Lecture Control system design: Stability in the frequency domain - Nguyễn Công Phương

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This chapter include all of the following content: Mapping contours in the s – plane, the nyquist criterio, relative stability and the nyquist criterion, time – domain performance criteria in the frequency domain, system bandwidth, the stability of control systems with time delays, pid controllers in the frequency domain, stability in the frequency domain using control design software.

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Nội dung Text: Lecture Control system design: Stability in the frequency domain - Nguyễn Công Phương

Nguyễn Công Phương CONTROL SYSTEM DESIGN Stability in the Frequency Domain
Contents I. Introduction II. Mathematical Models of Systems III. State Variable Models IV. Feedback Control System Characteristics V. The Performance of Feedback Control Systems VI. The Stability of Linear Feedback Systems VII. The Root Locus Method VIII.Frequency Response Methods IX. Stability in the Frequency Domain X. The Design of Feedback Control Systems XI. The Design of State Variable Feedback Systems XII. Robust Control Systems XIII.Digital Control Systems sites.google.com/site/ncpdhbkhn 2
Stability in the Frequency Domain 1. Mapping Contours in the s – Plane 2. The Nyquist Criterion 3. Relative Stability and the Nyquist Criterion 4. Time – Domain Performance Criteria in the Frequency Domain 5. System Bandwidth 6. The Stability of Control Systems with Time Delays 7. PID Controllers in the Frequency Domain 8. Stability in the Frequency Domain Using Control Design Software sites.google.com/site/ncpdhbkhn 3
Mapping Contours Ex. 1 in the s – Plane (1) jω jv s – plane F (s) F(s) – plane j2 D j2 A D A j1 j1 −2 −1 0 1 2 σ −2 −1 0 1 2 u − j1 − j1 C B − j2 − j2 B C u = 2σ + 1 F ( s) = 2 s + 1 = 2(σ + jω ) + 1 = (2σ + 1) + j2ω = u + jv →  v = 2ω As = 1 + j1 = σ + jω → AF = u + jv = (2σ + 1) + j (2ω ) = (2 × 1 + 1) + j (2 × 1) = 3 + j 2 Bs = 1 − j1 → BF = (2 × 1 + 1) + j[2 × ( −1)] = 3 − j 2 Cs = −1 − j1 → CF = [2 × ( −1) + 1] + j[2 × ( −1)] = −1 − j 2 Ds = −1 + j1 → DF = [2 × ( −1) + 1] + j (2 × 1) = −1 + j 2 sites.google.com/site/ncpdhbkhn 4
Mapping Contours Ex. 2 in the s – Plane (2) s F ( s) = s+2 s-plane F(s)-plane 1.5 1.5 D A D 1 1 0.5 0.5 A 0 0 jω jv B -0.5 -0.5 -1 -1 C B C -1.5 -1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 σ u sites.google.com/site/ncpdhbkhn 5
Mapping Contours in the s – Plane (3) s-plane F(s)-plane 2 2 1 F ( s ) = 2 s + 11 jω 0 0 jv -1 -1 -2 -2 Cauchy’s -2theorem: -1 0If a contour 1 2 Γ3s in the s-plane encircles -2 -1 Z zeros 0 and 1 P2 poles 3 of F(s) σ u and does not pass through any poles or zeros of F(s) and the traversal is in the clockwise directions-plane along the contour, the corresponding contour ΓF in the F(s)-plane F(s)-plane 1.5 encircles the origin of the F(s)-plane N = Z – P1.5times in the clockwise direction. 1 1 0.5 0.5 0 0 jω jv -0.5 s-0.5 F ( s) = -1 s + -12 -1.5 -1.5 -2 -1 0 1 -2 -1 0 1 σ sites.google.com/site/ncpdhbkhn u 6
Mapping Contours Ex. 3 in the s – Plane (4) s F ( s) = s + 0.5 s-plane F(s)-plane 6 0.15 4 0.1 2 0.05 jω 0 0 jv -2 -0.05 -4 -0.1 -6 -6 -4 -2 0 2 4 6 0.9 0.95 1 1.05 1.1 1.15 σ u If a contour Γs in the s-plane encircles Z zeros and P poles of F(s) and does not pass through any poles or zeros of F(s) and the traversal is in the clockwise direction along the contour, the corresponding contour ΓF in the F(s)-plane encircles the origin of the F(s)-plane N = Z – P times in the clockwise direction. sites.google.com/site/ncpdhbkhn 7
Mapping Contours Ex. 2 in the s – Plane (5) s s+z F ( s) = = = F ( s) ∠F ( s) = F ( s) ∠(φz − φ p ) s+2 s+ p If a contour Γs in the s-plane encircles Z zeros and P poles of F(s) and does not pass through any poles or zeros of F(s) and the traversal is in the clockwise direction along the contour, the corresponding contour ΓF in the F(s)-plane encircles the origin of the F(s)-plane N = Z – P times in the clockwise direction. s-plane F(s)-plane 1.5 1.5 1 1 0.5 φz 0.5 φz − φ p φp jω 0 0 jv -0.5 -0.5 -1 -1 -1.5 -1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 σ sites.google.com/site/ncpdhbkhn u 8
Stability in the Frequency Domain 1. Mapping Contours in the s – Plane 2. The Nyquist Criterion 3. Relative Stability and the Nyquist Criterion 4. Time – Domain Performance Criteria in the Frequency Domain 5. System Bandwidth 6. The Stability of Control Systems with Time Delays 7. PID Controllers in the Frequency Domain 8. Stability in the Frequency Domain Using Control Design Software sites.google.com/site/ncpdhbkhn 9
The Nyquist Criterion (1) • F(s) = 1 + L(s) = 0 • A feedback system is stable if and only jω s – plane if the contour ΓL in the L(s) – plane does not encircle the (–1, 0) point when r→∞ the number of poles of L(s) in the right – hand s – plane is zero (P = 0). 0 σ • (when the number of poles of L(s) in the Γs right – hand s – plane is other than zero) A feedback system is stable if and only if, for the contour ΓL , the number Nyquist contour of counterclockwise encirclements of the (–1, 0) point is equal to the number of poles of L(s) with positive real parts. sites.google.com/site/ncpdhbkhn 10
The Nyquist Criterion (2) Ex. 1 1 1 R(s) 1 1 Y ( s) . τ 1s + 1 τ 2 s + 1 τ 1s + 1 τ2s + 1 T ( s) = ( −) 1 1 1+ K . τ 1s + 1 τ 2 s + 1 K 1 A feedback system is 1stable if and only if the contour ΓL in the1L(s) – plane 1 does not encircle → 1 + K the (–1, 0)τ point . = 0 = 1 + L ( s ) → L ( s ) = K . 1s + 1 τ 2 s + 1 τ 1s–+hand 1 τ s2 s– +plane when the number of poles of L(s) in the right 1 is zero (P = 0). As = jω ω →∞ jω jv AL = L( jω ) ω →∞ = 0 s – plane A Bs = σ σ →∞ ω→∞ r→∞ BL = L(σ ) σ →∞ = 0 D D B A , B, C Cs = jω ω→−∞ 1 1 0 σ 0 u − − τ1 τ2 ω=0 CL = L( jω ) ω→−∞ = 0 Γs Ds = 0 + j 0 C DL = L(0) = K L(s) – plane sites.google.com/site/ncpdhbkhn 11
The Nyquist Criterion (3) Ex. 2 jω K s – plane L( s ) = D s(τ s + 1) C r→∞ As = jω ω
The Nyquist Criterion (4) K = 2, τ = 1 Ex. 2 jω 20 K s – plane 15 A L( s ) = D s(τ s + 1) 10 C r→∞ AL = − Kτ + j∞ B E 5 0 D, F B jv − 1 0 σ Bs = σ τ ε σ →0 A -5 -10 K F Γs BL = L(σ ) σ →0 = -15 σ (σ + 1) σ →0 -20 C =∞ 0 10 20 u K Cs = jω ω >0, ω →0 → CL = L( jω ) ω >0, ω →0 = = − Kτ − j∞ ( jω )( jτω + 1) ω >0, ω →0 K Ds = jω → DL = L( jω ) = =0 ω →∞ ω→∞ ( jω )( jτω + 1) ω→∞ K Fs = jω → FL = L( jω ) = =0 ω→−∞ ω→−∞ ( jω )( jτω + 1) ω→−∞ sites.google.com/site/ncpdhbkhn 13
The Nyquist Criterion (5) jω jv K s – plane L( s ) = A (τ1s + 1)(τ 2 s + 1) ω→∞ r→∞ D B A, B,C D • It is sufficient to 1 1 0 σ 0 u construct the contour − − ω=0 τ1 τ2 ΓL for the frequency range 0
jω s – plane The Nyquist Criterion (6) D Ex. 3 C r→∞ K B E L( s ) = σ s(τ 1s + 1)(τ 2 s + 1) 1 0 − ε τ A K L( jω ) = jω ( jωτ1 + 1)( jωτ 2 + 1) F Γs − K (τ1 + τ 2 ) − jK (1/ ω )(1 − ω 2τ1τ 2 ) = 1 + ω 2 (τ 12 + τ 22 ) + ω 4τ 12τ 22 K = ∠[− tan −1 (ωτ 1 ) − tan −1 (ωτ 2 ) − π / 2] ω 4 (τ 1 + τ 2 )2 + ω 2 (1 − ω 2τ 1τ 2 )2 K Cs = jω ω >0, ω →0 → CL = lim L( jω ) = lim ∠( −π / 2) ω→ 0 ω→ 0 ω (τ1 + τ 2 ) + ω (1 − ω τ 1τ 2 ) 4 2 2 2 2 K Ds = jω ω →∞ → DL = lim L( jω ) = lim ∠( −3π / 2) ω →∞ ω →∞ τ1τ 2ω 3 sites.google.com/site/ncpdhbkhn 15
jω s – plane The Nyquist Criterion (7) D Ex. 3 C r→∞ K B E L( s ) = σ s(τ 1s + 1)(τ 2 s + 1) 1 0 − ε τ A K K CL = ∠( −π / 2); DL = ∠(−3π / 2) 0 ∞ F Γs − K (τ1 + τ 2 ) K (1/ ω )(1 − ω 2τ 1τ 2 ) L( jω ) = −j 1 + ω (τ 1 + τ 2 ) + ω τ1 τ 2 2 2 2 4 2 2 1 + ω 2 (τ 12 + τ 22 ) + ω 4τ12τ 22 K = 1, τ1 = 5, τ2 = 5 6 K (1/ ω )(1 − ω 2τ 1τ 2 ) Im{L( jω )} = 0 → =0 1 + ω (τ1 + τ 2 ) + ω τ 1 τ 2 2 2 2 4 2 2 4 1 − Kτ 1τ 2 2 →ω = → Re( L) ω = 1 = τ 1τ 2 τ1τ 2 τ1 + τ 2 0 jv -2 − Kτ 1τ 2 τ +τ → ≥ −1 → K ≤ 1 2 τ1 + τ 2 τ 1τ 2 -4 -6 -6 -4 -2 0 2 4 sites.google.com/site/ncpdhbkhn u 16
The Nyquist Criterion (8) Ex. 3 K τ1 + τ 2 L( s ) = K≤ s(τ 1s + 1)(τ 2 s + 1) τ 1τ 2 K = 1, τ = 1, τ = 1 K = 2, τ = 1, τ = 1 K = 3, τ = 1, τ = 1 1 2 1 2 1 2 1.5 1.5 1.5 1 1 1 0.5 0.5 0.5 0 0 0 jv jv jv -0.5 -0.5 -0.5 -1 -1 -1 -1.5 -1.5 -1.5 -2 -1 0 1 -2 -1 0 1 -2 -1 0 1 u u u sites.google.com/site/ncpdhbkhn 17
jω s – plane The Nyquist Criterion (9) D Ex. 4 C r→∞ K B E L( s ) = 2 0 σ s (τ s + 1) 1 − ε τ A K K L( jω ) = = ∠[ −π − tan −1 (ωτ )] −ω 2 ( jωτ + 1) ω +τ ω 4 2 6 F Γs Cs = jω ω >0, ω →0 2 K → CL = lim L( jω ) = lim ∠( −π ) 1.5 ω →0 ω →0 ω2 1 Ds = jω ω →∞ 0.5 0 K jv → DL = lim L( jω ) = lim ∠( −3π / 2) ω →∞ ω →∞ τω 3 -0.5 -1 Bs = ε e jφ ε →0 -1.5 K → BL = lim L(ε e jφ ) = lim e −2 jφ -2 ε →0 ε →0 ε 2 -2.5 -2 -1.5 -1 -0.5 u 0 0.5 1 1.5 2 sites.google.com/site/ncpdhbkhn 18
The Nyquist Criterion (10) 2.5 Ex. 5 jω 2 R ( s) 1 1 Y ( s) s – plane K1 D 1.5 s −1 s 1 ( −) C r→∞ 0.5 B E 0 jv σ -0.5 0 1 K1 ε -1 K1 L( s ) = A s( s − 1) -1.5 F Γs -2 -2.5 -3 -2 -1 0 u Ex. 6 jω 2 s – plane R ( s) 1 1 Y ( s) D 1.5 K1 (− ) s −1 s 1 ( −) C r→∞ 0.5 K2 B E 0 jv − 1 0 1 σ -0.5 K2 ε K ( K s + 1) A -1 L( s ) = 1 2 s( s − 1) -1.5 F Γs -2 -5 -4 -3 -2 -1 0 sites.google.com/site/ncpdhbkhn u 19
The Nyquist Criterion (11) Ex. 6 jω 2 s – plane R ( s) 1 1 Y ( s) D 1.5 K1 (− ) s −1 s 1 ( −) C r→∞ 0.5 K2 B E 0 jv − 1 0 1 σ -0.5 K2 ε K ( K s + 1) A -1 L( s ) = 1 2 s( s − 1) -1.5 F Γs -2 -5 -4 -3 -2 -1 0 K1 ( K 2 jω + 1) − K1ω 2 ( K 2 + 1) K1ω (1 − K 2ω 2 ) u → L( jω ) = = + j jω ( jω − 1) ω +ω 4 2 ω4 + ω2 K1ω (1 − K 2ω 2 ) 1 Im{L( jω )} = 0 → = 0 → ω 2 = ω4 + ω2 K2 − K1ω 2 ( K 2 + 1) Re{L( jω )} ω 2 =1/ K = = − K1K 2 2 ω4 + ω2 ω 2 =1/ K 2 If − K1K 2 < −1 → K1K 2 > 1 → Z = N + P = −1 + 1 = 0 → stable sites.google.com/site/ncpdhbkhn 20