7. Control System Design by the Root-Locus Method

(1)

System Control

Professor Kyongsu Yi

Vehicle Dynamics and Control Laboratory Seoul National University

7. Control System Design by

the Root-Locus Method

(2)

Control system design

• Control System Design ; to perform specific tasks

• requirements ; performance specifications

- accuracy (steady state error)

- relative stability ( , damping ratio)

- speed of response ( , , : natural frequency)

• Given purpose  precise performance specifications  an optimal control system

• Controller ;

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

- hydraulic pneumatic

electronic (analog) mechanical

ζ

T e

⁻^ζωⁿ^t ωn

(3)

Design procedure

Yes

Given purpose for given system

Modeling

Design controller

Digital simulation

Evaluation

Prototype test / Gain tuning

END

1) Build up  test month to years Suspension

- spring, damper - tune the parameter

- test/evaluation (road tests)

2) Lab tests (simulator) week to months

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

48개월 개발기간 ; 80년대 초반

36개월 80년대 말

30개월 90년대 초반

24개월 2000년대

자동차

(4)

Preliminary Design Considerations

• Root – Locus approach ; - graphical method

- gain or parameter variations

• Effect of the addition of poles ; less stable

• Effect of the addition of zeros ; more stable

× σ jω

× × σ

jω

× ×

× σ jω

× ×

× σ jω

× ×



× σ jω

× ^ ×

× σ jω

× × ^

× σ jω

× ×



(5)

Lead/Lag Compensation

1

( ) 1

c c

s T G s K

s α T

= +

+

(6)

1 ( )

( ) 1

o

c i

E s K s T

E s s

α T

= +

+

1 1 2 2 4 1 3 2 c

T R C T R C K R C

R C α

=

(7)

1

( ) 1

c c

s T G s K

s αT

= +

+

When s= jω 1 1

( )

Gc j j j

T T

ω ω ω

α

   

=  + −  + 

   

  

Lead Compensator  Lag Compensator 

1 1

( ) 0 0 1 G jc

T T ω α

<α  > < <

1 1

( ) 0 1 G jc

T T ω α

>α  < <

σ jω

1

−T

×

1 αT

− σ 

jω

1

−T

×

1 αT

−



Lead Lag

(8)

Root Locus / Lead/Lag Compensator

• Root locus lead/lag compensator : powerful tool

when ; specification – time domain spec.

 • damping ratio • natural frequency

• dominant closed loop poles • rise time etc.

① Time domain spec

② unstable or

undesirable transient-response characteristics

 Lead Compensators ( relocate dominant closed loop poles)

(9)

0.5 2

2

n

Kv

ζ ω

 =

 =

 =



Example

Ex.

(

⁴

)

( ) 2

G s = s s +

2

( ) 4

( ) 2 4

C s

R s = s s

+ + , closed loop poles : _s= − ±₁ ₃_j

The static velocity error constant

0 2

0

( ) 1 ( )

1 ( )

1 1

( ) lim

1 ( )

1 1 1

lim 0.5

( ) 2

ss s

s v

E s R s

G s

e t s

G s s sG s K

→

 =

 +



 = +



= = = =



Desired : ^ωⁿ ⁼^4,^ζ ⁼^0.5 2 2 3

s j

⇒ = − ± σ

jω

2

×



×

3

current closed loop poles _{Root Locus}

Real Axis

Imaginary Axis

-1 -0.8 -0.6 -0.4 -0.2 0

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

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

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

(10)

Example

Desired : ^ωⁿ ⁼^4,^ζ ⁼^0.5 2 2 3

s j

⇒ = − ±

1) Angle of deficiency

φ

2 2 3

4 210

( 2) _s _j s s _{=− ±} = −

+ ^



1

30 , 30 , 0 1

1 s T K

s T

φ α

α

= = + = < <

+

  

 Lead Compensator 2) Choose such that ¹

αT 1 T

1 1 30 s T s αT

+ = +

 

가능한한

α

가크게  증가 (Good)

K

v

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

1 0.345 2.9

1 0.185 5.4

T αT

= =

α

(11)

Example

Pole-zero location for large

α

0 2

0

1 1

lim 1 ( ) ( )

1 1

lim ( ) ( )

ss s

c

s

c v

e s

G s G s s

sG s G s K

→

= +

= =

0

lim ( ) ( ) 1 lim 4

1 2

2

v c

s

K sG s G s s T

K s

s T K

α α

→

=

= +

+ +

= ⋅ ⋅

(12)

Example

Lead Compensator : pole-zero selection for large

P

α

σ jω

×

15^

4 D

−

P

15^

C B

2 2 φ

φ

1

−T 1

αT

−

: desired pole

2 2 3 j

− +

C D

: pole :

: zero : ¹

−T 1 αT

−

Lead Compensator

Fig 7-9 Ogata

(13)

Example

3) Magnitude Condition

2 2 3

2.9 4 ( ) ( )

5.4 ( 2)

1 ( ) ( ) 0

2.9 4

( ) ( ) 1

5.4 ( 2)

c c

c s j

c c

s j

G s G s K s

s s s 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

5.4 1 1 0.185 1

c c c

s s T Ts s

G s K K

s Ts s

s T

αα α

+ + + +

= = = =

+ + + +

 K_c =4.68

Analog controller

2 4 1 1

1 3 2 2

( ) 1

o i

E s R R R C s E s R R R C s

= +

+

1 2

3

10 10

C C F

R k

µ

= =

 = Ω Arbitrarily chosen 

Digital controller

1

1 1

1 2 2 2

( ) 0.185 1

, 0.185

c

u a

G s K

e s

u K e

u u u u u ae

u

= = +

+

 =

= +  + =

 >







(14)

Example

Root locus of the compensated system

× σ j ω

×  ×

− 5.4

− 2.9

− 2 0

2 2 3 j

• − +

2 2 3 j

• − −

( )

2.9 4

5.4 2

c

K s

s s s

+

+ +

 Comparison of step responses • original system

• compensated system

Root Locus

Real Axis

Imaginary Axis

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

-15 -10 -5 0 5 10 15

System: sysDG Gain: 1.08 Pole: -2 - 3.54i Damping: 0.492 Overshoot (%): 17 Frequency (rad/sec): 4.07 System: sysDG

Gain: 1.09 Pole: -2 + 3.58i Damping: 0.489 Overshoot (%): 17.2 Frequency (rad/sec): 4.1

(15)

Example

Root locus of the compensated system

( )

2.9 4

5.4 2

c

K s

s s s

+

+ +

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0 0.2 0.4 0.6 0.8 1 1.2 1.4

time [sec]

response

step response

uncompensated compensated

(16)

Example

4) Static velocity error constant ;

 가능한한 큰 가 증가에기여 , 가 lead Compensator로충분히증가하지못한 경우 Lag compensator ^v추가  Lead-Lag Compensator

K K_v

α

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 0.1984

2.52 2

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

→ → →

= = = =

+ + ⋅  ++  +

= =

×

5.04 ( 2)

v v

K = old K =

(17)

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

φ

3)

dc gain ;

; compensate angle deficiency 1

( ) 1

1 1

c c c

s T Ts

G s K K

s Ts T

αα α

+ +

= =

+ +

Kcα ,T

α

4) Tune to locate the closed – loop poles at the desired locations ; as large as possible

,T α α

−

5) Tune from the magnitude condition K_c

(18)

Lag Compensation

1

( ) ˆ , 1

c c 1

s T G s K

s T

β β

= + >

+

① satisfactory transient response

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

ii) open loop gain increase

 Lag compensator

β >1

Procedure

1) Draw the root-locus plot for the uncompensated system  locate the dominant closed loop poles

2)

calculate 1

( ) 1

c c

s T G s K

s βT

= +

+

c( )

G s

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 the static error constant is sufficiently large ② without altering the original root loci.

5) Draw a new root-locus plot

6) Adjust from the magnitude condition K^ˆ_c

1 1 1 s T

s βT +

+  Angle

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

( ) 50 G sc ≤



(19)

Lag Compensation

The static velocity gain ;

Uncompensated system

Compensated system

lim0 ( )

v s

K sG s

= →



0 0

ˆ_v lim _c( ) ( ) lim _c( ) _v ˆ_c _v

s s

K sG s G s G s K K β K

→ →

= = = ⋅ ⋅

Increased by a factor of ˆ_c 1

factor K

β =

≈

ˆc

K ⋅β

(20)

Example

Ex. ( )( )

( )( )

( ) 1.06

1 2

( ) 1.06

( ) 1 2 1.06

G s s s s

C s

R s s s s

= + +

= + + +

The dominant closed-loop poles ;

1

0.3307 0.5864 0.491

0.673 / sec 0.53sec

n v

s j

rad K

ζ ω

−

= − ±

=

Increase without change of the location of dominant poles  Lag compensator K_v →5sec⁻¹

1) ¹

( ) ˆ

c c 1

s T G s K

s βT

= +

+ ˆ ˆ

v c v

K =K βK

2) Let ^{10 ,} ¹ ^0.05, ¹ ^0.005

T T

β ⁼ ⁼ β ⁼

ˆ 0.05

( ) 0.005

c c

G s K s 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 ₌₋ _±

+ +

  

 No significant change

Root Locus

Real Axis

Imaginary Axis

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

-4 -3 -2 -1 0 1 2 3 4

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

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

(21)

Example

3) ^{G s G s}^c^{( ) ( )}⁼^K^ˆ^c ^s^s⁺⁺^0.005^0.05 ^⋅^{s s}

(

⁺^1.06¹

)(

^s⁺²

)

• dominant closed loop poles same damping ratio



s = − 0.31 ± j 0.55

4) Magnitude condition

( )( )

_0.31 _0.55

0.05 1.06

ˆ 1

0.005 1 2

c

s j

K s

s s s s ₌₋ _±

+ ⋅ =

+ + +

ˆ_c 0.9656

⇒ K =

× × ×

× 0

o

<

Root Locus

Real Axis

Imaginary Axis

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

-4 -3 -2 -1 0 1 2 3 4

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

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

(22)

Example

Verification ;

( )( )

0 2

0

1

,

1 1 1

ˆ lim 1 ( ) ( ) ˆ lim ( ) ( )

0.05 1.06 0.9656

0.005 1 2

5.12 sec

s c

v

v c

s

v old

s G s G s s K

K sG s G s s

s s s

βK

→

−

= +

=

= +

+ + +

= 

5)

s

₃

= − 2.326 s

₄

= − 0.0549

A long tail of small amplitude

(23)

Example

0 5 10 15 20 25 30 35 40 45 50

time [sec]

response

ramp response

0 5 10 15 20 25 30 35 40 45 50

0 0.2 0.4 0.6 0.8 1 1.2 1.4

time [sec]

response

step response

(24)

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

Lead Lag

s s

T T

G s K

s s

T T

γ β

γ

β

> >

 +  + 

  

=     +      +   

(25)

Lag-Lead Compensation

Spec. 1) dominant poles

2) (steady state velocity error constant) K_v Step 1. Lead Compensator

1) choose desired poles 2) angle condition

 determine the angle deficiency

3) magnitude condition

( ) ( )

1

( ) 180 2 1

c

s s

s T

K G s k

s T γ

=

 + 

 

  = +

 + 

 

 

 

s1

φ

( )

1

180 2 1 ( )

s s

s T

k G s

s T γ φ

=

+

= = + −

+

   (infinitely many ) ; choose one γ, T

1 1

1

( ) 1

c

s T

K G s

s T γ +

= +

 K_c

(26)

Step 2. Lag Compensator, spec

1)

2) choose such that i)

ii)

Kv

1 2

0 0 0

1 2

1 1

lim ( ) ( ) lim ( ) lim ( )

v c c 1 c

s s s

s s

T T

K sG s G s sK G s sK G s

s s

T T

β

γ γ

β

→ → →

 +  + 

  

= = =

 +  + 

  

  

 β

T2

Lag-Lead Compensation

2

1 1 1 s T s βT

+

+ 

2

1

5 0

1 s T s βT

+

− < <

+

  



(27)

Example

Ex

(

⁴

)

( ) 0.5

G s = s s +

The closed-loop poles ;

1

0.25 1.9843 0.125

2 / sec 8sec

n v

s j

rad K

ζ ω

−

= − ±

=

= Desired spec : ^0.5

5 80

n

Kv

ζ ω

=

 =

 =



 Lead – Lag compensator

1 2

1 1

( ) , 1, 1

c c

1

s s

T T

G s K

s s

T T

γ β

γ

β

 +  + 

  

= > >

 +  + 

  

  

(28)

Example

Step 1. Lead Compensator 1) choose desired poles 2) angle condition

 the angle deficiency = 55˚

choose , such that pole-zero cancelation happens

let

 angle condition

3) magnitude condition

1 2.50 4.33

s = − ± j

(

⁴^0.5

)

^s ^s¹ ²³⁵

s s = = −

+

 

( )

( ) ( ) 180 2 1

G sc + G s = − ^ k+

 

T1

1 1

1 0.5 2

s s T

+T = + ⇒ =

(

1

) (

1

)

10.04

0.5 0.5 55

s s

γ

=

+ + + = ^

 

( )

1

0.5 4

5.021 0.5 1

6.26

c

s s c

K s

s s s

K

=

+ =

+ +

⇒ =

(29)

Example

Step 2. Lag Compensator

1)

2) choose such that i)

ii)

( ) ( )

0 0

0

lim ( ) ( ) lim ( )

lim 6.26 4 4.988 80

10.04 0.5 16.04

v c c

s s

s

K sG s G s sK G s

s s s

β γ

β β

β

→ →

→

= =

= = =

+

⇒ =

T2

1

2

1 1 1

s s

s T s βT ₌

+

+ 

2

2 2

1

5 0

1

5 5

s T

T let T

β +

− < <

+

⇒ ≥ ⇒ =

 

1 1

5 ( ) 6.26 2

10.04 1

2 16.04 5

25.04( 0.2) ( ) ( )

( 5.02)( 0.01247)

c

s s G s

s s

G s G s s

s s s

 

 +   + 

 

=    

 

 +  +

   × 

= +

+ +

(30)

Example

Root locus of the compensated system

×

σ jω



×

Root Locus

Real Axis

Imaginary Axis

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

-0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25

System: sysDG Gain: 0 Pole: -5.02 Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 5.02

System: sysDG Gain: 0 Pole: -0.0125 Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 0.0125

System: sysDG Gain: 0 Pole: 0 Damping: -1 Overshoot (%): 0 Frequency (rad/sec): 0

(31)

Example

0 1 2 3 4 5 6 7 8 9 10

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

time [sec]

response

step response

0 1 2 3 4 5 6 7 8 9 10

time [sec]

response

ramp response

(32)

End of lecture note 7