Linear systems are described by \(A,B,C,D\). as in \begin{align*} x^{\prime } & =Ax+Bu\\ y & =Cx+Du \end{align*}
For non-linear systems we have\begin{align*} x^{\prime } & =f\left ( x,u\right ) \\ y & =g\left ( x,u\right ) \end{align*}
We assume \(f,g\) are smooth and solution exists (may be using numerical). We talked about equilibrium point \(\left ( x,u\right ) \). This can be stable or unstable equilibrium.
We want to linearize the equations above. Linearization must be done about a stable point. Do not linearize around an unstable equilibrium. The linearized system\[ \Delta x^{\prime }=\overbrace{\left . \frac{\partial f}{\partial x}\right \vert _{\left ( x,u\right ) _{eq}}}^A\Delta x+\overbrace{\left . \frac{\partial f}{\partial u}\right \vert _{\left ( x,u\right ) _{eq}}}^B\Delta u \] So solution, near \(x_{eq}\) is \[ x\left ( t\right ) =x_{eq}+\Delta x \] Output equation is \[ \Delta y=\overbrace{\left . \frac{\partial g}{\partial x}\right \vert _{\left ( x,u\right ) _{eq}}}^C\Delta x+\overbrace{\left . \frac{\partial g}{\partial u}\right \vert _{\left ( x,u\right ) _{eq}}}^D\Delta u \] And \[ y\left ( t\right ) =g\left ( x,y\right ) +\Delta y \] Reader: Distinction between domain of attraction and linear approximation.
Region of attraction: Domain of initial states that converges to \(x_{eq}\).
Transfer functions: Motivating example. Given \(H\left ( s\right ) =\frac{Y\left ( s\right ) }{U\left ( s\right ) }=\frac{5}{s^{2}+6s+7}\). Recall, \(\mathcal{L}\)\(\frac{d^{k}f\left ( t\right ) }{dt^{k}}=s^{k}F\left ( s\right ) \) with initial added. Example, \(\mathcal{L}\)\(\left \{ f^{\prime }\left ( t\right ) \right \} =sF\left ( s\right ) -f\left ( 0\right ) \) and \(\mathcal{L}\)\(\left \{ f^{\prime \prime }\left ( t\right ) \right \} =s^{2}F\left ( s\right ) -sf\left ( 0\right ) -f^{\prime }\left ( 0\right ) \).
Note: \(H\left ( s\right ) \) is derived assuming all initial conditions are zero.
Reader: Find \(H\left ( s\right ) \) for \(y^{\prime \prime }+6y^{\prime }+7y=5u\left ( t\right ) \) and find the state space realization.
Suppose we are given \(A,B,C,D\). There is SISO (single input, single output) and MIMO (multiple input, multiple output).
Consider \begin{align*} \,x^{\prime } & =Ax+Bu\\ y & =Cx+Du \end{align*}
Take Laplace transform results in \(X\left ( s\right ) =\left ( sI-A\right ) ^{-1}BU\left ( s\right ) \) and the output becomes \begin{align*} Y\left ( s\right ) & =C\left ( sI-A\right ) ^{-1}BU+DU\\ & =\left ( C\left ( sI-A\right ) ^{-1}B+D\right ) U \end{align*}
Hence\[ H\left ( s\right ) =\frac{Y}{U}=C\left ( sI-A\right ) ^{-1}B+D \] Reader:
Find SISO \(H(s)\) for the above. Writing loop equations, using \(v=Ri\) for voltage across resistor, results in\[ R i + L \frac{di}{dt} + \frac{1}{c} \int \limits _{0}^{t} i\, d\tau = u(t) \] Taking derivative gives \[ R \frac{di}{dt} + L\frac{d^2 i}{dt^2} + \frac{1}{c} i= u'(t) \] The output equation is \(y=\frac{1}{c} \int \limits _{0}^{t} i \, d\tau \)
Reader: Given \(A=\begin{pmatrix} 1 & -2\\ 3 & 4 \end{pmatrix} ,B=\begin{pmatrix} 2 & 1\\ -1 & 0 \end{pmatrix} ,C=\begin{pmatrix} 0 & 1\\ -1 & 2 \end{pmatrix} ,D=\begin{pmatrix} 2 & -1\\ 1 & 1 \end{pmatrix} \). Find MIMO \(H(s)\). Can solve using syms.
Given transfer function matrix, can we find state space realization? Remark on MIMO: Consider \(Y_{r\times 1}(s) = H_{r\times m}(s) U_{m\times 1}(s)\), so the \(i^{th}\) output is \(\sum \limits _{j=1}^{m} H_{ij}(s) U_{j}(s)\). So entry \(H_{ij}(s)\) in the matrix transfer function is the transfer function between the \(j^{th}\) input to the \(i^{th}\) output. \[ H_{ij}(s) = \frac{Y_i}{U_j} \] This suggests experimental method to find \(H_{ij}(s)\). Zero out all inputs except for one. Measure the output at the port of interest. This finds \(H(s)\) between the input which is not zero and the output port being measured.