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Stability of Closed-Loop Contol Systems

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Stability of Closed-Loop Contol Systems CHE302 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS Professor Dae Ryook Yang Fall 2001 Dept. of Chemical and Biological Engineering Korea University CHE302 Process Dynamics and Control Korea University 10-1 DEFINITION OF STABILITY • BIBO Stability – “An unconstrained 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 function are negative or have negative real parts (OLHP). Otherwise, the system is unstable. Þ What is the difference between the two definitions? – Open-loop stable/unstable – Closed-loop stable/unstable – Characteristic equation: 1+=GsOL ()0 – Nonlinear system stability: Lyapunov and Popov stability CHE302 Process Dynamics and Control Korea University 10-2 • Supplements for stability – For input-output model, • Asymptotic 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 occurring on the imaginary axis non-repeated. – If the imaginary pole is repeated the mode is tsin(wt) and it is unstable. – For state-space model, • Even though there are unstable poles and if they are cancelled by the zeros exactly (pole-zero cancellation), the system is BIBO stable. • Internally AS: if u(t)=0 for all time t implies that x(t) goes to zero with time for all initial conditions x(0). – Cancelled poles have to be in OLHP. CHE302 Process Dynamics and Control Korea University 10-3 EXAMPLES L G • Feedback control system L X1 + R + E P M + C G()sK= K G G G cc m c v p X 1 1 - 2 Gsv ()= Gsm ()= B 21s + s +1 Gm 1 G(s)==Gs() pL51s + Cs() KGGG = mcvp R(s)1+ GGGGcvpm Ksc (+1) = 32 10s+17s+81sK++ c – Using root-finding techniques, the poles can be calculated. – As Kc increases, the step response gets more oscillatory. – If Kc >12.6, the step response is unstable. CHE302 Process Dynamics and Control Korea University 10-4 • Simple Example 1 Gc(sK)=c, Gv(sK)=v, Gm(s)=1, Gp(s)=+Kspp/(t 1) Characteristic equation: 1+GOL(s)=1++KKcvKspp/(t 1) =0 ttps+(1+KKcvKp)=0 Þ s=-+(1KKKcvpp)/ \KKKcvp>-1 for stability – When Kp>0 and Kv>0, the controller should be reverse acting (Kc>0) for stability. • Simple example 2 Gc(sK)=c, Gv(s)=1/(2s+1), Gmp(s)=1, G(ss)=+1/(51) Characteristic equation: 1+Kc/[(2ss++1)(51)] =0 10s2 +7s+1+K=0 Þ sK=-éù7±-+4940(1)/20 ccëû \Kc >-1 for stability CHE302 Process Dynamics and Control Korea University 10-5 ROUTH-HURWITZ STABILITY CRITERION • From the characteristic equation of the form: nn-1 ans+ann-1s+L +a10s+aa=>0 (0) • Construct the Routh array aa ba=- nn- 2 / Row 11aa n- b1=-(an-1an-2anaann--31)/ nn--13 n aa sanaann--24L ba=- nn-4 / b2=-(an-1an-4anaann--51)/ 21n- n-1 aann--15 saaannn---135L n-2 M sbbb123 aa L cb=- nn--13/ n-3 11 c1=-(b1ann--3a1bb21)/ bb12 scc12L aann--15 c2=-(b1ann--5a1bb31)/ cb21=- / MM bb13 sz0 1 >0 M – A necessary condition for stability: all ai’s are positive – “A necessary and sufficient condition for all roots of the characteristic equation to have negative real parts is that all of the elements in the left column of the Routh array are positive.” CHE302 Process Dynamics and Control Korea University 10-6 • Example for Routh test – Characteristic equation 32 10s+17s+8sK+10+=c – Necessary condition 1+KKcc>01Þ>- • If any coefficient is not positive, stop and conclude the system is unstable. (at least one RHP pole, possibly more) – Routh array 3 17(8)-+10(1)K 108 c s bK1 ==-7.410.588 c 2 17 s 171+ Kc 17(0)-10(0) b ==0 1 2 17 s bb12 0 bK1(1+-c )17(0) s c1 cK1 ==+1 c b1 – Stable region b1 =7.41-0.588KKcc>0 and >-Þ1 -1<<K c 12.6 CHE302 Process Dynamics and Control Korea University 10-7 • Supplements for Routh test – It is valid only when the characteristic equation is a polynomial of s. (Time delay cannot be handled directly.) – If the characteristic equation contains time delay, use Pade approximation to make it as a polynomial of s. – Routh test can be used to test if the real part of all roots of characteristic equation are less than -c. • Original characteristic equation nn-1 ans+ann-1s+L +a10s+aa=>0 (0) • Modify characteristic equation and apply Routh criterion nn-1 ann(s+c)+a-1(s+c)+L +a10()s++ca ¢nn¢-1 ¢¢ =anns+a-1s+L +a10sa+=0 – The number of sign change in the 1st column of the Routh array indicates the number of poles in RHP. – If the two rows are proportional or the any of 1st column is zero, Routh array cannot be proceeded. CHE302 Process Dynamics and Control Korea University 10-8 • Remedy for special cases of Routh array – Only the pivot element is zero and others are not all zero • Replace zero with positive small number (e), and proceed. • If there is no sign change in the 1st column, it indicates there is a pair of pure imaginary roots (marginally stable ). If not, the sign change indicates the no. of RHP poles. – Entire row becomes zero (two rows are proportional) Im • It implies the characteristic polynomial is divided exactly by the Re polynomial one row above (always even-ordered polynomial). x x • Replace the row with the coefficients of the derivative (auxiliary polynomial) of the polynomial one row above and proceed. Im • This situation indicates at least either a pair of real roots x x Re symmetric about the origin (one unstable), and/or two complex x x pairs symmetric about the origin (one unstable pair). • If there is no sign change after the auxiliary polynomial, it Im indicated that the polynomial prior to the auxiliary polynomial x Re has all pure imaginary roots. x CHE302 Process Dynamics and Control Korea University 10-9 DIRECT SUBSTITUTION METHOD • Find the value of variable that locates the closed- loop poles at the imaginary axis (stability limit). • Example 32 – Characteristic equation: 1+GcGpc=+10s17s+8sK+10+= – On the imaginary axis s becomes j w . 3222 -10jww-178+jw+1+Kcm=(1+Kjcm -17w)+ww(8-=10)0 22 \(1+Kcm -17w)=0 and ww(8-=10)0 2 \ww=0 or =0.8ÞKKcm=-=1 or cm 12.6 – Try a test point such as Kc=0 10s32+17s+8s+1=(sss++1)(21)(5+=1)0 (All stable) \Stable range is -1<<Kc 12.6 CHE302 Process Dynamics and Control Korea University 10-10 ROOT LOCUS DIAGRAMS • Diagram shows the location of closed-loop poles (roots of characteristic equation) depending on the parameter value. (Single parametric study) • Find the roots as a function of parameter • Each loci starts at open-loop poles and approaches to zeros or ±¥ . For G(s)= N(s)/Ds() Stability limit lim(D(s)+KNc (s))==Ds()0 (poles) Kc ®0 lim(D(s)+KNc (s))==Ns()0 (zeros) Kc ®¥ Ex) (s+1)(s+2)(sK+3)+=20c Open-loop poles Þ<Kc 30 CHE302 Process Dynamics and Control Korea University 10-11 BODE STABILITY CRITERION • Bode stability criterion “A closed-loop system is unstable if the frequency response of the open-loop transfer function GOL=GcGvGpGm has an amplitude ratio greater than one at the critical frequency. Otherwise, the closed-loop system is stable.” – Applicable to open-loop stable systems with only one critical frequency – Example: 2K G = c OL (0.5s +1) 3 Kc AROL Classification 1 0.25 stable 4 1 Marginally stable 20 5 unstable CHE302 Process Dynamics and Control Korea University 10-12 GAIN MARGIN (GM) AND PHASE MARGIN (PM) • Margin: How close is a system to stability limit? • Gain Margin (GM) GM=1/AR()wc Critical or Phase – For stability, GM>1 crossover freq. • Phase Margin (PM) PM =fw(g )+°180 Gain crossover – For stability, PM>0 freq. • Rule of thumb – Well-tuned system: GM=1.7-2.0, PM=30° -45° – Large GM and PM: sluggish – Small GM and PM: oscillatory – If the uncertainty on process is small, tighter tuning is possible. CHE302 Process Dynamics and Control Korea University 10-13 EFFECT OF PID CONTROLLERS ON FREQUENCY RESPONSE • P – As Kc increases, AROL increases (faster but destabilizing) – No change in phase angle • PI – Increase AROL more at low freq. – As t I decreases, AROL increases (destabilizing) – More phase lag for lower freq. (moves critical freq. toward lower freq. => usually destabilizing) • PD – Increase AROL more at high freq. – As t D increases, AROL increases at high freq. (faster) – More phase lead for high freq. (moves critical freq. toward higher freq. => usually stabilizing) CHE302 Process Dynamics and Control Korea University 10-14 NYQUIST STABILITY CRITERION • Nyquist stability criterion “If N is the number of times that the Nyquist plot encircles the point (-1,0) in the complex plane in the clockwise direction, and P is the number of open-loop poles of GOL(s) that lies in RHP, then Z=N+P is the number of unstable roots of the closed-loop characteristic equation.” – Applicable to even unstable systems and the systems with multiple critical frequencies – The point (-1,0) corresponds to AR=1 and PA=-180°.
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