State-space equations
Control design using pole placement
Introducing the reference input
Observer design

Key Matlab commands used in this tutorial: acker, lsim, place, plot, rscale

Matlab commands from the control system toolbox are highlighted in red.
Non-standard Matlab commands used in this tutorial are highlighted in green.

State-space equations

There are several different ways to describe a system of linear differential equations. The state-space representation is given by the equations:

To introduce the state space design method, we will use the magnetically suspended ball as an example. The current through the coils induces a magnetic force which can balance the force of gravity and cause the ball (which is made of a magnetic material) to be suspended in midair. The modeling of this system has been established in many control text books (including Automatic Control Systems by B. C. Kuo, the seventh edition). The equations for the system are given by:

where h is the vertical position of the ball, i is the current through the electromagnet, V is the applied voltage, M is the mass of the ball, g is gravity, L is the inductance, R is the resistance, and K is a coefficient that determines the magnetic force exerted on the ball. For simplicity, we will choose values M = 0.05 Kg, K = 0.0001, L = 0.01 H, R = 1 Ohm, g = 9.81 m/sec^2 . The system is at equilibrium (the ball is suspended in midair) whenever h = K i^2/Mg (at which point dh/dt = 0). We linearize the equations about the point h = 0.01 m (where the nominal current is about 7 amp) and get the state space equations:

A = [  0  1     0
     980  0  -2.8
       0  0  -100];
B = [0
     0
     100];
C = [1   0   0];

One of the first things you want to do with the state equations is find the poles of the system; these are the values of s where det(sI - A) = 0, or the eigenvalues of the A matrix:

poles = eig(A)

poles =

   31.3050
  -31.3050
 -100.0000

To check out what happens to this unstable system when there is a nonzero initial condition, add the following lines to your m-file,

t = 0:0.01:2;
u = 0*t;
x0 = [0.005 0 0];
[y,x] = lsim(A,B,C,0,u,t,x0);
h = x(:,2);   %Delta-h is the output of interest
plot(t,h)

It looks like the distance between the ball and the electromagnet will go to infinity, but probably the ball hits the table or the floor first (and also probably goes out of the range where our linearization is valid).

Control design using pole placement

Let's build a controller for this system. The schematic of a full-state feedback system is the following:

Recall that the characteristic polynomial for this closed-loop system is the determinant of (sI-(A-BK)). Since the matrices A and B*K are both 3 by 3 matrices, there will be 3 poles for the system. By using full-state feedback we can place the poles anywhere we want. We could use the Matlab function place to find the control matrix, K, which will give the desired poles.

Before attempting this method, we have to decide where we want the closed-loop poles to be. Suppose the criteria for the controller were settling time < 0.5 sec and overshoot < 5%, then we might try to place the two dominant poles at -10 +/- 10i (at zeta = 0.7 or 45 degrees with sigma = 10 > 4.6*2). The third pole we might place at -50 to start, and we can change it later depending on what the closed-loop behavior is. Remove the lsim command from your m-file and everything after it, then add the following lines to your m-file,

p1 = -10 + 10i;
p2 = -10 - 10i;
p3 = -50;

K = place(A,B,[p1 p2 p3]);

lsim(A-B*K,B,C,0,u,t,x0);

The overshoot is too large (there are also zeros in the transfer function which can increase the overshoot; you do not see the zeros in the state-space formulation). Try placing the poles further to the left to see if the transient response improves (this should also make the response faster).

p1 = -20 + 20i;
p2 = -20 - 20i;
p3 = -100;
K = place(A,B,[p1 p2 p3]);
lsim(A-B*K,B,C,0,u,t,x0);

This time the overshoot is smaller. Consult your textbook for further suggestions on choosing the desired closed-loop poles.

Compare the control effort required (K) in both cases. In general, the farther you move the poles, the more control effort it takes.

K = acker(A,B,[p1 p2 p3])

Introducing the reference input

t = 0:0.01:2; 
u = 0.001*ones(size(t));
lsim(A-B*K,B,C,0,u,t)

The system does not track the step well at all; not only is the magnitude not one, but it is negative instead of positive!

Recall the schematic above, we don't compare the output to the reference; instead we measure all the states, multiply by the gain vector K, and then subtract this result from the reference. There is no reason to expect that K*x will be equal to the desired output. To eliminate this problem, we can scale the reference input to make it equal to K*x_steadystate. This scale factor is often called Nbar; it is introduced as shown in the following schematic:

Nbar=rscale(A,B,C,0,K)

lsim(A-B*K,B*Nbar,C,0,u,t)

and now a step can be tracked reasonably well.

Observer design

When we can't measure all the states x (as is commonly the case), we can build an observer to estimate them, while measuring only the output y = C x. For the magnetic ball example, we will add three new, estimated states to the system. The schematic is as follows:

The observer is basically a copy of the plant; it has the same input and almost the same differential equation. An extra term compares the actual measured output y to the estimated output ; this will cause the estimated states to approach the values of the actual states x. The error dynamics of the observer are given by the poles of (A-L*C).

First we need to choose the observer gain L. Since we want the dynamics of the observer to be much faster than the system itself, we need to place the poles at least five times farther to the left than the dominant poles of the system. If we want to use place, we need to put the three observer poles at different locations.

op1 = -100;
op2 = -101;
op3 = -102;

L = place(A',C',[op1 op2 op3])';

At = [A - B*K           B*K
     zeros(size(A))     A - L*C];
Bt = [    B*Nbar 
      zeros(size(B))];
Ct = [   C        zeros(size(C))];

lsim(At,Bt,Ct,0,zeros(size(t)),t,[x0 x0])

Responses of all the states are plotted below. Recall that lsim gives us x and e; to get we need to compute x-e.

The blue solid line is the response of the ball position , the blue dotted line is the estimated state ;
The green solid line is the response of the ball speed , the green dotted line is the estimated state ;
The red solid line is the response of the current, the red dotted line is the estimated state .

We can see that the observer estimates the states quickly and tracks the states reasonably well in the steady-state.

The plot above can be obtained by using the plot command.

State Space Examples

Cruise Control | Motor Speed | Motor Position | Bus Suspension | Inverted Pendulum Pitch Controller | Ball & Beam

Tutorials

Basics | Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control | Examples