National Instruments 370753C-01

Chapter 6 State-Space Design

Figure 6-4. Diagram of Plant for the Inverted Pendulum Problem

Figure 6-4 shows the pendulum at φ = 0 and φ > 0. The distance of the cart

from some initial reference point along the line of its motion is represented

by the state variable x. You can measure the angle φ and the distance x

easily—in fact, you will use measurements as your system outputs—but it

is more difficult to obtain accurate measurements of the rate at which x and

φ change.

Designating as the state vector, you can set up the system in

Xmath:

A = [0,1,0,0;21,0,0,0.8;0,0,0,1;0,0,0,-4];

B = [0,-2,0,1]';

C = [1,0,0,0;0,0,1,0];

D = [0,0]';

ipsys = system(A,B,C,D);

You design a regulator with the assumption that all four states are available.

Recalling that you defined the state vector as , you can decide

the weighting you want to associate with each state. Refer to the quadratic

performance index equation in the Linear Quadratic Regulator section for

more information. For this particular problem, your most important

performance goal is that the pendulum stay upright—that is, that φ be

tightly controlled to stay as close to zero as possible. You also might prefer,

though to a lesser extent, that you not have to move the cart over too great

a distance. Physical limitations, such as the size of the room in which the

experiment is conducted, should be considered.

If you are not particularly concerned about the speed of the cart across the

floor or that rate of change of the angle, you might define,

Rxx = diagonal([1,0,0.1,0]);

with the larger values in the matrix corresponding to states whose values

you care most about. Presuming you are not too worried about the size of

the input u you impart to the cart:

Ruu = 1e-5;

φφ

·xx

[]′

φφ

·xx

[]′