Transcript Routh-Hurwitz Criterion

CH. 6 Stability 穩定度
6.1 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
• Stable system 定義
Natural (transient) response → 0 as t →∞
• Unstable system:
Natural (transient) response → ∞ as t →∞
• Marginally stable system:
Natural (transient) response → neither decays nor
grows
由閉回路系統之極點位置判別是否為穩定系統
Figure 6.1 Closed-loop
poles and response:
a. stable system;
closed-loop poles at
LHP (stands for Left
Half Plane)
Underdamped responses
two complex poles:
由閉回路系統之極點位置判別是否為穩定系統
Figure 6.1 Closed-loop
poles and response:
b. unstable system
at least 1 closed-loop pole
at RHP (Right Half Plane)
2 closed-loop poles on imaginary axis → unstable system
C(t) = A tn cos(ωt+φ)
註: marginally stable
1 closed-loop pole on imaginary axis
T(s)
Closed-loop T.F.
Figure 6.2
Common cause of problems in finding closed-loop poles:
a. original system; b. equivalent system
由因式分解求closed-loop poles → 困難 (高階系統) → Routh-Hurwitz Criterion
• Closed-loop T.F. T(s) = N(s)/D(s)
• Stable system之條件
(1) D(s) = ( s + ai ) = ( s + a1 ) … ( s + an )=0
ai＞0 穩定系統之條件 → D(s) 係數符號需一致
註1: 若D(s) 缺項或係數之符號不同→不穩定系統
註2: D(s) 缺項, 如 s2+1 = 0 缺 s 項
(2) 若D(s) 無缺項且係數之符號相同
→仍無法確認是否為穩定系統
→ Routh-Hurwitz Criterion
• ( s + 1 ) ( s + 2) = s2 + 3s + 2
• ( s + 1 ) ( s + 2) ( s + 3)
= (s2 + 3s + 2) ( s + 3)
= s3 + 6s2 + 11s + 6
ai＞0 穩定系統 → 係數符號一致
→ 不缺項
6.2 Routh-Hurwitz Criterion
T(s)
Figure 6.3
Equivalent
closed-loop
transfer function
Table 6.1
Initial layout for
Routh table
Table 6.2 Completed Routh table
Intepret the Routh table: 由Routh table判讀closed-loop poles在RHP之數目
closed-loop poles在RHP之數目 = first column 符號變更之次數
如Routh table’s first column 無符號變更 →
stable system
• Ex. 6.1
Fig. 6.4
Table 6.3
Completed Routh
table
first column 符號變
更之次數 = 2
→ 2 closed-loop
poles at RHP
→ unstable system
6.3 Routh-Hurwitz Criterion: Special Cases
Case 1: Zero Only in the First Column 1/2
Example 6.2
Table 6.4 Completed Routh table
6.3 Routh-Hurwitz Criterion: Special Cases
Case 1: Zero Only in the First Column 2/2
Example 6.2
Table 6.5
Determining signs in first column of a Routh table with zero as first element in a row
2 sign changes → 2 closed-loop poles at RHP → unstable system
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 1/6
Example 6.4
Closed-loop T.F.
T(s) = 10/ D(s)
D(s) = s5+7s4+6s3+42s2+8s+56
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero
2/6
Example 6.4
D(s) = s5+7s4+6s3+42s2+8s+56
Table 6.7
Routh table
→
↓
Even polynomial: p(s) = s4+6s2+8
→ dP/ds = 4s3+12s+0 → 代入 s3 row
→ complete the Routh table
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 3/6
Example 6.4 Closed-loop T.F.
T(s) = 10/ D(s)
D(s) = s5+7s4+6s3+42s2+8s+56
→
Table 6.7
Completed Routh table
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 4/6
Example 6.4
poles 所在區域的判讀分兩部分
↑
→
↓
even
polynomials
Table 6.7
Completed Routh table
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 5/6
Figure 6.5 Root positions to
generate even polynomials: A , B,
C, or any combination
註:
p(s) 之根對稱於原點
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 6/6
↑
Example 6.4
→
Table 6.7 first column
of Routh table
↓
系統穩定度判別
註1:
檢驗 p(s) 以下, 第1列有無變號
→無變號→無closed-loop poles在RHP
→無closed-loop poles在LHP
→ 4 closed-loop poles在jω軸上
註2:檢驗 p(s) 以上, 第1列有無變號
1 closed-loop pole在LHP → unstable system
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero
Example 6.5 Closed-loop T.F.
1/4
T(s) = 10/ D(s)
D(s) = s8+s7+12s6+22s5+39s4+59s3+48s2+38s+20
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 2/4
Example 6.5
D(s) = s8+s7+12s6+22s5+39s4+59s3+48s2+38s+20
Table 6.8
Routh table
→
↓
Even polynomial: p(s) = s4+3s2+2
→ dP/ds = 4s3+6s+0 → 代入 s3 row
→ complete the Routh table
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 3/4
Example 6.5 Closed-loop T.F.
T(s) = 10/ D(s)
D(s) = s8+s7+12s6+22s5+39s4+59s3+48s2+38s+20
Table 6.8
Completed
Routh table
←
6.3 Routh-Hurwitz Criterion: Special Cases
Case 2: Entire Row is Zero 4/4
系統穩定度判別
←
Example 6.5
Table
6.8 first column of
Routh table
註: 檢驗 p(s) 以下, 第1列有無變號 → 無變號 →
無closed-loop poles在RHP →無closed-loop poles
在LHP → 4 closed-loop poles在jω軸上
註: 檢驗p(s) 以上, 第1列有無變號 → 2變號 →
2 closed-loop pole在RHP → 2 closed-loop pole在
LHP → unstable system
• 自修題: Ex. 6.6 - Ex. 6.8
• H.W.: Skill-Assessment Exercise 6.1
Skill-Assessment Exercise 6.2
Skill-Assessment Exercise 6.3
Example 6.9
Stability Design via Routh-Hurwitz
Figure 6.10 Feedback control system
T(s) = k/ D(s)
D(s) = s3+18s2+77s+k
For stable system:
No sign change in the first column
→ k＜1386 stable sys
→ k＞1386 unstable sys
→ k = 1386 unstable sys
(via even polynomial, 2 roots on jω
axis, 1 root at LHP, unstable system )
Example 6.10
Factoring via Routh-Hurwitz
Factor the polynomial s4+3s3+30s2+30s+200
Table 6.17
→
Even polynomial p(s) = s2+10
s4+3s3+30s2+30s+200 = (s2+10)(s2+3s+20)
Table 6.19 case study
Routh table for antenna control
求K範圍 such that system is stable.