8. Control System Design by the Root-Locus Method

(1)

System Control System Control

Professor Kyongsu Yi

Vehicle Dynamics and Control Laboratory Seoul National University

Seoul National University

(2)

Control system design

• Control System Design ; to perform specific tasks

• requirements ; performance specifications

- accuracy (steady state error)

- relative stability ( , damping ratio) 

 t

- speed of response ( , , : natural frequency)

• Given purpose  precise performance specifications

 ti l t l t

T e

^^ⁿ^t



n

 an optimal control system

• Controller ;

digital controller in many cases - digital controller , in many cases - microprocessor based controller - software; control algorithm

- hydraulic - hydraulic pneumatic

electronic (analog) mechanical

mechanical

(3)

Design procedure

1) Build up test

Suspension

- spring, damper - tune the parameterp

- test/evaluation (road tests) 2) Lab tests (simulator) week to months

3) Computer simulation (software) / test 검증(vehicle test/final tuning)

48개월 개발기간 ; 80년대 초반

36개월 80년대 말

자동차

30개월 90년대 초반

24개월 90년대 말

(4)

Preliminary Design Considerations

R t L h hi l th d

• Root – Locus approach ; - graphical method

- gain or parameter variations

• Effect of the addition of poles ; less stable

j



j



j

 



 

 

• Effect of the addition of zeros ; more stable

j j j

 

 

 

 



 

 ^ 

j ^j



 

  ^

  

_



(5)

Lead Compensation

• Fig 7 4

1 ( )

( ) 1

o

c

E s K s T E s

 

• Fig 7-4.

1 1 2 2

T R C T R C

R C





i( )

E s s

T

 ₄ ₁

3 2 c

K R C

 R C 1

( ) 0 1

s T

G s( )K ^ , 0  1

c c 1

G s K

s T



 



When s j 1 1

( )

G jc j j

T T

  



   

      

  

T T

   

Lead Compensator  Lag Compensator 

1 1

( ) 0 G jc

T T 

  

1 1

( ) 0 G jc

T T 

  

T T

j j

Lead Lag

1 

T



1

T



  1

T



1

 T



T T

T T



(6)

Root Locus / Lead/Lag Compensator

• Root loc s lead/lag compensator : po erf l tool

• Root locus lead/lag compensator : powerful tool when ; specification – time domain spec.

 • damping ratio

• natural frequency

• dominant closed loop poles

• rise time etc rise time etc.

① Time domain spec

② unstable or

undesirable transient-response characteristics undesirable transient response characteristics

 Lead Compensators ( relocate dominant closed loop poles)

(7)

Example

Ex 7 1 Ex. 7-1



⁴



( ) 2

G s  s s



( ) 4

C

0.5 2





 

 



2

( ) 4

( ) 2 4

C s

R s  s s

  , closed loop poles : s  1 3j

2

n

Kv



 

 The static velocity error constant

( ) 1 ( )

1 ( )

1 1

E s R s

G s

 

 

 current closed loop poles _{Root Locus}

0 2

0

1 1

( ) lim

1 ( )

1 1 1

lim 0.5

( ) 2

ss s

s v

e t s

G s s

sG s K



 

 



   



j



2 3 4 5

System: sysT Gain: 0 Pole: -1 + 1.73i Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/sec): 2

 v

 2

 

3

Imaginary Axis

-2 -1 0 1

System: sysT

D i d  ⁴ ^{0 5} ^-1 ^-0.8 ^{Real Axis}^-0.6 ^-0.4 ^-0.2 ⁰

-5 -4 -3

System: sysT Gain: 0 Pole: -1 - 1.73i Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/sec): 2

Desired : ⁿ ^4, ^0.5 2 2 3

s j

   

(8)

Example

Desired : ⁿ ^4, ^0.5 2 2 3

s j

   

1) Angle of deficiency



1) Angle of deficiency



2 2 3

4 210

( 2) _s _j s s _{ }  



 

1

30 , 30 , 0 1

1 s T K

s T

 



     



  

 Lead Compensator 2) Choose such that ¹ ¹

1 30 sT

 ^ ) 

T

T 30

s 1

T





가능한한



가 크게 증가(Good)

K

v

Text 방법 ploe = -5.4 zero = -2.9 =0.536 1 0.345

2.9 1 T  



1 0.185 T 5.4

  

(9)

Example

Pole-zero location for large



0 2

1 1

lim 1 ( ) ( )

1 1

ss s

c

e s

G s G s s

 



0

1 1

lim^s^ sG s G s_c( ) ( ) K_v

 

lim0 ( ) ( ) 1

v c

K s sG s G s

 

0

1 lim 4

1 2

2

s

s T

K s s

T K





 

 

  

(10)

Example

Lead Compensator : pole-zero selection for large



P

: desired pole

2 2 3 j

 

j P

2 



C

^{: pole :}

1 1

T







15^

4 D



15^

C B

2

1 1

 T

D

: zero : ¹

T

Lead Compensator

T

T Fig 7-9 Ogata

(11)

Example

3) Magnitude Condition

2.9 4 ( ) ( )

5.4 ( 2)

1 ( ) ( ) 0

c c

G s G s K s

s s s

G G

 

 

 _{2 2 3}

2 2 3

1 ( ) ( ) 0

2.9 4

( ) ( ) 1

5.4 ( 2)

c s j

c c

s j

G s G s G s G s K s

s s s

 

 

  

 

 Lead Compensator 1

2.9 1 0.345 1

( ) 4.68 2.51

1

s s T Ts s

G s  ^ K ^ K  ^  ^

 K_c 4.68

( ) 4.68 2.51

5.4 1 1 0.185 1

c c c

G s K K

s Ts s

s T



   

Analog controller

2 4 1 1

( ) 1

E so R R R C s₂ ₄ ₁ ₁ 

1 3 2 2

( )

( ) 1

o

E si  R R R C s



1 2

3

10 10

C C F

R k

  

   Arbitrarily chosen 

Digital controller ( ) 1

0.185 1

c

u a

G s K

e s

u K e

  



 ₁ ₁

1 2, 0.185 2 2

u K e

u u u u u ae

u

 

    

 





(12)

Example

Root locus of the compensated system

 

2.9 4

5.4 2

c

K s

s s s



 

Root Locus

j 

2 2 3 j

  

¹⁰

15

System: sysDG Gain: 1.09 Pole: -2 + 3.58i Damping: 0.489



  2.9  

nary Axis 0 5

p g

Overshoot (%): 17.2 Frequency (rad/sec): 4.1



 5.4  2 0

2 2 3 j

  

Imagin

-5 System: sysDG

Gain: 1.08 Pole: -2 - 3.54i Damping: 0.492

Real Axis

-6 -5 -4 -3 -2 -1 0

-15

-10 p g

Overshoot (%): 17 Frequency (rad/sec): 4.07

 Comparison of step responses

• original system

• compensated system

Real Axis

(13)

Example

Root locus of the compensated system

 

2.9 4

5.4 2

c

K s

s s s



 

1.4

step response

uncompensated t d

0.8 1 1.2

se

compensated

0.4 0.6 0.8

respons

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0 0.2

time [sec][ ]

(14)

Example

4) Static velocity error constant ;

 가능한한 큰 가 증가에 기여, 가lead Compensator로 충분히 증가하지 못한 경우Lag compensator 추가  Lead-Lag Compensator

Kv K_v



0 0 2 0

1 1 1 1 1

( ) lim ( ) lim lim

0.345 1 4

1 ( ) ( ) 1 ( ) ( ) ( ) ( )

2.51 0.185 1 ( 2) 1

ss s s s

c c c

e t s R s s

G s G s G s G s s sG s G s s

s s s s

  

   

      

1 0.1984 2.52 2

 



5 04 ( 2)

K

_v

5.04 ( ld K

_v

2)

K  old K 

(15)

The Procedure for design a Lead compensator

1) Performance Spec.

desired locations for the dominant closed-loop poles 2) Draw the root locus)

(1) Check whether or not the gain adjustment alone satisfactory results (2) If not calculate the angle deficiency



3)

1 ( ) 1

s T Ts

G K K

 

3)

dc gain ;

; compensate angle deficiency

( ) 1 1

c c T c

G s K K

s Ts T





 

 

Kc

T ; compensate angle deficiency ,T



4) Tune to locate the closed – loop poles at the desired locations

; as large as possible ,T





5) Tune from the magnitude conditionK_c

(16)

Lag Compensation

1

( ) ˆ , 1

c c 1

s T

G s K ^  

① satisfactory transient response

② unsatisfactory steady state characteristics i) closed loop poles : no change

 1

( ) 1

c c

s T





ii) open loop gain increase

 Lag compensator

 1

Procedure Procedure

1) Draw the root-locus plot for the uncompensated system

locate the dominant closed loop poles

2) ^s ¹

T )

calculate

( ) 1

c c T

G s K

s T





c( )

G s_c( )

3) Evaluate the particular static error constant specified ; The static velocity error constant

4) Determine pole and zero of the lag compensator s.t.

① th t ti t t i ffi i tl l ^s^ ¹ Angle ^^{G s}^c^{( )}^⁵⁰

① the static error constant is sufficiently large

② without altering the original root loci.

5) Draw a new root-locus plot

1 1 s T

s T



 

Angle

pole, zero close together and near the origin of the s-plane

c( )

6) Adjust from the magnitude conditionK^ˆ_c

(17)

Lag Compensation

The static velocity gain ;

Uncompensated system ^K^v ^^lim_s_₀^{sG s}^{( )}

Compensated system _

0 0

ˆ_v lim _c( ) ( ) lim _c( ) _v ˆ_c _v

s s

K sG s G s G s K K  K

 

    

Increased by a factor of ˆ_c 1

factor K

 



ˆc

K 

(18)

Example

Root Locus 4

Ex. 7-2

  

( ) 1.06

1 2

( ) 1.06

( ) 1 2 1 06

G s s s s

C s R

  

 s 1

2 3

System: sysT Gain: 0 Pole: -0.331 + 0.586i Damping: 0.491 Overshoot (%): 17 Frequency (rad/sec): 0.673

  

( ) 1 2 1.06

R s s s s 

The dominant closed-loop poles ; 0.3307 0.5864 0.491

s j



  



Imaginary Axi

-2 -1 0

System: sysT Gain: 0 Pole: -0.331 - 0.586i Damping: 0.491 Overshoot (%): 17

1

0.673 / sec 0.53sec

n v

rad K







Increase without change of the location of dominant poles K_v 5sec^¹ Lag compensator ^{Real Axis}

-6 -5 -4 -3 -2 -1 0 1 2

-4 -3

Overshoot (%): 17 Frequency (rad/sec): 0.673

1) ¹

( ) ˆ

c c 1

s T G s K

s T

 

 ˆ ˆ

v c v

K K K

2) Let ^{10 ,} ¹ ^0.05, ¹ ^0.005

T T

 ^ ^  ^

ˆ 0.05

( ) 0 005

c c

G s K s s

 

0.005 s

Angle contribution near a dominant closed loop poles

0 3307 0 5864

ˆ 0.05 4

0.005

c

s j

K s

s _ _



  

0.3307 0.5864

s j

 No significant change

(19)

Example

0 05 1 06

s

3) ^{G s G s}^c^{( ) ( )}^^K^ˆ^c ^s^s^^^0.005^0.05 ^^{s s}



^^1.06¹



^s^²



Root Locus

× × o ××

<

Root Locus

2 3 4

System: sysDG Gain: 1.01 Pole: -0 31 + 0 556i

× × ×

0 o ×

<

ginary Axis

0 1

2 Pole: 0.31 + 0.556i

Damping: 0.487 Overshoot (%): 17.4 Frequency (rad/sec): 0.636

• dominant closed loop poles same damping ratio



s   0 31  j 0 55

Imag

-2

-1 System: sysDG

Gain: 1.01 Pole: -0.31 - 0.556i Damping: 0.487 Overshoot (%): 17.4 Frequency (rad/sec): 0.636



s  0.31  j 0.55

4) Magnitude condition

0.05 1.06

ˆ s 1

K   ^{Real Axis}

-6 -5 -4 -3 -2 -1 0 1 2

-4 -3

Frequency (rad/sec): 0.636

  

_0.31 _0.55 ¹

0.005 1 2

c

s j

K s s s s _ _ 

  

ˆ

_c

0.9656



K



(20)

Example

Verification ;

0 2

1 1 1

ˆ

_v

lim

^s

s 1

_c

( ) ( ) G s G s s K

 ^



  

0

ˆ lim ( ) ( )

0.05 1.06 0.9656

0.005 1 2

v c

K

s

sG s G s s

s s s

 

 













1

5.12 sec

^

 K

_{v old},

 

5)

s

₃

  2.326 s

₄

  0.0549

A long tail of small amplitude

(21)

Example

50

ramp response

35 40 45 50

15 20 25 30

response

0 5 10 15 20 25 30 35 40 45 50

0 5 10

time [sec]

uncompensated compensated

1 1.2 1.4

step response

0.6 0.8 1

response

0 5 10 15 20 25 30 35 40 45 50

0 0.2 0.4

0 5 10 15 20 25 30 35 40 45 50

time [sec]

(22)

Lag-Lead Compensation

Lead compensator  speeds up, increase the stability

Lag compensator improves the steady state accuracy but reduce the speed

Lag-Lead compensators

Design procedure

1) Design lead compensator to relocate the dominant poles 2) Design lag compensator to improve the s.s. gain

  

1 2

1 1

( ) 1

c c

s s

T T

G s K

s  s

    

  

        

1 2

1 1

Lead Lag

s s

T T

 



 

 

  

      

(23)

Lag-Lead Compensation

Spec. 1) dominant poles

2) (steady state velocity error constant)K_v

Step 1. Lead Compensator

1) choose desired poles

2) l di i ^ ¹ ^

s1

2) angle condition

   

1

( ) 180 2 1

c

s s

s T

K G s k

s T





  

 

   

  

 

 

 

determine the angle deficiency



 

1

180 2 1 ( )

s T

k G s





  ^  

  (infinitely many  ^T ) ; choose one

3) magnitude condition

 

1

180 2 1 ( )

s s

k G s

s T

 



   



  (infinitely many ) ; choose one^, ^T

1 1

1

( ) 1

c

s T

K G s

s T











K

_c

(24)

Lag-Lead Compensation

Step 2. Lag Compensator, spec

1)

K

v

1 2

1 1

lim ( ) ( ) lim ( ) lim ( )

1

s s

T T

K sG s G s sK G s sK  G s

    

  

  

1)

0 0 0

1 2

lim ( ) ( ) lim ( ) lim ( )

v c c

1

c

s s s

K sG s G s sK G s sK G s

s s

T T

 



     

    

  

  





2) choose such that i)

T

2 2

1 1 1 s



T



ii)

2

s 1

 T



s 1



T

2

5 0

1 T s  T

  



  



(25)

Example

Ex. 7-3



⁴



( ) 0.5

G s  s s



The closed-loop poles ; 0.25 1.9843 0.125

2 /

s j

d



  



1

2 / sec 8sec

n v

rad K







 Desired spec : ^0.5

n 5





 

 



80

n

Kv

 



 Lead – Lag compensator

1 2

1 1

( ) , 1, 1

c c

1

s s

T T

G s K   

    

  

  

  

1 2

s s 1

T T



     

  

  

(26)

Example

Step 1. Lead Compensator

1) choose desired poles s₁  2.50  j4.33 2) angle condition



⁴^0.5



^s ^s¹ ²³⁵

s s   



 

the angle deficiency = 55˚

choose , such that pole-zero cancelation happens let

 

( ) ( ) 180 2 1

G sc  G s   ^ k

 

T1

1

1 0.5 2

s s T

T    

angle condition

1

T1

   

10.04

 

3) magnitude condition



s10.5







s10.5



55^

 

 

0.5 4

5 021 0 5 1

c

K s

 



1

5.021 0.5 6.26

c

s s c

s s s

K

  

 

(27)

Example

Step 2. Lag Compensator

1)

K

_

lim sG s G s ( ) ( )

_

lim sK  G s ( )

1)

   

0 0

0

lim ( ) ( ) lim ( )

lim 6.26 4 4.988 80

10.04 0.5

v c c

s s

s

K sG s G s sK G s

s s s



 

 



 

  



2) choose such that i)

16.04



 

T

2

)

1

2

1 1 1

s s

s T s  T

_





ii)

s s1

2

1

5 0

1 s T s



  



 

1 1

5 ( ) 6.26 2

10.04 1

c

s s G s

 

     

 

    

 

2

5

2

5

s T

T let T





   

2 16.04 5

25.04( 0.2) ( ) ( )

( 5.02)( 0.01247)

c

s s

G s G s s

s s s

 

   

    

 

 

( )( )

(28)

Example

Root locus of the compensated system

j











Root Locus Root Locus

0.1 0.15 0.2 0.25

System: sysDG Gain: 0 Pole: -5.02

D i 1

System: sysDG Gain: 0 Pole: -0.0125

D i 1

Imaginary Axis

-0.1 -0.05 0 0.05

Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 5.02

Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 0.012

System: sysDG Gain: 0 Pole: 0 Damping: -1 Overshoot (%): 0

Real Axis

-6 -5 -4 -3 -2 -1 0

-0.25 -0.2

-0.15 Frequency (rad/sec): 0

(29)

Example

step response

1.2 1.4 1.6 1.8

step response

0.6 0.8 1

response

0 1 2 3 4 5 6 7 8 9 10

0 0.2 0.4

time [sec]

8 9 10

ramp response

4 5 6 7 8

esponse

0 1 2 3

re 4

0 1 2 3 4 5 6 7 8 9 10

0

time [sec]