Linear Control Systems Engineering 线性控制系统工程

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 -1/τ -1/βτ ▽Transfer Function of Phase Lag Element • Phase lag networks

1/ 1/   90

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

Phase Lag Compensation Process • 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

60 40 20 0 0 -90 -180 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

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

60 40 20 0 0 -90 -180 Fig.20.4 Determining the new gain crossover frequency.

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

0.01 0.1 1.0 10 60 40db 40 Compensated magnitude 20 0 0 -90 Compensated phase -180 -270 Fig.20.5 Bode diagram of compensated system

2 1.8 1.6 1.4 1.2 Amplitude 1.0 0.8 0.6 Uncompensated 0.4 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time (s) (a) 1.2 1.0 Amplitude 0.8 0.6 compensated 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time (s) (b) Fig.20.6 Closed-loop step response of compensated system

Comment on Phase Lag compensation (P433) • 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.

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)

ω= 2.7时φo(2.7)= –133o 例6-4 设计校正网络使图示系统 OK

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

Transfer Function of Lead- Lag Element • 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. Buffer amplifier 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.

0 0 Fig.20.9 Bode diagram of lead-lag filter

The features of Bode diagram 1. The filter provides attenuation only and no gain (α< β ) 2. The low frequency gain is 1 while high frequency gain is

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.

Sample Problem 20.1 • 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%

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.

0.001 0.01 1.0 0.1 10 80 60 40 M=31db 20 0 -90 -180 -270 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

0.001 0.01 1.0 0.1 10 80 A 60 compensated 40 20 B 0 C D -90 G F H E -180 compensated -270 Fig.SP20.1.2

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。

例6-5图1 80 60 [-20] 40 0.5s+1 =58.25o, 20 0.01s+1 ω 3.5 0dB 0.1 1 10 3.5 100 -20 w w w a =1 [-60] =100, a b b ¢ ¢ w = -40 3.5 c -60 取 =45o,ts=2.7s, ∴可取 -80 取 =2 降阶采用滞后超前校正 26.8 a=50 由(6-8) ～(6-10)求得 L0(3.5)=26.8dB j0(3.5) = -180o

¢ ¢ 42.8o g = 嘿嘿ok! ¢ ¢ w = 3.29 c 180(s+1) G(s) = s(s/6+1)(50s+1)(0.01s+1) ¢ ¢ =27.7dB h 例6-5图2 √ ts=1.65s

Homework P461 20.4

Linear Control Systems Engineering 线性控制系统工程