Houjun Liu

NUS-MATH570 Circuits

We declare known battery voltage \(E(t)\).

Here are the \(y\) values.

\begin{equation} \begin{cases} \dv{x_1}{t} = y_{4}\\ \dv{x_2}{t} = y_{3}\\ \dv{x_3}{t} = y_{1}\\ \dv{x_4}{t} = y_{2}\\ \end{cases} \end{equation}

And here are some of the \(x\) values.

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{y_1}{t}=-\frac{1}{LC}x_2-\frac{E(t)}{C} \\ \dv{y_4}{t} = -\frac{R}{L}y_2-\frac{2E(t)}{L} \end{cases} \end{equation}

Right off the bat, we can see that we can make one substitution. That, given:

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{x_4}{t} = y_{2} \end{cases} \end{equation}

we have that:

\begin{equation} y_2 = -\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \end{equation}

This renders the last expression:

\begin{align} \dv{y_4}{t} &= -\frac{R}{L}y_2-\frac{2E(t)}{L} \\ &= -\frac{R}{L}\qty(-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R})-\frac{2E(t)}{L} \\ &= \qty(\frac{2}{LC}x_2+\frac{1}{LC}x_{3}+\frac{2E(t)}{L})-\frac{2E(t)}{L} \\ &= \frac{2}{LC}x_2+\frac{1}{LC}x_{3} \end{align}

So now, we have the final unused expressions:

\begin{equation} \begin{cases} \dv{x_1}{t} = y_4 \\ \dv{x_2}{t} = y_3 \\ \dv{x_{3}}{t} = y_1 \\ \dv{y_1}{t} = -\frac{1}{LC}x_2-\frac{E(t)}{C} \\ \dv{y_4}{t} = \frac{2}{LC}x_2+\frac{1}{LC}x_3 \end{cases} \end{equation}