Linear Systems 101 - Part 2 - (De)motivation for Continuous-time Systems

 So let us continue our great journey. For the ones who did not read the previous post yet, you can find it here. Here's a brief recap. We have studied an unforced first order scalar linear system with the form

\begin{equation}\dot x = -ax, x(0) = x_0\end{equation}

defined some useful terms, and found its time solution. Before proceeding, let us have a closer look on the procedure of finding this solution. There are quite a few methods of solving it, the one I used in the last post was by integration. Now, let us apply the very common method of guessing the solution:

\begin{equation}\dot x + ax = 0, \text{ for } x = Ce^{\lambda t}, t \geq 0\end{equation}

Why did I guess an exponential solution? Well, an explanation is that it is a function whose derivative leads to itself multiplied by a constant term. Besides, we have now two unknowns, C and λ. Thus, we need two equations for obtaining those values. The first one comes from 

\begin{equation}\dot x + ax = Ce^{\lambda t}(\lambda +a) =0 \end{equation}

and the second one comes from the initial condition, that is, evaluated at t = 0:

\begin{equation}Ce^{\lambda t} =  Ce^{0} = C = x_0.\end{equation}

so that, by substituting C on the first equation yields \begin{equation}x_0 e^{\lambda t}(\lambda +a) =0, t \geq 0\end{equation}

Let us briefly analyse this equation. Note that the system is not subject to any input. If the initial condition were to be zero, then we would have a trivial solution to the last equation. This is intuitive in a practical sense: our system would stay at rest if we would not apply any external force or initial condition. But, considering the more exciting situation that the initial condition is not zero, then the only way to have a solution for all t would be to set \begin{equation}x_0 e^{\lambda t}(\lambda +a) =0, \forall t \geq 0 \leftrightarrow   \lambda = -a,  \end{equation}

which yields the same solution we have previously found \begin{equation}x(t) = e^{-at}x_0, t \geq 0\end{equation}

Right........ Why did we bother do this different method just for obtaining the same solution we had already obtained by simple itnegration? Because it leads to an important tool and provide some useful insightswe are going to use in the next posts: the characteristic equation. It is an algebraic equation obtained from the original differential equation. Notice the similarities: \begin{equation}\dot x + ax = 0 \text{ and } \lambda + a = 0 \end{equation} We have substituted the first order differential by the unknown λ and the lowest, zero order, term by 1. By finding the root of this characteristic equation, we have found the argument of the exponential solution. That is, we can found the solution by solving this equation! 

Finally, we can also make the following inference

  • If a > 0, then the system is stable
  • If a < 0, then the system is unstable
which is the same to say that

  • If  λ < 0, then the system is stable
  • If λ > 0, then the system is unstable
This root λ is called the pole. It is a fundamental concept and an entity which describes several important characteristics of our system.

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