Chapter4 Time Domain Analysis of Control System Routh stability - - PDF document

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 1

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Chapter4 Time Domain Analysis of Control System

Routh stability criterion Steady state errors Transient response of the first-order system Transient response of the second-order system

• Time domain performance specifications
• The relationship between the performance specifications

and system parameters Transient response of higher-order systems

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Definition of characteristic equation ) ( ) ( ) (

1 1 1 1 1 1

s G a s a s a s a b s b s b s b s R s C

n n n n m m m m

= + + + + + + + + =

− − − −

L L The characteristic equation of the system is defined as

1 1 1

= + + + +

− −

a s a s a s a

n n n n

L

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 2

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Transfer function:

2

, 1

j k k k n n k

p j j σ ω ζ ω ω ζ − − ± =− ± −

poles of the closed loop

∏ ∏ ∏

= = =

− + + + + + = =

q j r k k k k k j m i i

j s j s p s z s k s R s C s

1 1 1

)] ( )][( ( [ ) ( ) ( ) ( ) ( ) ( G ω σ ω σ

zi : zeros of the closed loop

4 2 2 1 1

( ) ( 1 ) ( 1 )

q r j k k j k j k n n k n n k

a a s C s s s p s j s j α β ζ ω ω ζ ζω ω ζ

= =

+ = + + +     + + − + − −    

∑ ∑

) 1 sin( ) 1 sin 1 cos ( ) (

1 1 2 2 1 1 2 k q j r i k k t k t p j k k k q j r k k k k t t p j

t e D e a a t C t B e e a a t C

k k j k k j

ϕ ζ ω ζ ω ζ ω

ω ζ ω ζ

+ − + + = − + − + + =

∑ ∑ ∑ ∑

= = − − = = − −

s j s j s p s z s k s C

q j r k k k k k j m i i

1 )] ( )][( ( [ ) ( ) ( ) (

1 1 1

⋅ − + + + + + =

∏ ∏ ∏

= = =

ω σ ω σ

Step response:

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 3

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Consider that the characteristic equation of a LTI system

) (

1 1 1

= + + + + =

− −

a s a s a s a s F

n n n n

L

Where all the coefficients are real numbers. In order that there be no roots of the above equation with positive real parts, it is necessary but not sufficient that

• 1. All the coefficients of the polynomial have the same sign.
• 2. None of the coefficients vanishes.

4.1 Routh stability criterion

6 6 6 1 6 6 2 1 6 1 5 1 3 3 6 1 6 1 1 5 4 1 1 3 2 1 4 5 3 1 5 6 4 2 6

F E Fa s F E Ca ED s C A C a C A Ca E C AD BC s A a A D A a a Aa C A B a Aa s a a a a a B a a a a a A a a a a a s a a a s a a a a s × − = − = × − × = × − = − = × − × = − = − = × − = − = −

The Routh tabulation

) (

1 2 2 1 1

= + + + + + =

− − − n n n n n

a s a s a s a s a s F L

for n = 6

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 4

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The roots of the polynomial are all in the left half of the s-plane if all the elements of the first column of the Routh Array are

• f the same sign.

The necessary and sufficient condition for the stability of a system Rout-Hurwitz criterion If there are changes of signs in the elements of the first column, the number of sign changes indicates the number

• f roots with positive real parts.

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Example: The characteristic equation of a system is 50 10 4

2 3

= + + + s s s Determine the stability of the system using Routh criterion. Solution:

50 5 . 2 50 4 10 1

1 2 3

s s s s −

The system has two roots located in the right half of the s-plane. （2). Routh array is （1）. Check the necessary condition

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 5

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The first element in any one row of the Routh Array is zero but the other elements are not. We can replace the zero element in the Routh tabulation by an arbitrary small positive number ε and then proceed with the Routh test. Special case 1

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The characteristic equation of a system is Determine the stability of the system using Routh criterion. Example:

1 2 5 5

2 3 4 5

= + + + + + s s s s s

1 1 5 1 5 1 1 5 1 ) ( 1 5 1 2 5 1

2 1 2 3 4 5

s s s s s s − − − − ε ε ε ε ε ε

Solution: Routh array is There are two sign changes in the first column of the tabulation, the system has two roots located in the right half of the s-plane.

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 6

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The elements in one row of the Routh Array are all zero. Special case 2

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Example:

16 16

2 3

= + + + s s s

The characteristic equation of the system is: The Routh array is : The auxiliary equation is:

16 16 1 16 1

1 2 3

s s s s

16 ) (

2

= + = s s A

16 2 16 1 16 1

1 2 3

s s s s

The sign of the elements in the first column does not change, the system is stable .

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 7

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Example:

( 1) ( 1)(2 1) K s s Ts s + + + ( ) R s ( ) C s

+

−

Determine the value of K and T to make the closed loop be stable. Solution: the characteristic equation:

) 1 ( ) 2 ( 2

2 3

= + + + + + K s K s T Ts

2 2 ) 1 )( 2 ( 2 1 2

1 2 3

K s T TK K T s K T s K T s + − + + + + Routh array:

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, 2 , > > T K

2 ) 1 )( 2 ( > − + + TK K T

2 > +T

The condition for the stability is:

> T 2 2 − + < < T T K

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 8

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K T 2 2 4 4 6 6 8 2 2 T K T + < −

2 2 − + < < T T K

Stable

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Example:

2 10 8

2 3

= + + + s s s

The characteristic equation of the system is Determine the stability of the system. Analyze how many roots lie between the imaginary axis and the line .

1 − = s

Solution:

2 75 . 9 2 8 10 1

1 2 3

s s s s

the system is stable.

1

1 −

= s s

Let The characteristic equation becomes:

1 3 5

1 2 1 3 1

= − − + s s s

Routh array:

Im Re 1 −

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 9

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1 3 5

1 2 1 3 1

= − − + s s s

Routh array for the above equation:

1 8 . 2 1 5 3 1

1 1 1 2 1 3 1

− − − − s s s s

there is one root on the right side of the line s = -1.

Im Re 1 −

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4.6 Steady state errors ) ( ) ( ∞ − ∞ = ∞ r c e ） （

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 10

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( ) R s ( ) C s ( ) G s ( ) B s ( ) E s

+

−

( ) H s

) ( ) ( 1 ) ( ) ( s H s G s R s E + =

1 lim ( ) lim ( ) lim ( ) 1 ( ) ( )

s s t s s

e e t s E s s R s G s H s

→ ∞ → →

= = = +

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∏ ∏

+ = =

+ + = + + + + + + =

q v j j m i i r q m r

p s s z s K p s p s p s s z s z s z s K s G

1 1 2 1 2 1

) ( ) ( ) ( ) )( ( ) ( ) )( ( ) (

ν ν

L L

ν = 0, type 0 system ν = 1, type 1 system ν = 2, type 2 system the types of the control systems

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 11

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• a. position error constant

For a step input

) ( ) ( t u R t r ⋅ =

) ( ) ( 1 lim ) ( ) ( 1 / lim ) ( ) ( 1 ) ( lim ) ( lim ) ( lim s H s G R s H s G s R s s H s G s R s s sE t e e

s s s s t ss

+ = + = + = = =

→ → → → ∞ →

lim1 ( ) ( ) 1 lim ( ) ( )

ss s s

R R e G s H s G s H s

→ →

= = + + lim ( ) ( )

P s

G s H s K

→

=

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• b. velocity error constant

For a ramp input:

Rt t r = ) ( ) ( ) ( lim ) ( ) ( 1 / lim ) ( ) ( 1 ) ( lim ) ( lim ) ( lim

2

s H s sG R s H s G s R s s H s G s R s s sE t e e

s s s s t ss → → → → ∞ →

= + = + = = = lim ( ) ( )

v s

K sG s H s

→

=

v ss

K R e =

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 12

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• c. Acceleration error constant

for acceleration input

2

1 ( ) 2 r t Rt =

) ( ) ( lim ) ( ) ( 1 / lim ) ( ) ( 1 ) ( lim ) ( lim ) ( lim

2 3

s H s G s R s H s G s R s s H s G s R s s sE t e e

s s s s t ss → → → → ∞ →

= + = + = = =

) ( ) ( lim

2

s H s G s K

s a →

=

a ss

K R e =

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Steady state error

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 13

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Example:

+

−

( ) R s ( ) C s 10 ( 5) s s +

Determine the steady state errors, when the input is

) (t u

t

2

2 1 t

, and . Solution: ∞ = + = =

→ →

) 5 ( 10 lim ) ( ) ( lim s s s H s G K

s s p

2 ) 5 ( 10 lim ) ( ) ( lim = + = =

→ →

s s s s H s sG K

s s v

) 5 ( 10 lim ) ( ) ( lim

2 2

= + = =

→ →

s s s s H s G s K

s s a 26

2 1 1 = =

v ssv

K e

1 1 = + =

P ssp

K e

∞ = =

a ssa

K e 1

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 14

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4.1 The transient response of the first-order system 4.1.1 The math model

Ts s R s C + = 1 1 ) ( ) (

( ) R s +

−

( ) C s Ts 1 28

4.1.2 the step response r(t) = u(t) , R(s)=1/s

1 1 ( ) ( ) ( ) ( 1) 1 1 C s G s R s Ts s T s Ts = = ⋅ + = − +

T t

e t c

−

− =1 ) (

t ≥ 0

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 15

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1 0.632 ( ) c t t T 2T 3T 4T 5T 63.2% 86.5% 95% 98.2% 99.3%

slope

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Impulse response:

Ts s R Ts s C + =

• +

= 1 1 ) ( 1 1 ) (

T t

e T t c t g

−

= = 1 ) ( ) (

( ) R s +

−

( ) C s Ts 1

T 2T 3T 4T t ( ) c t 1 T

/

1 ( )

t T

c t e T

−

=

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 16

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Ramp response ：

T s T s T s s Ts s R s G s C / 1 1 1 1 1 ) ( ) ( ) (

2 2

+ + − = ⋅ + = =

( ) R s +

−

( ) C s Ts 1

T t

Te T t t c

/

) (

−

+ − =

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) (t r ) (t c t ) (t r ) (t c T ess =

T t

Te T t t c

/

) (

−

+ − = T t Te T t t r t c

T t t t

− = − + − = −

− ∞ → ∞ →

] [ lim )] ( ) ( [ lim

/

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 17

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4.2 the transient response of the second-order system 4.2.1 the mathematical model

( 1) K s Ts + ( ) R s ( ) C s

+ −

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( 1) K s Ts + ( ) R s ( ) C s

+ −

T K s T 1 s T K ) s ( R ) s ( C ) s (

2

+ + = = Φ

let T K

n = 2

ω

T

n

1 2 = ςω

2 n n 2 2 n

s 2 s ) s ( ω ζω ω Φ + + =

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 18

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The characteristic equation :

s 2 s

2 n n 2

= + + ω ζω

The roots of the characteristic equation are

1 s

2 n n 2 , 1

− ± − = ζ ω ζω

36 d

j s ω σ ± − =

2 , 1

1 s

2 n n 2 , 1

− ± − = ζ ω ζω

• 1. underdamped

) 1 ( < < ζ

If input is a unit step u(t):

2 2 2 2 2 2 2

) ( ) ( 1 1 ) 2 ( ) (

d n n d n n n n n

s s s s s s s s C ω ζω ζω ω ζω ζω ω ζω ω + + − + + + − = ⋅ + + =

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 19

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) t sin 1 t (cos e 1 ) t ( c

d 2 d t n

ω ζ − ζ + ω − =

ζω −

) sin cos 1 ( 1 1 1

2 2

t t e

d d t

n

ω ζ ω ζ ζ

ζω

+ − − − =

−

) t sin( 1 e 1

d 2 t n

θ ω ζ

ζω

+ − − =

−

ζ = θ arccos

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1

2

1 1 1 ζ + −

2

1 1 1 ζ − −

2

1 1

nt

e ζω ζ

−

+ − 1

n

T ζω = T 2T 3T 4T t ( ) c t

2

1 1

nt

e ζω ζ

−

− −

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 20

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1 s

2 n n 2 , 1

− ± − = ζ ω ζω

n

s ω − =

2 , 1 n n n n n n

s s s s s s s c ω ω ω ω ω ω + − + − = ⋅ + + = 1 ) ( 1 1 ) 2 ( ) (

2 2 2 2

) 1 ( 1 ) ( t e t c

n t

n

ω

ζω

+ − =

−

( ) c t ( ) r t 1 t ( ) c t

• 2. Critically damped response

1 = ζ

u(t)

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1 > ζ

1

2 2 , 1

− ± − = ζ ω ζω

n n

s 1 ] 1 1 ( 2 [ 1 )] 1 1 ( 2 [ 1 ) (

2 1 2 2 2 1 2 2

− + + − − + + − − + − − − + =

− −

ζ ω ζω ζ ζ ζ ζ ω ζω ζ ζ ζ

n n n n

s s s s c

) T e T e ( 1 2 1 1 ) t ( c

2 t 2 T 1 t 1 T 2

− − − − + = ζ

n n

T T ω ζ ζ ω ζ ζ ) 1 ( ) 1 (

2 2 2 1

− − − = − + − =

• 3. Over damped case

Let

1 1 2 2 2 1 2 2

] 1 1 ( 2 [ )] 1 1 ( 2 [ 1 ) ( T s T s s s c − − − + + − − − − + =

− −

ζ ζ ζ ζ ζ ζ

( ) r t t ( ) c t

c(t)

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 21

41 n

j s ω ± =

2 , 1

1 s

2 n n 2 , 1

− ± − = ζ ω ζω t cos 1 ) t ( c

n

ω − = ) t ( ≥

( ) c t 1 2 t

) ( = ζ

• 4. undamped case

2 2 2 2 2

1 1 ) ( ) (

n n n

s s s s s s c ω ω ω + − = ⋅ + =

) ( = ζ

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• 1. Rise time tr
• 2. The peak time tp

4.3 Time domain performance specifications

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 22

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• 3. Percentage overshoot σp

% 100 ) ( c ) ( c ) t ( c

P P

× ∞ ∞ − = σ

• 4. Steady-state error ess

( ) ( ) 100% ( )

ss

c r e r ∞ − ∞ = × ∞

44

• 5. Settling time ts

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 23

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4.4 The relationship between the performance specifications and system parameters

• 1. Rise time-system parameter

1 ) ( =

r

t c

According to definition:

1 ) sin( 1 1 ) (

2

= + − − =

−

θ ω ζ

ω ζ r d t r

t e t c

r n

π θ ω = +

r dt 2

1 ζ ω θ π ω θ π − − = − =

n d r

t ) 1 ( < < ζ

) sin( 1 1 ) (

2

θ ω ζ

ζω

+ − − =

−

t e t c

d t

n

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• 2. Peak time -system parameter

2

1 ζ ω π ω π − = =

n d P

t

( ) c t

r

t

p

t

s

t 1 0.1 0.9

p

σ t

2

2 2 2 1

P d n

t π π ω ω ζ = = −

(±0.05 or±0.02 ) × r (∞)

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 24

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• 3. Percentage overshoot - system parameter

) sin cos 1 ( 1 1 1 ) (

2 2

t t e t c

d d t

n

ω ζ ω ζ ζ

ζω

+ − − − =

−

d P

t ω π =

2

1 2

( ) 1 (cos sin ) 1

P

c t e

ζ π ζ

ζ π π ζ

− −

= − + −

2 2

1 1

n n n

P d n

t π π π ω ω ω ω ω ξ ξ = = = − −

2 2

1 ( ) 1 ( 1 cos sin ) 1

n p

t p d p d p

c t e t t

ζ ω

ζ ω ζ ω ζ

−

= − − + −

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2 2

1 1 2

( ) 1 (cos sin ) 1 1

P

c t e e

ζπ ζπ ζ ζ

ζ π π ζ

− − − −

= − + = + − ( ) ( ) ( ) 1 100% 100% ( ) 1

P P P

c t c c t c σ − ∞ − = × × ∞ ＝

2

1 P

e

ζπ ζ

σ

− −

=

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 25

49 2

1 ζ ζπ

σ

− −

= e

p

σp is only related to ζ.

0.5 1.0 1.5

2 2 2

( ) 2

n n n

s s s ω ζω ω Φ = + +

p

σ 20 40 60 80 100 % ζ

p

σ

Percentage overshoot 50

• 4. The settling time ts- - system parameter

) ( ) ( ) ( ∞ ⋅ ∆ = ∞ − c c t c

s

0.02 0.05

• r

∆ = ±

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 26

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1

2

1 1 1 ζ + −

2

1 1 1 ζ − −

2

1 1

nt

e ζω ζ

−

+ − 1

n

T ζω = T 2T 3T 4T t ( ) c t

2

1 1

nt

e ζω ζ

−

− −

∆ ζ

ζω

= −

− 2 s t n

1 e

2 n s

1 1 ln 1 t ζ ∆ ζω − =

2

1

n s

t

e ζω ζ

−

= ∆ −

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9 . < < ζ

n s

4 t ζω =

n s

3 t ζω =

02 . = ∆ 05 . = ∆

approximately: for for

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 27

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Example:

( ) R s ( ) C s

+

−

( 1) K s s + 1 K s +

The desired specifications are:

2 0 % , 1

P P

t s σ = =

What should the value of K and K0 be? Determine the value of ts and tr . Solution: The transfer function of the closed loop is K s KK s K s KK K s s K s R s C + + = + + + = ) 1 ( ) 1 ( ) ( ) (

2

＋

1 2 , KK K

n n

+ = = ζω ω

54

(1)

2 .

2

1

=

− − ζ ζπ

e 456 . = ζ

(2)

1 1

2 ＝

ζ ω π ω π − = =

n d P

t

ωn

s rad

n

/ 53 . 3 = ω

5 . 12 K

2 n =

= ω

n

KK ζω 2 1

0 =

+ 178 .

0 =

K

K s KK s K s KK K s s K s R s C + + = + + + = ) 1 ( ) 1 ( ) ( ) (

2

＋

ζ

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 28

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(3) tr

s 65 . 1 t

2 n r

= − − = ζ ω θ π

(4) ts

s t

n s

86 . 1 3 = = ζω

rad 1 . 1 1 tg

2 1

= − =

−

ζ ζ θ

for ∆=5%

56

s j s j s p s z s k s C

q j r k k k k k j m i i

1 )] ( )][( ( [ ) ( ) ( ) (

1 1 1

⋅ − + + + + + =

∏ ∏ ∏

= = =

ω σ ω σ

( )

( )

2 2 2 1 1

( ) 1

q r k k j j k j k k k k

s a a C s s s p s α β ζ ω ω ζ

= =

+ = + + + + + −

∑ ∑

• 1. Step response of a higher order system

4.5 Transient response of higher-order systems

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 29

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( ) c t ) (t r 1

t t

( ) c t

1 ) (t r ( ) c t t 1 ) (t r

2 2 1 1 2 1 1

( ) ( cos 1 sin 1 ) sin( 1 )

j k k j k k

q r p t t j k k k k k k j k q r p t t j k k k k j i

C t a a e e B t C t a a e D e t

ζ ω ζ ω

ω ζ ω ζ ω ζ φ

− − = = − − = =

= + + − + − = + + − +

∑ ∑ ∑ ∑

58

• 3. The dominant poles of higher-order systems

2 2 1 1 2 1 1

( ) ( cos 1 sin 1 ) sin( 1 )

j k k j k k

q r p t t j k k k k k k j k q r p t t j k k k k j i

C t a a e e B t C t a a e D e t

ζ ω ζ ω

ω ζ ω ζ ω ζ φ

− − = = − − = =

= + + − + − = + + − +

∑ ∑ ∑ ∑

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Automatic Control System 30

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σ j

[ ]

s 主导极点

Re[ ] 1 Re[ ] 5 poles of closed loop dominant poles <

there is no zeros of the closed loop near the dominant poles dominant closed loop poles