I recently learned about the standard continuous-time state-space model (wiki):
$$\begin{align}
\dot{x}(t) &= Ax(t) + Bu(t) \\
y(t) &= Cx(t) + Du(t).
\end{align}$$
It's known that these systems are LTI (linear time-invariant), but that's not entirely obvious to me. Take the linearity of some system $G$, for example, which stipulates among other things that $$\alpha u(t) \rightarrow G \rightarrow \alpha y(t).$$
I don't see how this is implied by the CT SSM above. Am I making a super simple mistake in my analysis below?
Let $C=1, D=0$ for simplicity so that $y(t) = x(t)$. This reduces the problem to solving the first differential equation, whose solution is the sum of the homogeneous solution and some particular solution, which comes out via variation of parameters to
$$y(t) = x_0 e^{At} + \int K e^{A(t-\tau)} \, Bu(\tau) \, d\tau.$$
From this, it doesn't seem like $y(t)$ should always be linear w.r.t. $u(t)$. What am I missing / what have I got wrong? Why then are these systems said to be linear?
Best Answer
Assuming that ${\bf D} = {\bf O}$, the response of the LTI system is
$$ {\bf y} (t) = {\bf C} e^{ (t - t_0) {\bf A}} {\bf x}_0 + \int_{t_0}^t {\bf C} e^{ (t - \tau) {\bf A}} {\bf B} {\bf u} (\tau) \, {\rm d} \tau $$
which is the sum of
the natural (or zero-input) response (dependent on ${\bf x}_0$), and
the forced (or zero-state) response (given by the convolution integral above).
Assuming that $\bf A$ is Hurwitz stable, after "a while", the initial state ${\bf x}_0$ is "forgotten" and what is left is mostly the forced response. This is when linearity applies. Say, double the input, and the output is also doubled. The reason linearity was not obvious to you is that you were not looking at the convolution integral alone. If you assume that the system was "switched on" a geological era ago, you don't need to worry about the natural response because the initial state has been totally "forgotten".