Chapter 7 Steady-State Errors 穩態誤差

7.1 Introduction 控制系統設計 3規格: Transient response 暫態反應 (Tp , Ts , Tr , %OS ) Stability 穩定度 Steady-state errors 穩態誤差, e(∞) System discussed: stable system only.

討論3類系統的控制誤差 位置控制； 等速度控制； 等加速度控制。 Figure 7.1 Test inputs for steady-state error analysis and design vary with target type

位置控制指令 等速度控制指令 等加速度控制指令 3 Inputs (即3種指令) : Step input(位置) ；Ramp input(等速度) ；Parabolic input(等加速度) 位置控制指令 等速度控制指令 等加速度控制指令 Table 7.1 Test waveforms for evaluating steady-state errors of position control systems

Figure 7. 2 Steady-state error e(∞) a Figure 7.2 Steady-state error e(∞) a. step input; output1: e(∞)=0 output2: e(∞)=constant b. ramp input output1: e(∞)=0 output2: e(∞)= constant output3: e(∞)= ∞ unstable

Error 定義: e(t) = Input (t) - Output (t) E(s) = R(s) – C(s) Steady-state error 定義: e(∞) = Input (∞) - Output (∞) at time domain e(∞) = lims→0 s E(s) (by final value theorem)

Figure 7. 4 e(∞) 由system configuration and input 決定: a Figure 7.4 e(∞) 由system configuration and input 決定: a. finite steady-state error for a step input; Csteady-state = K esteady-state K↑ esteady-state ↓ esteady-state = 0 → impossible b. zero steady-state error for step input esteady-state = 0 (∵ 系統具積分器)

7.2 Steady-State Error for Unity Feedback Systems (case 1) 1/2 E(s) = R(s) – C(s) where C (s) = G(s)E(s) → E(s) = R(s)/(1 + G(s)) 註：e(∞) 由system configuration and input 決定 G(s) R(s)

7.2 Steady-State Error for Unity Feedback Systems (case 1) 2/2 E(s) = R(s) – C(s) where C (s) = G(s)E(s) → E(s) = R(s)/(1 + G(s)) → e(∞) = lims→0 S E(s) or e(∞) = lims→0 S R(s)/(1 + G(s)) 註：Steady-State Error

7.2 Steady-State Error for Unity Feedback Systems (case 2) E(s) = R(s) – C(s) where C (s) = T(s)R(s) → E(s) = R(s) [1 － T(s)] → e(∞) = lims→0 S E(s) or e(∞) = lims→0 S R(s)[1 － T(s)] e(∞) 由system configuration and input 決定

Figure 7.8 Feedback control system for defining system type 定義 of System Type: n=0 Type 0 system n=1 Type 1 system n=2 Type 2 system

※ 求 esteady-state under 3 input signals 1/4 For step input R(s)=1/S e(∞) = lims→0 S R(s)/(1 + G(s)) ←公式 = lims→0 S (1/S)/(1 + G(s)) = lims→0 1/(1 + G(s)) = 1/(1 + lims→0 G(s)) if wish e(∞) = 0 → then lims→0 G(s) = ∞

※ 求 esteady-state under 3 input signals 2/4 For step input (續) e(∞) = 1/(1 + lims→0 G(s)) if wish e(∞) = 0 → then lims→0 G(s) = ∞ lims→0 G(s) = ∞ if n≧1 for n≧1 stands for 1 integrator in the forward path i.e. system type ≧1 to derive e(∞) = 0

※ 求 esteady-state under 3 input signals 3/4 For ramp input R(s)=1/S2 e(∞) = lims→0 S R(s)/(1 + G(s)) = lims→0 S (1/S2)/(1 + G(s)) = lims→0 1/S(1 + G(s)) = 1/ lims→0 SG(s) if wish e(∞) = 0 → then lims→0 SG(s) = ∞ lims→0 sG(s) = ∞ if n≧2 for n≧2 stands for 2 integrators in the forward path i.e. system type ≧2 to derive e(∞) = 0

※ 求 esteady-state under 3 input signals 4/4 For parabolic input R(s)=1/S3 e(∞) = lims→0 S R(s)/(1 + G(s)) = lims→0 S (1/S3)/(1 + G(s)) = lims→0 1/ S2(1 + G(s)) = 1/ lims→0 s2G(s) if wish e(∞) = 0 → then lims→0 s2G(s) = ∞ lims→0 s2G(s) = ∞ if n≧3 for n≧3 stands for 3 integrators in the forward path i.e. system type ≧3 to derive e(∞) = 0

公式彙總 指令不同 求 esteady-state 公式不同 For step input R(s)=1/S e(∞) = 1/(1 + lims→0 G(s)) For ramp input R(s)=1/S2 e(∞) = 1/ lims→0 SG(s) For parabolic input R(s)=1/S3 e(∞) = 1/ lims→0 s2G(s)

Example 7. 2 不同指令下 求 esteady-state Figure 7 Example 7.2 不同指令下 求 esteady-state Figure 7.5 Feedback control system for system with no integrator type 0 系統 R(s) = 5u(t) = 5/S e(∞) = 5/(1 + lims→0 G(s)) = 5/21 R(s) = 5tu(t) = 5/S2 e(∞) = 5/ lims→0SG(s) = 1/0 = ∞ R(s) = 5t2u(t) = 10/S3 e(∞) = 5/ lims→0 S2G(s) = 1/0 = ∞ type 0系統 只能執行位置控制 產生有限誤差; 無法執行速度及加速度控制

Example 7.3 Figure 7.6 Feedback control system for system with no one integrator type1 系統 R(s) = 5u(t) = 5/S e(∞) = 5/(1 + lims→0 G(s)) = 0 R(s)= 5tu(t) = 5/S2 e(∞) = 5/ lims→0SG(s) = 1/20 = finite R(s)= 5t2u(t) = 10/S3 e(∞) = 10/ lims→0 S2G(s) = 1/0 = ∞ type 1系統 執行位置控制 無誤差產生; 執行速度控制 產生有限誤差; 無法執行加速度控制 H.W.: Skill-Assessment Exercise 7.1

7.3 Static Error Constants and System Type 定義: Static Error Constants Kp Kv Ka For step input R(s) = 1/s e(∞) = 1/(1 + lims→0 G(s)) = 1/1+Kp Kp = lims→0 G(s) position error constant For ramp input R(s) =1/s2 e(∞) = 1/ lims→0SG(s) = 1/Kv Kv = lims→0 SG(s) velocity error constant For parabolic input R(s) = 1/s3 e(∞) = 1/ lims→0 s2G(s) = 1/Ka Ka = lims→0 S2G(s) acceleration error constant

Example 7. 4 利用Static Error Constants 求解 Figure 7 Example 7.4 利用Static Error Constants 求解 Figure 7.7 Feedback control systems 求3系統之steady-state error? 1/3 Type 0 system For step input: R(s) = 1/s Kp = lims→0G(s)= 5.208 e(∞) = 1/ /(1+Kp) = 0.161 For ramp input: R(s) =1/s2 Kv = lims→0 sG(s) = 0 e(∞) = 1/Kv = ∞ For parabolic input: R(s) = 1/s3 Ka = lims→0s2G(s)=0 e(∞) = 1/Ka = ∞

Example 7.4 Figure 7.7 Feedback control systems 求3系統之steady-state error? 2/3 (b) Type 1 system For step input: R(s) = 1/s Kp = lims→0G(s)= ?/0 = ∞ e(∞) = 1/(1+Kp) = 0 For ramp input: R(s) =1/s2 Kv = lims→0 sG(s) = 30000/960 =31.25 e(∞) = 1/Kv = 0.032 For parabolic input: R(s) = 1/s3 Ka = lims→0s2G(s)=0*?=0 e(∞) = 1/Ka = ∞

Example 7.4 Figure 7.7 Feedback control systems 求3系統之steady-state error? 3/3 (c) Type 2 system For step input: R(s) = 1/s Kp = lims→0 G(s)= ?/0 = ∞ e(∞) = 1/ (1+Kp) = 0 For ramp input: R(s) =1/s2 Kv = lims→0 sG(s) = ?/0 = ∞ e(∞) = 1/Kv = 0 For parabolic input: R(s) = 1/s3 Ka = lims→0 s2G(s) = 875 e(∞) = 1/Ka = 0.00114

Table 7.2 Relationships between input, system type, static error constants, and steady-state errors Static Error Constants: Kp Kv Ka 決定系統之 e(∞) ; 其可為steady-state error 之規格 H.W.: Skill-Assessment Exercise 7.2

7.4 Steady-State Error Specifications Example 7.5 Given Kv=1000 → draw ?? conclusions 1. Stable system 2. Ramp input 3. Type 1 system

Example 7.6 Find K=? → e(∞) = 10% Type 1 system (已知) 有限的e(∞) → Ramp input (已知) e(∞) = 1/ Kv = 0.1 → Kv = 10 Kv = lims→0 SG(s) = k*5 / 6*7*8 = 10 → k = 672 自修 Skill-Assessment Exercise 7.3

7.5 Steady-State Error for Disturbances 1/3 Figure 7.11 Feedback control system showing disturbance 2 inputs R(s) & D(s) C(s) =﹝E(s)G1(s) ＋ D(s) ﹞G2(s) = E(s)G1(s)G2(s) ＋ D(s)G2(s) E(s) = R(s) – C(s) → R(s) –E(s) = E(s)G1(s)G2(s) ＋ D(s)G2(s) E(s)G1(s)G2(s) ＋ E(s) = R(s) – D(s)G2(s) E(s)(1+G1(s)G2(s)) = R(s) – D(s)G2(s)

7.5 Steady-State Error for Disturbances 2/3 E(s)(1+G1(s)G2(s)) = R(s) – D(s)G2(s)

7.5 Steady-State Error for Disturbances 3/3 DC gain of G1(s) eD(∞)↓ (分母變大) if DC gain of G1(s)↑ or DC gain of G2(s)↓

D(s) = step disturbance Example 7.7 Fig. 7.13 如下 自修 D(s) = step disturbance Find eD(∞) = ? H.W. : Skill-Assessment Exercise 7.4

Figure 7.12 Figure 7.11 system rearranged to show disturbance as input and error as output, with R(s) = 0 -C(s) = E(s)

7.6 Steady-State Error for Nonunity Feedback Systems

7.7 Sensitivity H.W. : Skill-Assessment Exercise 7.6 Defination Examples: 7.10 7.11 7.12 H.W. : Skill-Assessment Exercise 7.6

Se:a = ﹝a/(a/k)﹞﹝δ(a/k)/δa﹞= 1 Example 7.11 Figure 7.19 Find Se:a = ? Se:k = ? R(s) = Ramp input = 1/s2 → e(∞) = 1/kv = 1/(k/a) = a/k Se:a = ﹝a/(a/k)﹞﹝δ(a/k)/δa﹞= 1 Se:k = ﹝k/(a/k)﹞﹝δ(a/k)/δk﹞= -1