2.5a Internal Stability

Internal (Asymptotic) Stability

A system is internally stable if, in the absence of any input (e.g. with $x(n)=0$  for all $n$), every state trajectory $q(t)$ converges to the zero vector (i.e. $q(t)\to \vec{0}$) as $t\to \infty$, regardless of the value of its initial state $q(0)$.

Additionally, any discrete-time LTI system is internally stable if, for every eigenvalue ${\lambda }_i\in \{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N\}\in \mathbb{C}$ of the state matrix $A\in {\mathbb{R}}^{N\times N}$ defining the system’s state evolution equation,
$|{\lambda }_i|<1.$

Similarly, any continuous-time LTI system is internally stable if
$\text{Re}\{{\lambda }_i\}<0.$

Note (Marginal Stability) $|{\lambda }_i|=1$ in the discrete-time, and $\text{Re}\{{\lambda }_i\}=0$ in the continuous-time), the system is said to be marginally stable. These are the conditions at which $q(t)=q(0)$ $\forall t$.

If one or more eigenvalues meet the border conditions (i.e.

Proof

1. Discrete-Time Proof
   We will complete the proof by observing that any LTI state trajectory $q(n)$ with input $x(n)=0$ can be expressed as a product of its initial state $q(0)$, such that
   $q(n)=A^{n}q(0).$
   This is easily observed when we understand that $q(2)=Aq(1)=A(Aq(0))=A^{2}q(0)$. We infer that $q(n)\to 0$ as $n\to 0$ only if $A^n\to 0$. We then observe that any diagonalizable matrix $A$ can be expressed using eigendecomposition as
   $A^n=V{\Lambda }^n{V}^{-1}.$
   Finally, we observe that $q(0)$ can be expressed as a linear combination of any eigenbasis in its dimension, such that
   $q(0)={\alpha }_1{\vec{v}}_1+\dots +{\alpha }_N{\vec{v}}_N=V\alpha$
   Substituting the two latter equations into the first one, we get

q(n) &= V\Lambda^n(V^{-1}V)\alpha \\ &=V\Lambda^n\alpha. \end{align}$$ We see that if $\Lambda^n \to 0$ as $n \to \infty$ (every eigenvalue is strictly less than one), then $q(n) \to 0$. Then, all 2. **Continuous-Time Proof** We proceed in the same manner by observing that $$q(t) = e^{At}q(0)$$ which is the matrix solution of the scalar differential equation $\dot{q}(t)=aq(t) \iff q(t) = e^{at}q(0)$. Since by its Taylor series expansion $e^{At} = I + \frac{t}{1!}A + \frac{t^2}{2!}A^2 + \ldots$, then by decomposing $q(0)$ into a linear combination of an eigenbasis and expressing $A^n=V\Lambda^n V^{-1}$ as defined above, we get $$\begin{align} q(t)&=e^{At}V\alpha \\ &= V\alpha + \frac{t}{1!}V\Lambda (V^{-1}V)\alpha+\frac{t^2}{2!}V\Lambda^2(V^{-1}V)\alpha+\ldots \\ &= V\alpha+tV\Lambda\alpha + \frac{t^2}{2!}V\Lambda^2\alpha+\ldots \\ &= \sum^{\infty}_{k=0}\frac{t^k}{k!}V\Lambda^k\alpha \\ &= Ve^{\Lambda t}\alpha \end{align}$$ This means that as $t \to \infty$, $q(t) \to 0$ only if $e^{\Lambda t} \to 0$. This is only a possibility if for all $\lambda_{i} \in \{ \lambda_{1},\lambda_{2},\ldots,\lambda_{N} \} \in \mathbb{C}$, $\text{Re}\{\lambda_{i}\} < 0$. # Examples >[!example] Stability of an RLC Circuit > >In a previous note (2.2f), we found that the state matrix of a series RLC circuit is $A=\begin{bmatrix}-\frac{R}{L} & 1 \\ \frac{1}{LC} & 0\end{bmatrix}$.