Transcript R(s)

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
• 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. 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. 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
定義 of System Type:
n=0 Type 0 system
n=1 Type 1 system
n=2 Type 2 system
type
※ 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/
if wish
lims→0 SG(s)
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 7.5 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 7.7 Feedback control systems 求3系統之steady-state error?
(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 = ∞
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?
(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
3/3
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
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)
1/3
7.5 Steady-State Error for Disturbances
E(s)(1+G1(s)G2(s)) = R(s) – D(s)G2(s)
2/3
7.5 Steady-State Error for Disturbances
DC gain of G1(s)
• eD(∞)↓ (分母變大)
if DC gain of G1(s)↑
or DC gain of G2(s)↓
3/3
• 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
• 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