Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc.

Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc. Electrical Engineering Department University of Indonesia

2 Steady State Error How well can a system track a standard input step, ramp, parabola based on final value theorem of Laplace Transforms Steady state value of ( ) ( ) et () is { } lim{ ()} e = lim e t = se s t s 0

Table 7.1 Test waveforms for evaluating steady-state errors of position control systems Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright 2004 by John Wiley & Sons. All rights reserved.

5 Steady State Error Consider a unity feedback system note that any closed loop system may be reduced to a unity gain system R(s) E(s) C(s) + - G(s)

6 Steady State Error R(s) + E(s) C(s) - G(s) Steady state error e Cs () Es () + Rs () = Rs () Rs () Gs () = 1 + Gs () { se s } Es () Gs () 1 = 1 = Rs () 1 + Gs () 1 + Gs () ss = lim ( ) s 0

7 Steady State Error R(s) + E(s) C(s) - G(s) E() s 1 = Es () = Rs () Rs () 1 + Gs () e = ss sr() s lim 0 1 + Gs () s 1 + 1 Gs ()

8 Error Constants Error constants give figure of merit for steady state error The bigger the error constant, the better the steady state error tracking of input signal defined in terms of response to standard inputs step (position), ramp(velocity), parabola(acceleration)

9 System types Systems classified into how many open loop poles lie at the origin e.g. type 0 open loop systems have no poles at the origin, type 1 systems have one pole at the origin, and so on

10 Static Position Error Constant The steady state error of a unity feedback system for a unit step input is 1 1 ess = s s + Gs s = 1 lim 0 1 () 1+ G() 0 Define static position error constant K = lim G( s) = G( 0) p s 0 { }

11 Static Position Error Constant 1 ess = 1 + K Note for a type 0 system K is finite, e is finite e ss p for a type 1 or higher system it is infinite is zero p If zero steady state error is required for a step input a first or higher order system is needed. ss

12 Static Velocity Error Constant Steady state response fo system to unit ramp input 1 1 ess = s s + Gs s = 1 lim lim 0 1 () 2 s 0 sg() s Define static velocity error constant K v = e = ss lim sg( s) s 0 { } 1 K v

13 Static Velocity Error Constant For a type 0 system K = 0 e is infinite Type 1 system K is finite e is finite v v Type 2 or higher system K is infinite e is zero v ss ss ss

Static Acceleration Error 14 Constant Steady state response fo system to unit parabola input e = lim s s 1 + Define static acceleration error constant K ss a = s 0 e = ss { 2 s G s } lim ( ) 1 K a 1 1 Gs s = 1 lim () s s G() s 0 3 0 2

Static Acceleration Error 15 Constant For a system types 0 &1 Type 2 system K is finite e is finite a K a is infinite Type 3 or higher system K is infinite e is zero v = 0 ss e ss ss

Table 7.2 Relationships between input, system type, static error constants, and steady-state errors Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright 2004 by John Wiley & Sons. All rights reserved.

Steady State Error - A 21 Summary SYSTEM TYPE Step Input Ramp Input Parabolic Input 1 0 1 + K 1 0 2 0 0 p 1 K v 1 K a

Figure 7.15 Forming an equivalent unity feedback system from a general nonunity feedback system Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright 2004 by John Wiley & Sons. All rights reserved.

29 Example 1 Taken from Ogata, p287 Consider liquid level control system. Determine the steady state effect of disturbance of size D 0 if proportional control is used and alternatively if integral control is used.

30 Example 1 Block Diagram D(s) G(s) X(s) E(s) H(s) + - G c (s) G(s) + + controller Disturbance

31 Example 1 R Gs () =, RCs + 1 Kp Proportional control Gc () s = K Integral control s D Disturbance Ds () = 0 s

32 Example 1 Taking variation in setpoint as zero X( s) = 0 KpR R Hs () = Es () + Ds () RCs + 1 RCs + 1 KpR R Es () = Hs () = Es () Ds () RCs + 1 RCs + 1

33 Example 1 R Es () = RCs + + K R Ds () 1 D Ds () = 0 s R Es () = RCs + 1 + p K R p D s 0

34 Example 1 RD = + + + RD Es () 0 1 0 1 1 K R 1 K R p p 1 + K s pr s RC RD e( ) = lim { se( s) } = 0 s 0 1 + K R p To get good steady state error need high value of K p

35 Example 1 if controller is integral G () s Rs Es () = RCs s KR Ds () 2 + + Rs e( ) = lim { se( s) } = lim s s 0 s 0 2 RCs + s + KR c = K s D s = 0 0 Integrator eliminates steady state error due to step disturbance

Homeworks Ogata Chapter Nise chapter 7: 4, 12,16, 35, 38, 45