Linear Control Systems Engineering

线性控制系统工程

Module 20

Phase Lag and Lead- Lag Compensation

(相位滞后和滞后 -超前校正 )

Module 20 Phase Lag and Lead- Lag Compensation

▽▽Transfer Function of Phase Lag ElementTransfer Function of Phase Lag Element

• Phase lag networks

-1/βτ-1/τ

2

212 ,

1

1

R

RRCR

s

ssG

v

vc

i

o

1R

Civ ov

2R

11

2

2

tantan

1

1

M

1/ 1/

90

β >1, the designing of phase lag compensation element include the specification of τ and β.

The maximum phase lag occurs at

Note: The phase part of Bode diagram are no consequence at all in the design of the compensator

The attenuation at high frequency

1

m

10log20M

1/

1/

90

Example:

(1) The velocity error for unit ramp is to be no more than 0.32% K > = 316

(2) Maximum percentage overshoot is to be no more than 20 %. k <= 6.3

Plot the Bode diagram for K=316

Phase Lag Compensation ProcessPhase Lag Compensation Process

ss

KsGH

2.01

5 /rad s

45PM

60

40

20

0

0

-90

-180

dbM

( / )rad s1 10 100

Fig.20.3 Uncompensated system with K=316.

The required PM = 45º occur at ω = 5 rad/s

1 Phase lag compensation: use a phase lag element to reduce

the uncompensated system gain to 0db at ω = 5 rad/s.

2 During phase lag compensating, it is assumed that no modification of composite phase

45º + 10% = 50º ω = 3 rad/s 3 New gain crossover frequency ωgc’=3 rad/s 4 At ωgc’= 3 rad/s, the magnitude must reduced from 40db to 0db

100 40log20 10

occurs. But a safety margin 10% is added to phase margin.

60

40

20

0

0

-90

-180

dbM

( / )rad s1 10 100

Fig.20.4 Determining the new gain crossover frequency.

50PM

gc

40db

Determine τ : the two break frequencies are at a lower frequency than ωgc’=3. This is determine by:

(1) The higher break frequency 1/ τ should at a frequency that the residual phase is no more than 10% for gain crossover frequency. Usually 1/ τ = ωgc’/10.

(2) The lower break frequency should not be too small to keep the bandwidth. Usually 1/ τ = ωgc’/10.

The compensator transfer function

The compensated open loop transfer fucntion

s

ssGc 3341

34.31

sss

ssGH

33412.01

34.31316

334

34.3

sradgc / 3.01.01

0.01 0.1 1.0 1060

40

20

0

0

-90

-180

-270

dbM

Compensated

magnitude

Compensated phase

40db

gc

50PM

Zmp

Fig.20.5 Bode diagram of compensated system

( / )rad s

1.8

2

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

00.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

00 1 2 3 4 5 6 7 8 9 10

0.2

0.4

0.6

0.8

1.0

1.2

Time (s) (a)

Time (s) (b)

Fig.20.6 Closed-loop step response of compensated system

compensated

Uncompensated

Am

plitu

deA

mpl

itude

1. Phase lag compensation provides a form of integral control (lag)

2. Phase lag compensation provide the necessary damping ratio to limit the overshoot.

3. The filter design process is simpler than phase lead because the zero and pole break frequencies is not too critical.

Comment on Phase Lag compensationComment on Phase Lag compensation (P433)(P433)

4. The phase lag compensation reduces the open- ( close-) loop bandwidths.

5. The settling time is also increased considerably.

6. Phase lag compensation may change the phase margin by more than 90º (Phase lag < 90º Fig. 19.7)

例 6-4

9.77c 17.2o

(0.1 )(0.2 1)

k

s s s

设计校正网络使图示系统30, 40 , dB 10dB, 2.3o

v ck h

30, 40 , dB 10dB, 2.3ov ck h

ω= 2.7 时φo(2.7)= –133o

3.7 1( )41 1csG ss

2.39c 45.1o

14.2dBh OK

Combined lead-lag compensation may allow more requirements to be met than by using either lead or lag compensation alone.

Lead- Lag CompensationLead- Lag Compensation

Lead-lag compensator may be constructed by connecting together a phase lead and a phase lag network in series. The value of α, β, τ, T are independent.

Transfer Function of Lead- Lag Transfer Function of Lead- Lag ElementElement

sTs

sTs

v

vsG

i

oc

11

11

1Riv ov2R

1C Buffer amplifier

2C3R

4R

Fig.20.7 Lead-lag from separate elements

The compensator can replaced by

But the value α, β, τ, T are not independent, so it is not applied.

It is assumed that α< β the break frequencies of lag portion are lower that that of lead portion.

1Riv ov

2R

1C

2C

3R

1

1

1

T1

T

10log

10log

1020log

Fig.20.9 Bode diagram of lead-lag filter

dbM

0

0

The features of Bode diagram1. The filter provides attenuation only and no

gain

(α< β )

2. The low frequency gain is 1 while high

frequency gain is

10log20M

3. The phase angle first lag, then leads, but high- and low- frequency phases are both zeros.

4. The maximum phase lag and phase lead occur between their respective filter break frequencies.

Problem: Consider the system with open loop transfer function

Design a phase lag compensation element that enforces the following performance specificities on the closed loop system:

(1) A damping ratio of = 0.5

(2) A velocity error of no more than 10%

Sample Problem 20.1

21 ss

KsGH

Solution: For (2)

Plot the Bode diagram for K=10: PM=0,

system is unstable.

For (1) , PM=50º. The PM required

The above PM occur at ω = 0.27, and the

corresponding magnitude is 31db.

101.0

11

ssv eKK

55%1050

5.35 31log20 10 dbM

dbM

0.001 0.01 0.1 1.0 10

-270

-180

-90

0

20

40

60

80

( / )rad s

10K

M=31db

55PM

0.27 gc

Fig.SP20.1.1

The high break frequency is set one decade below the new gain crossover frequency

ω = 0.27.

The compensated open loop transfer function

srad

srad

p

z

/ 00077.01

/ 027.01

s

s

s

ssGc 12961

371

1

1

sss

ssGHGc

129611

371102

0.001 0.01 0.1 1.0 10

( / )rad s

dbM

-270

-180

-90

0

20

40

60

80 A

compensated

compensated

E

gc

Fig.SP20.1.2

F

B

C D

HG

例 6-5 设未校正系统开环传递函数如下，试设计校正网络使： 1 ）在最大指令速度为 180/s 时， 位置滞后误差不超过 1o; 2) 相角裕度为 45o±3o; 3) 幅值裕度不低于 10dB; 4) 动态过程调节时间 ts 不超过 3秒。

( )( / 6 1)( / 2 1)

kG s

s s s

0dB

20

40

60

80

-20

-40

-60

-80

0.1 1 10 100

ω

[-20]

12.6c 55.5o

[-60]

取=45o,ts=2.7s, 由 (6-8) ～ (6-10) 求得

3.5

c

L0(3.5)=26.8dB

采用滞后超前校正

3.5

b取 =2 降阶ba =100, a=50

0.5s+1

0.01s+1=58.25o,

3.5∴ 可取

a =1

例 6-5图1

26.8

例 6-5图2

G(s) =180(s+1)

s(s/6+1)(50s+1)(0.01s+1)

3.29c 42.8o

h=27.7dB

ts=1.65s√

HomeworkHomework P461 20.4 P461 20.4