kevin Silken Nov 14 '19 at 6:33

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