机电控制技术考试试卷及答案解析

Mechatronic Control Systems

Spring 2013

Dr. Bin Yao

FINAL EXAM

April 30, 2013

INSTRUCTIONS:

1.This is a Closed book exam. You are allowed one help sheet of hand-written

summary.

2.Your exams must be stapled.

3.Circle your final answers.

4.Be neat and clear.

PROBLEM 1 (20Points)

Consider the following feedback system:

where

()3

()

(1)

s

P s

s s

+

=

-

.

You are required to design a controller to meet the following performance specifications: (P1).Zero steady-state error for ramp type reference input ()

r t and constant disturbance ()

d t

(P2).The resulting closed-loop system should not have excessive transient responses for step reference input ()

R s, i.e., your design should avoid either excessive large overshoot or large undershoot in the step responses.

To solve this problem, you are required to follow the following procedure:

a)Determine the correct controller structure that is needed to meet the performance

requirement P1. To receive full credit, you need to justify your answer as well.

b)Determine the suitable desired pole locations of the closed-loop system so that the

performance requirement P2 can be satisfied. Again, to receive full credit, you need to justify your answer as well.

c)Determine the unknown controller parameters to meet the above performance

requirements.

Solutions:

由上面两条定理可以得到结论：

1、

a). As the plant has an integrator, to satisfy (P1), the controller only needs one integrator, i.e.,

()()

()

(),with order()order()1

()

C

C C

C

N s

C s N s

D s

sD s

=≤+(1) With the controller (1), the closed-loop output is given by

()

()()

()

()()

22

3()3()

()()()

1()3()1()3()

C C

C C C C

s N s s sD s

Y s R s d s

s s D s s N s s s D s s N s

++

=+

-++-++

(2)

Y(s)

-

()

R s

()

d s

()

C s()

P s

Controller Plant

Thus, for ramp type reference input (i.e., 2()/c R s r s =) and constant disturbance (i.e., ()/d d s A s =), the system output tracking error is

()()()()()()()()()()2222

()()()

1()

3() ()()1()3()1()3()

1()3() 1()3()

C C C C C C C c C d

C C E s R s Y s s s

D s s sD s R s d s s s D s s N s s s D s s N s s D s r s D s A s s D s s N s =--+=--++-++--+=-++ (3)

So as long as the CL system is stable (i.e., the denominator in (3) has all roots in LHP), the condition for applying FVT is satisfied. By FVT, you can easily show that the steady-state error in (3) is zero.