MM4 System s Poles and Feedback Characteristics
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1 MM4 System s Poles ad Feedback Characteristics Readigs: Sectio 3.3 (resose & ole locatios ); Sectio 4.1 (basic roerties of feedback ); Extra readigs (feedback characterisitcs) 9/6/011 Classical Cotrol 1
2 What have we talked i MM3? The reose aalysis Performace secificatio Numerical simulatio 9/6/011 Classical Cotrol
3 9/6/011 Aalog ad Digital Cotrol 3 MM3: Time Resose Aalysis (I) d(t)=0 Tyical iut u(t) Time resose y(t) TF model ) (0 0) m( m ) (0) (0) ( f e a m m m T T v f Lalace Tras Iv Lalace T. i i i m i i i s a s b s G s U s G s Y 0 0 ) ( ) ( ) ( ) (
4 MM3: Time Resose Aalysis (II) Tyical iuts: imulse ste ad ram sigals 1st d ad high-order (LTI) systems G( s) G( s) time k s c s 1 domai : ole : ole : 1 time costat time costat 1 : : g( t) ke t or g( t) c e 1 t Time resose = excitatio resose + iitial coditio resose (free resose) 9/6/011 Classical Cotrol 4
5 MM3: Performace Secificatio t t r s M t % 16 % 35 % d d /6/011 Classical Cotrol 5
6 MM3: Numerical Simulatio Imulse resose: imulse(sys) Ste resose: ste(sys) ltiview(sys) Sublot(m1) EXAMPLE: sys1: Sys: um1=[1]; um=[1 ]; de1=[1 1]; de=[1 3]; imulse(tf(um1de1)'r'tf(umde)'b') ste(tf(um1de1)'r'tf(umde)'b') 9/6/011 Process Cotrol 6
7 Goals for this lecture (MM4) System oles vs. time resoses Poles ad zeros Time resoses vs. Pole locatios Feedback characteristics Characteristics A simle feedback desig Block diagram decomositio (simulik) 9/6/011 Classical Cotrol 7
8 System Poles First-order system k c 1 G( s) ole : G( s) ole : s s 1 Secod-order system G( s) s oles : 1 1 real (differet) oles : real (idetical) oles : if 1 if ξ 1 High-order system comlex oles : comlex oles : 1 1 j j 1 if 0 1 if 0 G ( s ) m i 0 i 0 b a i i s s i i ( s z ( s 1 1 )( s )( s z ) ( s ) ( s z m 9/6/011 Classical Cotrol 8 ) )
9 Pole vs Time Resose: First-order System (I) 1 G( s) s 1 1 ole: Imulseresose: assume 0 time costat: 1 y(t) L( ) e s 1 1 Steresose: y(t) L( s( s 1) 1 t ) 1e 1 t 63% Time costat why? 9 Time resose is determied by the time costat System ole is the egative of iverse time costat 9/6/011 Classical Cotrol
10 Pole vs Time Resose: First-order System (II) A desig roblem how to use this kowledge Seed-u the resose Elimiate the steady-state error S1=tf(0.95[10 1]); S=tf(1[1 1]); Ste(s1s) Seed-u the resose Shorte the time costat!... How? 9/6/011 Classical Cotrol 10
11 Time Resose: d-order System (I) 9/6/011 Classical Cotrol 11
12 Time Resose: d-order System (II) G( s) s assume 0 0 oles: 1 1 real(differet ) oles: real(idetical ) oles: if 1 if ξ 1 comlex oles: comlex oles: 1 1 j j 1 if 0 1 if 0 9/6/011 Classical Cotrol 1
13 d-order System: Poles vs Performace Pole locatios Time resose 9/6/011 Classical Cotrol 13
14 d-order System Pole Locatios (I) Examle: How is this figure geerated? How to iterretate the iformatio? s1=tf(1[1 1]); s=tf(1[ ]); s3=tf(1[ ]); s4=tf(1[1 0 1]); zma(s1ss3s4) sgrid 9/6/011 Process Cotrol 14
15 SGRID geerates a grid over a existig cotiuous s-lae root locus or ole-zero ma. Lies of costat damig ratio (zeta) ad atural frequecy (W) are draw. zma(sys); ZGRID geerates a grid over a existig discrete z-lae root locus or ole-zero ma. Lies of costat damig factor (zeta) ad atural frequecy (W) are draw i withi the uit Z-lae circle. 9/6/011 Process Cotrol 15
16 d-order System Pole Locatios (II) G( s) s assume 0 0 oles: 1 1 real(differet ) oles: real(idetical ) oles: if 1 if ξ 1 comlex oles: comlex oles: 1 1 j j 1 if 0 1 if 0 9/6/011 Process Cotrol 16
17 d-order System Pole Locatios (III) Execise: (1)Figure out the damig ratio ad atural frequecy of the followig systems ()Sketch all ole locatios i the s-lae accordig to the iformatio you get from (1) (3)Sketch ad comare the ste resoses of all systems ad exlai the results s1=tf(1[1 1]); s=tf(1[ ]); s3=tf(1[ ]); s4=tf(1[1 0 1]); zma(s1ss3s4) sgrid 9/6/011 Process Cotrol 17
18 MM3: Performace Secificatio t t r s M t % 16 % 35 % d d /6/011 Classical Cotrol 18
19 zma(s1ss3s4) ste(s1ss3s415); grid; 9/6/011 Classical Cotrol 19
20 Summary of Pole vs Performace (I) 9/6/011 Aalog ad Digital Cotrol 0
21 Summary of Pole vs Performace (II) Pole locatios Time resose 9/6/011 Aalog ad Digital Cotrol 1
22 Goals for this lecture (MM4) System oles vs. time resoses Poles ad zeros Time resoses vs. Pole locatios Feedback characteristics Characteristics A simle feedback desig Block diagram decomositio (simulik) 9/6/011 Classical Cotrol
23 See chater of Goodwi et al Material ca be dowloaded from course webage 9/6/011 Classical Cotrol 3
24 Revisit of examle : First-order System (I) A desig roblem how to use this kowledge Seed-u the resose Elimiate the steady-state error S1=tf(0.95[10 1]); S=tf(1[1 1]); Ste(s1s) Seed-u the resose Shorte the time costat!... How? 9/6/011 Classical Cotrol 4
25 Revisit of examle : First-order System (II) A desig solutio Dowload feedbackexamle.mdl 9/6/011 Classical Cotrol 5
26 Summary of Feedback Characteristics System errors ca be made less sesitive to disturbace with feedback tha they are i oeloo systems I feedback cotrol the error i the cotrolled quatity is less sesitive to variatios i the system gai/arameters Desig tradeoff betwee gai ad disturbace 9/6/011 Classical Cotrol 6
27 Goals for this lecture (MM4) System oles vs. time resoses Poles ad zeros Time resoses vs. Pole locatios Feedback characteristics Characteristics A simle feedback desig Block diagram decomositio (simulik) See blackboard 9/6/011 Classical Cotrol 7
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