State Equations

Please can you help me in finding the state equations for the circuit below? My problem is the two capacitors. How will we cater for the second capacitor? I only know problems with 1 capacitor and 1 inductor.

Please can you help me and find my attached solution which I attempted half way through?

This is my attempt for the solution. Is it correct? (mathjax transcript below)

editor's note: missing parts of the second equation have been added in $\color{red}{red}$. They reappear in subsequent equations. \begin{align} C_1 \frac{dv_{c1}}{dt} &= i_1 - i_L = \frac{v_s-v_{c1}}{R_1} - i_L & \text{(using KCL at node 1)}\\ \frac{dv_{c1}}{dt} &= \color{red}{\frac{1}{C_1}} \left( \frac{v_s}{R_1} - \frac{v_{c1}}{R\color{red}{_1}} - i_L\right) & \text{(divide by }C_1\text{)} \\ \frac{dv_{c1}}{dt} &= \frac{1}{R_1C_1}v_s - \frac{1}{R_1C_1}v_{c1} - \frac{1}{C_1}i_L\\ \frac{dv_{c1}}{dt} &= - \frac{1}{R_1C_1}v_{c1}+\frac{1}{R_1C_1}v_s - \frac{1}{C_1}i_L\\ \\ &\text{using KVL on inductor}\\ \\ C_2 \frac{dv_{c2}}{dt} &= i_L - g_mv_{c1} & \text{(using KCL node 2)}\\ \frac{dv_{c2}}{dt} &= \frac{-g_m}{C_2}v_{c1} + \frac{1}{C_2}i_L & \text{(dividing by }C_2\text{)}\\ \\ \\ L \frac{di_L}{dt} &= v_{c1} - v_{c2} & \text{(dividing by }L\text{)}\\ \frac{di_L}{dt} &= \frac{1}{L}v_{c1} - \frac{1}{L}v_{c2}\\ \end{align}

$$\left[ \begin{matrix} \frac{dv_{c1}}{dt}\\ \frac{dv_{c2}}{dt}\\ \frac{di_L}{dt}\\ \end{matrix} \right] = \left[ \begin{matrix} -\frac{1}{R_1C_1} & 0 & - \frac{1}{C_1}\\ -\frac{g_m}{C_2} & 0 & \frac{1}{C_2}\\ \frac{1}{L} & -\frac{1}{L} & 0\\ \end{matrix} \right] \left[ \begin{matrix} v_{c1}\\ v_{c2}\\ i_L\\ \end{matrix} \right] + \left[ \begin{matrix} \frac{1}{R_1C_1}\\ 0\\ 0\\ \end{matrix} \right] v_s$$

• Welcome on EE.SE. Please, post what have you tried. We don't give complete solutions for homeworks, unless one shows us real effort. Visit the help center to learn about this site netiquette. – Lorenzo Donati Jul 16 '15 at 13:48
• Transient or permanent? – MathieuL Jul 16 '15 at 16:39
• Are you allowed to use impedance to resolve the problem? – MathieuL Jul 17 '15 at 14:31
• I got much the same using charge instead of capacitor voltage. I think it's fine. – Chu Jul 19 '15 at 14:11