The idea Automate estim at ion and co nt rol Self-tuning regulator Deterministic sTR Process parameters 1. The idea 2. Pole placement desig n parameters ndirect STR Reference Controller Process 4. Continuo us time str Output 5. Direct stR 6. Distur bances wit h known properties The separat io n principle 7. Conclusio ns Estimate parameters Desig n co nt roller as if the estimates were correct.(The Cert ainty Equivalence Principle) lers Direct and indirect co nt rolle Process mode Pole placement design . Process Model A(q)y(t)=B(q)(u(+)+v(t) Model Following rely e Causality Co ndit io ns A(q-4)y(t)=B"(q-1)(u(t-d)+v(t-d) ● xam ples Interpret at io n of poly Relative degree do= deg A(g)-deg B Relations to Model followi Same notat io n for co ntinuo us time systems Ay(t=B(ut+u(t) C K.J. Astrom and BWittenmark
Deterministic STR 1. The idea 2. Pole placement design 3. Indirect STR 4. Continuous time STR 5. Direct STR 6. Disturbances with known properties 7. Conclusions The idea Automate estimation and control Process parameters Controller design Estimation Controller Process Controller parameters Reference Input Output Specification Self-tuning regulator The separation principle Estimate parameters Design controller as if the estimates were correct. (The Certainty Equivalence Principle) Direct and indirect controllers Pole Placement Design Process Model Model Following Causality Conditions Examples Interpretation of polynomial Ao Relations to Model Following Summary Process Model A(q)y(t) = B(q) (u(t) + v(t)) Alternatively A (q1 )y(t) = B (q1 ) (u(t d0) + v(t d0)) Relative degree d0 = deg A(q) deg B Same notation for continuous time systems Ay(t) = B (u(t) + v(t)) c K. J. Åström and B. Wittenmark 1
Closed Loop system Model Following Process Closed loop response Ay(t)=B(u(t)+u(t) yt AR+ Bsc(t) Co nt roller Desired respo nse Ru(t)=Tue(t)-Sy(t) Amym(t)=Bm(t) Controller Process Perfect model foll B BT Bn Cancellat ions in LHS,B=B+B Closed loop system Bm=bb BR Ac=A.+ y()=AR+ BS W(t)+ AR+Bs u(t) R=RB+ BS u(t) H AR+ B ()-AE+B50( Closed loop charact eristic polyno mial AR+ B-S=A.Am=A AR+BS=A Causality Conditions Example 3.1 Cancellation of Zero Tuc-S Sampling wit hh=0.5 Hence H(Q)=609+b1 0.1065g+0.0902 R 0.1761q q2+am1q+am2q2-1.32065g+04966 deg tdeg b (4) CK.J. Astrom and BWittenmark
Closed Loop System Process Ay(t) = B (u(t) + v(t)) Controller Ru(t) = T uc(t) Sy(t) Controller u Process y B A uc Ru = Tuc − Sy v Σ Closed loop system y(t) = BT AR + BS uc(t) + BR AR + BS v(t) u(t) = AT AR + BS uc(t) BS AR + BS v(t) Closed loop characteristic polynomial AR + BS = Ac Model Following Closed loop response y(t) = BT AR + BS uc(t) Desired response Amym(t) = Bmuc(t) Perfect model following BT AR + BS = BT Ac = BmAm Cancellations in LHS, B = B+B Bm = BB0m Ac = AoAmB+ R = R0B+ (1) Hence AR0 + B S = AoAm = A0c T = AoB0m Causality Conditions Controller Ru = T uc Sy Hence deg S deg R deg T deg R Diophantine Equation AR + BS = Ac has many solutions R = R0 + QB S = S0 QA We have deg A > deg B deg Ac 2 deg A 1 Example 3.1 Cancellation of Zero G(s) = 1 s(s + 1) Sampling with h = 0:5 H(q) = b0q + b1 q2 + a1q + a2 = 0:1065q + 0:0902 q2 1:6065q + 0:6065 Bm(q) Am(q) = bm0q q2 + am1q + am2 = 0:1761q q2 1:3205q + 0:4966 B+ (q) = q + b1=b0 B (q) = b0 B0m(q) = bm0q=b0 (2) Choose Ao = 1 (q 2 + a1q + a2) 1 + b0(s0q + s1) = q 2 + am1q + am2 a1 + b0s0 = am1 a2 + b0s1 = am2 (3) s0 = am1 a1 b0 s1 = am2 a2b0 (4) c K. J. Åström and B. Wittenmark 2
Example 3.2-No Cance lation H()=a+a1q+a2=-16059+06065 Example 3. 3-Continuous time We have b+=1 andb-=B Bn(0)=B+b=4+一 G(s ose Dio phantine equation (q2+a19+a2)(q+71)+(bq+b1)(s0q+s1) Ao(s=s+a, Put q =-b1/bo solve for T1 B Xob+x1b061+ Dio phantine eq s(s+a(s+r1)+b(sos+s1) X1 Then solve fo Example 3. 3- Cont Identification of coefficients of powers of aT1+ bso =w+2cwao Inte rpre tation of poly nomial Ao What happens if we solve the pro blem wit h fb≠ state space met ho ds? What is the interpret ation of pole-zero cancel- T1=2Cw+ao-a C K.J. Astrom andB.Wittenmark
Example 3.2 - No Cancellation H(q) = b0q + b1 q2 + a1q + a2 = 0:1065q + 0:0902 q2 1:6065q + 0:6065 We have B+ = 1 and B = B = b0q + b1 Hm(q) = b0q + b1 q2 + am1q + am2 = bm0q + bm1 q2 + am1q + am2 Diophantine equation (q 2 + a1q + a2)(q + r1)+(b0q + b1)(s0q + s1) = (q 2 + am1q + am2)(q + ao) Put q = b1=b0 solve for r1 r1 = X0b 2 0 + X1b0b1 + X2b 2 1 b 2 1 a1b0b1 + a2b 2 0 where X0 = aoam2 X1 = a2 am2 aoam1 X2 = ao + am1 a1 (5) Then solve for s0 and s1. Example 3.3 - Continuous time G(s) = b s(s + a) Choose Ao (s) = s + ao and Bm(s) Am(s) = !2 s2 + 2!s + !2 Diophantine equation s(s + a)(s + r1) + b(s0s + s1) = (s 2 + 2!s + !2 )(s + ao ) Example 3.3 - Cont. Identi cation of coecients of powers of s a + r1 = 2! + ao ar1 + bs0 = !2 + 2!ao bs1 = !2ao (6) If b 6= 0 r1 = 2! + ao a s0 = ao2! + !2 ar1 b s1 = !2ao b (7) Interpretation of Polynomial A0 What happens if we solve the problem with state space methods? What is the interpretation of pole-zero cancellations? c K. J. Åström and B. Wittenmark 3
Relations to model followi The control law is Ru= Tuc-Sy № tice that Ro R ABm SBm Su mma ry Pole place me nt give s line ar e quations When do solutions e xist? T S ABm SBm S R Am Whe n do the y nat eXIst? ABmuc- R ()· What is the key difficulty? BAm E summation Proce ss model no distubances an(t-n bou(t-d (9) trod uang n bo Indirect ST r 1。 Estimation g(t-1)=(-y(t-1)…-y(t-n) (10) 2. An alo orithm The model be cores a re g re ssion y(t)="(t-1)9 Estimate S g iven by e(t)=6t-1)+K(t(t) c(t)=y(t)-g2(t-1)(t-1) P(t-1)p(t-1) K(t)=X+"(t-1)Pt-1)(t-(1) ()=÷(I-K(t)(t-1)Pt-1) C K.J. Astrom and BWittenmark
Relations to Model Following The control law is Ru = T uc Sy Notice that T R = AoB0m R = (AR0 + B S)B0m AmR = ABm BAm + SBm RAm Hence u = T R uc S R y = ABm BAm uc + SBm RAm uc S R y = ABm BAm uc S R (y ym) (8) u y Σ Σ u c A B S R B A − 1 ym B m A m Summary Pole placement gives linear equations When do solutions exist? When do they not exist? What is the key diculty? Indirect STR 1. Estimation 2. An Algorithm 3. Examples Estimation Process model (No disturbances!) y(t) = a1y(t 1) a2y(t 2) ::: any(t n) + b0u(t d0) + ::: + bmu(t d0 m) (9) Introducing T = ( a1 a2 ::: an b0 ::: bm ) ' T (t 1) = (y(t 1) y(t n) u(t d0) : : :u(t d0 m)) (10) The model becomes a regression y(t) = 'T (t 1) Estimates given by ^ (t) = ^ (t 1) + K(t)"(t) "(t) = y(t) 'T (t 1)^ (t 1) K(t) = P (t 1)'(t 1) + 'T (t 1)P (t 1)'(t 1) P (t) = 1 I K(t)'T (t 1) P (t 1) (11) c K. J. Åström and B. Wittenmark 4
t STR EXam ple 3. 4-Cancellation of ze ro 1. Data Aoa nd amand n 2. Est imate para mEt ers of modd by Rls Process mode 3. Obtain pdy nomials R and sby sov,ing Diopha n ine Eq uation y(t)+a1y(t-1)+a2y(t-2) 4. Compute Pdy nomial T= AoBm bou(t-1)+b1u(t-2) 5. Compute cont rd sigNal from Contrd law R (t)+miu(t-1)=touct-soy(t)-siy(t-1) 6. Goto 2 latio n EXam ple 3.5- No Cancellat io n Process mode y()+a1y(t-1)+a2y(t-2) bou(t-1)+b1u(t-2) u(t)+riu(t-1)= touc(t)-soy(t C K.J. Astrom and BWittenmark
An Indirect STR 1. Data Ao and Am and n 2. Estimate parameters of model by RLS 3. Obtain polynomials R and S by solving Diophantine Equation 4. Compute Polynomial T = AoB0m 5. Compute control signal from Ru = T uc Sy 6. Go to 2 Example 3.4 - Cancellation of Zero Process model y(t) + a1y(t 1) + a2y(t 2) = b0u(t 1) + b1u(t 2) Control law u(t) + r1u(t 1) = t0uc(t) s0y(t) s1y(t 1) Simulation 0 20 40 60 80 100 −1 0 1 0 20 40 60 80 100 −4 −2 0 2 Time Time uc y u 0 5 10 15 20 −2 −1 0 0 5 10 15 20 0.0 0.1 0.2 Time Time ^a2 ^a1 ^ b1 ^ b0 Example 3.5 - No Cancellation Process model y(t) + a1y(t 1) + a2y(t 2) = b0u(t 1) + b1u(t 2) Control law u(t) + r1u(t 1) = t0uc(t) s0y(t) s1y(t 1) c K. J. Åström and B. Wittenmark 5
Simulation Simulation P, G(s ao-2 polynomial Ads)=s+aowit h Controller struct ure u(t) y(t)+ to(p+ao (12) So 14 (15)
Simulation 0 20 40 60 80 100 −1 0 1 0 20 40 60 80 100 −4 −2 0 2 Time Time uc y u 0 100 200 300 400 500 −2 −1 0 0 100 200 300 400 500 0.0 0.1 0.2 Time Time ^a2 ^a1 ^ b1 ^ b0 Simulation Process G(s) = b s(s + a) Desired response Gm(s) = !2 s2 + 2!s + !2 Observer polynomial Ao(s) = s + ao with ao = 2. Controller structure u(t) = s0p + s1 p + r1 y(t) + t0(p + ao) p + r1 uc(t) where r1 = 2! + ao a (12) s0 = ao2! + !2 ar1 b (13) s1 = !2ao b (14) t0 = !2 b (15) c K. J. Åström and B. Wittenmark 6
