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How do I choose the least number of state variables for the following circuit? I started out by assuming currents $$i_{1}, i_{2}, i_{3}$$ in each of the loops and applied KVL, which fetched me three differential equations: $$ i_{1} + 2\frac{di_{1}}{dt} - 2\frac{di_{2}}{dt} = v_{i}(t) $$ $$ 2\frac{di_{1}}{dt} - 3\frac{di_{2}}{dt} + \frac{di_{3}}{dt} = 3i_{2} $$ $$ \frac{di_{2}}{dt} - \frac{di_{3}}{dt} - i_{3} = v_{o}(t) $$

But when it comes to choosing the state variables, while keeping the number to minimum, I am unable to proceed. Can anyone help me?enter image description here

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  • \$\begingroup\$ Show your work. \$\endgroup\$ – Eugene Sh. Aug 20 '18 at 14:34
  • \$\begingroup\$ Added the differential equations. \$\endgroup\$ – MaxFrost Aug 20 '18 at 15:35
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The way this is generally done is to get the equations into a matrix form (or solve them.

\$\begin{bmatrix} \dot{x} \end{bmatrix}= \begin{bmatrix} Ax \end{bmatrix}+ \begin{bmatrix}Bf\end{bmatrix}\$

or this notation since the state would be \$\dot{x}=\frac{di}{dt}\$

\$\begin{bmatrix} \frac{di}{dt} \end{bmatrix}= \begin{bmatrix} Ax \end{bmatrix}+ \begin{bmatrix}Bf\end{bmatrix}\$

The matrix coefficients are the same for fly speck or differential notation.

$$ 2\frac{di_{1}}{dt} - 2\frac{di_{2}}{dt} + 0\frac{di_{3}}{dt} = -i_{1} + 0i_{2} +0i_{3} +v_{i}(t) $$ $$ 2\frac{di_{1}}{dt} - 3\frac{di_{2}}{dt} + \frac{di_{3}}{dt} = 0i_{1}+ 3i_{2}+0i_{3} $$ $$ 0\frac{di_{1}}{dt} + \frac{di_{2}}{dt} - \frac{di_{3}}{dt} = 0i_{1}+0i_{2} + i_{3} + v_{o}(t) $$

You can represent a matrix form from the above equations:

\$\begin{bmatrix} 2 & -2 & 0 \\ 2&-3&1 \\ 0&1&-1 \end{bmatrix}= \begin{bmatrix} -1& 0& 0 \\0&3&0 \\0 &0&1 \end{bmatrix}+ \begin{bmatrix} v_{i}(t) \\ 0\\v_{o}(t) \end{bmatrix}\$

You need to reduce the system of equations until you get this (all the differential equations are reduced, this form makes it easy to work with or use it for simulation)

\$\begin{bmatrix} 1 & 0 & 0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}= \begin{bmatrix} Blah \end{bmatrix}+ \begin{bmatrix}vin blah\end{bmatrix}\$

if you substitued the variables back in it would look like this:

\$\begin{bmatrix} \frac{di_1}{dt} & 0 & 0 \\ 0&\frac{di_2}{dt}&0 \\ 0&0&\frac{di_3}{dt} \end{bmatrix}= \begin{bmatrix} Blah \end{bmatrix}+ \begin{bmatrix}vin blah\end{bmatrix}\$

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A good rule of thumb I picked up from some guy on YouTube, is to look at every element that stores energy in your system. In this case you have two inductors and one capacitor. Energy in inductors is calculated with the current flowing through them. \begin{equation} E = (1/2)i_L^2, \end{equation} Where as the energy in a capacitor is calculated with the voltage developed on the cap. \begin{equation} E = (1/2)v_c^2, \end{equation}

Following this the state equations would be \begin{equation} i_1,i_2,v_0 \end{equation} where the currents are for the first two loops from the left of the schematic. You have the added benefit that the voltage out at the capacitor is also a state.

Hope this Helps

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