RLC Series Circuit: State Equations

[Dustinsfl]

Using the capacitor voltage \(v_c(t)\) and the inductor current \(i_L(t)\) as states, write the state equations for the RLC series circuit shown in the figure.

We can write that \(e(t) = iR + \frac{di}{dt} + \frac{1}{C}\int i(t)dt\). I am not sure with what it wants when it says to write it as states.

 

[Ackbach]

Shouldn't it be $e(t) = iR + L \frac{di}{dt} + \frac{1}{C}\int i(t)dt?$ I think when they say "state", they mean that the DE should be in that unknown variable. So if $i_{L}$ is the state variable, write the DE with $i_{L}$ as the unknown function.

 

[chisigma]

Using the capacitor voltage \(v_c(t)\) and the inductor current \(i_L(t)\) as states, write the state equations for the RLC series circuit shown in the figure.

We can write that \(e(t) = iR + \frac{di}{dt} + \frac{1}{C}\int i(t)dt\). I am not sure with what it wants when it says to write it as states.

In electric egeneering are called states of a network somewhat is related to the 'energy' contained in the reactive elements, i.e. the voltage across a capacitors and the current in inductors. In Your case the network has two states, i.e.the two unknown variables are $v_{c} (t)$ and $i_{L} (t)$. The equation governing the network are...

$\displaystyle v_{c} (t) + R\ i_{L}(t) + i_{L}^{\ '} (t) = e(t)$

$\displaystyle v_{c}^{\ '} (t) = C\ i_{L} (t)\ (1)$

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

I have determined my state transition matrix as
\[
\Phi(s) =
\frac{1}{(s + 1)(s + 2)}
\begin{pmatrix}
s & -1\\
2 & s + 3
\end{pmatrix}\Rightarrow
\varphi(t) = e^{-t}
\begin{pmatrix}
2e^{-t} - 1 & e^{-t} - 1\\
2(1 - e^{-t}) & 2 - e^{-t}
\end{pmatrix}.
\]
Then
\[
\begin{pmatrix}
i_L(s)\\
v_c(s)
\end{pmatrix} = \Phi(s)\mathbf{B}\mathbf{U}(s)
\]
What is \(\mathbf{B}\) and \(\mathbf{U}(s)\).
Additionally, I have that
\[
\begin{pmatrix}
\frac{di_L(t)}{dt}\\
\frac{dv_c(t)}{dt}
\end{pmatrix} =
\begin{pmatrix}
\frac{-R}{L} & \frac{-1}{L}\\
\frac{1}{C} & 0
\end{pmatrix}
\begin{pmatrix}
i_L(t)\\
v_c(t)
\end{pmatrix}
+
\begin{pmatrix}
\frac{1}{L}\\
0
\end{pmatrix}e(t) \qquad (*)
\]
where \(R = 3\), \(L = 1\), and \(C = \frac{1}{2}\).

The solution has
\[
\Phi(s)\mathbf{B}\mathbf{U}(s) =
\begin{pmatrix}
\frac{s}{(s+1)(s+3)}\\
\frac{2}{(s+1)(s+3)}
\end{pmatrix}
\frac{2}{s}
\begin{pmatrix}
e^{-s} & -e^{-2s}
\end{pmatrix}
\]
Therefore, \(\mathbf{U}(s) = \frac{2}{s}
\begin{pmatrix}
e^{-s} & -e^{-2s}
\end{pmatrix}\) where does this come from?

If we take the Laplace Transform of \((*)\), we end up with the factor \(\frac{2}{s}\). That is, we have
\begin{gather}
I(s) = \frac{E(s) - V(s)}{s + 3}\\
V(s) = \frac{2}{s}I(s)
\end{gather}
but that still leaves the question where does \(U(s)\) come from?