Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

Transcription

1 Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter θ D dc amplifier Control Transformer Prof. Dr. Y. Samim Ünlüsoy Prof. Dr. Y. Samim Ünlüsoy 1

2 COURSE OUTLINE I. INTRODUCTION & BASIC CONCEPTS II. MODELING DYNAMIC SYSTEMS III. CONTROL SYSTEM COMPONENTS IV. STABILITY V. TRANSIENT RESPONSE VI. STEADY STATE RESPONSE CH VI VII. DISTURBANCE REJECTION VIII. BASIC CONTROL ACTIONS & CONTROLLERS IX. FREQUENCY RESPONSE ANALYSIS X. SENSITIVITY ANALYSIS XI. ROOT LOCUS ANALYSIS Prof. Dr. Y. Samim Ünlüsoy 2

3 STEADY STATE RESPONSE - OBJECTIVES In this chapter : Open loop transfer functions of feedback control systems will be classified. Steady state error of feedback control systems due to step, ramp, and parabolic inputs will be investigated. Selection of controller parameters for a specified steady state error and the measures to be taken to reduce steady state error will be examined. Prof. Dr. Y. Samim Ünlüsoy 3

4 STEADY STATE RESPONSE Steady State Response is the response of a system as time goes to infinity. It is particularly important since it provides an indication of the accuracy of a control system when its output is compared with the desired d input. If they do not agree exactly, then a steady state error exists. Prof. Dr. Y. Samim Ünlüsoy 4

5 STEADY STATE RESPONSE The Steady State Response of a system is judged by the steady state error due to step, ramp, and parabolic (acceleration) inputs. These inputs may be associated with the ability of a control system to : Position itself relative to a stationary target, t Follow a target moving at constant speed, and Track an object that is accelerating. Prof. Dr. Y. Samim Ünlüsoy 5

6 STEADY STATE RESPONSE Note that for the steady state response to exist, the system must be stable. Therefore before going into steady state analysis it would be good practise to check the stability of the system. Prof. Dr. Y. Samim Ünlüsoy 6

7 STEADY STATE RESPONSE The steady state response and error can be obtained by using the final value theorem. The final value theorem : The final value of a time signal can be found from the Laplace transform of the signal in the s-domain. lim y(t)=lim sy(s) t s o Prof. Dr. Y. Samim Ünlüsoy 7

8 STEADY STATE RESPONSE The final value theorem : In the case of the output of a system C(s)=G(s)R(s) )R( ) lim c(t)=lim sg(s)r(s) t s 0 Prof. Dr. Y. Samim Ünlüsoy 8

9 OPEN LOOP TRANSFER FUNCTION The open loop transfer function of a general feedback control system is given by : B(s) = R(s) G(s)H(s) R(s) + E a (s) C(s) G(s) - Remove B(s) H(s) Prof. Dr. Y. Samim Ünlüsoy 9

10 OPEN LOOP TRANSFER FUNCTION The open loop transfer function of a unity feedback control system is given by : B(s) = R(s) G(s) R(s) + E(s) C(s) G(s) Remove - B(s) Prof. Dr. Y. Samim Ünlüsoy 10

11 STEADY STATE ERROR Nise Ch , 7.5, Dorf & Bishop Section 4.5, 5.7 The Laplace transform of actuating error, (s), for a general closed loop system : E a R(s) + E a (s) C(s) G(s) - B(s) H(s) E (s)=r(s)-h(s)c(s) a =R(s)-H(s)G(s)E (s) a 1 E(s)= a R(s) 1+G(s)H(s) Prof. Dr. Y. Samim Ünlüsoy 11

12 R(s) E(s)= a 1+G(s)H(s) STEADY STATE ERROR H(s)=1 R(s) E(s)= 1+G(s) R(s) + E(s) G(s) C(s) - Note that the actuating and actual errors will be identical only for a unity feedback system, i.e. with an ideal sensor. For a non-unity feedback system, the actual error may not be zero when the actuating error is zero. Prof. Dr. Y. Samim Ünlüsoy 12

13 STEADY STATE RESPONSE Here, with a systematic approach, it will be shown that the steady state error for a unity feedback system due to a certain type of input depends on the type of its open loop transfer function. For a non-unity feedback system, the analysis is somewhat more involved* and will not be covered here. * See Nise, Section 7.6, and Kuo, pp Prof. Dr. Y. Samim Ünlüsoy 13

14 SYSTEM TYPE Nise 7.3, Dorf & Bishop 5.7 The open loop transfer function of a unity feedback control system can be written in the general form : K ( 1+T s) p P p=1 T(s)= M Q 2 N s s s ( 1+τms) 1+2ξ q + 2 m=1 q=1 ωn q ω nq A unity feedback control system with this open loop transfer function is called a type N system. s N K : N poles at the origin (free integrators), : open loop gain. Prof. Dr. Y. Samim Ünlüsoy 14

15 SYSTEM TYPE EXAMPLE 1a R(s) s 2 s ( s+3 ) s 3 ( s+2 ) _ C(s) G(s)= 2 s s+1 s+1 ( s+3) s ( s+2) s( s+2)( s+3) = N=1 : type 1 system Prof. Dr. Y. Samim Ünlüsoy 15

16 SYSTEM TYPE EXAMPLE 1b R(s) + _ s 2 s+1 2 ( s+3 ) s 3 ( s+2) C(s) 1 ( s+1 ) s+1 G(s)= = s( 2) s+1 ( 3) s+1 s s+1 s Open loop gain : 1/18 Prof. Dr. Y. Samim Ünlüsoy 16

17 E(s)= R(s) 1+G(s) STEADY STATE ERROR Using the final value theorem : e =lim e(t) ss t =lim sr(s) lims 0sE(s)=lim s 0 1+G(s) Note that t the steady state t error depends on the input and the open loop transfer function of the system. Prof. Dr. Y. Samim Ünlüsoy 17

18 STEADY STATE ERROR Step Input Step Input : r(t)=r R(s)= R s R s s e ss =lims 0 1+G(s) R = 1+lim G(s) s 0 Prof. Dr. Y. Samim Ünlüsoy 18

19 R e ss = 1+lims 0 G(s) STEADY STATE ERROR Define the position error constant Then K =lim s s 0 G(s) e = ss R 1+K s Prof. Dr. Y. Samim Ünlüsoy 19

20 K s =G(0) R e ss = 1+K s STATE ( 1 )( 2 ) ( m ) ( τ )( τ ) ( τ ) K T s +1 T s T s +1 G(s)= N s 1s+1 2s+1... ns+1 STEADY STATE ERROR K s = K for type 0 systems for type 1 or higher systems e = ss R 1+K 0 for type 0 systems for type 1 or higher systems Prof. Dr. Y. Samim Ünlüsoy 20

21 STEADY STATE ERROR Ramp Input Ramp Input : r(t)=rt { } L t n! = s n n+1 R R(s)= s 2 R s 2 s e ss =lims 0 1+G(s) =lim s 0 s 0 R s+sg(s) R = lim sg(s) Prof. Dr. Y. Samim Ünlüsoy 21

22 R e ss = lims 0 sg(s) STEADY STATE ERROR Define the velocity error constant Then K =lim v s 0 sg(s) e = ss R K v Prof. Dr. Y. Samim Ünlüsoy 22

23 K v =lims 0sG(s) ss R ( 1 )( 2 ) ( m ) ( τ )( τ ) ( τ ) K T s +1 T s T s +1 G(s)= N s 1s+1 2s+1... ns+1 e = K STEADY v STATE ERROR v 0 K = K for type 0 systems for type 1 systems for type 2 or higher systems R for type 0 systems e = ss for type 1 systems K for type 2 or higher systems K 0 Prof. Dr. Y. Samim Ünlüsoy 23

24 STEADY STATE ERROR Acceleration Input Parabolic (acceleration) Input : R 2 R r(t)= t s 3 2 s e ss =lim s 0 1+G(s) { } n! L t n = s n+1 R =lims 0 s 2 +s 2 G(s) R(s)= R 3 s R = lim s 2 G(s) s 0 Prof. Dr. Y. Samim Ünlüsoy 24

25 R e = ss 2 lim s 0s G(s) STEADY STATE ERROR Define the acceleration error constant K =lim 2 a s 0s G(s) Then e ss R = K a Prof. Dr. Y. Samim Ünlüsoy 25

26 2 K a =lims 0s G(s) ss R ( 1 )( 2 ) ( m ) ( τ )( τ ) ( τ ) K T s +1 T s T s +1 G(s)= N s 1s+1 2s+1... ns+1 e = K STEADY a STATE ERROR a 0 K = K for type 0 and 1 systems for type 2 systems for type 3 or higher systems R for type 0 and 1 systems e = ss for type 2 systems K for type 3 or higher systems K 0 Prof. Dr. Y. Samim Ünlüsoy 26

27 STEADY STATE ERROR - Summary System Type Step Input Ramp Input Parabolic Input R 0 1+K 1 0 R K R K Prof. Dr. Y. Samim Ünlüsoy 27

28 STEADY STATE ERROR Multiple Inputs For linear systems, the steady state error for the simultaneous application of two or more inputs will be the superposition p of the steady state errors due to each input applied separately. Prof. Dr. Y. Samim Ünlüsoy 28

29 STEADY STATE ERROR For example, for an input of 3t r(t)=1+2t the steady state error is given as the superposition of the steady responses to each of the inputs e ss = K K K s v a Prof. Dr. Y. Samim Ünlüsoy 29

30 STEADY STATE ERROR Therefore the steady state error of the system subjected to the composite input will be : e ss = Ks Kv Ka = + + = 1+K 0 0 for type 0systems = K 0 = for type1systems = + + = 1+ K K for type 2 systems = + + = 0 1+ for type 3 or higher systems Prof. Dr. Y. Samim Ünlüsoy 30

31 STEADY STATE ERROR - Observations It is observed that : i) When e ss is finite, increasing the open loop gain decreases the steady state error. Step : R 1+K Ramp : Acceleration : R K K : open loop gain Prof. Dr. Y. Samim Ünlüsoy 31

32 STEADY STATE ERROR - Observations ii) As the type of the system is increased, e ss decreases. Therefore one may attempt to improve the steady state response by including an integrator in the controller. This, however, may cause stability problems which become critical for type 3 or higher systems. Prof. Dr. Y. Samim Ünlüsoy 32

33 STEADY STATE ERROR - Observations iii) The approach to the minimization of the steady state error in control systems may thus be : Determine the maximum possible open loop gain and check the maximum allowable open loop gain that will not result in instability. Choose the smaller of the two open loop gain values. Prof. Dr. Y. Samim Ünlüsoy 33

34 STEADY STATE ERROR Example 1a Determine the steady state error for r(t)=2+3t R(s) + _ Amplifier gain 10 ss ( ) C(s) G(s)= = = s(s+5) s(5)(0.2s+1) s(0.2s+1) N=1 : type 1 system, K=200 Prof. Dr. Y. Samim Ünlüsoy 34

35 STEADY STATE ERROR Example 1b G(s) = 200 s(0.2s+1) N=1 : type 1 system, K=200 e ss = 0 for r(t)=2 (step input) R 3 e ss = = =0.015 for r(t)= 3t (ramp input) K 200 Hence e = =1.5 % ss It is obvious that if you increase the value of the open loop gain K, say by increasing amplifier gain, steady state error will decrease. Prof. Dr. Y. Samim Ünlüsoy 35

36 STEADY STATE ERROR Example 1c Let us check the maximum value of the amplifier gain without causing instability. R(s) + _ Amplifier gain K a 10 ss+5 ( ) C(s) 10K a ( ) C(s) ss+5 = = R(s) 10Ka 1+ ss+5 ( ) 10K a 2 s +5s+10Ka Prof. Dr. Y. Samim Ünlüsoy 36

37 STEADY STATE ERROR Example 1d Thus the characteristic polynomial is given by : 2 D(s)= s +5s+10K a It is obvious that it passes Hurwitz test. Application of Routh s stability criterion will result in only one condition on stability : s K a s s 0 10K a K > 0 a Prof. Dr. Y. Samim Ünlüsoy 37

38 STEADY STATE ERROR Example 1e As an alternative solution, let us calculate the steady state error using the final value theorem C(s) s(s+5) 1000 = = R(s) 1000 s(s+5) s(s+5) r(t)=2+3t 2 3 R(s)= + s s E(s)=R(s)-C(s)=R(s)- R(s) s(s+5)+1000 s(s+5) s(s +5) 2 3 E(s)= R(s)= s +5s+1000 s +5s s s Prof. Dr. Y. Samim Ünlüsoy 38

39 STEADY STATE ERROR Example 1f Thus with the error expression in Laplace domain : s+5 2s+3 E(s)= 2 s +5s+1000 s e =lim se(s) ss s 0 2s+3 s+5 =lim s s 0 s s 2 +5s = = =1.5 % 1000 It is obvious that this method requires considerably higher effort. Prof. Dr. Y. Samim Ünlüsoy 39

40 STEADY STATE ERROR Example 2a Determine the value of amplifier gain K a such that t the steady state t error for a ramp input r(t)=3t is going to be at most 2 %. R(s) 300 K a + ss+5 ( )( s+2 ) _ Amplifier gain C(s) 300Ka 300Ka 30Ka s(s+2) s+5 s 2 (5)(0.5s+1) 5s+1) 0.2s+1 s(0.5s+1)(0.2s+1) G(s)= = ( ) ( ) ( ) = N=1 : type 1 system, K=30K a Prof. Dr. Y. Samim Ünlüsoy 40

41 STEADY STATE ERROR Example 2b G(s) = 30K a s(0.5s+1)(0.2s+1) N=1 : type 1 system, K=3K ; a R=3 R 3 ess = = 0.02 for r(t)= 3t (ramp input) K 30K 3 Hence Ka = a ( ) Now, the limit on K a with respect to stability should be checked. Prof. Dr. Y. Samim Ünlüsoy 41

42 STEADY STATE ERROR Example 2c Let us check the maximum value of the amplifier gain without causing instability. R(s) + _ K a Amplifier gain 30 ss+5 s+2 ( )( ) C(s) 30K a ( ) C(s) s(s+2) s+5 = = R(s) 30Ka 1+ s(s+2) s+5 ( ) 30K a 3 2 s +7s +10s+30Ka Prof. Dr. Y. Samim Ünlüsoy 42

43 STEADY STATE ERROR Example 2d Thus the characteristic polynomial is given by : 3 2 D(s)= s +7s +10s +30K a It is obvious that it passes Hurwitz test. Application of Routh s stability criterion will result in : s s 2 a > 7 30K a 70-30K a 7 s 1 0 s 0 30K a Ka > 7 0 K 0 K 70 a < = Prof. Dr. Y. Samim Ünlüsoy 43

44 STEADY STATE ERROR Example 2e It is obvious that the value of K required a for the specified steady state error will make the system unstable. Therefore, the minimum possible steady state error will be higher than the desired value. Can you determine the minimum possible steady state t error for this system? Prof. Dr. Y. Samim Ünlüsoy 44