Bài giảng Control system design - Chapter VII: The Root Locus Method - Nguyễn Công Phương

Nguyễn Công Phương CONTROL SYSTEM DESIGN The Root Locus Method 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 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 3 The Root Locus Concept (1) R( s ) Y( s ) Ys() KGs () K G( s ) T( s ) = = R() s 1+ KG () s (− ) 1+KG ()0 s = →KG( s ) = − 1 →KG() s ∠ [ KG ()] s =∠ 1180 o  KG( s )= 1 →  ∠[KG ()] s =+ 180o k 360; o k =±± 0, 1, 2,... The root locus is the path of the roots of the characteristic equation traced out in the s-plane as a system parameter varies from zero to infinity. sites.google.com/site/ncpdhbkhn 4 The Root Locus Concept (2) Ex. 1 R( s ) 1 Y( s ) K + K − s( s 2) 1+KG ()1 s =+ = 0 ( ) s( s + 2) →∆=++=+2 2ζω + ω 2 = ()2ss sKs 2n s n 0 → =−ζωωζ ±2 −=−± ωζ 2 − s1,2 n n1 1 n 1 Ex. 2 R( s ) 1 Y( s ) 10 s( s+ a ) (− ) sites.google.com/site/ncpdhbkhn 5 The Root Locus Concept (3) N ∆=−+ − + (s ) 1 ∑ Ln ∑ LL nm ∑ LLL nmp ... n=1 nm , nmp ,, nontouching nontouching =1 + F ( s ) ∆()0s =→ F () s =− 1 Kszszsz(+ )( + )( + )...( sz + ) F( s ) = 1 2 3 M + + + + (spspsp1 )( 2 )( 3 )...( sp n )  Ks+ z s + z ... F( s )=1 2 = 1  + + →  s ps1 p 2 ... ∠ =∠+ +∠+ +−∠+ +∠+ + =o + o  Fs()[( sz12 ) ( sz )...][( sp 1 ) ( sp 2 )...]180 k 360 sites.google.com/site/ncpdhbkhn 6 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 7 The Root Locus Procedure (1) 1. Prepare the root locus sketch. 2. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. 3. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. 4. Determine the points at which the locus crosses the imaginary axis (if it does so). 5. Determine the breakaway point on the real axis (if any). 6. Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. 7. Complete the root locus sketch. sites.google.com/site/ncpdhbkhn 8 The Root Locus Procedure (2) Step 1 1+F () s = 0 →+1KPs ( ) = 0, 0 ≤≤∞ K M + ∏ (s z i ) → +i=1 = 1K n 0 + ∏ (s p j ) j=1 n M nM → ++ +=↔1 ++ += ∏(spKszj ) ∏ ()0 i ∏ ()()0 sp ji ∏ sz j=1 i = 1K ji == 11 n = → + = K0∏ ( s p j )0 j=1 M →∞→ + = K∏( s z j ) 0 j=1 The locus of the roots of the characteristic equation 1 + KP (s) = 0 begins at the poles of P(s) and ends at the zeros of P(s) as K increases from zero to infinity. sites.google.com/site/ncpdhbkhn 9 The Root Locus Procedure (3) 1. Prepare the root locus sketch. 2. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols: the root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros 3. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. 4. Determine the points at which the locus crosses the imaginary axis (if it does so). 5. Determine the breakaway point on the real axis (if any). 6. Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. 7. Complete the root locus sketch. sites.google.com/site/ncpdhbkhn 10 The Root Locus Procedure (4) Ex. 1 Step 2 2(s + 2) 1+K = 0 s( s + 4) + s1 2 s1 − − s2 4 2 s1 0 + s1 4 The locus of the roots of the characteristic equation 1 + KP (s) = 0 begins at the poles of P(s) and ends at the zeros of P(s) as K increases from zero to infinity. - The number of separate loci is equal to the number of poles. - The root loci must be symmetrical with respect to the horizontal real axis. sites.google.com/site/ncpdhbkhn 11 The Root Locus Procedure (5) 1. Prepare the root locus sketch. 2. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. 3. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. 4. Determine the points at which the locus crosses the imaginary axis (if it does so). 5. Determine the breakaway point on the real axis (if any). 6. Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. 7. Complete the root locus sketch. sites.google.com/site/ncpdhbkhn 12 The Root Locus Procedure (6) Step 3 M + ∏ (s z i ) + =+i=1 = ≤≤∞ 1KPsK ()1n 0,0 K + ∏ (s p j ) j=1 n M − − − ∑(pj ) ∑ () z i polesofPs ( )− zerosof Ps ( ) σ =∑ ∑ = j=1 i = 1 A nM− nM − 2k + 1 φ =180o ,k = 0,1,2,...,( n −− M 1) A n− M sites.google.com/site/ncpdhbkhn 13 The Root Locus Procedure (7) Ex. 2 Step 3 (s + 1) 1+K = 0 s( s+ 2)( s + 4) 2 polesofPs ()− zerosof Ps ( ) σ = ∑ ∑ A n− M [(− 2) +− ( 4) +− ( 4)] −− ( 1) = = − 3 4− 1 2k + 1 − − − 0 φ =180o ,k = 0,1,2,...,( n −− M 1) 4 2 1 A n− M 2k + 1 =180o =+ (2k 1)60, o k = 0,1,2. 4− 1 = →φ = o k 0A 60 = →φ = o k 1A 180 = →φ = o k 2A 300 sites.google.com/site/ncpdhbkhn 14 The Root Locus Procedure (8) 1. Prepare the root locus sketch. 2. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. 3. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. 4. Determine the points at which the locus crosses the imaginary axis (if it does so). 5. Determine the breakaway point on the real axis (if any). 6. Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. 7. Complete the root locus sketch. sites.google.com/site/ncpdhbkhn 15 The Root Locus Procedure (9) Step 5 ∆()s =→ 0 psK () = dp( s ) breakaway point := 0 ds Ex. 3 (s + 1) 1+K = 0 s( s+ 2)( s + 3) −s( s + 2)( s + 3) →K = = p( s ) s +1 −++3 +++ 2 →dpdss =( 2)( s 3) = 2 s 8 s 10 s 6 ds dt s +1 (s + 1) 2 dp =→0 2sss3 + 8 2 + 10 +=→=− 60 s 2.46, s =− 0.77 ± j 0.79 ds 1 2,3 sites.google.com/site/ncpdhbkhn 16 The Root Locus Procedure (10) 1. Prepare the root locus sketch. 2. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. 3. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. 4. Determine the points at which the locus crosses the imaginary axis (if it does so). 5. Determine the breakaway point on the real axis (if any). 6. Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion: The angle of locus departure from a pole is the difference between the net angle due to all other poles and zeros and the criterion angle of ±180 °(2k + 1). 7. Complete the root locus sketch. sites.google.com/site/ncpdhbkhn 17 The Root Locus Procedure (11) Step 6 θ 1 The angle of locus departure from a θ pole is the difference between the 3 net angle due to all other poles and 0 zeros and the criterion angle of ±180 °(2k + 1). θ 2 θθθ++=o →= θ o −+ θθ 123180 1 180 ( 23 ) sites.google.com/site/ncpdhbkhn 18 The Root Locus Procedure (12) Ex. 2 5 4 p +K = 1 14 3 2 0 s+12 s + 64 s + 128 s 3 K →1 + = 0 2 ss(+ 4)( s ++ 4 js 4)( +− 4 j 4) = − + 1 p1 4 j 4 = − − p2 4 j 4 0 = − p3 p4 p3 4 -1 = p4 0 -2 1.VVV Prepare the root locus sketch. 2.VVV Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. -3 3. The loci proceed to the zeros at infinity along asymptotes centered at σ and with -4 A p2 angle ϕA. 4. Determine the points at which the locus -5 crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1 5. Determine the breakaway point on the real poles ofPs ( )− zeros of Ps ( ) axis (if any). σ = ∑ ∑ 6. Determine the angle of locus departure A from complex poles and the angle of locus n− M arrival at complex zeros, using the phase criterion. −−−44j 44 −+ j 4 = = − 3 7. Complete the root locus sketch. 4− 0 sites.google.com/site/ncpdhbkhn 19 The Root Locus Procedure (13) Ex. 2 5 4 p +K = 1 14 3 2 0 s+12 s + 64 s + 128 s 3 2k + 1 2 φ =180o ,k = 0,1,2,...,( n −− M 1) A − n M 1 2k + 1 0 =180o ,k = 0,1,2,3 p p − 3 4 4 0 -1 -2 1.VVV Prepare the root locus sketch. 2.VVV Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. -3 3.VVV The loci proceed to the zeros at infinity along asymptotes centered at σ and with -4 A p2 angle ϕA. 4. Determine the points at which the locus -5 crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1 5. Determine the breakaway point on the real = →φ = o axis (if any). k 0A 45 6. Determine the angle of locus departure = →φ = o from complex poles and the angle of locus k 1A 135 arrival at complex zeros, using the phase criterion. = →φ = o k 2A 225 7. Complete the root locus sketch. = →φ = o sites.google.com/site/ncpdhbkhn k 3A 315 20 The Root Locus Procedure (14) Ex. 2 5 4 p +K = 1 14 3 2 0 s+12 s + 64 s + 128 s 3 s4 1 64 K 2 3 s 12 128 0 1 s2 b 0 1 K 0 1 p3 p4 s c1 0 0 -1 s0 K -2 1.VVV Prepare the root locus sketch. 2.VVV Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. -3 3.VVV The loci proceed to the zeros at infinity along asymptotes centered at σ and with -4 A p2 angle ϕA. 4. Determine the points at which the locus VVV -5 crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1 5. Determine the breakaway point on the real ×− ×− axis (if any). =12 64 128 = = 53.33 128 12 K 6. Determine the angle of locus departure b1 53.33; c 1 from complex poles and the angle of locus 12 53.33 arrival at complex zeros, using the phase c>0 → K < 568.89 criterion. 1 7. Complete the root locus sketch. 2 + =→ =± 53.33s 568.89 0 s1,2 j 3.27 sites.google.com/site/ncpdhbkhn 21 The Root Locus Procedure (15) 5 θ Ex. 2 1 K 4 1+ = 0 4 3 2 p1 s+12 s + 64 s + 128 s 3 2 →=−+K( s4 12 s 3 + 64 s 2 + 128 sps ) = ( ) dp 1 → =−(4s3 + 36 s 2 + 128 s + 128) θ θ 4 3 ds 0 p p dp 3 4 =→=−0s 1.58; s =− 3.71 ± 2.55 -1 ds 1 2,3 -2 1.VVV Prepare the root locus sketch. 2.VVV Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. -3 θ 3.VVV The loci proceed to the zeros at infinity 2 along asymptotes centered at σA and with -4 angle ϕA. p2 4. Determine the points at which the locus VVV -5 crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1 5. Determine the breakaway point on the real VVV θ+ θ + θ + θ = o axis (if any). 1 2 3 4 180 6.VVV Determine the angle of locus departure from complex poles and the angle of locus →=θo − θθθ ++ arrival at complex zeros, using the phase 1180( 234 ) criterion. 7. Complete the root locus sketch. =180oooo − (90 + 135 + 90) =− 135 o = 225 o sites.google.com/site/ncpdhbkhn 22 The Root Locus Procedure (16) Ex. 2 5 K 4 1+ = 0 4 3 2 p1 s+12 s + 64 s + 128 s 3 2 1 0 p3 p4 -1 -2 1.VVV Prepare the root locus sketch. 2.VVV Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. -3 3.VVV The loci proceed to the zeros at infinity along asymptotes centered at σA and with -4 angle ϕA. p2 4. Determine the points at which the locus VVV -5 crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1 5.VVV Determine the breakaway point on the real axis (if any). 6.VVV Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. 7.VVV Complete the root locus sketch. sites.google.com/site/ncpdhbkhn 23 The Root Locus Procedure (17) Ex. 2 K K 1+ = 0 →1 + = 0 s4+12 s 3 + 64 s 2 + 128 s ss(+ 4)( s ++ 4 js 4)( +− 4 j 4) 5 4 p1 3 2 1 s 1 s 0 2 s 3 p3 p4 s -1 4 -2 -3 -4 p2 -5 -7 -6 sites.google.com/site/ncpdhbkhn-5 -4 -3 -2 -1 0 1 24 The Root Locus Procedure (18) Ex. 3 K Given the characteristic 1 + = 0. Find K so that s = –1 + j2? s4+12 s 3 + 64 s 2 + 128 s K 5 + + → = − 1 s1 4 j 4 ss(+ 4)( s ++ 4 js 4)( +− 4 j 4) 4 p1 →=Kss +4 s ++ 44 js +− 44 j 3 2 +=2 + 2 = s1 4 2 3 3.61 s1 1 s + 1 s1 4 2 2 s s++4 j 4 = 3 + 6 = 6.71 0 2 1 s 3 p3 p4 s s+4 − j 4 -1 4 1 +− =2 + 2 = s1 4 j 4 2 3 3.61 -2 =2 + 2 = s1 1 2 2.24 -3 →K =×××3.61 6.71 3.61 2.24 -4 s=−1 + j 2 p2 -5 = 195 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 sites.google.com/site/ncpdhbkhn 25 The Root Locus Procedure (19) Ex. 4 K( s2 + 16 s + 113) Given the characteristic equation 1 + = 0 . Find K so that: s2 ( s + 16) 1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds. 1 −ζω s2+2ζωω s += 2 0;()1 rt =→=− yt ()1 ent sin( ωβθ t + ) n n β n 1.8 ζ = 0.1 1.6 ζ = 0.2 ζ = 0.5 1.4 ζ = 0.7 ζ = 1.1 1.2 ζ = 2.0 1 0.8 0.6 0.4 0.2 0 -0.2 0 0.5 1 1.5 2 2.5 3 sites.google.com/site/ncpdhbkhn 26 The Root Locus Procedure (20) Ex. 4 K( s2 + 16 s + 113) Given the characteristic equation 1 + = 0 . Find K so that: s2 ( s + 16) 1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds. Ks(++ 8 js 7)( +− 8 j 7) 1+ = 0 s2 ( s + 16) 8 poles ofPs ( )− zeros of Ps ( ) 6 σ = ∑ ∑ A n− M 4 − −−− +−+ = (0 16) [( 8j 7) ( 8 j 7)] 2 3− 2 0 = 0 -2 2k + 1 φ =180o ,k = 0,1,...,( n −− M 1) A n− M -4 2k + 1 -6 φ =180,o k = 0 A 1 -8 o = 180 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 sites.google.com/site/ncpdhbkhn 27 The Root Locus Procedure (21) Ex. 4 K( s2 + 16 s + 113) Given the characteristic equation 1 + = 0 . Find K so that: s2 ( s + 16) 1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds. 3 s 1 16 8 s2 0 0 θ 1 6 1 s b1 0 4 s c1 2 θ θ −1 1 16 3, 4 b = θ 1 5 0 0 0 0 3 2 s+16 s -2 K= − = p( s ) s2 +16 s + 113 -4 dp = 1 -6 θ ds 2 θ+ θ + θ + θ + θ = o -8 1 2 3 4 5 180 →θ = o -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 3 90 sites.google.com/site/ncpdhbkhn 28 The Root Locus Procedure (22) Ex. 4 K( s2 + 16 s + 113) Given the characteristic equation 1 + = 0 . Find K so that: s2 ( s + 16) 1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds. 2 2 +ζω + ω = s2n s n 0 s 8 1 2 →=−ζω ± ω ζ − s s1,2 n j n 1 2 s * 6 s 3 4 Im{s } ω ζ 2 −1 1 k =1 =n =−1 ζω ζ 2 2 Re{s1 } n 0 ζ =0.5 →k = 1.73 -2 →s* =− 4.5 + j 8.0 -4 -6 4 4 →T = == 0.89 second -8 settling ζω n 4.5 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 sites.google.com/site/ncpdhbkhn 29 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 30 Parameter Design by the Root Locus Method (1) 1+KP ( s ) = 0, 0 ≤≤∞ K n+ n −1 +++= asn as n −1... asa 1 0 0 a s →1 +1 = 0 n+ n −1 ++ 2 + asn as n −1... asa 2 0 s3+ s 2 +β s + α = 0  α β=0 →s3 + s 2 + α = 01 → + = 0 → α *  2 +  s( s 1) β α*32→ + + βα + * → +s = → β  s s s 1 0  s3+ s 2 + α * sites.google.com/site/ncpdhbkhn 31 Parameter Design by the Root Ex. Locus Method (2) Design the system to satisfy the following R( s ) K1 Y( s ) specifications: G( s ) = s( s + 2) 1. Steady – state error for a ramp input is less (− ) (− ) than 35% of input slope. = 2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2 2. The settling time is less than 3 seconds. R/ s 2 R/ s2 R s+( K K + 2) E( s ) = = = 1 2 G( s ) K s 2 + + + 1+ 1 s( KK1 2 2) sK 1 1+ G () s H () s s( s + 2) 1+ K 1+ 1 K s s( s + 2) 2 R + + + = = =sKK(12 2) = KK 12 2 esteady state lim e () t lim sE () s lim s R t→∞ s → 0 s→0 2 + + + ss( KK1 2 2) sK 1 K 1 e K K + 2 ss ≤ 0.35 →1 2 ≤ 0.35 R K1 sites.google.com/site/ncpdhbkhn 32 Parameter Design by the Root Ex. Locus Method (3) Design the system to satisfy the following R( s ) K1 Y( s ) specifications: G( s ) = s( s + 2) 1. Steady – state error for a ramp input is less (− ) (− ) than 35% of input slope. = 2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2 2. The settling time is less than 3 seconds. 2+ζωω +=→=−± 2 ζωωζ 2 − ss2nn 0 s1,2 nn j 1 ζ = 0.707 Im{s } ω ζ 2 −1 1 k =1 =n =−1 ζω ζ 2 Re{s1 } n 1.33 ζ ≥0.707 →k ≤ 1 4 T = ≤→=3σ ζω ≥ 1.33 settlingζω n n sites.google.com/site/ncpdhbkhn 33 Parameter Design by the Root Ex. Locus Method (4) Design the system to satisfy the following R( s ) K1 Y( s ) specifications: G( s ) = s( s + 2) 1. Steady – state error for a ramp input is less (− ) (− ) than 35% of input slope. = 2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2 2. The settling time is less than 3 seconds. s2 +( KK ++= 2) sK 0 1 2 1 5 5 2 s s →s +2 s +β s += α 0 4 1 4 1 s s 2 2 β=→0s2 + 2 s + α = 0 3 3 s * α 2 2 →1 + = 0 1 1 s( s + 2) ζ = 0.707 0 0 α= = →+++=2 β 20K1 s 2 s s 200 -1 -1 β s -2 -2 →1 + = 0 s2 +2 s + 20 -3 -3 = − + →β = = -4 -4 s* 3.15 j 3.15 4.3 K1 K 2 -5 -5 → = -6 -4 -2 0 -6 -4 -2 0 K2 0.215 sites.google.com/site/ncpdhbkhn 34 Parameter Design by the Root Ex. Locus Method (5) Design the system to satisfy the following R( s ) K1 Y( s ) specifications: G( s ) = s( s + 2) 1. Steady – state error for a ramp input is less (− ) (− ) than 35% of input slope. = 2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2 2. The settling time is less than 3 seconds. = = 5 K120; K 2 0.215 s 4 1 s 2 × + 3 20 0.215 2 s * e = =0.315 ≤ 0.35 2 ss 20 1 ζ = 0.707 0 ζ = 0.707 -1 -2 4 4 -3 T == =1.27 < 3 seconds settling σ 3.15 -4 -5 -6 -4 -2 0 sites.google.com/site/ncpdhbkhn 35 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 36 Sensitivity and the Root Locus (1) ∂lnT ∂ T / T The logarithmic sensitivity: ST = = K ∂lnK ∂ K / K ∂r ∂ r ∆ r The root sensitivity: S ri =i = i ≈ i K ∂lnK ∂ KK / ∆ KK / sites.google.com/site/ncpdhbkhn 37 Sensitivity and the Root Locus Ex. (2) R( s ) K Y( s ) G( s ) = K 2 2 + β 1+ =↔++==++ 0sβ sK 0 s β s α (− ) s( s ) s( s + β ) αα= ±∆ αββ, = ±∆ β 0 0 0.8 s β =→+K = 1 0 1 1 0 s s( s + 1) 2 0.6 0.1 = r1 K 0.6 = → =− ± = K0.5 r1,2 0.5 j 0.5 r1 K 0.5 0.4 r−0.1 K = 0.4 = →=−±0.1 1 K0.6 (20%) r1,2 0.5 j 0.59 0.2 = →−0.1 =−± K0.4(20%) r1,2 0.5 j 0.39 0 ∆ −+ −−+ r1 r1 ( 0.5 j 0.59) ( 0.5 j 0.5) S + = = K ∆K/ K 0.1/0.5 -0.2 = = ∠ o −0.1 = j0.45 0.45 90 -0.4 r2 K 0.4 r2 K = 0.5 ∆r( −+ 0.5 j 0.39) −−+ ( 0.5 j 0.5) -0.6 0.1 = r1 1 r K 0.6 S − = = 2 K ∆K/ K 0.1/ 0.5 -0.8 o =−j0.55 = 0.55 ∠− ( 90 ) -1 -0.8 -0.6 -0.4 -0.2 0 sites.google.com/site/ncpdhbkhn 38 Sensitivity and the Root Locus Ex. (3) R( s ) K Y( s ) G( s ) = K 2 2 + β 1+ =↔++==++ 0sβ sK 0 s β s α (− ) s( s ) s( s + β ) αα= ±∆ αββ = ±∆ β 0, 0 1 2 (∆β ) s +∆β β=→1s + (1 +∆ β ) s + α = 0 →1 + = 0 -∆β 0 s2 + s + α 0.8 2 ∆β = − 0.2 −+∆(1β ) ± (1 +∆ β ) − 4 α 0.6 r = 1,2 2 ∆β = 0 0.4 ∆β = + ∆=+β → =−± 0.2 0.2(20%)r1,2 0.6 j 0.37 ∆=−β → =−± 0.2 0.2(20%)r1,2 0.4 j 0.58 ∆ −+ −−+ 0 r1 r1 ( 0.6 j 0.37) ( 0.5 j 0.5) Sβ + = = ∆β/ β 0.2/1 -0.2 o =0.82 ∠ ( − 127.6 ) -0.4 ∆r( −+ 0.4 j 0.58) −−+ ( 0.5 j 0.55) -0.6 S r1 =1 = β − ∆β β / 0.2 /1 -0.8 o =0.64 ∠ 38.7 -1 -1 -0.8 -0.6 -0.4 -0.2 0 sites.google.com/site/ncpdhbkhn 39 Sensitivity and the Root Locus Ex. (4) R( s ) K Y( s ) G( s ) = + β (− ) s( s ) r1 = = ∠ o SK+ j 0.45 0.45 90 r1 =− = ∠− o SK− j 0.55 0.55 ( 90 ) r1 o Sβ + =0.82 ∠ ( − 127.6 ) r1 o Sβ − =0.64 ∠ 38.7 sites.google.com/site/ncpdhbkhn 40 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 41 PID Controllers (1) K K s2 + K s + K PIDcontroller:GsK ( ) = +I + Ks = DPI c Ps D s K K s+ K PIcontroller:G ( s ) = K +I = P I c P s s = + PDcontroller:Gsc ( ) K P Ks D sites.google.com/site/ncpdhbkhn 42 PID Controllers (2) Ex. 1 2 R( s ) s2 +6 s + 10 1 Y( s ) s+6 s + 10 1 K K D (s+ 2)( s + 4) D s( s+ 2)( s + 4) (− ) s T( s ) = s2 +6 s + 10 1 1+ K D s( s+ 2)( s + 4) K( s2 + 6 s + 10) = D →+sK3( + 6) s 2 + (6 K ++ 8) sK 10 = 0 3+ + 2 + ++ D D D sK(D 6) s (6 K D 8) sK 10 D 2 s 1 s 1.5 2 s 3 1 0.5 0 -0.5 -1 -1.5 -2 -5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 sites.google.com/site/ncpdhbkhn 43 PID Controllers (3) 1 −ζω − =−±−→=ζωωζ2 nt ωζ −+2 1 ζ s1,2 n j n 1 yt () e sin(1n t cos) y( t ) 1 −ζ 2 1.6 M pt M− final value Percent overshoot = pt 100% 1.4 final value Overshoot 2 = −ζπ/ 1 − ζ 1.2 100 e 1.0 + δ 1 0.9 1.0 −δ 0.8 0.6 π Peak time T = p 2 0.4 ω1− ζ n 4 Rise time T Settling time T = r s ζω 0.2 n Tr1 0.1 t 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 PID Gain Percent Overshoot Settling Time Steady – State Error K Increasing P Increases Minimal impact Decreases K Increasing I Increases Increases Zero steady – state error K Increasing D Decreases Decreases No impact sites.google.com/site/ncpdhbkhn 44 PID Controllers (4) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. Step response with K = 885.5, K = 0, and K = 0 P I D 2 1.5 1 Amplitude 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) Root locus with K = 0, and K = 0 I D 10 = s KP 885.5 5 1 s 2 0 s 3 -5 -10 -15 -10 -5 0 sites.google.com/site/ncpdhbkhn 45 PID Controllers (4) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. Step response with K = 0, and K = 0 I D 2 K = 885.5 1.8 P K = 442.75 P 1.6 K = 370 P 1.4 1.2 1 Amplitude 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) sites.google.com/site/ncpdhbkhn 46 PID Controllers (5) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. Root locus with K = 0, and K = 370 I P 25 s 1 20 s 2 s 15 3 10 = −ζπ/ 1 − ζ 2 Percent overshoot100 e 5 4 0 Settling Time = ζω n -5 -10 -15 -20 -25 -14 PID-12 Gain-10 Percent-8 Overshoot-6 Settling-4 Time -2 0 K Increasing D Decreases Decreases sites.google.com/site/ncpdhbkhn 47 PID Controllers (6) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. 60 40 20 Percent Overshoot 0 = −ζπ/ 1 − ζ 2 0 10 20 30 40 50 Percent overshoot100 e K D 4 4 Settling Time = ζω n 3 2 Settling Time 1 0 10 20 30 40 50 K D PID Gain Percent Overshoot Settling Time K Increasing D Decreases Decreases sites.google.com/site/ncpdhbkhn 48 PID Controllers (7) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. Root locus with K = 370, K = 0 P D 100 80 60 ) -1 40 −ζπ − ζ 2 20 Percent overshoot=100 e / 1 0 4 -20 Settling Time = ζω n -40 Imaginary Axis ImaginaryAxis (seconds -60 -80 -100 -120 -100 -80 -60 -40 -20 0 20 40 60 Real Axis (seconds -1 ) PID Gain Percent Overshoot Settling Time K Increasing I Increases Increases sites.google.com/site/ncpdhbkhn 49 PID Controllers (8) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. 90 80 70 60 Percent Overshoot 50 −ζπ − ζ 2 Percent overshoot=100 e / 1 0 100 200 300 400 500 600 K I 4 20 Settling Time = ζω n 15 10 Settling Time 5 0 0 100 200 300 400 500 600 K I PID Gain Percent Overshoot Settling Time K Increasing I Increases Increases sites.google.com/site/ncpdhbkhn 50 PID Controllers (9) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. 4 60 3 40 2 20 Settling Time 1 Percent Overshoot 0 0 0 20 40 60 0 20 40 60 K K D D 20 90 15 80 10 70 Settling Time 5 60 Percent Overshoot 0 50 0 200 400 600 0 200 400 600 K K I sites.google.com/site/ncpdhbkhn I 51 PID Controllers (9) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. Step response with K =370, K = 100, and K = 60 P I D 1.4 1.2 1 0.8 Amplitude 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time (s) sites.google.com/site/ncpdhbkhn 52 PID Controllers (10) Ziegler-Nichols PID Controller Gain Tuning Using Closed-loop Concepts Controller Type KP KI KD P 0.5 KUltimate PI 0.45 KUltimate 0.54 KUltimate /TUltimate PID 0.6 KUltimate 1.2 KUltimate /TUltimate 0.6 KUltimate TUltimate /8 sites.google.com/site/ncpdhbkhn 53 PID Controllers (11) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. Step response with K = 885.5, K = 0, and K = 0 P I D 2 T = 0.83s 1.5 1 Amplitude 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) Root locus with K = 0, and K = 0 I D 10 = s KP 885.5 5 1 s 2 0 s 3 -5 -10 -15 -10 -5 0 sites.google.com/site/ncpdhbkhn 54 PID Controllers (12) Ex. 2 R( s ) 1 Y( s ) +KI + KP K D s + + ζω − s ss( b )( s 2n ) K=885.5; T = 0.83 ( ) U U β= ζ = ω = 10, 0.707,n 4. Ziegler-Nichols PID Controller Gain Tuning Using Closed-loop Concepts Controller Type KP KI KD P 0.5 KUltimate PI 0.45 KUltimate 0.54 KUltimate /TUltimate PID 0.6 KUltimate 1.2 KUltimate /TUltimate 0.6 KUltimate TUltimate /8 = =× = KP0.6 K U 0.6 885.5 531.3 =KU =885.5 = KI 1.2 1.2 1280.2 TU 0.83 K T 885.5× 0.83 K =0.6U U = 0.6 = 55.1 D 8 8 sites.google.com/site/ncpdhbkhn 55 PID Controllers (13) Ex. 2 R( s ) K 1 Y( s ) K+I + K s P D ss(+ b )( s + 2ζω ) (− ) s n β= ζ = ω = 10, 0.707,n 4. 1.6 Step response with the manual tuning 1.4 Step response with the Ziegler - Nichols PID tuning 1.2 1 0.8 Amplitude 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) sites.google.com/site/ncpdhbkhn 56 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 57 Negative Gain Root Locus Ex. s − 20 1+K = 0 s2 +5 s − 50 ) Root Locus -1 20 10 0 -10 -20 -20 -10 0 10 20 30 40 50 Imaginary Axis Axis Imaginary (seconds Real Axis (seconds -1 ) ) Root Locus -1 20 10 K = − 5.0 K = − 2.5 0 -10 -20 -20 -10 0 10 20 30 40 50 Imaginary Axis Imaginary Axis (seconds Real Axis (seconds -1 ) sites.google.com/site/ncpdhbkhn 58 The Root Locus Method 1. The Root Locus Concept 2. The Root Locus Procedure 3. Parameter Design by the Root Locus Method 4. Sensitivity and the Root Locus 5. PID Controllers 6. Negative Gain Root Locus 7. The Root Locus Using Control Design Software sites.google.com/site/ncpdhbkhn 59 The Root Locus Using Control Ex. 1 Design Software (1) −s + 20 1+K = 0 s2 +5 s − 50 • rlocus • rlocfind sites.google.com/site/ncpdhbkhn 60 The Root Locus Using Control Ex. 1 Design Software (2) s + 20 G( s )= = 0 s2 +5 s + 20 • step • impulse sites.google.com/site/ncpdhbkhn 61