The Performance of Feedback Control System

2017年10月24日 1

The Performance of Feedback Control System

• Introduction

• Performance of a Second-order System

• The S-Plane Root Location and The Transient Response

• The Concept of Stability

• The Routh-Hurwitz Stability Criterion

– Case1:No element in the first column is zero.

– Case 2: Zeros in the first column while some other elements of the row containing a zero in the first column are nonzero.

– Case 3: Zeros in the first column, and the other element of the row containing the zero are also zero.

– Case 4: Repeated roots of the characteristic equation on the imaginary axis.

Introduction

2017年10月24日 3

Second-order system parameters

system gain: DC gain

s s s s

s K

M K M M

s K M s B

s M T gain

DC 1

1 1

lim )

( lim

0 2

0  







  

If the transfer function is stable, the DC gain as follows:

Damping ratio,

Characteristic equation

Second-order standard form:

1 ) ( :

2 ,

) (

) ) (

( _{2 2}

2 



 

 DC gain T s

s s

s R

s s Y

T _s

n n

n

s  



0 2

)

(

F

 

The roots of ₁

2

4 ) 2 ( , 2

- : )

( ²

2 2

2

1       



      



 ^{n n n} _{n n}

s F

Case I: for ,the roots are negative real (over-damping) ^{ 1}

2 2 1

1 2

1 2

) )(

) (

(    



 

 



 

s A s

A s

s s

T_s ⁿ

If (unit step function)

s s

R 1

)

( 





t y

s A s

A s

s s

s s Y

2 1

2 1

1

2 2 1

1 2

1 2 2

1 ) ( )

(

1 , )

)(

( ) 1

(



















   



 

 

 



 







 

t

1 S-plane

Re Im

1 2 





2017年10月24日 5

Case II: for ,they are double roots, (critical-damping) ^{ 1} 







S-plane

Re Im





t

n y t Ae t te

s s s R

s

T ^  ^



 _{     }

  1, ( ) 1 )

( ) ,

) (

( ₂

2

t

1

1 1 

 

Case III: for , the roots are complex conjugate pair (under-damping) 0  1

2 2

1,    1 

    

 _n j _n

S-plane

Re Im

1

1 1 

 

1 0 

j 





Definition: damping factor:

delay (oscillation ) frequency:

n

 

1 2



_d  _n 



 





 

 2

1 2

2

tan 1 ),

1 sin(

1 1

)

( 





 



 e

t

t

y

nt 



12

1

2017年10月24日 7

) (t y

(sec) t

ymax

tp t_s

td

1

0 1 . 0

5 . 0

9 . 0

t2

t1

Overshoot

t_r=t₂-t₁

% 5

% 2 or

 

Typical unit-step response of a second-order control system 1 2

1 



 e^ ^nt 1 2

1 



 e^ ^nt

d

% 100

% 100 .

1 1

1

1

1

0 1

sin

t 0

0 1

sin 1

) (

2 2

1 max /

1 / max

2 2

2 2

2 2





 









 





 























 



















































 



y e y O y

P e y

M t

t t n n

t

n t t

e

t dt e

t dy

fin fin p

n p

p n

n n

t

n n t

n

n

Percent over-shoot:

2017年10月24日 9

Settling Time:

02 . 0 for

, 4 )

1 02 . 0 1 ln(

02 . 0 1

1

2 2











 



ss n

n s

t

e t

e ^{n s}

 







Factor n

Damping :  

2 2

1 1

1 1

tan cos 1





















 _

 



 



 







 ^{ }

n n n

d d

tr

Case IV: for (zero-damping)  ⁰

2 2

2 2

1, , ( )

n n

n T s s

j 

 



     



y ( t )  1  cos 

t

t

1

CaseV: for (negative-damping) _{1 0}

2



 





 

 2

1 2

2

tan 1 ),

1 sin(

1 1

)

( 





 



 e

t

t

y

nt

 0



S-plane

Re Im

2017年10月24日 11 0>>-1

0>>-1 0<<1

0<<1

=-1

=0

=1

>1 <-1

=0

 -

 

_n is held constant while the damping ratio  is varied

2 1

2 ,

1    

 s _n _n 

  >1 : Overdamped

s  n

 _1,2

  =1 : Critically damped

2 2

,

1     1

 s _n j _n

 0< <1 : Underdamped

j n

s   

 _1,2

  =0: Undamped

2 2

,

1     1

 s _n j _n

  <0 : Negatively damped

1 S - 平面

0

 j

1

0 t

c t( )

1

0 t

c t( )

1 S - 平面

 j

0

1

0 t

c t( )

0  1 S - 平面

 j

0

1

0 t

c t( )

0 S - 平面

 j

0

1

0 t

c t( )

  1  0 S - 平面

 j

0

1

0 t

c t( )

 1 S - 平面

 j

0

2017年10月24日 13





)

(s p

s k

 )

(s

E C(s)

) (s

5

, 100 ) ,

( ) (

2

 



  k p

k ps s

k s

R s C

% 44

% 100 .

, 25 . 2 0

, 5 10

100     ^{1 2}  

 ^{ }



 

 PO e

n n

sec / 6825 . 9 1

, 3245 . 0 1

sec, 6 .

4 1 ₂

2 rad

t

t _{d n}

n p

n

s    

 



   









sec 648 . 1 0 ,

2 :

Re   

T f f mark _d 

T

0.3245 1.6 1.44

1

sec

 j

1 

n ⁿ²

3

n 1

2

3 n n

n  

  

constant _n



 j

0



1

2

3 1 2

3  

  

constant 



 j

2 ₃

1

0

0

3

1 2















 j

1

d^d2 1 

2 d

d 

 

2017年10月24日 15

The S-plane Root Location and the Transient Response

























1











n d

S-plane A

B

C

Second-order step response

System A

2017年10月24日 17

ζ P.O 0.1

0.2 0.3 0.4 0.5 0.6 0.7 0.8





) (s G

) (s H ) (s E )

(s

R C(s)

2 2.4

t(msec) t1

2 t1=40 ,k=? P=?

2 2.4

t(msec) T

3 T=40 ,k=? P=?

2 2.4

t(msec) ts

1 P.o=?

Home Work1

2017年10月24日 19

The Concept of Stability

 Stability  Absolute stability

 Relative stability

 The methods for the determination of stability of linear continuous-time systems:

 Routh-Hurwitz criterion

 Nyquist criterion

 Bode diagram

 j



S-plane

stable

stable

unstable

unstable

A stable system is a dynamic system with a bounded response to a bounded input

S-plane

2017年10月24日 21

2017年10月24日 23

 Bounded-input bounded-output (BIBO) stability:

System

) (t

u y(t)

)

(t h(t)



















d h t

u t

y d

h t

u t

y

d h t

u t

y

) ( ) (

) ( )

( ) (

) (

) ( ) (

) (

0 0

0

































M t

u( ) 

If u(t) is bounded, ^ y⁽t^{) } M







 N t

y )( Thus, if y(t) is to be bounded, or

























Q d

h

N d

h M









0 0

) (

) (

number positive

finite are

Q and N M,

 The Routh-Hurwitz Stability Criterion

Consider the characteristic equation of a linear SISO system

0 )

(   ^{1 1}  ¹  ⁰ 

n n

n

n n n

a s a a s a

a s a

s

q 

ime ree at a t s taken th

f the root products o

e o at a tim s taken tw

f the root products o

all roots

a a

a a

a a

n n

n n

n n

3 2 1











  









0 ...

) 1 ( ....

...) (

...) (

) ...

(

0 ) )...(

)(

( )

(

2 1 3

4 2 1 3 2 1

2 3

2 2 1 1

2 1

2 1















































r r a

s r

r r r r r a

s r

r r r a s

r r

r a s

a

r s r

s r s a s

q

n n

n n

n n

n n n

n n

n n

2017年10月24日 25

b1

c1

. . . .

The necessary condition to guarantee that all roots of q(s)=0 with negative real part are:

 All the coefficients of the equation have the the same sign.

 None of the coefficients vanishes.

Necessary but not sufficient!

0 ...

)

(s a s a _1s ^1 a₁sa₀ 

q _n ⁿ _n ⁿ

0 3 2

7 5

3 1

1

6 4

2

. .

....

...

s s s s

a a

a a

s

a a

a a

s

n n

n n

n n

n

n n

n n

n





















1 3 2

1 1







 



n n n n n

a a a a b a

b2

1 5 4

1 2







 



n n n n n

a a a a b a

b3

1 2 1 3 1

1 b

b a a

c b ^n  ^n

The Routh-Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array

Case1:No element in the first column is zero Second-order system

0 )

(s a₂s² a₁sa₀  q

0 1 2

s s s

0

1

0 2

a

a a

0 1

2 0 1 1

0 a a

a a

b a  



The system is stable if all the coefficients have the same sign.

Third-order system

0 )

(s  a₃s³a₂s² a₁sa₀  q

0 1 2 3

s s s s

0 2

1 3

a a

a a

0

2 0 3 1 2

1 a

a a a

b a 

 a0

set

The system is stable if

0

~ ₀

3 a 

a

0 3 1

2a a a

a 

2017年10月24日 27

Case 2: Zeros in the first column while some other elements of the row containing a zero in the first column are nonzero.

0 10 11

4 2

2 )

(s  s⁵  s⁴  s³  s²  s  q

0 1 2 3 4 5

s s s s s s

10 0 2 1

1 1

d c

0 0 10

6 4 2

0 0 0 0 10

11



) (

) 10 11

4 2

2 1 (

10 11

4 2

2 )

(

5 5

1 2

3 4

1 5

1

5 4

3 2

1 1

1 1

1 1

1

x F

x x

x x

x s

F

x x

x x

x x

s _x x





























 







 

  

 12 12 4

c1

10 6 6

1 1

1   

c

d c 

 0



Case 3: Zeros in the first column, and the other element of the row containing the zero are also zero.

This condition occurs when the polynomial contains singularities that are symmetrically located about the origin of the s-plane.

S-plane S-plane S-plane

2017年10月24日 29

0 8 4

2 )

(

0 1 2 3

0 0

8 2

4 1

s s s s

j2 s

0 4 s

0 8 2

) (

2 2















 s s

A

8 2s

8 2s

4s s

8 4 2

s 4 s

2 s

) 1 (

2 2 3

2 3

2

















s s

0 ) 2 )(

2 )(

2 (

)

(

q

The system has not any pole on the RHP, but system is marginally stable.

) 2

( ) 2 (

ds s s dA 

8

0

8 2

4 1

0 1 2 3

s s s s

2

0 0 0 s

0 1 1

0 4 4

1 2 1

1 2 1

1 2 3 4 5

s s s

s s s

ds s

dA( )  4 ₃ 4

1 2s s

1 2

s

s 2s

1 2

2 s

1 2

1

2 4

2 4

3 5

2 3

4 5 2

4





























s s

s s s

s s

s

s

.

0 0 0

1 2 1

1 2 1

0 1 2

2 )

(

2 3 4 5

2 3 4 5

s s s s

s s s s

s s

q       

j j j j s

s s

s s A



















, , ,

0 ) 1 ( 1 2 )

( ^{4 2 2 2}

(2)

(1)

j j j j s

s s

s s A



















, , ,

0 ) 1 ( 1 2 )

( ^{4 2 2 2}

The system has not any root on the RHP, but the system is unstable.

Repeated roots on the imaginary axis

Case 4: Repeated roots of the characteristic equation on the imaginary axis.

2017年10月24日 31

0 4

2 )

(

k s

s k

k s

s

0 2

8

2

4 1

0 1 2 3



0

) 2 (

8

) 1 (



 k k





8

0

Ans:0< k<8, system is stable.

0 4 )

2 (

3 )

(s  s³  ks²  k  s  q

Ans: k>0.528, system is stable.

2017年10月24日 33

ka s s

ka b

s

k s

ka s

c

0 ) 10 (

8

17

1

0 1 1

1 2 3 4



0 )

10 (

17 8

)

( s  s

 s

 s

 k  s  ka  q

0 64 ) 126 )(

10 (

0 126

8 ) 10 (

and 8 , 126

1 1

1 1















 

 

ka k

k ka k

b

ka k

c b b k

2017年10月24日 35