Reader: Recall the mass, spring, damper with \(k,c,m\) and we generated the second order ODE for it. Let \(u=0\) be the input. Initial conditions are \(y(0)=1,y'(0)=0\).
Experiment with various values of the parameters \(k,m,c\) and observe variety of responses. Also consider \(u(t)\) as unit step.
Reader: Consider discrete time state equation \begin{align*} x(k+1) & = A x(k) +B u(k) \\ y(k) & = Cx(k) + D u(k) \end{align*}
with zero initial conditions. Solve for \(x(N)\) and \(y(N)\).
\begin{align*} x(1) &= A x(0) + B u(0) \\ x(2) &= A ( Ax(0) + B u(0) ) +B u(1) \\ &= A^2 x(0) + A B u(0) +B u(1) \\ x(3) &= A( A^2 x(0) + A B u(0) +B u(1)) + Bu(2) \\ &= A^3 x(0) +A^2 Bu(0) +A B u(1) + B u(2) \\ &\vdots \end{align*}
Hence
\[ x(N) = A^{N} x(0) + \sum \limits _{k=0}^{N-1} A^{N-1-k} B u(k) \]
In solving continuous time state equations, the solution will contain terms such as \(\sin /\cos \) and exponential and \(t\) multipliers. These are the only things that come up in linear system theory.
Sometimes an LTI system is stable and sometimes it is not stable.
Stable vs. not stable: if \(A\) has all its eigenvalues with real part negative, then it is stable, else not stable. But even if one one eigenvalue had real part which is positive, it might still be stable. This depends if the initial conditions activate the mode with this eigenvalue. For Example
\(A= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, B=0\), then \(x_{1}' = x_1\) and \(x_{2}' = -x_2\). Then \(x_1=e^t x_1(0)\) and \(x_2 = e^{-t} x_2(0)\). Now if the initial conditions are \(x_1(0)=0\) and \(x_2(0)=1\), then \(x_1(t)=0\) even though the eigenvalue was positive.
Now we will talk about non-linear systems. Consider
\begin{align*} x_{1}' & =x_{1}x_{2}+ux_{3}\\ x_{2}' & =x_{2}+2x_{2}x_{3}\\ x_{3}' & =x_{2}+x_{3} \end{align*}
More generally, \(x' = f(x,u)\). There are no \(A,B,C,D\) in nonlinear system. there is only \(f(x,u)\) and \(g(x,u)\). To linearize this, we need to talk about equilibrium. There is stable and there is unstable equilibrium. Always linearize around the stable equilibrium point. To find \((\bar{x},\bar{u}) \) solve \(x'=0\). i.e. it is when \(f(\bar{x},\bar{u})=0\).
How does linearization work? Since we assume \(f(x,u)\) is smooth function, we expand in Taylor series around equilibrium \((\bar{x},\bar{u})\)
\begin{align*} f_{1}\left ( \bar{x}+\Delta x\right ) & =\overset{zero\text{ at equilibrium}}{\overbrace{f\left ( x_{eq}\right ) }}+\left . \frac{\partial f_{1}}{\partial x_{1}}\right \vert _{\left ( \bar{x},\bar{u}\right ) }\Delta x_{1}+\frac{1}{2}\left . \frac{\partial ^{2}f_{1}}{\partial x_{1}^{2}}\right \vert _{\left ( \bar{x},\bar{u}\right ) }\left ( \Delta x_{1}\right ) ^{2}+\cdots \\ & +\left . \frac{\partial f_{1}}{\partial x_{2}}\right \vert _{\left ( \bar{x},\bar{u}\right ) }\Delta x_{2}+\frac{1}{2}\left . \frac{\partial ^{2}f_{1}}{\partial x_{2}^{2}}\right \vert _{\left ( \bar{x},\bar{u}\right ) }\left ( \Delta x_{2}\right ) ^{2}+\cdots \\ & +\left . \frac{\partial f_{1}}{\partial x_{3}}\right \vert _{\left ( \bar{x},\bar{u}\right ) }\Delta x_{3}+\frac{1}{2}\left . \frac{\partial ^{2}f_{1}}{\partial x_{3}^{2}}\right \vert _{\left ( \bar{x},\bar{u}\right ) }\left ( \Delta x_{3}\right ) ^{2}+\cdots \end{align*}
Similarly for each \(f_{i}\left ( x,u\right ) \). For small \(\Delta x\) we obtain, after dropping all higher order terms
\[ f\left ( \bar{x}+\Delta x\right ) =\begin{pmatrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} & \cdots & \frac{\partial f_{1}}{\partial x_{n}}\\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}} & \cdots & \frac{\partial f_{2}}{\partial x_{n}}\\ \vdots & \vdots & \vdots & \vdots \\ \frac{\partial f_{n}}{\partial x_{1}} & \frac{\partial f_{n}}{\partial x_{2}} & \cdots & \frac{\partial f_{n}}{\partial x_{n}}\end{pmatrix} \]
Two roles for the small \(\Delta x\): approximates linear behavior, and remain around domain of influence so system returns to \(x_{eq}\).
Reader Argue that incremental dynamics are now \(\Delta x^{\prime }=\overbrace{\left . \frac{\partial f}{\partial x}\right \vert _{x_{eq}}}^{A}\Delta x\)
Reader: Generalize to \(f\left ( x,u\right ) \) instead of just \(f\left ( x\right ) \): Define \(\left ( \bar{x},\bar{u}\right ) \) s.t. \(f\left ( \bar{x},\bar{u}\right ) =0\). So that above become \(f\left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) \) then
\[ B\left ( \bar{u}+\Delta u\right ) =\begin{pmatrix} \frac{\partial f_{1}}{\partial u_{1}} & \frac{\partial f_{1}}{\partial u_{2}} & \cdots & \frac{\partial f_{1}}{\partial u_{m}}\\ \frac{\partial f_{2}}{\partial u_{1}} & \frac{\partial f_{1}}{\partial u_{2}} & \cdots & \frac{\partial f_{1}}{\partial u_{m}}\\ \vdots & \vdots & \vdots & \vdots \\ \frac{\partial f_{n}}{\partial u_{1}} & \frac{\partial f_{n}}{\partial u_{2}} & \cdots & \frac{\partial f_{n}}{\partial u_{m}}\end{pmatrix} \]
Can also introduce the output equation \(y=g\left ( x,u\right ) \) and now obtain linearized \(C,D\) as we did above for \(A,B\), but now using \(g\left ( x,u\right ) \) in place of \(f\left ( x,u\right ) \)