SECTION 7: STEADY-STATE ERROR. ESE 499 Feedback Control Systems
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1 SECTION 7: STEADY-STATE ERROR ESE 499 Feedback Control Systems
2 2 Introduction
3 Steady-State Error Introduction 3 Consider a simple unity-feedback system The error is the difference between the reference and the output EE ss = RR ss YY ss The input to the controller, DD ss Consider a case where: Reference input is a step Plant has no poles at the origin finite DC gain Controller is a simple gain block In steady state, the forward path reduces to a constant gain:
4 Steady-State Error Introduction 4 In steady state, we d like: Output to be equal to the input: yy ssss = rr ssss Zero steady-state error: ee ssss = 0 Is that the case here? ee ssss = rr ssss yy ssss = rr ssss ee ssss KK ee ssss = rr ssss + KK No, if rr ssss 0, then ee ssss 0 Non-zero steady-state error to a step input for finite steady-state forward-path gain Finite DC gain implies no poles at the origin in DD ss or GG ss
5 Steady-State Error Introduction 5 Now, allow a single pole at the origin An integrator in the forward path Now the error is For a step input EE ss = RR ss EE ss KK ss ss EE ss = RR SS ss + KK EE ss = ss ss ss + KK = ss + KK Applying the final value theorem gives the steady-state error ss ee ssss = lim ssss ss = lim ss + KK = 0 Zero steady-state error to a step input when there is an integrator in the forward path
6 Steady-State Error Introduction 6 Next, consider a ramp input to the same system Now the error is rr tt = tt tt and RR ss = ss 2 EE ss = ss 2 The steady-state error is ss ss + KK = ss ss + KK ee ssss = lim ssss ss = lim ss ss ss + KK = KK Non-zero, but finite, steady-state error to a ramp input when there is an integrator in the forward path
7 Steady-State Error Introduction 7 Two key observations from the preceding example involving unity-feedback systems: Steady-state error is related to the number of integrators in the open-loop transfer function Steady-state error is related to the type of input We ll now explore both of these observations more thoroughly First, we ll introduce the concept of system type
8 8 System Type and Steady-State Error
9 System Type 9 System Type The degree of the input polynomial for which the steadystate error is a finite, non-zero constant Type 0: finite, non-zero error to a step input Type : finite, non-zero error to a ramp input Type 2: finite, non-zero error to a parabolic input For the remainder of this sub-section, and the one that follows, we ll consider only the special case of unity-feedback systems
10 System Type Unity-Feedback Systems 0 For unity-feedback systems, system type is determined by the number of integrators in the forward path Type 0: no integrators in the open-loop TF, e.g.: GG ss = ss + 4 ss + 6 ss 2 + 4ss + 8 Type : one integrator in the open-loop TF, e.g.: GG ss = 5 ss ss 2 + 3ss + 2 Type 2: two integrators in the open-loop TF, e.g.: ss + 5 GG ss = ss 2 ss + 3 ss + 7
11 Types of Inputs When characterizing a control system s error performance we focus on three main inputs: Step Ramp Parabola We will derive expressions for the steady-state error due to each Step: rr tt = tt RR ss = ss For a positioning system, this represents a constant position
12 Types of Inputs 2 Ramp: rr tt = tt tt RR ss = ss 2 For a positioning system, this represents a constant velocity Parabola: rr tt = 2 tt2 tt RR ss = ss 3 For a positioning system, this represents a constant acceleration
13 Steady-State Error Unity-Feedback 3 For unity-feedback systems steady-state error can be expressed in terms of the open-loop transfer function, GG ss EE ss = RR ss YY ss = RR ss EE ss GG ss EE ss = RR ss + GG ss Steady-state error is found by applying the final value theorem ee ssss = lim ssss ss = lim ssrr ss + GG ss We ll now consider this expression for each of the three inputs of interest
14 Steady-State Error Step Input 4 For a step input rr tt = tt RR ss = ss Steady-state error to a step input is ee ssss = lim ssss ss = lim ee ssss = lim + GG ss ss ss + GG ss ee ssss = + lim GG ss
15 Steady-State Error Step Input 5 ee ssss = + lim GG ss In order to have ee ssss = 0, as we d like, we must have lim GG ss = That is, the DC gain of the open-loop system must be infinite If GG ss has the following form then GG ss = ss + zz ss + zz 2 ss + pp ss + pp 2 lim GG ss = zz zz 2 pp pp 2 and we ll have non-zero steady-state error
16 Steady-State Error Step Input 6 However, consider GG ss of the following form where nn GG ss = ss + zz ss + zz 2 ss nn ss + pp ss + pp 2 That is, GG ss includes nn integrators It is a type nn system lim GG ss = and ee ssss = 0 A type or greater system will exhibit zero steadystate error to a step input
17 Steady-State Error Ramp Input 7 For a ramp input rr tt = tt tt RR ss = ss 2 Steady-state error to a ramp input is ee ssss = lim ssss ss = lim ee ssss = lim ss + ssgg ss ss ss 2 + GG ss ee ssss = lim ssss ss
18 Steady-State Error Ramp Input 8 ee ssss = lim ssgg ss In order to have ee ssss = 0, the following must be true lim ssss ss = If there are no integrators in the forward path, then and lim ssss ss = lim ss ss + zz ss + zz 2 ss + pp ss + pp 2 = 0 ee ssss = A type 0 system has infinite steady-state error to a ramp input
19 Steady-State Error Ramp Input 9 If there is a single integrator in the forward path, i.e. a type system GG ss = ss + zz ss + zz 2 ss ss + pp ss + pp 2 then and lim ssss ss = zz zz 2 pp pp 2 ee ssss = pp pp 2 zz zz 2 A type system has non-zero, but finite, steadystate error to a ramp input
20 Steady-State Error Ramp Input 20 If there are two or more integrators in the forward path, i.e. a type 2 or greater system then and GG ss = ss+zz ss+zz 2 ss nn ss+pp ss+pp 2, nn 2 lim ssss ss = lim ss ss + zz ss + zz 2 ss nn ss + pp ss + pp 2 = ee ssss = 0 A type 2 or greater system has zero steady-state error to a ramp input
21 Steady-State Error Parabolic Input 2 For a Parabolic input rr tt = tt2 2 tt RR ss = ss 3 Steady-state error to a parabolic input is ee ssss = lim ssss ss = lim ee ssss = lim ss 2 + ss 2 GG ss ss ss 3 + GG ss ee ssss = lim ss2 GG ss
22 Steady-State Error Parabolic Input 22 ee ssss = lim ss2 GG ss In order to have ee ssss = 0, the following must be true lim ss2 GG ss = If there are no integrators in the forward path, then and lim ss2 GG ss = lim ss 2 ss + zz ss + zz 2 ss + pp ss + pp 2 = 0 ee ssss = A type 0 system has infinite steady-state error to a parabolic input
23 Steady-State Error Parabolic Input 23 If there is a single integrator in the forward path, i.e. a type system GG ss = ss + zz ss + zz 2 ss ss + pp ss + pp 2 then and lim ss2 GG ss = lim ss zz zz 2 pp pp 2 = 0 ee ssss = A type system has infinite steady-state error to a parabolic input
24 Steady-State Error Parabolic Input 24 If there are two integrators in the forward path, i.e. a type 2 system then and GG ss = ss + zz ss + zz 2 ss 2 ss + pp ss + pp 2 ss + zz ss + zz 2 lim ss2 GG ss = lim ss + pp ss + pp 2 = zz zz 2 pp pp 2 ee ssss = pp pp 2 zz zz 2 A type 2 system has non-zero, but finite, steady-state error to a parabolic input
25 Steady-State Error Parabolic Input 25 If there are three or more integrators in the forward path, i.e. a type 3 or greater system then and GG ss = ss+zz ss+zz 2 ss nn ss+pp ss+pp 2, nn 3 lim ss2 GG ss = lim ss 2 ss + zz ss + zz 2 ss nn ss + pp ss + pp 2 = ee ssss = 0 A type 3 or greater system has zero steady-state error to a parabolic input
26 26 Static Error Constants
27 Static Error Constants Unity-Feedback 27 We ve seen that the steady-state error to each of the inputs considered is Step: ee ssss = Ramp: ee ssss = Parabola: ee ssss = +lim GG ss lim ssgg ss lim ss2 GG ss The limit term in each expression is the static error constant associated with that particular input: Position constant: Velocity constant: Acceleration constant: KK pp = lim GG ss KK vv = lim ssss ss KK aa = lim ss 2 GG ss
28 Steady-State Error vs. System Type 28 Steady-state error vs. input and system type System Type Input Step Ramp Parabola 0 + KK pp 2 0 KKKK Note that the given steady-state error is for inputs of unit magnitude Actual error is scaled by the magnitude of the reference input KKKK
29 29 Non-Unity-Feedback Systems
30 Non-Unity-Feedback Systems 30 So far, we ve focused on the special case of unityfeedback systems System type determined by # of integrators in the forward path i.e., # of open-loop poles at the origin Steady-state error determined using static error constants Static error constants determined from the open-loop transfer function
31 Non-Unity-Feedback Systems 3 More general approach to determining steady-state error is to use the closed-loop transfer function Applicable to non-unity-feedback systems, e.g.: TT ss = GG ss GG 2 ss + GG 2 ss HH ss The error is EE ss = RR ss YY ss = RR ss RR ss TT ss EE ss = RR ss TT ss
32 Non-Unity-Feedback Systems 32 Apply the final value theorem to determine the steady-state error: ee ssss = lim ssss ss = lim ssrr ss TT ss Here, system type is determined by using the more general definition: System type is the degree of the input polynomial for which the steady-state error is a finite, non-zero constant
33 Non-Unity-Feedback Systems 33 Alternatively, find steady-state error by converting to a unity-feedback configuration, e.g.: Add and subtract unity-feedback paths:
34 Non-Unity-Feedback Systems 34 Combine the two upper parallel feedback paths: Collapsing the inner feedback form leaves a unityfeedback system Can now apply unityfeedback error analysis techniques
35 35 Steady-State Error Examples
36 Steady-State Error Example 36 What is the steady-state error to a constant reference input, rr tt = 3 cccc tt, for the following feedback positioning system? A type 0 system Non-zero error to a constant reference Position constant: KK pp = lim GG ss = 0 Steady-state error: ee ssss = rr ssss + KK pp = 3 cccc + 0 ee ssss = 0.27 cccc
37 Steady-State Error Example 37 ee ssss = 0.27 cccc
38 Steady-State Error Example 38 What is the same system s steady-state error to a unit ramp input, rr tt = tt tt? A type 0 system, so error to a ramp reference will be infinite Verify using closed-loop transfer function TT ss = GG ss + GG ss = 0 ss + Steady-state error is ee ssss = lim ssss ss TT ss = lim ss ss 2 0 ss + ee ssss = lim ss ss + ss + =
39 Steady-State Error Example 39 ee ssss =
40 Steady-State Error Example 2 40 Design the controller, DD ss, for error of 0.05 to a unit ramp input Plant is type 0 Forward path must be type for finite error to a ramp input DD ss must be type, so one very simple option is: DD ss = KK ss Forward-path transfer function is KK ss + 2 DD ss GG ss = ss ss + ss + 5
41 Steady-State Error Example 2 4 The velocity constant is Steady-state error is For error of 0.05: KK vv = lim ssss ss = lim ss ee ssss = KK vv = 5 2KK ee ssss = 0.05 = 5 2KK KK = 50 KK ss + 2 ss ss + ss + 5 = 2KK 5
42 Steady-State Error Example 2 42 ee ssss = 0.05
43 Steady-State Error Example 3 43 Next, consider a non-unity-feedback system: Determine controller gain, KK, to provide a 2% steady-state error to a constant reference input First, convert to a unityfeedback system Combine forward-path blocks Simultaneously add and subtract unity-feedback paths
44 Steady-State Error Example 3 44 Combine the top two parallel feedback paths Simplifying the inner feedback form leaves a unity-feedback system
45 Steady-State Error Example 3 45 Steady-state error for this type 0 system is ee ssss = + KK pp where For 2% steady-state error The controller gain is KK pp = lim GG ss = ee ssss = 0.02 = KK = KK 0 = 2 KK + 2 KK
46 Steady-State Error Example 3 46 Note that the controller gain has been set to satisfy a steady-state error requirement only Closed loop poles are very lightlydamped Dynamic response is likely unacceptable ee ssss = 0.02
47 Steady-State Error Example 4 47 Now, consider a unity-feedback system with a disturbance input where DD ss = KK and GG ss = ss+5 Determine the controller gain, KK, such that error due to a constant disturbance is % of WW ss For this value of KK, what is the steady-state error to a constant reference input?
48 Steady-State Error Example 4 48 The total error is given by EE ss = RR ss YY ss = RR ss EE ss DD ss GG ss + WW ss GG ss EE ss + DD ss GG ss = RR ss WW ss GG ss EE ss = RR ss + DD ss GG ss WW ss GG ss + DD ss GG ss Substituting in controller and plant transfer functions gives EE ss = RR ss ss + 5 ss KK WW ss ss KK
49 Steady-State Error Example 4 49 Error due to a constant disturbance can be found by applying the final value theorem ee ssss,ww = lim ss WW ss ss KK ee ssss,ww = lim ss ss ss KK = 5 + KK We can calculate the required gain for % error ee ssss,ww = 0.0 = 5 + KK KK = 95 At this gain value, the error due to a constant reference is ee ssss,rr = lim ss ss ss + 5 ss KK = %
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