LECTURETHREE-3P-I-Denginespeedgovernors1LEARNINGOBJECTIVESTounderstand single-input-single-output control concepts:Todefine and understand benefits &limitations of on/off control.To define the full PID control law in its main formsTo understand P/PI/PID /PD-control benefits and issues·To be able to perform basic steady-state error analysis of linear.systems using the Laplace Transform's Final Value TheoremTo be able to perform perturbation analysis of marine propulsion.engines in order to determine the rpm regulation problem·To apply robust control theory and poleplacement todesigndisturbance rejection PI and PID speed governors·To applyrobustness analysis for both parametric uncertainty andneglected dynamicsTo use available signals shipboard in order to improve robustness.of PIDenginecontrollers21
1 LECTURE THREE – 3 P-I-D engine speed governors 1 LEARNING OBJECTIVES • To understand single-input-single-output control concepts • To define and understand benefits & limitations of on/off control • To define the full PID control law in its main forms • To understand P / PI / PID / PD-control benefits and issues • To be able to perform basic steady-state error analysis of linear systems using the Laplace Transform’s Final Value Theorem • To be able to perform perturbation analysis of marine propulsion engines in order to determine the rpm regulation problem • To apply robust control theory and pole placement to design disturbance rejection PI and PID speed governors • To apply robustness analysis for both parametric uncertainty and neglected dynamics • To use available signals shipboard in order to improve robustness of PID engine controllers 2
PIDControlFundamentalsNik.XirosSingle-Input-Single-Output system controlDisturbancedReferencesignalResponseProcess(setpoint)Error(or output)inputHG(s)7y1C+K(s)uProcessControllerControl(or plant)signalFeedbackpath (orbranch)42
2 PID Control Fundamentals Nik. Xiros 3 4 r e d u y ∆ιαταραχή Ελεγκτής Σήµα ελέγχου Σφάλµα Έξοδος ∆ιεργασία Είσοδος διεργασίας Κλάδος ανατροφοδοτήσεως Σήµα αναφοράς K(s) G(s) Process input Process (or plant) Disturbance Control signal Controller Error Reference signal (setpoint) Response (or output) Feedback path (or branch) Single-Input-Single-Output system control
Single-Input-Single-Output system controlY(s) =G(s) [U(s)+ D(s)]U(s) = K(s) E(s) = K(s) [R(s) -Y(s)]]G(s)G(s)K(s)=Y(s)=D(s)R(s)+1+G(s)K(s)1+G(s)K(s)ON/OFFControl+A, e≥+Zd[A·sgn(e), le|≥ Za0,el<Zdu =0,lel<Za-A, e≤-Zdu+1e3
3 [ ] [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 ( ) ( ) 1 ( ) ( ) Y s G s U s D s U s K s E s K s R s Y s G s K s G s Y s R s D s G s K s G s K s = ⋅ + ⇒ = ⋅ = ⋅ − ⇒ = ⋅ + ⋅ + + 5 Single-Input-Single-Output system control , sgn( ), 0, 0, , d d d d d A e Z A e e Z u e Z e Z A e Z + ≥ + ⋅ ≥ = = < < − ≤ − 6 ON/OFF Control e u +1 -1
ON/OFFControl-EXAMPLEA*sgn(u)5S+1ReferenceResponseON/OFFProcesssignalcontrollawu+1ON/OFFControl-EXAMPLEAE0.1F1.0A = 15.0The controlsignal is notplottedbecauseitoscillatesathigh frequencybetyen-15and+15!!-
4 7 ON/OFF Control - EXAMPLE Σήµα αναφοράς A*sgn(u) Ελεγκτής ON/OFF 1 5s+1 ∆ιεργασία y Έξοδος ON/OFF Process Response control law Reference signal e u +1 -1 8 ON/OFF Control - EXAMPLE 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 Σήµα ελέγχου Απόκριση (έξοδος) A = 0.1 Control signal Response (output) 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 A = 1.0 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 A = 2.0 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 A = 15.0 Σήµα ελέγχου ταλαντώνεται µεταξύ -15 και +15!! The control signal is not plotted because it oscillates at high frequency between –15 and +15 !!
ON/OFFControl-EXAMPLE70.0Zo912009A*sgn(u)5s+1ReferenceResponseON/OFFProcesssignalcontrol lawTheP-I-D ControllerControlTrackingSetpointProportionalactionerrorvaluegainCkiOverallIntegralIntegratorgaingainActual★du/dtKdvalueDifferentialDerivative4gaindu(0)=K,-e(0)+K, Je(5)dE+Ka-e(t)dt2K,+K,-$+K, -s?FU(s)= K(s)-E(s)=+K.:sE(s)=-E(s)sS105
5 9 ON/OFF Control – EXAMPLE Σήµα αναφοράς A*sgn(u) Ελεγκτής ON/OFF 1 5s+1 ∆ιεργασία y Έξοδος ON/OFF Process Response control law Reference signal 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Zd = 0.03 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Zd = 0.3 The P-I-D Controller 10 Tracking error e Actual value y Setpoint value Control action u Kp Proportional gain K Overall gain s 1 Integrator Ki Integral gain Kd Differential gain du/dt Derivative 0 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) t p i d i i p d p d d u t K e t K e d K e t dt K K K s K s U s K s E s K K s E s E s s s = ⋅ + ⋅ + ⋅ ξ ξ + ⋅ + ⋅ = ⋅ = + + ⋅ ⋅ = ⋅ ∫
Forms of thePIDControlLawAnalyticformdu(t)=K, e(t)+K, Je(5)d$+Kde(t)dt0↑K,+K,s+K,-s?K;U(s)= K(s)-E(s) =L+KasK.E(s) =.E(s)sSPracticalform[e(c)d5-T.% 0)u(t)= K,-y(t)+dt11Proportionalcontrol[uo +K,-eo, e(t)>eou(t)=uo +Kp-e(t), -eo ≤e(t)≤eouo-Kpeo, e(t)<-eu+1e126
6 - Analytic form 0 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) t p i d i i p d p d d u t K e t K e d K e t dt K K K s K s U s K s E s K K s E s E s s s = ⋅ + ⋅ + ⋅ ξ ξ + ⋅ + ⋅ = ⋅ = + + ⋅ ⋅ = ⋅ ∫ 0 1 ( ) ( ) ( ) ( ) t p d i d u t K y t e d T y t T dt ξ ξ = ⋅ − + ⋅ − ⋅ ∫ 11 Forms of the PID Control Law - Practical form 0 0 0 0 0 0 0 0 0 , ( ) ( ) ( ), ( ) , ( ) P P P u K e e t e u t u K e t e e t e u K e e t e + ⋅ > = + ⋅ − ≤ ≤ − ⋅ < − 12 Proportional control e u +1 -1
Proportionalcontrol-EXAMPLE1Static plant equation y=Kss u=Kss (uo +K,e)Kso+Kk,=e=KsoClosed-loopequation y=1+KssK,1+KssK,Steady-state erroruo=r/KssK,→813Proportional control-EXAMPLE 2KssIst-order plantG(s)=Ts+1Closed-loopequationKssK,KssY(s) =D(s)R(s)+ts+1+KsK,Ts+1+KssK,147
7 Static plant equation 0 ( ) SS SS p y K u K u K e = ⋅ = ⋅ + ⋅ Closed-loop equation 0 0 1 1 SS SS p SS SS p SS p K u K K r r K u y e K K K K ⋅ + ⋅ − ⋅ = ⇒ = + + Steady-state error 0 / SS p u r K K = → ∞ 13 Proportional control – EXAMPLE 1 1 st-order plant ( ) 1 KSS G s τ s = + Closed-loop equation ( ) ( ) ( ) 1 1 SS p SS SS p SS p K K K Y s R s D s τ τ s K K s K K = ⋅ + ⋅ + + + + 14 Proportional control – EXAMPLE 2
Proportional control-Final Value Theoremd(t)=ustep(t) D(s)=1/ suKssD(s)lim(y(t)) = lim(sY(s))= limSTs+1+KsK,5-→0=Kss1lim (y(t) = lim1+KssK,Ts+1+KssK,50s15Proportional-Integralcontrolu(0)= K, e(t)+K, f'e(5)ds↑K,-s+K,K.i.E(s)U(s)=IK+·E(s) =sS168
8 ( ) ( ) ( ) 0 0 0 ( ) ( ) ( ) 1/ lim ( ) lim ( ) lim ( ) 1 1 lim ( ) lim 1 1 step SS t s s SS p SS SS t s SS p SS p d t u t D s s K y t sY s s D s s K K K K y t s s K K s K K τ τ →∞ → → →∞ → = ⇔ = ⇓ = ⋅ ⋅ + + ⇓ = ⋅ ⋅ = + + + = 15 Proportional control – Final Value Theorem 0 ( ) ( ) ( ) ( ) ( ) ( ) t p i p i i p u t K e t K e d K K s K U s K E s E s s s = ⋅ + ⋅ ξ ξ ⋅ + = + ⋅ = ⋅ ∫ 16 Proportional-Integral control
Proportional-Integral controlG(s)= P,(s)Generic linearplantPo(s)(K,+K,s)·p,(s)Closed-loop equation Y(s)=R(s)+s-po(s)+(K, +K, s)-p,(s)s-p,(s)D(s)+S-po(s)+(K, +K,-s)-p,(s)Closed-loop Characteristic PolynomialP.(s)=S·po(s)+(K,+K,s)-P,(s)17Proportional-Integral control-EXAMPLEClosed-loopequationfor1storderopen-loopplantKss '(K, +K,s)Y(s) =·R(s) +s-(1+ts)+ Kss (K,+K,s)Kss*sD(s)一s-(1+ts)+K (K,+K,s)Steady-state error using the Final Value Theoremd(t)=ustep (t) - D(s)=1 / sUKss-0(t→0)=lm(s.Y(0)= 0.(1+T:0)+kg (K,+, 0)=0189
9 Generic linear plant 0 ( ) ( ) ( ) n p s G s p s = Closed-loop equation 0 0 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) i p n i p n n i p n K K s p s Y s R s s p s K K s p s s p s D s s p s K K s p s + ⋅ ⋅ = ⋅ + ⋅ + + ⋅ ⋅ ⋅ + ⋅ ⋅ + + ⋅ ⋅ 17 Closed-loop Characteristic Polynomial 0 ( ) ( ) ( ) ( ) c i p n p s s p s K K s p s = ⋅ + + ⋅ ⋅ Proportional-Integral control Closed-loop equation for 1st order open-loop plant ( ) ( ) ( ) (1 ) ( ) ( ) (1 ) ( ) SS i p SS i p SS SS i p K K K s Y s R s s s K K K s K s D s s s K K K s τ τ ⋅ + = ⋅ + ⋅ + + ⋅ + ⋅ + ⋅ ⋅ + + ⋅ + 18 Proportional-Integral control – EXAMPLE Steady-state error using the Final Value Theorem ( ) 0 ( ) ( ) ( ) 1/ 0 ( ) lim ( ) 0 0 (1 0) ( 0) step SS s SS i p d t u t D s s K y t s Y s → τ K K K = ⇔ = ⇓ ⋅ → ∞ = ⋅ = = ⋅ + ⋅ + ⋅ + ⋅
Proportional-Integral-Differential controlFullPID controllerequationK,+K,s+Kas?+K+K,s)E(s)=U(s)=/ K, +.E(s)SClosed-loop equation for 1st order open-loop plantKs(K, +K,S+K,s*) R(s)+ Ksss·D(s)Y(s) =(t+KssKa)s?+(1+KssK,)s+KssK,Proportional-DifferentialcontrolClosed-loop equationfor1storderopen-loopplantKss(K, +Kas)Y(s)=R(s)(t+KssK)s+(1+KssK,)KssD(s)(T+KssKa)s+(1+KssK,Steady-state error analysis using Final Value TheoremKssy(t→0)=1+Ksk,2010
10 Closed-loop equation for 1st order open-loop plant ( ) ( ) 2 2 ( ) ( ) ( ) ( ) 1 SS i p d SS SS d SS p SS i K K K s K s R s K s D s Y s τ K K s K K s K K + + ⋅ + ⋅ = + + + + 19 Proportional-Integral-Differential control Full PID controller equation 2 ( ) ( ) ( ) i i p d p d K K K s K s U s K K s E s E s s s + + = + + ⋅ = ⋅ Closed-loop equation for 1st order open-loop plant ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 ( ) 1 SS p d SS d SS p SS SS d SS p K K K s Y s R s K K s K K K D s K K s K K τ τ + = ⋅ + + + + ⋅ + + + 20 Proportional-Differential control Steady-state error analysis using Final Value Theorem ( ) 1 SS SS p K y t K K → ∞ = +