Bài giảng Control system design - Chapter V: The Performance of Feedback Control Systems - Nguyễn Công Phương

Nguyễn Công Phương CONTROL SYSTEM DESIGN The Performance of Feedback Control Systems 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 Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 3 Introduction Performance Performance measure, M1 measure, M2 pmin Parameter, p sites.google.com/site/ncpdhbkhn 4 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 5 Test Input Signals (1) • If the system is stable, the response to a specific input signal will provide several measures of the performance. • But because the actual input signal of a system is usually unknown, a standard test input signal is normally chosen. • Using a standard input allows the designer to compare several competing designs. • Many control systems experience input signals that are very similar to the standard test signals. • 4 types: – Unit impulse, – Step, – Ramp, – Parabolic. sites.google.com/site/ncpdhbkhn 6 Test Input Signals (2) r() t r() t   A  2 2 0 t 0 t 1    , t  ,  0 A, t  0 A r( t ) 2 2 ; R ( s )  1 r();() t R s   0,t  0 s 0, otherwise Unit impulse Step Ramp Parabolic r() t r() t A 0 t 0 t At, t  0 A At2, t  0 2A r();() t R s  2 r();() t R s  3 0,t  0 s 0,t  0 s sites.google.com/site/ncpdhbkhn 7 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 8 Performance of Second – Order Systems (1) R() s 2 Y() s G() s  n G() s s( s  2n ) Y()() s R s () 1 G ( s ) 2 n  2 2 R() s s2n s   n 2 1 n R()() s  Y s  2 2 s s( s 2n s   n ) 1 nt 2 1 y( t )  1  e sin(n 1   t  cos  ) 1  2 1 1 ent sin(  t   )  n sites.google.com/site/ncpdhbkhn 9 Performance of Second – Order Systems (2) R() s 2 Y() s G() s  n s( s  2 ) () n 1nt2 1 1   n t r( t ) 1  y ( t )  1  e sin(n 1   t  cos  )  1  e sin(  n  t   ) 1  2  1.8 y() t  = 0.1 1.6  = 0.2  = 0.4 1.4  = 0.7  = 1.1 1.2  = 2.0 1 0.8 0.6 0.4 0.2 t 0 0 0.5 1 1.5 2 2.5 3 sites.google.com/site/ncpdhbkhn 10 Performance of Second – Order Systems (3) R() s 2 Y() s G() s  n G() s s( s  2n ) Y()() s R s () 1 G ( s ) 2 n  2 2 R() s s2n s   n 2 n R( s ) 1  Y ( s )  2 2 s2n s   n n nt 2 y( t )  e sin(n 1   t ) 1  2 1 1  ent sin(  t )  n sites.google.com/site/ncpdhbkhn 11 Performance of Second – Order Systems (4) R() s 2 Y() s G() s  n s( s  2 ) () n n nt2 1   n t r( t ) ( t )  y ( t )  e sin( n 1   t )  1  e sin(  n  t ) 1  2  4 y() t  = 0.10  = 0.25 3  = 0.50  = 1.1 2 1 0 -1 -2 t -3 0 0.5 1 1.5 2 2.5 3 sites.google.com/site/ncpdhbkhn 12 Performance of Second – Order Systems (5) 1 1 r( t ) 1  y ( t )  1  ent sin( 1  2 t  cos 1  )  1  e   n t sin(   t   ) y() t n n 1  2  1.6 M pt M final value Percent overshoot  pt 100% 1.4 final value Overshoot / 1   2 1.2 100e 1.0   1 0.9 1.0   0.8 0.6  Peak time T  p 2 0.4 1  n 4 Rise time Tr Settling time Ts  0.2 n Tr1 0.1 t 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 sites.google.com/site/ncpdhbkhn 13 Performance of Second – Order Systems (5) 120 Percent overshoot Peak time 100 80 60 40 20 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Damping ratio,  sites.google.com/site/ncpdhbkhn 14 Performance of Second – Order Systems (6) 1.6  = 10rad/s n 1.4  = 1rad/s n 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 15 Performance of Second – Order Ex. Systems (7) Find K & p so that the transient response to a step should be as R() s K Y() s fast as attainable while retaining an overshoot of less than 5%, s() s p and the settling time should be less than 4 seconds. () K G() ss() s p K 2 T() s     n 1 G ( s ) K 2 2 2 1 s ps  K s 2n s   n s() s p 2 2 Percent overshoot100 e/ 1   s1,2  n  j  n 1   2 100e(1/ 2) / 1(1/  2) 4 Ts   4 n 1  4.32% n n 1 s 1  j 1  1,2  2 n 1   1  p 2n  2(1/ 2) 2  2   1/ 2 2 2   K n ( 2)  2 n  2 sites.google.com/site/ncpdhbkhn 16 Performance of Second – Order Ex. Systems (8) Find K & p so that the transient response to a step should be as R() s K Y() s fast as attainable while retaining an overshoot of less than 5%, s() s p and the settling time should be less than 4 seconds. () K2; p  2 1.4 1.2 1 0.8 y(t) 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 17 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 18 Effects of a Third Pole & a Zero on the Second – Order System Response (1) 1 T() s  (s2  2 s  1)(  s  1) 1.4 1.2 1 0.8 y(t) 0.6 0.4  = 2.25  = 1.5  = 0.9 0.2  = 0.4  = 0.05  = 0.001 0 0 2 4 6 8 10 12 14 Time (s) sites.google.com/site/ncpdhbkhn 19 Effects of a Third Pole & a Zero on the Second – Order System Response (2) 1 T() s  (s2  2 s  1)(  s  1) 1 Percent Settling   overshoot time 2.25 0.444 0 9.63 1.50 0.666 3.9 6.30 0.90 1.111 12.3 8.81 0.40 2.50 18.6 8.67 0.05 20.0 20.5 8.37   0.45 sites.google.com/site/ncpdhbkhn 20 Effects of a Third Pole & a Zero on the Second – Order System Response (3) 2 (n /a )( s a ) T() s  2 2 s2n s   n 3.5 a/ = 5 n a/ = 2 n 3 a/ = 1 n a/ = 0.5 n 2.5 2 y(t) 1.5 1 0.5 0 0 1 2 3 4 5 6 7 8 9 10 Time (s) sites.google.com/site/ncpdhbkhn 21 Effects of a Third Pole & a Zero on the Second – Order System Response (4) 2 (n /a )( s a ) T() s  2 2 s2n s   n Percent Settling Peak a /n overshoot time time 5 23.1 8.0 3.0 2 39.7 7.6 2.2 1 89.9 10.1 1.8 0.5 210.0 10.3 1.5   0.45 sites.google.com/site/ncpdhbkhn 22 Effects of a Third Pole & a Zero on the Second Order System Response (5) Ex. – 62.5(s 2.5) 10( s  2.5) 4 3 T1();() s2 T 2 s  2 (s 6 s  25)( s  6.25) s  6 s  25 2 1 2 0 -1 ImaginaryPart -2 1.6 -3 T 1 -4 1.4 T -8 -6 -4 -2 0 2 2 Real Part 1.2 1 0.8 y(t) 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 Time (s) sites.google.com/site/ncpdhbkhn 23 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 24 The s – Plane Root Location & the Transient Response 1 MNMNA B s C Y( s ) i  k k  y ( t )  1  A eit  D e   k t sin( t   )  2 2 2 i  k k k si1 s  i k  1s2k s  (  k   k ) i  1 k  1 j 1 1 10 1 0 0 0 0 -1 -1 -1 -10 0 5 10 0 5 10 0 5 10 0 5 10 1 1 10 1 0 0 0 0 -1 -1 -1 -10 0 5 10 0 5 10 0 5 10 0 5 10 1 1 2 10  0.5 0.5 1 5 0 0 0 0 0 5 10 0 5 10 0 5 10 0 5 10 sites.google.com/site/ncpdhbkhn 25 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 26 The Steady – State Error of Feedback Control Systems (1) R() s Y() s 1 G() s G() s E()() s R s c 1 Gc ( s ) G ( s ) () Controller Process 1 lime ( t ) esteady state  lim s R ( s ) t s 0 1 Gc ( s ) G ( s ) A r()() t A  R s  s 1 AA ess lim s  s0 1Gc ()() s G s s 1lim  G c ()() s G s s0 Kp lim G c ( s ) G ( s ) s0 A ess  1 K p sites.google.com/site/ncpdhbkhn 27 The Steady – State Error of Feedback Control Systems (2) R() s Y() s 1 G() s G() s E()() s R s c 1 Gc ( s ) G ( s ) () Controller Process 1 lime ( t ) esteady state  lim s R ( s ) t s 0 1 Gc ( s ) G ( s ) A r()() t At  R s  s2 1 AA ess lim s  lim s02 s  0 1 Gc ( s ) G ( s )s sG c ( s ) G ( s ) Kv lim sG c ( s ) G ( s ) s0 A ess  Kv sites.google.com/site/ncpdhbkhn 28 The Steady – State Error of Feedback Control Systems (3) R() s Y() s 1 G() s G() s E()() s R s c 1 Gc ( s ) G ( s ) () Controller Process 1 lime ( t ) esteady state  lim s R ( s ) t s 0 1 Gc ( s ) G ( s ) At2 A r()() t  R s  2 s3 1 AA ess lim s  lim s03 s  0 2 1 Gc ( s ) G ( s ) s s Gc ()() s G s 2 Ka lim s G c ( s ) G ( s ) s0 A ess  Ka sites.google.com/site/ncpdhbkhn 29 The Steady – State Error of Feedback Control Systems (4) R() s Y() s 1 G() s G() s E()() s R s c 1 Gc ( s ) G ( s ) () Controller Process 2 Kplim GsGsK c ()(); v  lim sGsGsK c ()(); a  lim sGsGs c ()() s0 s  0 s  0 Number of Input Integrations Step, r(t) = A Ramp Parabola in Gc(s)G(s), Type Number R()/ s A s At,/ A s2 At2/ 2, A / s 3 A 0 ess  Infinite Infinite 1 K p A 1 ess  0 Infinite Kv A 2 ess  0 0 Ka sites.google.com/site/ncpdhbkhn 30 The Steady – State Error Ex. 1 of Feedback Control Systems (5) R() s K2 K Y() s Gc () s K1  G() s  () s  s 1 A R() s A  ess  1 K p KK2   Kplim G c ( s ) G ( s )  lim K1      s0 s  0 s  s 1  ess  0 A R() s At  ess  Kv KK2   Kplim sG c ( s ) G ( s )  lim s K1     KK2 s0 s  0 s  s 1  A ess  KK2 sites.google.com/site/ncpdhbkhn 31 The Steady – State Error Ex. 1 of Feedback Control Systems (6) R() s K K Y() s R( s ) A  e  0 G() s K  2 G() s  ss c 1 s  s 1 A () R() s At  ess  KK2 1 0.8 0.6 Step input Output 0.4 Error 0.2 0 0 1 2 3 4 5 6 7 8 1 Ramp input Output 0.5 Error 0 -0.5 -1 0 1 2 3 4 5 6 7 8 sites.google.com/site/ncpdhbkhn 32 The Steady – State Error Ex. 1 of Feedback Control Systems (7) R() s K K Y() s R( s ) A  e  0 G() s K  2 G() s  ss c 1 s  s 1 A () R() s At  ess  KK2 1 K = 0.2 2 K = 2 0.5 2 K = 10 2 0 K = 20 2 Error of Step -0.5 -1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.4 K = 0.2 2 0.3 K = 2 2 K = 10 0.2 2 K = 20 0.1 2 Error of Ramp 0 -0.1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 sites.google.com/site/ncpdhbkhn 33 The Steady – State Error of Feedback Control Systems (8) R() s Y() s K1 Gc () s G() s () Controller Process H() s Sensor K H() s  1  s 1 1 s [1  K G ( s ) G ( s )] E()() s  1 c R s  s1  K1 Gc ( s ) G ( s ) 1 ess lim sE ( s )  s0 1 K1 lim Gc ( s ) G ( s ) s0 sites.google.com/site/ncpdhbkhn 34 The Steady – State Error Ex. 2 of Feedback Control Systems (9) R() s 1 Y() s K 40 1 s  5 () Controller 3 Process Step input 2 Output 2 Error 0.1s  1 1 Sensor 0 1 ess  -1 1 K1 lim Gc ( s ) G ( s ) s0 -2 0 1 2 3 4 5 6 7 8 1  1.5 1 Ramp input 1 2lim40 1 Output s0 s  5 Error 0.5 0.059  5.9% 0 -0.5 -1 -1.5 0 1 2 3 4 5 6 7 8 sites.google.com/site/ncpdhbkhn 35 The Steady – State Error Ex. 3 of Feedback Control Systems (10) R() s 1 Y() s Find K so that the steady – state error to a step K input is minimize? () Controller s  5 Process G()() s G s K( s  4) 2 T() s c  s  4 1Gc ( s ) G ( s ) H ( s ) ( s  2)( s  4)  2 K Sensor 1.2 Es()()()()()() Rs  Ys  Rs  TsRs 1 [1  T ( s )] R ( s ) 0.8 1 R( s )  ess  lim sE ( s ) 0.6 s s0 Step input Output 1 0.4 Error lims [1  T ( s )] s0 s 1  T (0) 0.2 0 4K ess 0  T (0)   1  K  4 -0.2 8 2K 0 0.5 1 1.5 2 2.5 3 sites.google.com/site/ncpdhbkhn 36 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 37 Performance Indices (1) • A performance index is a quantitative measure of the performance of a system and is chosen so that emphasis is given to the important system specifications. • A system is considered an optimum control system when the system parameters are adjusted so that the index reaches an extremum, commonly a minimum value. • A performance index must be a number that is always positive or zero. • Then the best system is defined as the system that minimizes this index. sites.google.com/site/ncpdhbkhn 38 Performance Indices (2) 1 e(t)=r(t)-y(t) 0.5 0 -0.5 -1 The Integral of the Square of the Error 0 5 10 15 20 T 2 1 ISE e ( t ) dt 2 0 0.8 e (t) 0.6 0.4 0.2 0 0 5 10 15 20 2 100 Input r(t) 80 1.5 Output y(t) 60 1 40 0.5 20 ISE 0 0 0 5 10 15 20 0 5 10 15 20 sites.google.com/site/ncpdhbkhn 39 Performance Indices (3) T Integral of Absolute magnitude of Error, IAE IAE e ( t ) dt 200 0 150 T ITAE t e ( t ) dt 100 0 50 T 0 ITSE te2 ( t ) dt 0 5 10 15 20 0 Integral of Time multiplied by Absolute Error, ITAE 1 3000 e(t)=r(t)-y(t) 0.5 2000 0 1000 -0.5 -1 0 0 5 10 15 20 0 5 10 15 20 Integral of Time multiplied by Squared Error, ITSE 100 250 80 200 60 150 40 100 20 50 ISE 0 0 0 5 10 15 20 0 5 10 15 20 sites.google.com/site/ncpdhbkhn 40 Performance Indices (4) Ex. 1 1 1 R() s 1 1 Y() s T() s  2  2 s  2 s2 s  1 s2  0.75 s  1 () s 8 ISE ITAE 7 ITSE 6 5 4 Indices 3 2 1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2  sites.google.com/site/ncpdhbkhn 41 T() s Performance Indices (5) d Ex. 2 R() s K1 K2 X() s Y() s Find K3 to minimize the effect of the disturbance? () s s () P  Y() s k k k  K3 Td () s  K p N  1 LLLLLLn   n m   n m p  ... n1 n , m n , m , p nontouching nontouching = 1 – (sum of all different loop gains) + (sum of the gain products of all combinations of two nontouching loops) – (sum of the gain products of all combinations of three nontouching loops) + ... KKK1   1 2   1  KK3     p   s   s s   KKK 1 K1  K 1 2  1  K K s1  K K K s  2 3sp s s 1 3 1 2 p sites.google.com/site/ncpdhbkhn 42 T() s Performance Indices (6) d Ex. 2 R() s K1 K2 X() s Y() s Find K3 to minimize the effect of the disturbance? () s s () P  Y() s k k k  K3 Td () s  K p 1  2  1 K1 K 3 s  K 1 K 2 Kp s Pk: gain of kth path from input to output Δk (cofactor): the determinant Δ with the loop(s) touching the kth path removed. P1 1 1 1  2 1  K 1 K 3 s Loop ofK1 K 2 Kp s removed 1 Y() s 1(1K1 K 3 s ) s ( s  K 1 K 3 )  1  2  2 Td () s 1KKs1 3  KKKs 1 2p s  KKs 1 3  KKK 1 2 p sites.google.com/site/ncpdhbkhn 43 T() s Performance Indices (7) d Ex. 2 R() s K1 K2 X() s Y() s Find K3 to minimize the effect of the disturbance? () s s () Y() s s() s K K  1 3 2 K3 Td () s s K1 K 3 s  K 1 K 2 K p K p KKKK10.5; 1 2 p  2.5 Td ( s ) 1/ s s( s 0.5 K3 ) 1 Y(). s  2 s0.5 K3 s  2.5 s 10 0.25K3 t    2 y() t  e sin t    ,   10(/2)  K3  2     210 0.5K3 t 2   1 I y( t ) dt  e sin  t   dt   0.1K3 0  0 2  2  K3 dI 2  K3 0.1  0 K3 10  3.2 dK3 sites.google.com/site/ncpdhbkhn 44 T() s Performance Indices (8) d Ex. 2 R() s K1 K2 X() s Y() s Find K3 to minimize the effect of the disturbance? () s s () Y( s ) s ( s  1.6)  2 K3 Td () s s1.6 s  2.5 K 1.2 p 1 0.8 0.6 t (t) d y(t) 0.4 0.2 0 -0.2 0 1 2 3 4 5 6 7 8 9 10 sites.google.com/site/ncpdhbkhn 45 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 46 The Simplification of Ex. 1 Linear Systems (1) KK /10 T(),() s T s  1s( s 2)( s  10) 2 s ( s  2) 1 0.8 0.6 T 1 0.4 0.2 0 0 1 2 3 4 5 6 1 0.8 0.6 T 2 0.4 0.2 0 0 1 2 3 4 5 6 sites.google.com/site/ncpdhbkhn 47 The Simplification of Linear Systems (2) m m1 p am s a m1 s ...  a 1 s  1 cp s...  c1 s  1 GHL()(), s Kn n1  G s  K g p  g  n bn s b n1 s ...  b 1 s  1 d g s  ...  d 1 s  1 G()() s M s H  GL ()() s s d k M()k ()() s M s dsk d k ()k ()()s   s dsk 2q ( 1)k qMM( k ) (0) (2 q  k ) (0) M2q  , q  0,1,2,... k0 k!(2 q k )! 2q ( 1)k q ( k ) (0)  (2 q  k ) (0) 2q  ,q  0,1,2,... k0 k!(2 q k )! M2q  2 q ( q  1,2,...)  c , d sites.google.com/site/ncpdhbkhn 48 The Simplification of Ex. 1 Linear Systems (3) 10 1 GHL()() s3 2  G s  2 s10 s  16 s  10 d2 s  d 1 s  1 1 G() s  H 0.1s3 s 2  1.6 s  1 2 GH ( s ) d2 s d 1 s  1 M ( s ) 3 2  GL ()() s0.1s s  1.6 s  1  s (0) 2 (0) (0) 3 2 (0) M( s ) d2 s  d 1 s  1  M (0)  1 (s )  0.1 s  s  1.6 s  1   (0)  1 (1) (1) (1) 2 (1) M( s ) 2 d2 s  d 1  M (0)  d 1 (s )  0.3 s  2 s  1.6   (0)  1.6 (2) (2) (2) (2) M( s ) 2 d2  M (0)  2 d 2 (s )  0.6 s  2   (0)  2 M(3)( s ) 0  M (3) (0)  0 (3)(s )  0.6   (3) (0)  0.6 sites.google.com/site/ncpdhbkhn 49 The Simplification of Ex. 1 Linear Systems (4) 10 1 GHL()() s3 2  G s  2 s10 s  16 s  10 d2 s  d 1 s  1 (0) (1) (2) (3) M(0)1, M (0)  d1 , M (0)  2 d 2 , M (0)0  (0)(0)  1,  (1) (0)  1.6,  (2) (0)  2,  (3) (0)  0.6 2q ( 1)k qMM( k ) (0) (2 q  k ) (0) M 2q   k0 k!(2 q k )! ( 1)01MMMM (0) (0) (20)  (0) (  1) 11  (1) (0) (21)  (0) q1  M   2 0!(2 0)! 1!(2  1)! ( 1)2 1MM (2) (0) (2  2) (0)  2!(2 2)! ( 1)MMMMMM(0) (0) (2) (0) (1) (0) (1) (0) (  1) (2) (0) (0) (0)    2 1 2 1  2d d d  2 d  1 2  1 1  2  2d  d 2 2 1 2 2 1 sites.google.com/site/ncpdhbkhn 50 The Simplification of Ex. 1 Linear Systems (5) 10 1 GHL()() s3 2  G s  2 s10 s  16 s  10 d2 s  d 1 s  1 (0) (1) (2) (3) 2 M(0)1, M (0)  d1 , M (0)2,  d 2 M (0)0,  M 2   2 d 2  d 1 (0)(0)  1,  (1) (0)  1.6,  (2) (0)  2,  (3) (0)  0.6 2q ( 1)k q ( k ) (0)  (2 q  k ) (0) 2q   k0 k!(2 q k )! ( 1)01(0)  (0)  (20)  (0) (  1) 11(1)   (0)  (21)  (0) q 1     2 0!(2 0)! 1!(2  1)! ( 1)2 1  (2) (0)  (2  2) (0)  2!(2 2)! ( 1) (0) (0)  (2) (0)  (1) (0)  (1) (0) (  1)  (2) (0)  (0) (0)    2 1 2 1  2 1.6  1.6  2  1     0.56 2 1 2 sites.google.com/site/ncpdhbkhn 51 The Simplification of Ex. 1 Linear Systems (6) 10 1 GHL()() s3 2  G s  2 s10 s  16 s  10 d2 s  d 1 s  1 (0) (1) (2) (3) 2 M(0)1, M (0)  d1 , M (0)2,  d 2 M (0)0,  M 2   2 d 2  d 1 (0) (1) (2) (3) (0)  1,  (0)  1.6,  (0)  2,  (0)  0.6, 2  0.56 2 M2  2  2 d 2  d 1  0.56 d1 1.4864 MMMMMM(0)(0) (4) (0) (1) (0) (3) (0) (2) (0) (2) (0) q2  M    4 0!4! 1!3! 2!2! MMMM(1)(0) (3) (0) (0) (0) (4) (0)    d 2 3!1! 4!0! 2 (0)(0)  (4) (0)  (1) (0)  (3) (0)  (2) (0)  (2) (0)     4 0!4! 1!3! 2!2! (1)(0)  (3) (0)  (0) (0)  (4) (0)    0.68 3!1! 4!0! 2 M4  4  d 2  0.68 d2  0.8246 sites.google.com/site/ncpdhbkhn 52 The Simplification of Ex. 1 Linear Systems (7) 10 1 1 GHL()() s3 2  G s  2  2 s10 s  16 s  10 d2 s  d 1 s  1 0.8246s 1.4864 s  1 1 0.8 0.6 G H 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 1 0.8 0.6 G L 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 sites.google.com/site/ncpdhbkhn 53 The Simplification of Ex. 1 Linear Systems (8) 10 1 1 GHL()() s3 2  G s  2  2 s10 s  16 s  10 d2 s  d 1 s  1 0.8246s 1.4864 s  1 G G H L 1 3 0.8 0.6 2 0.4 1 0.2 3 2 0 0 -0.2 ImaginaryPart -1 ImaginaryPart -0.4 -2 -0.6 -3 -0.8 -1 -8 -7 -6 -5 -4 -3 -2 -1 0 1 -1 -0.5 0 0.5 1 Real Part Real Part sites.google.com/site/ncpdhbkhn 54 The Performance of Feedback Control Systems 1. Introduction 2. Test Input Signals 3. Performance of Second – Order Systems 4. Effects of a Third Pole & a Zero on the Second – Order System Response 5. The s – Plane Root Location & the Transient Response 6. The Steady – State Error of Feedback Control Systems 7. Performance Indices 8. The Simplification of Linear Systems 9. System Performance Using Control Design Software sites.google.com/site/ncpdhbkhn 55 System Performance Using Ex. Control Design Software R() s s  2 1 Y() s Gc () s  G() s  () s 0.1s  1 sites.google.com/site/ncpdhbkhn 56