Notes on feedback control

The materials in this part heavily stems from MAE433 lecture note taught at Princeton University

State-space systems

Question: Why does if \(\dot{x} = Ax\) then the solution is \(x(t) = e^{At}x_0\)

Before talking about the matrix \(A\), let us first look at a simpler case where we have \(\dot{x} = ax\), with \(a\) a scalar value.

We have the following derivation:

\[\begin{split} \begin{align*} &\dot{x} = ax \\ &\leftrightarrow \frac{\dot{x}}{x} = a \\ &\leftrightarrow \int_{x_0}^{x(t)} \frac{\dot{x}}{x} \,dx = \int_{0}^{t} a \,dt && \text{taking integral both sides} \\ &\rightarrow \log{x(t)} - \log{x_0} = at \\ &\leftrightarrow \log{\frac{x(t)}{x_0}} = at \\ &\leftrightarrow \frac{x(t)}{x_0} = e^{at} \\ &\leftrightarrow x(t) = e^{at} x_0 \end{align*} \end{split}\]