《自动控制原理》课程教学资源（PPT课件讲稿）Module 6 Second-Order System Disturbance Rejection and Rate Feedback

Module 6 Second-Order System Disturbance Rejection and Rate feedback I hour

Second – Order System : Disturbance Rejection and Rate Feedback ( 1 hour ) Module 6

6. 1 Open-and Closed-loop Disturbance Rejection Open-loop control system with disturbance D D RR G1○G, D C=GG2R+G2D=CDesired +ACO Open When r=o, CoD.(ACOpen=g,Do D)

6.1 Open- and Closed-loop Disturbance Rejection • Open-loop control system with disturbance D: R R C D G1 G2 H C D C = G1 G2 R + G2 D = CDesired + COpen 0, . ( ) When R = C  D COpen = G2 D  D

Closed-loop control system with disturbance D RKC RE ○G1○1G H=1 H GG C R+-2D=C Desired +△C 1+gg 1+G,g Closed △C Closed △C 1+G,G Closed loop Open loop sired 1+GG2 →Kc·C+m=C

• Closed-loop control system with disturbance D: R KC D C G1 G2 H E * R Closed COpen G G C  +  = 1 1 2 1 D C Desired CClosed G G G R G G G G C = +  + + + = * 1 2 * 2 1 2 1 2 1 1 C D Closed loop Open loop 1 2 * * 1 G G C C Desired + = * * KC C Desired = C H=1

6.2 Effect of Velocity Feedback( Rate Feedback) 6. 2. 1 Second-Order System without Velocity Feedback 只AE K K K S(S+1) R S(TS+D)+KKH TS+S+KKH K The characteristic Equation is 11 TS+s+KK=o 2√TKKB 1 KK KK Or:s2+-s+-h=0 H T

6.2 Effect of Velocity Feedback (Rate Feedback) • 6.2.1 Second-Order System without Velocity Feedback E C s(Ts +1) K KH R H KKH Ts s K s Ts K K K R C + + = + +  = 2 ( 1) The characteristic Equation is: 0 2 Ts + s + KKH = or: 0 2 1 + + = T KK s T s H TKKH 1 2 1  = T KKH n =

The characteristic roots are 2√TKKB 4TKK ()K,Kn,7↑→54→ overshoot/kKy 2T 2T H disadvantage (2)K个→e(a) advantage 2T By changing K, we can only move the poles along the dashed line, i.e. cannot speed up the system by moving the poles to the left

2T 1 − The characteristic roots are: TKKH T T s 1 4 2 1 2 1 1,2 = −  − TKKH 1 2 1  = T KKH n = (1) K, KH , T     overshoot  –– disadvantage (2) K   e()  –– advantage

Fig. 6.10 Step responses for varying gain K

6.2.2 Second-Order System with Velocity Feedback R E、E1 K S(S+1 K K K K Open-loop Transfer E S(Ts+1)+KKS S[Ts+1+KK, Function Compare with the open-loop transfer function without velocity feedback K E S(TS+1)

• 6.2.2 Second-Order System with Velocity Feedback C s(Ts +1) R E K KH K sv E1 ( 1) [ 1 ] v Ts KKv s K s Ts KK s K E C + + = + + = Open-loop Transfer Function Compare with the open-loop transfer function without velocity feedback: ( +1) = s Ts K E C

We can find that 5,=(1+K)|→ overshoot where without velocity feedback with velocity feedback Ir The characteristic Equation Is TS +(+KKys+KKH=t Re KK no change

We can find that:  v = (1+ KKv ) where,  – without velocity feedback v  – with velocity feedback T KKH n = no change The characteristic Equation is: (1 ) 0 2 Ts + + KKv s + KKH = − n overshoot

6.2.3 Other forms of velocity feedback system R E K S(7S+1 K K R、EE1K R E K Ts+1 S(/s+1) K K +K H

• 6.2.3 Other forms of velocity feedback system C s(Ts +1) R E K KH K sv E1 R E C KH E1 Ts +1 K Kv s 1 C s(Ts +1) R E K K K s H + v

Sample problem:(P101 Fig 6.11) potentiometer Amplifier Motor Gearbox K Ko Js+c Tachometer K K Potentiometer Fig. 6. 11 Position control system with velocity feedback

Sample Problem: (P101 Fig.6.11)