Modern Control Engineering

850 Chapter 10 / Control Systems Design in State Space

where

Obtain the unit-step response of the system designed.

Solution.A MATLAB program to determine is given in MATLAB Program 10–34. The result is

k 1 =-188.0799, k 2 =-37.0738, k 3 =-26.6767, k 4 =-30.5824, kI=-10.0000

Kˆ

Q=E

100

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

U, R=0.01

MATLAB Program 10–34

% Design of quadratic optimal control system

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];

D = [0];

Ahat = [A zeros(4,1);-C 0];

Bhat = [B;0];

Q = [100 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1];

R = [0.01];

Khat = lqr(Ahat,Bhat,Q,R)

Khat =

-188.0799 -37.0738 -26.6767 -30.5824 10.0000

Unit-Step Response. Once we have determined the feedback gain matrix Kand the integral gain

constantkI,we can determine the unit-step response of the designed system. The system equation

is

(10–178)

[Refer to Equation (10–35).] Since

Equation (10–178) can be written as follows:

(10–179)

The output equation is

y=[C 0 ]B

x

j

R +[ 0 ]r

B

x#

j

#R =BA-BK

• C

BkI

0

RB

x

j

R + B

0

1

Rr

u=-Kx+kI j

B

x#

j

#R= B

A

- C

0

0

RB

x

j

R+ B

B

0

Ru+ B

0

1

Rr

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