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.

Figure 7.1

Test inputs for steady-state error analysis and design vary with target type

•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。

• 3 Inputs ( 即 3 種指令 ) : Step input( 位置 ) ； Ramp input( 等速度 ) ； Parabolic input( 等加速度 )

Table 7.1 Test waveforms for evaluating steady-state errors of position control systems

位置控制指令

等速度控制指令

等加速度控制指令

Figure 7.2Steady-state errore(∞)a. step input;output1: e(∞)=0output2: e(∞)=constant

b. ramp inputoutput1: e(∞)=0output2: e(∞)= constantoutput3: 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. 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 configurationsystem configuration and inputinput 決定

G(s)G(s) R(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 systemType 0 systemn=1 Type 1 systemType 1 systemn=2 Type 2 systemType 2 system

※ 求 esteady-state under 3 input signals 1/4

• For step inputFor step input R(s)=1/SR(s)=1/S

e(∞) = lims→0 S R(s)/(1 + G(s)) ←← 公式公式

= lims→0 S (1/S)/(1 + G(s))

= lims→0 1/(1 + G(s))

= 1/(1 + lims→0 G(s))

if wish e(∞) = 0e(∞) = 0 → then limlims→0s→0 G(s) = ∞ 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 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/ limlimss→0→0 s s22G(s)G(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

• 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/ limlimss→0→0 s s22G(s)G(s)

指令不同 求 指令不同 求 eesteady-state steady-state 公式不同公式不同

公式彙總公式彙總

Example 7.2 不同指令下 求 不同指令下 求 eesteady-state steady-state

Figure 7.5 Feedback control system for system with no integrator

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

type 0 系統

Example 7.3 Figure 7.6 Feedback control system for system with no one integrator

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

type1 系統

7.3 Static Error Constants and System Type

•定義 : Static Error Constants 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

(a) 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 利用 Static Error Constants Static Error Constants 求解 Figure 7.7 Feedback control systems 求 3 系統之 steady-state error? 1/3

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 7.2Relationships 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)

• eD(∞)↓ ( 分母變大 ) if DC gain of G1(s)↑

or DC gain of G2(s)↓

DC gain of G1(s)

7.5 Steady-State Error for Disturbances 3/3

• Example 7.7 Fig. 7.13 如下 自修

• H.W. : Skill-Assessment Exercise 7.4

D(s) = step disturbance

Find eD(∞) = ?

Figure 7.12Figure 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• Defination

Examples: 7.10 7.11 7.12

H.W. : Skill-Assessment Exercise 7.6

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﹝ ﹞﹝ ﹞