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CH. 6 Stability 穩定度. 6.1 Introduction. 控制系統設計 3 規格 : Transient response 暫態反應 ( T p , T s , T r , %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 →∞.
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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 → 0as t →∞
Unstable system: Natural (transient) response → ∞as t →∞
Marginally stable system: Natural (transient) response → neither decays nor grows
Figure 6.1 Closed-loop poles and response:a.stable system; closed-loop poles at LHP (stands for Left Half Plane) 由閉回路系統之極點位置判別是否為穩定系統 Underdamped responses two complex poles:
Figure 6.1Closed-loop poles and response:b.unstable systemat least1closed-looppole at RHP (Right Half Plane) 由閉回路系統之極點位置判別是否為穩定系統 2 closed-loop poles on imaginary axis→unstable system C(t) = A tn cos(ωt+φ)
Figure 6.2Common 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 T(s) Closed-loop T.F.
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 6.3Equivalent closed-looptransfer function Table 6.1Initial 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
Table 6.3Completed Routh table first column 符號變更之次數 = 2→ 2 closed-loop poles at RHP→ unstable system • Ex. 6.1Fig. 6.4
6.3 Routh-Hurwitz Criterion: Special CasesCase 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 6.4Closed-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 6.4poles 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 6.4Table 6.7first 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1closed-loop pole在LHP → unstable system
6.3 Routh-Hurwitz Criterion: Special Cases Case 2: Entire Row is Zero 1/4 Example 6.5 Closed-loop T.F. 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 6.5D(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 →2closed-loop pole在RHP → 2closed-loop pole在LHP → unstable system ← Example 6.5 Table 6.8first column of Routh table
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-HurwitzFigure 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 studyRouth table for antenna control 求K範圍such that system is stable.
