Bài giảng Control system design - Chapter IX: Stability in the frequency domain - Nguyễn Công Phương

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

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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 +j ω )1 + =(2σ + 1) + j 2 ω =u + jv →  v = 2ω =+ =σ + ω → =+ =σ ++ ω = ×++ ×=+ As 1 j 1 j AujvF (2 1) j (2 ) (2 1 1)j (2 1) 3 j 2 =− → = ×++ ×− =− BjBs1 1 F (211) j [2(1)]3 j 2 =−− → = ×−++ ×− =−− CjCs1 1 F [2(1)1] j [2(1)] 1 j 2 =−+ → = ×−++ ×=−+ DjDs1 1 F [2(1)1] j (21) 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 ω 0 jv 0 j 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 + 1 1 ω jv j 0 0 -1 -1 -2 -2 Cauchy’s -2theorem-1 : 0If a contour1 2 Γ3s in the s-plane encircles-2 -1 Z zeros0 and1 P2 poles3 of F(s) and does not pass throughσ any poles or zeros of F(s) and the traversalu is in the clockwise direction s-planealong the contour, the corresponding contourF(s)-plane ΓF in the F(s)-plane encircles1.5 the origin of the F(s)-plane N = Z – P1.5times in the clockwise direction. 1 1 0.5 0.5 ω jv j 0 0 -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 ω jv j 0 0 -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 ) = = =Fs() ∠= Fs () Fs ()( ∠−φ φ ) s + 2 s+ p z 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 φ φ− φ 0.5 z 0.5 z p ω 0 jv 0 j φ p -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 the number of poles of L(s) in the right r → ∞ – hand s – plane is zero ( P = 0). 0 σ • (when the number of poles of L (s) in the Γ right – hand s – plane is other than s 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 ) . τs+1 τ s + 1 τ s +1 τ s + 1 T( s ) = 1 2 (− ) 1 2 1 1 1+ K . τ+ τ + 1s1 2 s 1 K 1 1 Γ 1 1 →+A feedback1K system . is stable ==+ if 01() and only L if s the→ contourL( s ) = L Kin the L(s) . – plane does not encircle the (–1, 0)τ point+ when τ the + number of poles of L(s) in the rightτ –+ hand τ s – + plane is zero ( P = 0). 1s1 2 s 1 1s1 2 s 1 = ω As j ω→∞ jω jv A= L( j ω ) = 0 s – plane L ω→∞ A = σ Bs σ →∞ ω → ∞ r → ∞ B= L (σ ) = 0 L σ →∞ D B A, B,C D = ω 0 σ 0 u Cs j ω→−∞ − 1 − 1 τ τ ω = 0 =ω = 1 2 CL L( j )ω→−∞ 0 Γ = + C s Ds 0 j 0 = = 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 → ∞ A= j ω B s ω<0, ω → 0 E 0 σ − 1 ε τ A= L( j ω ) A L ω<0, ω → 0 Γ F s = K ω τω + (j )( j 1) ω<0, ω → 0 Kτ K  = − − j  =−Kτ + j ∞ τω22+1 ωτω ( 22 + 1)   ω<0, ω → 0 Cauchy’s theorem : If a contour Γs in the s-plane encircles Z zeros and P poles of L(s) and does not pass through any poles or zeros of L(s) and the traversal is in the clockwise direction along the contour, the corresponding contour ΓF in the L(s)-plane encircles the origin of the L(s)-plane N = Z – P times in the clockwise direction. sites.google.com/site/ncpdhbkhn 12 The Nyquist Criterion (4) K = 2, τ = 1 Ex. 2 ω 20 j A K s – plane 15 L( s ) = D s(τ s + 1) 10 C r → ∞ A=− Kτ + j ∞ 5 L B E jv 0 0 σ D, F B − 1 ε B = σ τ s σ →0 A -5 Γ -10 K F s B= L (σ ) = -15 L σ →0 σ( σ + 1) C σ →0 -20 = ∞ 0 10 20 K u C= j ω →=CLj(ω ) = =−−∞ Kj τ s ω>0, ω → 0 L ω>0, ω → 0 ω τω + (j )( j 1) ω>0, ω → 0 K D= j ω →=D L( j ω ) = = 0 s ω→∞ L ω→∞ ω τω + (j )( j 1) ω→∞ K F= j ω →=F L( j ω ) = = 0 s ω→−∞ L ω→−∞ ω τω + (j )( j 1) ω→−∞ sites.google.com/site/ncpdhbkhn 13 The Nyquist Criterion (5) jω jv K s – plane L( s ) = A (τs+ 1)( τ s + 1) ω → ∞ 1 2 r → ∞ D B A, B,C D • It is sufficient to 0 σ 0 u − 1 − 1 construct the contour τ τ ω = 0 Γ for the frequency 1 2 L Γ range 0< ω<∞ in order C s to investigate the τ stability. K = 2, = 1 jω 20 s – plane A D 15 • The magnitude of L(s) 10 as s = re jϕ and r →∞ C r → ∞ 5 will normally approach B E zero or a constant. jv 0 D, F B − 1 0 ε σ τ -5 A -10 K Γ L( s ) = F s -15 s(τ s + 1) C -20 sites.google.com/site/ncpdhbkhn 0 10 20 14 u jω s – plane The Nyquist Criterion (6) D Ex. 3 C r → ∞ B = K E L( s ) 0 σ τ+ τ + − 1 ε s(1 s 1)( 2 s 1) τ A K L( j ω ) = ω ωτ+ ωτ + Γ j( j1 1)( j 2 1) F s −+−ττ ω − ωττ2 = K(12 ) jK (1/ )(1 12 ) +ωτ22 + τ 2 + ωττ 422 1 (1 2 ) 12 K − − = ∠−−−[tan(1ωτ ) tan( 1 ωτ ) π /2] ωτ4+ τ 22 + ω − ωττ 22 1 2 (12 ) (1 12 ) =→==ω ω K ∠− π Cjsω> ω → C L lim()lim Lj (/2) 0, 0 ω→0 ω → 0 ωτ4+ τ 22 + ω − ωττ 22 (12 ) (1 12 ) K Dj=ω →= Dlim( Lj ω ) = lim ∠− (3/2) π sω→∞ L ω→∞ ω →∞ τ τ ω 3 1 2 sites.google.com/site/ncpdhbkhn 15 jω s – plane The Nyquist Criterion (7) D Ex. 3 C r → ∞ B = K E L( s ) 0 σ τ+ τ + − 1 ε s(1 s 1)( 2 s 1) τ K K A C=∠−(π /2); D =∠− (3/2) π L L ∞ Γ 0 F s −ττ + ωωττ − 2 ω =K(1 2 ) − K (1/ )(11 2 ) L( j ) 22 2 422 j 22 2 422 +ωτ ++ τ ωττ + ωτ ++ τ ωττ K = 1, τ = 5, τ = 5 1()12 12 1() 12 12 1 2 6 K(1/ω )(1− ωττ2 ) Im{()}0L j ω = →1 2 = 0 +ωτ22 + τ 2 + ωττ 422 4 1 (1 2 ) 12 − τ τ 2 →ω = 1 → = K 1 2 Re(L ) ω= 1 τ τ τ τ τ+ τ 1 2 1 2 1 2 jv 0 -2 −Kττ ττ + →12 ≥−→1 K ≤ 12 ττ+ ττ -4 12 12 -6 -6 -4 -2 0 2 4 sites.google.com/site/ncpdhbkhn u 16 The Nyquist Criterion (8) Ex. 3 K τ+ τ L( s ) = K ≤ 1 2 τ+ τ + τ τ s(1 s 1)( 2 s 1) 1 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 jv jv 0 jv 0 0 -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 ) = 0 σ 2 τ + − 1 ε s( s 1) τ A K K − L() j ω= = ∠−− [tan()] π1 ωτ 2 4 2 6 Γ −ω(j ωτ + 1) ω+ τ ω F s C= j ω s ω>0, ω → 0 2 →=ω =K ∠− π 1.5 CL lim( L j ) lim ( ) ω→0 ω → 0 ω2 1 = ω 0.5 Ds j ω→∞ jv 0 →=ω =K ∠− π DL lim L ( j ) lim3 (3/2) ω→∞ ω →∞ τω -0.5 -1 B= ε e jφ s ε→0 -1.5 → =ε jφ = K −2 j φ BL lim Le ( ) lim e -2 ε→0 ε → 0 ε 2 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 sites.google.com/site/ncpdhbkhn u 18 The Nyquist Criterion (10) Ex. 5 2.5 jω 2 Y( s ) s – plane R( s ) 1 1 1.5 K1 D s −1 s 1 − ( ) C r → ∞ 0.5 B E jv 0 0 σ -0.5 ε 1 K -1 K1 L( s ) = 1 A s( s − 1) -1.5 Γ -2 F s -2.5 -3 -2 -1 0 u Ex. 6 jω 2 Y( s ) s – plane R( s ) 1 1 D 1.5 K1 − 1 (− ) s 1 s − r → ∞ ( ) K C 0.5 2 B E jv 0 − 1 0 1 σ -0.5 K2 ε K( K s + 1) -1 L( s ) = 1 2 A − -1.5 s( s 1) Γ F s -2 -5 -4 -3 -2 -1 0 sites.google.com/site/ncpdhbkhn u 19 The Nyquist Criterion (11) Ex. 6 jω 2 Y( s ) s – plane R( s ) 1 1 D 1.5 K1 − 1 (− ) s 1 s − r → ∞ ( ) K C 0.5 2 B E jv 0 − 1 0 1 σ -0.5 K2 ε K( K s + 1) -1 L( s ) = 1 2 A − -1.5 s( s 1) Γ F s -2 -5 -4 -3 -2 -1 0 u KKj(ω+− 1) KK ω2 ( + 1) K ωω (1 − K 2 ) →=L( jω ) 12 = 12 + j 12 jωω( j − 1) ωω42 + ωω 42 + Kω(1− K ω 2 ) 1 Im{()}0L j ω=→1 2 =→= 0 ω 2 ω4+ ω 2 K2 −ω2 + ω =K1( K 2 1) = − Re{Lj ( )} ω2 = 4 2 KK1 2 1/ K2 ω+ ω ω2 = 1/ K2 − →=+=−+=→ IfKK12 1 KK 12 1 ZNP 110stable sites.google.com/site/ncpdhbkhn 20 The Nyquist Criterion (12) Ex. 6 jω 2 Y( s ) s – plane R( s ) 1 1 D 1.5 K1 − 1 (− ) s 1 s − r → ∞ ( ) K C 0.5 2 B E jv 0 − 1 0 1 σ -0.5 K2 ε > A -1 K1 K 2 1 -1.5 Γ F s -2 -5 -4 -3 -2 -1 0 u K = 0.5, K = 3, K K = 1.5 K = 0.5, K = 2, K K = 1 K = 0.5, K = 1, K K = 0.5 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 1 1 1 jv jv 0 jv 0 0 -1 -1 -1 -2 -2 -2 -4 -2 0 -4 -2 0 -4 -2 0 u u sites.google.com/site/ncpdhbkhn u 21 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 22 Relative Stability Ex. 1 and the Nyquist Criterion (1) τ = 1; τ = 1 1 2 K 0.5 L( s ) = τ+ τ + s(1 s 1)( 2 s 1) K L( j ω ) = ωτω+ τ ω + j(1 j 1)( 2 j 1) 0 −ω2 τ + τ = K (1 2 ) ωτ4(+ τ ) 22 + ω (1 − ττω 22 ) jv τ τ 12 12 K 1 2 2 Kω(1− ττω ) -0.5 τ+ τ − j 1 2 1 2 ωτ4+ τ 22 + ω − ττω 22 (12 ) (1 12 ) K = 2.5 2 ω− ττω K = 1.2 ω = →K (11 2 ) = Im{L ( j )} 04 22 22 0 K = 0.3 ωτ(+ τ ) + ω (1 − ττω ) -1 12 12 -1.5 -1 -0.5 0 u →ω = 1 τ τ 1 2 −Kωττ2 ( + ) − K ττ ω =12 = 12 Re{L ( j )} ω= τ τ 4 22 22 1/ 1 2 ωττ++− ω ττω ττ + (12 ) (1 12 ) ω= τ τ 12 1/ 1 2 sites.google.com/site/ncpdhbkhn 23 Relative Stability and the Nyquist Criterion (2) τ = 1; τ = 1 1 2 0.5 Phase difference before instability 0 α jv d = = 1 Gain marginGM 20log d -0.5 K = 2.5 K = 1.2 = Φ = α K = 0.3 Phase margin M -1 -1.5 -1 -0.5 0 u Gain difference before instability sites.google.com/site/ncpdhbkhn 24 Relative Stability Ex. 2 and the Nyquist Criterion (3) 6 0.5 Given L( s ) = (s+ 2)( s2 + 2 s + 2) Phase Find its gain & phase margin? difference before 6 0 L( j ω ) = instability α (jω+ 2)( −+ ω2 2 j ω + 2) jv −ω 2 d = 24(1 ) 22 2 22 16(1−ω ) + ω (6 − ω ) -0.5 6(6ω− ω 2 ) − j 16(1−ω22 ) + ω 2 (6 − ω 22 ) 2 6(6ω− ω ) -1 Im{L }= =→= 0ω 2.45 rad/s -1.5 -1 -0.5 0 16(1−ω22 ) + ω 2 (6 − ω 22 ) u Gain difference before instability 24(1−ω 2 ) d=Re{ L } = = 0.3 ω=2.45 −ω22 + ω 2 − ω 22 16(1 ) (6 ) ω =2.45 1 1 Gain margin==G 20log = 20log = 10.46 dB M d 0.3 sites.google.com/site/ncpdhbkhn 25 Relative Stability Ex. 2 and the Nyquist Criterion (4) 6 0.5 Given L( s ) = (s+ 2)( s2 + 2 s + 2) Phase Find its gain & phase margin? difference before ω = 0 L( j ) 1 instability α 2 24(1−ω 2 )  → jv 22 2 22  d 16(1−ω ) + ω (6 − ω )  2 -0.5 6(6ω− ω 2 )  +  = 1 →ω = 1.253 rad/s 16(1−ω22 ) + ω 2 (6 − ω 22 )    → ∠ω = Im{L } L( j )ω = atan   1.253   -1 Re{L } ω=1.253  -1.5 -1 -0.5 0 u 6(6ω− ω 2 )  Gain difference before instability   16(1−ω22 ) + ω 2 (6 − ω 22 ) =atan  = − 112.3 o 24(1−ω 2 )  −ω22 + ω 2 − ω 22  16(1 ) (6 ) ω=1.253  → =Φ==−α o −− o = o PhasemarginM 112.3 ( 180) 67.7 sites.google.com/site/ncpdhbkhn 26 Relative Stability and the Nyquist Criterion (5) τ = 1; τ = 1 1 2 0.5 = = 1 Gain marginGM 20log Phase d difference The gain margin is the increase in the before 0 system gain when phase = –180 ° that instability α will result in a marginally stable jv d system with intersection of the –1 + j0 point on the Nyquist diagram. -0.5 K = 2.5 K = 1.2 K = 0.3 Phase margin = Φ = α -1 M -1.5 -1 -0.5 0 u The phase margin is the amount of phase Gain difference before instability shift of the L(jωt) at unity magnitude that will result in a marginally stable system with intersection of the – 1 + j0 point on the Nyquist diagram. sites.google.com/site/ncpdhbkhn 27 Relative Stability Ex. 2 and the Nyquist Criterion (6) 6 Gain margin=G = 10.46 dB,ω = 2.45 rad/s Given L( s ) = M (s+ 2)( s2 + 2 s + 2) Phasemargin=Φ= 67.7,o ω = 1.253rad/s Find its gain & phase margin? M Bode Diagram 5 0 -5 GM -10 Magnitude (dB) Magnitude -15 -20 -25 -45 1 1.5 2 2.5 3 3.5 4 -90 -135 Φ M Phase (deg) Phase -180 -225 1 1.5 2 2.5 3 3.5 4 Frequency (rad/s) sites.google.com/site/ncpdhbkhn 28 Relative Stability Ex. 3 and the Nyquist Criterion (7) 1 1 Compare Ls()= &() Ls = . 1 ss(+ 1)(0.2 s + 1)2 ss ( + 1) 2 Bode Diagram 10 0 GM 2 -10 GM 1 -20 Magnitude (dB) Magnitude > -30 GM1 G M 2 Φ > Φ -40 M1 M 2 -90 L 1 L -135 2 Φ M 1 -180 Φ M 2 Phase (deg) Phase -225 -270 0.5 0.6 0.7 0.8 0.9 1 2 3 4 Frequency (rad/s) sites.google.com/site/ncpdhbkhn 29 Relative Stability and the Nyquist Criterion (8) ω 2 ω2   ω  L( j ω ) = n =n ω2 +4 ζ 22 ω ∠−− 90 o atan ω ω+ ζω ωn   ζω  j( j 2n )   2 n  ω o →Φ =−90 − atan 1 M ζω 2 n Exact 0.9 Linear approximation ω 2 n ω2+ ζ 2 ω 2 = 0.8 c4 n 1 ω 0.7 c ζ ω2 0.6 c 4 2 → =4ζ +− 1 2 ζ 0.5 ω2 n 0.4 Damping ratio, Damping → Φ = 2 0.3 M atan 4+1/ζ 4 − 2 0.2 0.1 0 ζ= Φ ζ ≤ 0 10 20 30 40 50 60 70 80 90 100 0.01 , 0.7 Phase margin, Φ ( o ) M M sites.google.com/site/ncpdhbkhn 30 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 31 Time – Domain Performance Criteria in the Frequency Domain (1) R( s ) Y( s ) Y( j ω ) G( jω )( Gj ω ) Gc ( s ) G( s ) T( j ω ) = = c ω+ ω ω (− ) Rj()1 GjGjc ()() φ ω = M(ω ) e j ( ) H( s ) ω ω = + Gjc ( )( Gj ) u jv ω ω 1.5 Gc ( j )( Gj ) = M = 0.7 →M (ω ) = M 1.5 M =1 + ω ω 1 1Gc ( j )( Gj ) ujv+ uv2 + 2 0.5 = = = 1++ujv (1 ++ u ) 2 v 2 0 M 2 M = 0.5 -0.5 -1 2 2 2 M  2  M  →−u  += v   -1.5 1−M2   1 − M 2  -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 sites.google.com/site/ncpdhbkhn 32 Time – Domain Performance Criteria in the Frequency Domain (2) 1 4.5 K M 1 1 K M 2 2 4 0.5 3.5 0 M1 3 2.5 jv -0.5 Magnitude 2 M2 -1 1.5 K> K 1 2 1 -1.5 0.5 K1 K2 -2 0 -2.5 -2 -1.5 -1 -0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 u ω Polar plot of Frequency response of ω ω ωω= ω = ω Gc ( j )( Gj ) GjGjc ()() Lj () KPj () T( j ω ) = 1+ G ( jω )( Gj ω ) sites.google.com/site/ncpdhbkhn c 33 Time – Domain Performance Criteria in the Frequency Domain (3) 6 R( s ) Y( s ) φ = o 10 Gc ( s ) G( s ) (− ) 4 φ = o 20 H( s ) 2 Y( j ω ) G( jω )( Gj ω ) φ ω Tj()ω= =c = Me () ω j ( ) o ω+ ω ω φ = 30 Rj()1 GjGjc ()() 1.5 0 = M = 0.7 M 1.5 M =1 φ = − 30 o 1 -2 0.5 0 M = 2 M = 0.5 φ = − 20 o -4 -0.5 -1 φ = − 10 o -6 -1.5 -4 -3 -2 -1 0 1 2 3 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 2 2 2 2 2 2 1   111    M  2  M  u+  +− v  = 1 +  u−  + v =   2   2φ  4  φ 2  1−M2   1 − M 2  sites.google.com/site/ncpdhbkhn 34 Time – Domain Performance Criteria Ex. in the Frequency Domain (4) R( s ) Y( s ) =0.47 = G( s ) G( s ) GsGsc ()() ,()1 Hs c s( s2 + s + 1) (− ) H( s ) Nichols Chart 20 1 dB 15 3 dB 10 6 dB 5 0 -5 Open-Loop Gain (dB) GainOpen-Loop -10 -15 -20 -270 -225 -180 -135 -90 Open-Loop Phase (deg) sites.google.com/site/ncpdhbkhn 35 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 36 System Bandwidth (1) Ex. 1 0 T =1 = 1 1 Ts1() , Ts 2 () T s+1 5 s + 1 -3 2 -6 -8 -10 20log|T|, dB 20log|T|, -18 0 0.2 1 2 ω Step Response 1 0.8 0.6 Amplitude 0.4 T 1 0.2 T 2 0 0 2 4 6 8 10 Time (seconds) sites.google.com/site/ncpdhbkhn 37 System Bandwidth (2) Ex. 2 100 900 Ts()= ,() Ts = 1ss2++10 100 2 ss 2 ++ 30 900 10 Step Response T 1.2 1 T T 1 2 0 T 2 -3 1 -10 0.8 -20 -30 0.6 20log|T|, dB 20log|T|, Amplitude -40 0.4 -50 0.2 -60 -70 0 0 1 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 10 10 10 ω sites.google.com/site/ncpdhbkhn Time (seconds) 38 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 39 The Stability of Control Systems with Time Delays (1) • Time delay : the time interval between the start of an even at one point in a system and its resulting action at another point in the system. • A pure time delay, without attenuation: –sT Gd(s) = e • A pure time delay does not change the magnitude of the transfer function. • The Nyquist criterion remains valid for a system with a time delay. sites.google.com/site/ncpdhbkhn 40 The Stability of Control Systems T Ex. with Time Delays (2) −s + 1 e−sT ≈ 2 31.5 − T( s ) = e s T + +2 ++ s +1 (s 1)(30 ss 1)[( / 3) s / 3 1] 2 6 4 2 0 -2 20log10|T|, dB 20log10|T|, -4 -6 -0.3 -0.2 -0.1 0 0.1 10 10 10 10 10 ω -100 With time delay Without time delay -150 ) o ( -180 Φ φ without time delay -200 Φ with time delay -250 -0.3 -0.2 -0.1 0 0.1 10 10 10 10 10 ω sites.google.com/site/ncpdhbkhn 41 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 42 PID Controllers Ex. in the Frequency Domain (1) Td ( s ) Kω 2 R( s ) K ng Y( s ) K+I + K s Ps D (τs+ 1)( s2 + 2 ζω s + ω 2 ) (− ) ng n g H( s ) K = –7000, τ = 5s. Design the PID controller so that o o GM ≥ 6dB, 30 ≤ ΦM ≤ 60 , rise time Tr < 4s, time to peak TP < 10s. s2 +(/ KKs )(/ + KK ) Ls( ) = Kω 2 K PD ID ng D ss(τ+ 1)( s2 + 2 ζω s + ω 2 ) ng n g ζ = Φ= ×o = 0.01M 0.01 30 0.3 2.16ζ + 0.6 2.16 × 0.3 + 0.6 T = = 4ω 0.31 r ω ω n n n 2.16× 0.3 + 0.6 π π ω =→=0.4T == 3s, T = = 8s n r 0.4 P ω− ζ 2 − 2 n 1 0.4 1 0.3 sites.google.com/site/ncpdhbkhn 43 PID Controllers Ex. in the Frequency Domain (2) K = –7000, τ = 5s. Design the PID controller so that o o GM ≥ 6dB, 30 ≤ ΦM ≤ 60 , rise time Tr < 4s, time to peak TP < 10s. s2 +(/ KKs )(/ + KK ) = ω2 PD ID ζ= ω = Ls( ) Kn K D , 0.3,n 0.4 g ss(τ+ 1)( s2 + 2 ζω s + ω 2 ) ng n g Bode Diagram 20 10 0 GM -10 -20 Magnitude (dB) Magnitude -30 -40 -45 -90 -135 Phase (deg) Phase Φ -180 M -225 -1 0 1 10 10 10 Frequency (rad/s) sites.google.com/site/ncpdhbkhn 44 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 45 Stability in the Frequency Domain Using Control Design Software • nyquist • nichols • margin • pade • ngrid sites.google.com/site/ncpdhbkhn 46

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