State Variables
System Modeling and Control · Chapter 14
Consider a mass–spring-damper system.
Equations of motion: \[m\frac{d^{2}y}{dt^{2}}=-ky-b\frac{dy}{dt}+u(t).\] Let \[\begin{aligned} x_{1} &=&y \\ x_{2} &=&\dot{y} \end{aligned}\] so that \[\begin{aligned} \frac{dx_{1}}{dt} &=&x_{2} \\ \frac{dx_{2}}{dt} &=&-\frac{k}{m}x_{1}-\frac{b}{m}x_{2}+\frac{1}{m}u. \end{aligned}\] This is called the statespace form.
\[\begin{aligned} \frac{dx_{1}}{dt} &=&x_{2} \\ \frac{dx_{2}}{dt} &=&-\frac{k}{m}x_{1}-\frac{b}{m}x_{2}+\frac{1}{m}u. \end{aligned}\]
Statespace form
Left-side of the equation has only first-order derivatives.
Right-side has no derivatives.
As these equations are linear, we can write them in matrix form as \[\frac{d}{dt}\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] =\underset{A}{\underbrace{\left[ \begin{array}{cc} 0 & 1 \\ -k/m & -b/m \end{array} \right] }}\underset{x}{\underbrace{\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] }}+\underset{b}{\underbrace{\left[ \begin{array}{c} 0 \\ 1/m \end{array} \right] }}u\]
DC Motor
Equations of motion: \[\begin{aligned} V_{a} &=&Ri+L\frac{di}{dt}+V_{b},\text{ }V_{b}=K_{b}\omega \\ J\frac{d\omega }{dt} &=&\tau _{m}-f\omega -\tau _{L},\text{ }\tau _{m}=K_{T}i \\ \frac{d\theta }{dt} &=&\omega \end{aligned}\] or \[\begin{aligned} L\frac{di}{dt} &=&-Ri-K_{b}\omega +V_{a} \\ J\frac{d\omega }{dt} &=&K_{T}i-f\omega -\tau _{L} \\ \frac{d\theta }{dt} &=&\omega . \end{aligned}\]
DC Motor (continued)
Rewrite \[\begin{aligned} L\frac{di}{dt} &=&-Ri-K_{b}\omega +V_{a} \\ J\frac{d\omega }{dt} &=&K_{T}i-f\omega -\tau _{L} \\ \frac{d\theta }{dt} &=&\omega \end{aligned}\] in statespace form as \[\begin{aligned} \frac{di}{dt} &=&-\frac{R}{L}i-\frac{K_{b}}{L}\omega +\frac{1}{L}V_{a} \\ \frac{d\omega }{dt} &=&\frac{K_{T}}{J}i-\frac{f}{J}\omega -\frac{1}{J}\tau _{L} \\ \frac{d\theta }{dt} &=&\omega . \end{aligned}\]
DC Motor (continued)
Matrix notation: \[\frac{d}{dt}\left[ \begin{array}{c} i \\ \omega \\ \theta \end{array} \right] =\underset{A}{\underbrace{\left[ \begin{array}{ccc} -R/L & -K_{b}/L & 0 \\ K_{T}/J & -f/J & 0 \\ 0 & 1 & 0 \end{array} \right] }}\underset{x}{\underbrace{\left[ \begin{array}{c} i \\ \omega \\ \theta \end{array} \right] }}+\underset{b}{\underbrace{\left[ \begin{array}{c} 1/L \\ 0 \\ 0 \end{array} \right] }}V_{a}+\underset{p}{\underbrace{\left[ \begin{array}{c} 0 \\ -1/J \\ 0 \end{array} \right] }}\tau _{L}.\]
Control input is \(u\triangleq V_{a}.\)
Disturbance input is \(\tau _{L}.\)
State variables are \(i,\omega ,\theta .\)
System state is \(x\triangleq \left[ \begin{array}{c} i \\ \omega \\ \theta \end{array} \right] \!.\)
In compact notation: \[\frac{d}{dt}x=Ax+bu+p\tau _{L}.\]
The statespace model \[\begin{aligned} dx_{1}/dt &=&x_{2} \\ dx_{2}/dt &=&-2x_{1}-3x_{2}+u \\ y &=&x_{1}. \end{aligned}\] is equivalent to the block diagram
The state variables \(x_{1},x_{2}\) are the outputs of the integrators.
In matrix notation: \[\begin{aligned} \frac{d}{dt}\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] &=&\underset{A}{\underbrace{\left[ \begin{array}{rr} 0 & 1 \\ -2 & -3 \end{array} \right] }}\underset{x}{\underbrace{\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] }}+\underset{b}{\underbrace{\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] }}u \\ y &=&\underset{c}{\underbrace{\left[ \begin{array}{cc} 1 & 0 \end{array} \right] }}\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] \!. \end{aligned}\]
Consider transfer function \[G(s)=\frac{Y(s)}{U(s)}=\frac{b_{2}s^{2}+b_{1}s+b_{0}}{ s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}.\]
Let \[\frac{X_{1}(s)}{U(s)}\triangleq \frac{1}{s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}.\]
Draw the simulation diagram for \(X_{1}(s)/U(s).\)
Some block diagram reduction shows \(X_{1}(s)/U(s)=\dfrac{1}{ s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}.\)
\(X_{1}(s)=\dfrac{1}{s}X_{2}(s)\Longrightarrow X_{2}(s)=sX_{1}(s)\) and \(X_{2}(s)=\dfrac{1}{s}X_{3}(s)\Longrightarrow X_{3}(s)=sX_{2}(s)=s^{2}X_{1}(s)\)
From previous slide: \(X_{2}(s)=sX_{1}(s),\) \(X_{3}(s)=sX_{2}(s)=s^{2}X_{1}(s).\)
As \[X_{1}(s)\triangleq \frac{1}{s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}U(s)\]
and \[Y(s)=\left( b_{2}s^{2}+b_{1}s+b_{0}\right) X_{1}(s)=\frac{ b_{2}s^{2}+b_{1}s+b_{0}}{s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}U(s)\] we have \[Y(s)=b_{2}s^{2}X_{1}(s)+b_{1}sX_{1}(s)+b_{0}X_{1}(s)=b_{2}X_{3}(s)+b_{1}X_{2}(s)+b_{0}X_{1}(s).\]
From the simulation diagram we have \[\begin{aligned} \frac{dx_{1}}{dt} &=&x_{2} \\ \frac{dx_{2}}{dt} &=&x_{3} \\ \frac{dx_{3}}{dt} &=&-a_{0}x_{1}-a_{1}x_{2}-a_{2}x_{3}+u \\ y &=&b_{0}x_{1}+b_{1}x_{2}+b_{2}x_{3} \end{aligned}\]
Transfer Function Realization
Let \[\frac{Y(s)}{U(s)}=\frac{10s+8}{s^{2}+5s+6}\] and \[\frac{X_{1}(s)}{U(s)}=\frac{1}{s^{2}+5s+6}.\]
Draw the simulation diagram for \(X_{1}(s)/U(s).\)
Then use the outputs of the two integrators to form \[Y(s)=10sX_{1}(s)+8X_{1}(s).\]
Transfer Function Realization (continued)
\[\begin{aligned} \frac{dx_{1}}{dt} &=&x_{2} \\ \frac{dx_{2}}{dt} &=&-6x_{1}-5x_{2}+u \\ y &=&8x_{1}+10x_{2}. \end{aligned}\] \[\begin{aligned} \frac{d}{dt}\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] &=&\underset{A}{\underbrace{\left[ \begin{array}{rr} 0 & 1 \\ -6 & -5 \end{array} \right] }}\underset{x}{\underbrace{\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] }}+\underset{b}{\underbrace{\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] }}u \\ y &=&\underset{c}{\underbrace{\left[ \begin{array}{cc} 8 & 10 \end{array} \right] }}\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] \end{aligned}\]
Block Diagram Reduction
Take “ \(10\)” outside the feedback loop to obtain
Block Diagram Reduction (continued)
From the previous slide
Simplify the inner feedback loop to obtain
Block Diagram Reduction (continued)
From the previous slide
Simplify the feedback loop and the output.
which shows \[\frac{Y(s)}{U(s)}=\dfrac{10s+8}{s^{2}+5s+6}.\]
Transfer Function Realization
\[G(s)=\frac{s^{2}+10s+5}{s^{2}+5s+6}\] This is proper, but not strictly proper. \[G(s)=\frac{s^{2}+5s+6-(5s+6)+10s+5}{s^{2}+5s+6}=\frac{s^{2}+5s+6+5s-1}{ s^{2}+5s+6}=1+\underset{G_{sp}}{\underbrace{\frac{5s-1}{s^{2}+5s+6}}}.\] We let \[G_{sp}(s)=\frac{Y_{sp}(s)}{U(s)}\triangleq \frac{5s-1}{s^{2}+5s+6}\] Do a realization of \(G_{sp}(s)\).
Transfer Function Realization (continued)
Including the feedforward signal from the input \(U(s)\) to the output \(Y(s)\): \[Y(s)=G_{sp}(s)U(s)+U(s).\]
Statespace Realization: \[\begin{aligned} \frac{dx_{1}}{dt} &=&x_{2} \\ \frac{dx_{2}}{dt} &=&-6x_{1}-5x_{2}+u \\ y &=&-x_{1}+5x_{2}+u \end{aligned}\]
Implementing a Feedback Controller
Consider the feedback system:
\(G_{c}(s)\) is proper, but not strictly proper. \[G_{c}(s)=\dfrac{b_{1}s+b_{0}}{s+a_{0}}=\dfrac{b_{1}(s+a_{0})-b_{1}a_{0}+b_{0} }{s+a_{0}}=b_{1}+\dfrac{b_{0}-b_{1}a_{0}}{s+a_{0}}\]
Statepace equations: \[\begin{aligned} \frac{dx(t)}{dt} &=&-a_{0}x(t)+e(t) \\ u(t) &=&(b_{0}-b_{1}a_{0})x(t)+b_{1}e(t) \\ e(t) &=&r(t)-c(t). \end{aligned}\]
Implementing a Transfer Function
Let \[\frac{U(s)}{E(s)}=G_{c}(s)=\frac{b_{c2}s^{2}+b_{c1}s+b_{c0}}{ s^{3}+a_{c2}s^{2}+a_{c1}s+a_{c0}}.\] A state space realization of \(G_{c}(s)\) is given by \[\begin{aligned} \frac{d}{dt}\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right] &=&\underset{A_{c}}{\underbrace{\left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -a_{c0} & -a_{c1} & -a_{c2} \end{array} \right] }}\underset{x}{\underbrace{\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right] }}+\underset{b_{c}}{\underbrace{\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] }}e \\ u &=&\underset{c_{c}}{\underbrace{\left[ \begin{array}{ccc} b_{c0} & b_{c1} & b_{c2} \end{array} \right] }}\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right] . \end{aligned}\]
