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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!

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    \$\begingroup\$ Think about what asymptotic stability means for the position of the poles of a system. \$\endgroup\$ – jramsay42 Feb 21 '18 at 22:28
  • \$\begingroup\$ I know for BIBO stability, we check whether the poles of the transfer function are in the OLHP. Does that apply for asymptotic stability as well? Or do we do it in terms of Y(s), which I don't know in this case. \$\endgroup\$ – ce1 Feb 21 '18 at 22:40
  • \$\begingroup\$ RLC circuits have 2 well defined equations for Q and ωo. What else do you need to know? Step or impulse response? \$\endgroup\$ – Tony Stewart EE75 Feb 22 '18 at 4:23
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    \$\begingroup\$ For linear circuits, there is no need to differentiate between the different forms of stability (Lyapunov / asymptotic / exponential / BIBO) as long as there are no poles with zero real part. The relationships between asymptotic and BIBO stability for linear systems are well explained at math.stackexchange.com/a/1525609/645472. \$\endgroup\$ – Koen Tiels Feb 24 '19 at 8:12
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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:

enter image description here

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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.

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