# Control Systems/State-Space Stability

## State-Space Stability

If a system is represented in the state-space domain, it doesn\'t make
sense to convert that system to a transfer function representation (or
even a transfer matrix representation) in an attempt to use any of the
previous stability methods. Luckily, there are other analysis methods
that can be used with the state-space representation to determine if a
system is stable or not. First, let us first introduce the notion of
unstability:

Also, a key concept when we are talking about stability of systems is
the concept of an **equilibrium point**:

The definitions below typically require that the equilibrium point be
zero. If we have an equilibrium point *x~e~ = a*, then we can use the
following change of variables to make the equilibrium point zero:

$$\bar{x} = x_e - a = 0$$

We will also see below that a system\'s stability is defined in terms of
an equilibrium point. Related to the concept of an equilibrium point is
the notion of a **zero point**:

### Stability Definitions

The equilibrium *x = 0* of the system is stable if and only if the
solutions of the zero-input state equation are bounded. Equivalently, *x
= 0* is a stable equilibrium if and only if for every initial time t~0~,
there exists an associated finite constant *k*(t~0~) such that:

$$\operatorname{sup}_{t \ge t_0}\|\phi(t, t_0)\| = k(t_0) < \infty$$

Where *sup* is the **supremum**, or \"maximum\" value of the equation.
The maximum value of this equation must never exceed the arbitrary
finite constant *k* (and therefore it may not be infinite at any point).

Uniform stability is a more general, and more powerful form of stability
than was previously provided.

A time-invariant system is asymptotically stable if all the eigenvalues
of the system matrix A have negative real parts. If a system is
asymptotically stable, it is also BIBO stable. However the inverse is
not true: A system that is BIBO stable might not be asymptotically
stable.

For linear systems, uniform asymptotic stability is the same as
**exponential stability**. This is not the case with non-linear systems.

### Marginal Stability

Here we will discuss some rules concerning systems that are marginally
stable. Because we are discussing eigenvalues and eigenvectors, these
theorems only apply to time-invariant systems.

1.  A time-invariant system is marginally stable if and only if all the
    eigenvalues of the system matrix A are zero or have negative real
    parts, and those with zero real parts are simple roots of the
    minimal polynomial of A.
2.  The equilibrium *x = 0* of the state equation is *uniformly stable*
    if all eigenvalues of A have non-positive real parts, and there is a
    complete set of distinct eigenvectors associated with the
    eigenvalues with zero real parts.
3.  The equilibrium *x = 0* of the state equation is *exponentially
    stable* if and only if all eigenvalues of the system matrix A have
    negative real parts.

## Eigenvalues and Poles

A Linearly Time Invariant (LTI) system is stable (asymptotically stable,
see above) if all the eigenvalues of A have negative real parts.
Consider the following state equation:

$$x' = Ax(t) + Bu(t)$$

We can take the Laplace Transform of both sides of this equation, using
initial conditions of x~0~ = 0:

$$sX(s) = AX(s) + BU(s)$$

Subtract AX(s) from both sides:

$$sX(s) - AX(s) = BU(s)$$

$$(sI - A)X(s) = BU(s)$$

Assuming (sI - A) is nonsingular, we can multiply both sides by the
inverse:

$$X(s) = (sI - A)^{-1}BU(s)$$

Now, if we remember our formula for finding the matrix inverse from the
adjoint matrix:

$$A^{-1} = \frac{\operatorname{adj}(A)}{|A|}$$

We can use that definition here:

$$X(s) = \frac{\operatorname{adj}(sI - A)BU(s)}{|(sI - A)|}$$

Let\'s look at the denominator (which we will now call D(s)) more
closely. To be stable, the following condition must be true:

$$D(s) = |(sI - A)| = 0$$

And if we substitute λ for s, we see that this is actually the
characteristic equation of matrix A! This means that the values for s
that satisfy the equation (the poles of our transfer function) are
precisely the eigenvalues of matrix A. In the S domain, it is required
that all the poles of the system be located in the left-half plane, and
therefore all the eigenvalues of A must have negative real parts.

## Impulse Response Matrix

We can define the **Impulse response matrix**, *G*(t, τ) in order to
define further tests for stability:

$$G(t, \tau) = \left\{\begin{matrix}C(t)\phi(t, \tau)B(\tau) & \mbox{ if } t \ge \tau \\0 & \mbox{ if } t < \tau\end{matrix}\right.$$

The system is *uniformly stable* if and only if there exists a finite
positive constant *L* such that for all time *t* and all initial
conditions t~0~ with $t \ge t_0$ the following integral is satisfied:

$$\int_0^t \|G(t, \tau)\|d\tau \le L$$

In other words, the above integral must have a finite value, or the
system is not uniformly stable.

In the time-invariant case, the impulse response matrix reduces to:

$$G(t) = \left\{\begin{matrix}Ce^{At}B & \mbox{ if } t \ge 0 \\0 & \mbox{ if } t < 0\end{matrix}\right.$$

In a time-invariant system, we can use the impulse response matrix to
determine if the system is uniformly BIBO stable by taking a similar
integral:

$$\int_0^\infty \|G(t)\|dt \le L$$

Where *L* is a finite constant.

## Positive Definiteness

These terms are important, and will be used in further discussions on
this topic.

-   f(x) is **positive definite** if f(x) \> 0 for all x.
-   f(x) is **positive semi-definite** if $f(x) \ge 0$ for all x, and
    f(x) = 0 only if x = 0.
-   f(x) is **negative definite** if f(x) \< 0 for all x.
-   f(x) is **negative semi-definite** if $f(x) \le 0$ for all x, and
    f(x) = 0 only if x = 0.

A Hermitian matrix X is positive definite if all its principle minors
are positive. Also, a matrix X is positive definite if all its
eigenvalues have positive real parts. These two methods may be used
interchangeably.

Positive definiteness is a very important concept. So much so that the
Lyapunov stability test depends on it. The other categorizations are not
as important, but are included here for completeness.

## Lyapunov Stability

### Lyapunov\'s Equation

For linear systems, we can use the **Lyapunov Equation**, below, to
determine if a system is stable. We will state the Lyapunov Equation
first, and then state the **Lyapunov Stability Theorem**.

$$MA + A^TM = -N$$

Where A is the system matrix, and M and N are *p* × *p* square matrices.

Notice that for the Lyapunov Equation to be satisfied, the matrices must
be compatible sizes. In fact, matrices A, M, and N must all be square
matrices of equal size. Alternatively, we can write:

If the matrix M can be calculated in this manner, the system is
asymptotically stable.

