Section 10–4 / Design of Servo Systems 749
MATLAB Program 10–6
A = [0 1 0 0; 20.601 0 0 0; 0 0 0 1; -0.4905 0 0 0];
B = [0;-1;0;0.5];
C = [0 0 1 0];
Ahat = [A zeros(4,1); -C 0];
Bhat = [B;0];
J = [-1+jsqrt(3) -1-jsqrt(3) -5 -5 -5];
Khat = acker(Ahat,Bhat,J)
Khat =
-157.6336 -35.3733 -56.0652 -36.7466 50.9684
Thus, we get
and
Unit Step-Response Characteristics of the Designed System. Once we determine the feed-
back gain matrix Kand the integral gain constant kI,the step response in the cart position can be
obtained by solving the following equation, which is obtained by substituting Equation (10–49)
into Equation (10–35):
(10–53)
The output y(t)of the system is x 3 (t),or
(10–54)
Define the state matrix, control matrix, output matrix, and direct transmission matrix of the
system given by Equations (10–53) and (10–54) as AA, BB, CC,andDD,respectively. MATLAB
Program 10–7 may be used to obtain the step-response curves of the designed system. Notice
that, to obtain the unit-step response, we entered the command
[y,x,t] = step(AA,BB,CC,DD,1,t)
Figure 10–10 shows curves x 1 versust, x 2 versust, x 3 (=outputy)versust, x 4 versust, andx 5
(=j)versust.Notice that y(t)C=x 3 (t)Dhas approximately 15%overshoot and the settling time
is approximately 4.5 sec.j(t)C=x 5 (t)Dapproaches 1.1. This result can be derived as follows: Since
or
D
0
0
0
0
T =D
0
20.601
0
- 0.4905
1
0
0
0
0
0
0
0
0
0
1
0
TD
0
0
r
0T + D
0
- 1
0
0.5
Tu(q)
x#(q)= 0 =Ax(q)+Bu(q)y=[0 0 1 0 0]B
x
jR +[0]r
B
x#
j#R= B
A-BK
- C
BkI
0RB
x
jR +B
0
1
Rr
kI=-50.9684K=Ck 1 k 2 k 3 k 4 D=[-157.6336 - 35.3733 - 56.0652 - 36.7466]