Analyse the following system

A Capacitor(C) and Inductor(L) are connected in series across a battery which provides variable supply voltage ($e_r$) which varies with the current as given by the following equation. The Capacitor and Inductor are respectively rated at $\frac{1}{2}$ farad and 1 Henry. Analyse the system for its stability and response. Supply voltage ($e_r$):-

$$\begin{array}{ccccccc} e_r &=& -3\dot q && && \vert{\dot q}\... ...\dot q > 1 \\ &=& 3\dot q + 6 && && \dot q < -1 \end{array}$$ (1.6)

The system can be mathematically expressed as :-

$$L{\ddot q} + e_r + \frac{q}{C} = 0$$ (1.7)

For the state space representation, state space variables are:-

$$\begin{array}{ccccccc} x_1 &=& q \\ x_2 &=& \dot q \end{array}$$ (1.8)

For $\vert\dot q\vert \leq 1$ i.e. $\vert x_2\vert \leq 1$, we have,

$$\dot x_1 = x_2$$ (1.9)

$$\dot x_2 = -2x_1 + 3x_2$$ (1.10)

and the state space representation is :-

$$\begin{pmatrix} \dot x_1 \\ \dot x_2 \end{pmatrix} = \begin{... ...matrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ (1.11)

the equillibrium point is (0,0), the eigenvalues are $1 , 2$ and hence the equillibrium point is an unstable node. For $\dot q > 1$ i.e. $x_2 > 1$, we have,

$$\dot x_1 = x_2$$ (1.12)

$$\dot x_2 = -2x_1 - 3x_2 + 6$$ (1.13)

and the state space representation is :-

$$\begin{pmatrix} \dot x_1 \\ \dot x_2 \end{pmatrix} = \begin{... ...matrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix}$$ (1.14)

the equillibrium point is (3,0), the eigenvalues are $-1 , -2$ and hence the equillibrium point is a stable node. But this equillibrium point does not lie in the range over which the function is defined, hence the equillibrium point is a virtual equillibrium point and the trajectories will never converge to this stable node. For $\dot q < -1$ i.e. $x_2 < -1$, we have,

$$\dot x_1 = x_2$$ (1.15)

$$\dot x_2 = -2x_1 - 3x_2 - 6$$ (1.16)

and the state space representation is :-

$$\begin{pmatrix} \dot x_1 \\ \dot x_2 \end{pmatrix} = \begin{... ...atrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ -6 \end{pmatrix}$$ (1.17)

the equillibrium point is (-3,0), the eigenvalues are $-1 , -2$ and hence the equillibrium point is a stable node. But this equillibrium point does not lie in the range over which the function is defined, hence the equillibrium point is a virtual equillibrium point and the trajectories will never converge to this stable node. The solution trajectory (phase portrait) is given below : Trajectory2