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enter image description here

I am confused with the above problem. Should I take a new state variable for rightmost capacitor or 3 variables are sufficient?

Please help me in writing the state space equations.

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  • \$\begingroup\$ You should take the output state. There is a general rule for this relating to capacitive loop and inductive cutset in the link below. books.google.co.kr/… \$\endgroup\$ – anhnha Mar 13 '18 at 17:53
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You will need 4 variables and there are 4 equations

$$V_{\text{C1}}-V_{\text{C2}}=L_4 i_1'$$ $$i_1=\frac{V_i-V_{\text{C1}}}{R_2}-C_3 V_{\text{C1}}'$$ $$i_1=C_4 \left(V_0'-V_{\text{C2}}'\right)+\frac{V_{\text{C2}}}{R_5}$$ $$\frac{V_{\text{C2}}-2 V_0}{R_6}=C_4 V_0'$$

And there are 4 unknowns \$\{i_1',V_{\text{C1}}',V_{\text{C2}}',V_0'\}\$ from which you can solve for the \$A\$ and \$B\$ matrices.

$$ A=\left( \begin{array}{cccc} 0 & \frac{1}{L_4} & -\frac{1}{L_4} & 0 \\ -\frac{1}{C_3} & -\frac{1}{C_3 R_2} & 0 & 0 \\ -\frac{1}{C_4} & 0 & -\frac{-R_5-R_6}{C_4 R_5 R_6} & -\frac{2}{C_4 R_6} \\ 0 & 0 & \frac{1}{C_4 R_6} & -\frac{2}{C_4 R_6} \\ \end{array} \right),B=\left( \begin{array}{c} 0 \\ \frac{1}{C_3 R_2} \\ 0 \\ 0 \\ \end{array} \right) $$

I am also attaching the computations in Mathematica enter image description here

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  • \$\begingroup\$ Thanks for such detailed explanation. My instructor combined the capacitor in series and then left the saying that you guys can do from here. \$\endgroup\$ – Nikhil Kashyap Mar 14 '18 at 14:03
  • \$\begingroup\$ I think your equations are wrong. Please correct them. \$\endgroup\$ – Nikhil Kashyap Mar 14 '18 at 16:21
  • \$\begingroup\$ All 4? Could you be more specific. \$\endgroup\$ – Suba Thomas Mar 14 '18 at 16:59

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