Finding asymptotic stability of RLC circuits

Question

Given the state model for an RLC circuit, how can I show that it is asymptotically stable? So my state assignment is as follows:

x1 = Vc

x2 = iL

And using the KVL and KCL equations I can get a state model for the first derivatives and can get the corresponding matrices and transfer function.

How can I find the asymptotic stability now?

Any help is appreciated!

Answer 1

Well, according to 'Faraday's law' in a series RLC-circuit:

$$\text{V}_{\space\text{C}}\left(t\right)+0+\text{V}_{\space\text{R}}\left(t\right)-\text{V}_{\space\text{in}}\left(t\right)=-\text{V}_{\space\text{L}}\left(t\right)\tag1$$

Now, we know a few things:

• $$\text{I}_{\space\text{C}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{C}}'\left(t\right)\cdot\text{C}\tag2$$
• $$\text{V}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)\cdot\text{R}=\text{I}_{\space\text{in}}\left(t\right)\cdot\text{R}\tag3$$
• $$\text{V}_{\space\text{L}}\left(t\right)=\text{I}_{\space\text{L}}'\left(t\right)\cdot\text{L}=\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{L}\tag4$$

So, we get:

$$\text{V}_{\space\text{C}}'\left(t\right)+0+\text{V}_{\space\text{R}}'\left(t\right)-\text{V}_{\space\text{in}}'\left(t\right)=-\text{V}_{\space\text{L}}'\left(t\right)\space\Longleftrightarrow\space$$ $$\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}+0+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}-\text{V}_{\space\text{in}}'\left(t\right)=-\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$ $$\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$ $$\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\tag5$$

Assuming that the initial conditions are equal to \$0\$, and using Laplace transform:

$$\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)=\text{s}^2\cdot\text{i}_{\space\text{in}}\left(\text{s}\right)\cdot\text{L}+\text{s}\cdot\text{i}_{\space\text{in}}\left(\text{s}\right)\cdot\text{R}+\text{i}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space$$ $$\text{i}_{\space\text{in}}\left(\text{s}\right)=\frac{\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag6$$

Using the 'Convolution Theorem' of the Laplace transform:

$$\text{I}_{\space\text{in}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{v}_{\space\text{in}}\left(\text{s}\right)\right]_{\left(\text{y}\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\space\text{d}\text{y}=$$ $$\int_0^t\text{V}_{\space\text{in}}\left(\text{y}\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\space\text{d}\text{y}\tag7$$

And \$\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\$, equals:

Answer 2

Besides @Jan's equations, maybe there's a way that's a bit more intuitive, or not. In simple terms, the three ways a system can respond to an impulse response is with an exponentially decaying output (oscillatory, or not), an stable oscillation, or a divergent output (oscillatory, or not). Since we're talking about a passive system, there is no divergence, because there's no way you can have a positive real part of a pole in there. So there can be only a decaying output, or an oscillation. In order for the latter to happen, you must have ideal L and C, and no R. Since this is no case here, it can only be that the system will decay, sometime, to a stable solution.

Or, in mathematical terms, an ideal LC (lowpass) has the transfer function:

$$H(s)=\frac{1}{LCs^2+1}\equiv\frac{\omega^2}{s^2+\omega^2}$$

i.e. no damping. As soon as an R comes into play (series with L):

$$H(s)=\frac{1}{LCs^2+RCs+1}\equiv\frac{\omega^2}{s^2+2\zeta\omega s+\omega^2}$$

you get the \$\zeta\$, which represents the damping. This is valid for any configuration, R represents the damping. And it makes sense, if you consider that a(n ideal) resistor is only dissipating power.