Linear Time-Invariant Systems
Brief introduction to linear time-invariant systems
Linear systems form a cornerstone of mathematical modelling of dynamical systems. Indeed a system or some aspects of it can be modelled using a linear model. Furthermore, non-linear systems can be linearized around a certain mode of operation. In this section, we give a brief overview of linear time-invariant systems. The major advantage of linear systems is that in terms of analysis a far simpler. Moreover, understanding the dynamics and thus stability of the system is easier.
Let's consider the following system
$$\frac{d \mathbf{x}}{dt} = \mathbf{f}(\mathbf{x}, \mathbf{u}), ~~ \mathbf{y} = \mathbf{g}(\mathbf{x}, \mathbf{u})$$
As before, $\mathbf{x}$ represents the state of the modelled system whilst $\mathbf{u}$ represents some control input to the system. Note that the right-hand side terms do not depend explicitly on the time variable $t$.
If $\mathbf{f}$ or $\mathbf{g}$ or both are non-linear, then the system is non-linear. In this case, we can linearize the system i.e. its dynamics, using a Taylor series expansion near a fixed point $(\bar{\mathbf{x}}, \bar{\mathbf{u}})$. Recall that at a fixed point is a point where
$$\mathbf{f}(\bar{\mathbf{x}}, \bar{\mathbf{u}}) = \mathbf{0}$$
The linear, or linearized, dynamics can be written in the following matrix form (assuming no errors)
$$\frac{d\mathbf{x}}{dt} = \mathbf{A}\mathbf{x} + \mathbf{B}\mathbf{u}, ~~ \mathbf{y} = \mathbf{C}\mathbf{x} + \mathbf{D}\mathbf{u}$$
When $\mathbf{u} = \mathbf{0}$ and when there are no measurement errors i.e. $\mathbf{y} = \mathbf{x}$. The system reduces to
$$\frac{d\mathbf{x}}{dt} = \mathbf{A}\mathbf{x}$$
The solution to this ODE is [1]
$$\mathbf{x}(t) = e^{\mathbf{A}t}\mathbf{x}(0)$$
Thus $\mathbf{x}(t)$ depends or is determined entirely by the matrix $\mathbf{A}$. The stability of the unforced system therefore, can be understood via the eigenvalues and eigenvectors of $\mathbf{A}$. In particular we have the following cases
- All the eigenvalues $\lambda$ satisfy $Re(\lambda) < 0$. Then the system is stable and all solutions decay to $\mathbf{u} = \mathbf{0}$ as $t \rightarrow \infty$
- There exists at least one eigenvalue $\lambda$ with $Re(\lambda) > 0$ then the system is unsatble and will diverge from the fixed point along the corresponding unstable eigenvector direction.
Now let's assume that $\mathbf{u} \neq \mathbf{0}$ and that $\mathbf{x}(0) = \mathbf{0}$. In this case the solution up to time $t$ is given by [1]
$$\mathbf{x}(t) = \int_{0}^t e^{\mathbf{A}(t - \tau)} \mathbf{B}\mathbf{u}(\tau)d \tau $$
This integral is nothing more than a convolution. Thus, we can write
$$\mathbf{x}(t) = e^{\mathbf{A}t} \mathbf{B} * \mathbf{u}(t)$$
- Steven L. Brunton, J. Nathan Kutz,
Data-Driven Science and Engineering. Machine Learning, Dynamical System and Control, Cambridge University Press.