Controle e Observabilidade de Sistemas

Prévia do material em texto

Problem 7.35PP
Consider a system with the transfer function
W =
(a) Find (Ao, Bo. Co) for this system in observer canonicai fonri.
(b) Is (Ao, Bo) controllable?
(c) Compute K so that the closed-loop poles are assigned to s = -3 ± 3/.
(d) Is the closed-loop system of part (c) observable?
(d) Is the closed-loop system of part (c) observable?
(e) Design a full-order estimator with estimator error poles at s = -12±12/
(f) Suppose the system is modified to have a zero;
G,(i) = 9(5-H )
j2 - 9 ■
Prove that if u = -K x + r, there is a feedback gain K that makes the closed-loop system 
unobservable. [Again assume an observer canonical realization for G1 (s).]
Step-by-step solution
step 1 of 9
(a)
Consider the numerator part of the gain.
b {s )= b ^ ^ + b 2 S ^ ^ + .. .+ b , ...... (1)
Consider the denominator part of the gain. 
a { s ) = s " + 0,$ '^ + ...+a^ (2)
Write the general canonical form of the equation.
Substitute equations (1) and (2) in equation (3).
...... m
Write the observer canonical form of the state space equations.
= ...... (5)
Write the output equation.
y = C ^ x ^ + D ...... (6)
Step 2 of 9
Write the state description matrices in observer canonical form,
-o , 1 0 ••• 0
A .=
B .=
- a , 0 1 0 0
= 0 0 1 0 
■ 0 0 0 1 
- a , 0 ........... 0
(7)
(8)
C ,= [ l 0 0 ••• 0 ] ......(9)
£), =Seperate valuefiomnumeratoraiiddeiiominatorelse 0 (10
Consider the value of gain G {s)-
(11)
Write the state equation in observer canonical fonn from equations (1)to (11).
R ]-[:
• fo n r®i« h?+ « w ......[9 Oj [9 j
Write the output equation in observer canonical form from equations (1) to (11). 
y = C x*D 
-̂ = [1 0 ]* (13)
Step 3 of 9
Write the value from equations (5) and (12).
Write the value from equations (5) and (12).
f o l
■(15)
Write the C , value from equations (9) and (13). 
C ,= [ l 0] (16)
Hence, the value of A„ =
0
9 0 j| 9 |C .= [1 0]|
step 4 of 9
(b)
Write the general formula for controllability matrix. C ■ 
C = [ b , A .B. ... A - ’B ,] (17)
Calculate A.B^from equations (14) and (15).
* • “ • = [ 9 0.
......
Substitute equations (15) and (18) in equation (17).
^̂=[9 3
Determine the value of det|C(from equation (19). 
det|C|»-81
Hence, value of |det|C| = -^ I| - This value is not equal to zero. So this is controllable system. 
Step 5 of 9
(c)
Consider the law of the form. 
a = -[Ar, + r (20)
Find -f B ,K ) from Equations (14) and (15)
( . i - A . . B , K ) = , [ ;
( r f -A ,+ B ,K ) =
(r f -A .+ B ,K ) =
( , i - A . B K ) = [ _ , ; , ^ ^
s 0 
p s 
s 0 
0 s
0 I
9 0
0 r
9 0 . ]
0
0 0 
9K, 9K,
(21)
Find det(5l-A-l-BK)sO from equation (21). 
det(jI-A-fBK)=»*-i-9A:,s-9-F9A:,......(2 2 )
Consider the closed loop systems satisfy the det(5l-A *F B K ) = 0 condition. 
Find desired characteristics polynomial of the system 
The two poles are placed at -3+ 3y ’ and - 3 —3y-
(i-F3-3y)(s-F3-F3y)=i’ - f6 j-H 8 ...... (23)
Compare the coefficients of s from equations (22) and (23)
9 A j= 6
A , = | ...... (24)
Compare the coefficients of constant from equations (22) and (23) 
-9 + 9 A .-1 8
A, = 3 ...... (25)
The design of state-feedback gain matrix, from Equations (14). (25) are AT, = 3 and K - IA . - 3
Step 6 of 9
Write the MATLAB program to verify the value of K: 
A= [0 1 ;9 0]:
B=[0;9]:
C=[1 0]:
D=0;
Po=10;
Ts=1;
z=(-log(Po/100))/(sqrt(pi'‘2-^log(Po/100)'‘2)):
wn=4/(z*Ts);
[numt.dent]=ord2(wn.z): 
r=roots(dent): 
poles=[-3■^3*i -3-3*1];
K=acker(A,B,poles)
The output of MATLAB program:
K =
3.0000 0.6667
Step 7 of 9
(d)
Determine the system observability Q-
Write the observable condition from description matrices.
‘’ • [ c l ] "
Calculate C^A^from equations (14) and (16).
c A - ( i • ( ; ; ]
C .A .- [0 1] (27)
Substitute equations (16) and (27) in equation (26).
‘ ’ = [ 0 “]
Determine the determination ^m atrix from equation (28). 
det|0|=l
Hence, value of l«‘» t|o |= o | . So this is [obsCTvable system|.
(e)
Write the MATLAB program to verify the value of full order estimator L at —12±12y poles. 
A= [0 1 ;9 Oj:
B=[0;9]:
C=[1 Oj:
D=0;
Po=10;
Ts=1;
z=(-log(Po/100))/(sqrt(pi'‘2-^log(Po/100)'‘2)):
wn=10*wn;
[numt.dent]=ord2(wn.z): 
r=roots(dent): 
poles=[-12■^12*i-12-12*1];
L=acker(A',C'.poles)'
The output of MATLAB program:
L =
24
297
Thus the values of full order estimator at —12±12ypolesare =24|and \L^ s297 |.
Step 8 of 9
(f)
Consider the value of gain. ^ ( 5)-
. ’ (4 + 1)c(*)= (29)
i ' - 9
Write the state description matrices in observer canonical form from equations (4). (5). (7), and 
(29).
- [ : ;i
.(30)
Write the state description matrices in observer canonical form from equations (4). (5). (8) and 
(29).
B .= (31)
Write the C , value from equations (9) and (29).
C , = [ l 0 ] (32)
Calculate the value of A , —B^Kfrom equations (30) and (31).
• - • “ ■ K '- J a
Calculate the value of C , ( A . - B^K)from equations (32) and (33).
c . K - . u n - | . . 1̂ , - 3
c . ( a . - b . k ) = [ ^ a:, 1-9AT,] (34)
Modify observable condition in equation (26).
’ [ c . ( A . - B . K ) ]
...... (35)
Write the observability matric from equations (34) and (35).
o J ' “ 1 ...... (35)
[-9AT, 1-91CJ
Calculate det|0|from equation (35).
det|0| = I - 9 A : j ...... (36)
If detjOjis zero, the system is unobservable.
Consider the value of det|0| is zero.
i - 9 a: , « o
■(37)
Determine de t(5 l-A ^■ !‘ B ,K )from equations (30)and (31).
det(*I-A,+B,K) = det[.[j ®]-[® 
=det['
L"
5+9AT, - l+ 9 A j
L-9-f9AT, S-I-9J:,
= ( i ■ ^9^ :, ) ( * -I-9AT, ) - ( ^ + 9 A , ) ( - l + 9AT, )
= * ’ ■l■9A:, -̂^9 :̂, -̂t-SlA.^, -9+818T, ■l■9A:, -818T,8r,
d e t ( j I -A ,- i -B ,K )= j ’ -f9/:,»-f9A:,i-9-F81Ar,-F9A:,...... (38)
Substitute equation (37) in equation (38).
d e t ( s l- A ,+ B .K )= « '+9^5+98:,i - 9 + 8 l | + 9 i r ,
= j ' - i-j -i-98:,j - 9 - i-9-f9A:,
= « ’ +*-F9 ir,»+9X ,
d e t ( j I -A ,- fB .K )= ( j- i-9 A r , ) ( * -H ) ...... (39)
Step 9 of 9
From equations (39), the reason for the unobservable system is 
| - l v a lu eo f pole is cancelled by zero o f thegain G{s)\-