CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS. Professor Dae Ryook Yang

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CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 208 Dept. of Chemial and Biologial Engineering 0-

Road Map of the Leture X Stability of losed-loop ontrol system Definition General stability BIBO stanility Stability riteria Routh-Hurwitz stability riterion Diret substitution method Root lous method Bode stability riterion Gain margin and Phase margin Nyquist stability riterion Robustness 0-2

DEFINITION OF STABILITY BIBO Stability An unonstrained linear system is said to be stable if the output response is bounded for all bounded inputs. Otherwise it is said to be unstable. General Stability A linear system is stable if and only if all roots (poles) of the denominator in the transfer funtion are negative or have negative real parts (OLHP). Otherwise, the system is unstable. What is the differene between the two definitions? Open-loop stable/unstable Closed-loop stable/unstable Charateristi equation: GOL ( s) 0 Nonlinear system stability: Lyapunov and Popov stability 0-3

Supplements for stability For input-output model, Asymptoti stability (AS): For a system with zero equilibrium point, if u(t)=0 for all time t implies y(t) goes to zero with time. Same as General stability : all poles have to be in OLHP. Marginally stability (MS): For a system with zero equilibrium point, if u(t)=0 for all time t implies y(t) is bounded for all time. Same as BIBO stability: all poles have to be in OLHP or on the imaginary axis with any poles ourring on the imaginary axis non-repeated. If the imaginary pole is repeated the mode is tsin(wt) and it is unstable. For state-spae model, Even though there are unstable poles and if they are anelled by the zeros exatly (pole-zero anellation), the system is BIBO stable. Internally AS: if u(t)=0 for all time, it implies that x(t) goes to zero with time for all initial onditions x(0). Canelled poles have to be in OLHP. 0-4

EXAMPLES Feedbak ontrol system G() s K Gv () s Gm () s 2s Gp() s GL() s 5s Cs () KmGGG v p Rs () GGGG v p m s K ( s) s s s K 3 2 0 7 8 Using root-finding tehniques, the poles an be alulated. As K inreases, the step response gets more osillatory. If K >2.6, the step response is unstable. R + K m - E G P B G v G m M L G L G p + X 2 X + C 0-5

Simple Example G () s K, G () s K, G () s, G () s K /( s) v v m p p p Charateristi equation: G ( s) K K K /( s) =0 OL v p p s ( KKK )=0 s( KKK ) / p v p v p p KKK v p for stability When K p >0 and K v >0, the ontroller should be reverse ating (K >0) for stability. Simple example 2 G () s K, G () s /(2s), G () s, G () s /(5s) v m p Charateristi equation: K / (2s)(5s) =0 2 0s 7s K=0 s 7 49 40( K) / 20 K for stability 0-6

ROUTH-HURWITZ STABILITY CRITERION From the harateristi equation of the form: as a s as a a n n n n 0 0 ( n 0) Construt the Routh array Row s a a a n n n2 n4 n s an an3 an5 n2 s b b2 b3 n3 s 2 s 0 z >0 A neessary ondition for stability: all a i s are positive A neessary and suffiient ondition for all roots of the harateristi equation to have negative real parts is that all of the elements in the left olumn of the Routh array are positive. b ( a a a a )/ a b ( a a a a )/ a n n2 n n3 n 2 n n4 n n5 n ( ba a b )/ b ( ba a b )/ b n3 n 2 2 n5 n 3 b b a a n n2 n an an3 a a / a / a n n4 2 n an an5 a a / b n n3 b b2 a a / b n n5 2 b b3 0-7

Example for Routh test Charateristi equation 3 2 0s 7s 8sK 0 Neessary ondition K 0 K If any oeffiient is not positive, stop and onlude the system is unstable. (at least one RHP pole, possibly more) Routh array s s s s 3 2 0 0 8 7 K b b2 Stable region 7(8) 0( K ) b 7.40.588K 7 7(0) 0(0) b2 0 7 b ( K ) 7(0) K b b 7.40.588K 0 and K K 2.6 0-8

Supplements for Routh test It is valid only when the harateristi equation is a polynomial of s. (Time delay annot be handled diretly.) If the harateristi equation ontains time delay, use Pade approximation to make it as a polynomial of s. Routh test an be used to test if the real part of all roots of harateristi equation are less than -. Original harateristi equation as a s as a a n n n n 0 0 ( n 0) Modify harateristi equation and apply Routh riterion n a ( s) a ( s) a ( s) a n n n 0 n n as n a ns as a 0 0 The number of sign hange in the st olumn of the Routh array indiates the number of poles in RHP. If the two rows are proportional or any element of st olumn is zero, Routh array annot be proeeded. 0-9

Im x Im x x Im x x Remedy for speial ases of Routh array x x x Re Re Re Only the pivot element is zero and others are not all zero Replae zero with positive small number (e), and proeed. If there is no sign hange in the st olumn, it indiates there is a pair of pure imaginary roots (marginally stable ). If not, the sign hange indiates the no. of RHP poles. Entire row beomes zero (two rows are proportional) It implies the harateristi polynomial is divided exatly by the polynomial one row above (always even-ordered polynomial). Replae the row with the oeffiients of the derivative (auxiliary polynomial) of the polynomial one row above and proeed. This situation indiates at least either a pair of real roots symmetri about the origin (one unstable), and/or two omplex pairs symmetri about the origin (one unstable pair). If there is no sign hange after the auxiliary polynomial, it indiated that the polynomial prior to the auxiliary polynomial has all pure imaginary roots. 0-0

DIRECT SUBSTITUTION METHOD Find the value of variable that loates the losedloop poles at the imaginary axis (stability limit). Example 3 2 Charateristi equation: GG p 0s 7s 8sK 0 On the imaginary axis s beomes j. 3 2 2 2 0 j 7 8 j Km ( Km 7 ) j(8 0 ) 0 2 2 ( Km 7 ) 0 and (8 0 ) 0 2 0 or 0.8 Km or Km 2.6 Try a test point suh as K =0 3 2 0s 7s 8s ( s)(2s)(5s) 0 (All stable) Stable range is K 2.6 0-

ROOT LOCUS DIAGRAMS Diagram shows the loation of losed-loop poles (roots of harateristi equation) depending on the parameter value. (Single parametri study) Find the roots as a funtion of parameter Eah loi starts at open-loop poles and approahes to zeros or. For Gs () Ns ()/ Ds () lim( Ds () KNs ()) Ds () 0 (poles) K 0 lim( Ds () KNs ()) Ns () 0 (zeros) K Stability limit Ex) ( s)( s2)( s3) 2K 0 30 K Open-loop poles 0-2

BODE STABILITY CRITERION Bode stability riterion A losed-loop system is unstable if the frequeny response of the open-loop transfer funtion G OL =G G v G p G m has an amplitude ratio greater than one at the ritial frequeny. Otherwise, the losed-loop system is stable. Appliable to open-loop stable systems with only one ritial frequeny Example: G OL 2K (0.5s ) K AR OL Classifiation 0.25 stable 3 4 Marginally stable 20 5 unstable 0-3

GAIN MARGIN (GM) AND PHASE MARGIN (PM) Margin: How lose is a system to stability limit? Gain Margin (GM) GM / AR( ) For stability, GM> Phase Margin (PM) PM ( g ) 80 For stability, PM>0 Rule of thumb Well-tuned system: GM=.7-2.0, PM=30-45 Large GM and PM: sluggish Critial or Phase rossover freq. Small GM and PM: osillatory Gain rossover freq. If the unertainty on proess is small, tighter tuning is possible. 0-4

P PI PD EFFECT OF PID CONTROLLERS ON FREQUENCY RESPONSE As K inreases, AR OL inreases (faster but destabilizing) No hange in phase angle Inrease AR OL more at low freq. I As dereases, AR OL inreases (destabilizing) More phase lag for lower freq. (moves ritial freq. toward lower freq. => usually destabilizing) Inrease AR OL more at high freq. D As inreases, AR OL inreases at high freq. (faster) More phase lead for high freq. (moves ritial freq. toward higher freq. => usually stabilizing) 0-5

NYQUIST STABILITY CRITERION Nyquist stability riterion If N is the number of times that the Nyquist plot enirles the point (-,0) in the omplex plane in the lokwise diretion, and P is the number of open-loop poles of G OL (s) that lies in RHP, then Z=N+P is the number of unstable roots of the losed-loop harateristi equation. Appliable to even unstable systems and the systems with multiple ritial frequenies The point (-,0) orresponds to AR= and PA=-80. Negative N indiates the enirlement of (-,0) in ounterlokwise diretion. 0-6

Some examples K (2s)( s) Gs () st order lag 2nd order lag s(20s)(0 s)(0.5s) Pure time delay Im Im Im Im Re Re Re Re Stable 3rd order lag + P ontrol GM and PM Im Im KK GM Re Re - - K stable unstable PM 0-7

CLOSED-LOOP FREQUENCY RESPONSE Closed-loop amplitude ratio and phase angle M Y ( j ) R ( j ) Y( j ) R( j ) For set point hange, M should be unity as 0. (No offset) M should maintain at unity up to high frequeny as possible. (rapid approah to a new set point) A resonant peak (M p ) in M is desirable but not greater than.25. (large p implies faster response to a new set point) Large bandwidth ( bw ) indiates a relatively fast response with a short rise time. 0-8

Definition ROBUSTNESS Despite the small hange in the proess or some inauraies in the proess model, if the ontrol system is insensitive to the unertainties in the system and funtions properly. The robust ontrol system should be, despite the ertain size of unertainty of the model, Stable Maintaining reasonable performane Unertainty (onfidene level of the model): Proess gain, Time onstants, Model order, et. Input, output If unertainty is high, the performane speifiation annot be too tight: might ause even instability 0-9