Lecture 13: Internal Model Principle and Repetitive Control


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1 ME 233, UC Berkeley, Spring 2014 Xu Chen Lecture 13: Internal Model Principle and Repetitive Control Big picture review of integral control in PID design example: 0 Es) C s) Ds) + + P s) Y s) where P s) = 1 ms + b, C s) = k p + k i 1 s + k ds, k p,k i,k d > 0 the integral action in PID control perfectly rejects asymptotically) constant disturbances D s) = d o /s): E s) = P s) 1 + P s)c s) D s) = d o m + k d )s 2 + k p + b)s + k i e t) 0 Lecture 13: Internal Model Principle and Repetitive Control ME
2 Big picture review of integral control in PID design 0 Es) C s) Ds) + + P s) Y s) the structure of the reference/disturbance is built into the integral controller: controller: 1 C s) = k p + k i s + k ds = 1 kd s 2 ) + k p s + k i s constant disturbance: d t) = d o L {d t)} = 1 s d o Lecture 13: Internal Model Principle and Repetitive Control ME General case: internal model principle IMP) Theorem Internal Model Principle) Rs) + Es) C s) = B cs) Ds)= B d s) A d s) + A c s) P s) = B ps) + A p s) Y s) Assume B p s) = 0 and A d s) = 0 do not have common roots. If the closed loop is asymptotically stable, and A c s) can be factoried as A c s) = A d s)a c s), then the disturbance is asymptotically rejected. Lecture 13: Internal Model Principle and Repetitive Control ME
3 General case: internal model principle IMP) Rs) + Es) C s) = B cs) Ds)= B d s) A d s) + A c s) P s) = B ps) + A p s) Y s) Proof: The steadystate error response to the disturbance is E s) = = P s) 1 + P s)c s) D s) = B p s)a c s) B d s) A p s)a c s) + B p s)b c s) A d s) B p s)a c s)b d s) A p s)a c s) + B p s)b c s) where all roots of A p s)a c s) + B p s)b c s) = 0 are on the left half plane. Hence e t) 0 Lecture 13: Internal Model Principle and Repetitive Control ME Internal model principle discretetime case: Theorem Discretetime IMP) R 1 ) + E 1 ) B c 1 ) D 1 )= B d 1 ) A d 1 ) + d B p 1 ) A c 1 ) + A p 1 ) Y 1 ) Assume B p 1 ) = 0 and A d 1 ) = 0 do not have common eros. If the closed loop is asymptotically stable, and A c 1 ) can be factoried as A c 1 ) = A d 1 ) A c 1 ), then the disturbance is asymptotically rejected. Proof: analogous to the continuoustime case. Lecture 13: Internal Model Principle and Repetitive Control ME
4 Internal model principle the disturbance structure: R 1 ) + E 1 ) B c 1 ) D 1 )= B d 1 ) A d 1 ) + d B p 1 ) A c 1 ) + A p 1 ) Y 1 ) example disturbance structures: d k) A d 1 ) constant d o 1 1 cosω 0 k) and sinω 0 k) cosω 0 ) + 2 shifted ramp signal d k) = αk + β periodic: d k) = d k N) 1 N Lecture 13: Internal Model Principle and Repetitive Control ME Internal model principle R 1 ) + E 1 ) B c 1 ) D 1 )= B d 1 ) A d 1 ) + d B p 1 ) A c 1 )A d 1 ) + A p 1 ) Y 1 ) observations: the controller contains a counter disturbance generator highgain control: the openloop frequency response P e jω) C e jω) = edjω B p e jω ) B c e jω ) A p e jω )A c e jω )A d e jω ) is large at frequencies where A d e jω ) = 0 D 1) = B d 1 ) /A d 1 ) means d k) is the impulse response of B d 1 ) /A d 1 ) : A d 1 ) d k) = B d 1 ) δ k) Lecture 13: Internal Model Principle and Repetitive Control ME
5 Outline 1. Big Picture review of integral control in PID design 2. Internal Model Principle theorems typical disturbance structures 3. Repetitive Control use of internal model principle design by pole placement design by stable poleero cancellation Lecture 13: Internal Model Principle and Repetitive Control ME Repetitive control Repetitive control focus on attenuating periodic disturbances with A d 1 ) = 1 N Control structure: R 1 ) + E 1 ) B c 1 ) D 1 )= B d 1 ) A d 1 ) + d B p 1 ) A c 1 )A d 1 ) + A p 1 ) Y 1 ) It remains to design B c 1 ) and A c 1 ). We discuss two methods: pole placement partial) cancellation of plant dynamics: prototype repetitive control Lecture 13: Internal Model Principle and Repetitive Control ME
6 1, Pole placement: prerequisite Theorem Consider G ) = β) α) = β 1 n1 +β 2 n2 + +β n. α ) and β ) are n +α 1 n1 + +α n coprime no common roots) iff S Sylvester matrix) is nonsingular: S = β α β 2 β α 1 1 β n α n1 α 1 β n β 1 α n β n... β 2 0 α n α n1.... β n β n α n β n 2n1) 2n1) Lecture 13: Internal Model Principle and Repetitive Control ME , Pole placement: prerequisite Example: G ) = β 1 n1 + β 2 n2 + + β n n + α 1 n1 + + α n = n1 + α 1 n2 + + α n1 n + α 1 n1 + + α n1 + 0 i.e. β 1 = 1 β i = α i1 i 2 α n = 0 α ) and β ) are not coprime, and S is clearly singular. Lecture 13: Internal Model Principle and Repetitive Control ME
7 1, Pole placement: big picture R 1 ) + E 1 ) B c 1 ) D 1 )= B d 1 ) A d 1 ) + d B p 1 ) A c 1 )A d 1 ) + A p 1 ) Y 1 ) Disturbance model: A d 1 ) = 1 N Pole placement assigns the closedloop characteristic equation: d B p 1 ) B c 1 ) + A p 1 ) A c 1 ) A d 1 ) = 1 + η η η q q }{{} η 1 ) which is in the structure of a Diophantine equation. Design procedure: specify the desired closedloop dynamics η 1) ; match coefficients of i on both sides to get B c 1 ) and A c 1 ). Lecture 13: Internal Model Principle and Repetitive Control ME , Pole placement: big picture R 1 ) + E 1 ) B c 1 ) D 1 )= B d 1 ) A d 1 ) + d B p 1 ) A c 1 )A d 1 ) + A p 1 ) Y 1 ) Diophantine equation in Pole placement: d B p 1 ) B c 1 ) + A p 1 ) A c 1 ) A d 1 ) = 1 + η η η q q }{{} η 1 ) Questions: what are the constraints for choosing η 1)? how to guarantee unique solution in Diophantine equation? Lecture 13: Internal Model Principle and Repetitive Control ME
8 Design and analysis procedure General procedure of control design: Problem definition Control design for solution current stage) Prove stability Prove stability robustness Case study or implementation results Lecture 13: Internal Model Principle and Repetitive Control ME , Pole placement: details Theorem Diophantine equation) Given η 1) = 1 + η η η q q α 1) = 1 + α α n n β 1) = β β β n n The Diophantine equation α 1) σ 1) + β 1) γ 1) = η 1) can be solved uniquely for σ 1) and γ 1) σ 1) = 1 + σ σ n1 n1) γ 1) = γ 0 + γ γ n1 n1) if the numerators of α 1) and β 1) are coprime and deg η 1)) = q 2n 1 Lecture 13: Internal Model Principle and Repetitive Control ME
9 1, Pole placement: details Proof of Diophantine equation Theorem key ideas): α 1) σ 1) +β 1) γ 1) = η 1) }{{}}{{} unknown unknown Matching the coefficients of i gives see one numerical example in course reader) σ 1 σ 2 S. σ n1 γ 0. γ n1 + α 1 α 2. α n 0. 0 = η 1 η 2. η n1 η n. η 2n1 The coprime condition assures S is invertible. deg η 1) 2n 1 assures the proper dimension on the right hand side of the equality. Lecture 13: Internal Model Principle and Repetitive Control ME , Prototype repetitive control: simple case D 1 )= B d 1 ) A d 1 ) R 1 ) + E 1 ) + C 1) d B p 1 ) + A p 1 ) Y 1 ) A d 1 ) = 1 N If all poles and eros of the plant are stable, then prototype repetitive control uses C 1) = k r N+d A p 1 ) ) 1 N B p 1 ) Theorem Prototype repetitive control) Under the assumptions above, the closedloop system is asymptotically stable for 0 < k r < 2 Lecture 13: Internal Model Principle and Repetitive Control ME
10 2, Prototype repetitive control: stability Proof of Theorem on prototype repetitive control: From 1 + k r N+d A p 1 ) ) d B p 1 ) 1 N B p 1 ) A p 1 ) the closedloop characteristic equation is = 0 A p 1 ) B p 1 )[ 1 1 k r ) N] = 0 roots of A p 1 ) B p 1 ) = 0 are all stable roots of 1 1 k r ) N = 0 are 1 k r 1 N e j 2πi N, i = 0,±1,..., for 0 < kr 1 1 k r 1 N e j 2πi N + π N ), i = 0,±1,..., for 1 < kr which are all inside the unit circle Lecture 13: Internal Model Principle and Repetitive Control ME , Prototype repetitive control: stability robustness Consider the case with plant uncertainty dk) rk)+ k r N+d A p 1 ) + d B p 1 ) 1 N )B p 1 ) + A p 1 ) )) yk) N openloop poles on the unit circle Root locus example: N = 4, 1 + 1) = q/ p) Imaginary Axis Root Locus Real Axis k r > 0, the closedloop system is now unstable! Lecture 13: Internal Model Principle and Repetitive Control ME
11 2, Prototype repetitive control: stability robustness dk) rk)+ + C 1) d B p 1 ) + A p 1 ) )) yk) To make the controller robust to plant uncertainties, do instead C 1) = k rq, 1 ) N+d A p 1 ) 1 q, 1 ) N) B p 1 ) 1) q, 1 ) : lowpass filter. e.g. erophase low pass α α 0 + α 1 α 0 + 2α 1 which shifts the marginary stable openloop poles to be inside the unit circle: A p 1 ) B p 1 )[ 1 1 k r )q, 1 ) N] = 0 Lecture 13: Internal Model Principle and Repetitive Control ME , Prototype repetitive control: extension D 1 )= B d 1 ) A d 1 ) R 1 ) + E 1 ) + C 1) d B p 1 ) + A p 1 ) Y 1 ) If poles of the plant are stable but NOT all eros are stable, let B p 1 ) = Bp 1 )B p + 1 ) [Bp 1 ) the uncancellable part] and C 1) = k r N+µ A p 1 )B p ) µ 1 N )B + p 1 ) d b, b > max ω [0,π] B p e jω ) 2 2) Similar as before, can show that the closedloop system is stable inclass exercise). Lecture 13: Internal Model Principle and Repetitive Control ME
12 2, Prototype repetitive control: extension Exercise: analye the stability of D 1 )= B d 1 ) A d 1 ) R 1 ) + E 1 ) + C 1) d B p 1 ) + A p 1 ) Y 1 ) C 1) = k r N+µ A p 1 )Bp ) µ 1 N )B p + 1 ) d b, b > max ω [0,π] B p e jω ) 2 3) Key steps: B p e jω )Bp e jω ) b < 1; k r Bp e jω )Bp e jω ) b 1 < 1; all roots from [ kr N Bp )Bp 1 ) ] = 0 b are inside the unit circle. Lecture 13: Internal Model Principle and Repetitive Control ME , Prototype repetitive control: extension dk) rk)+ + C 1) d B p 1 ) + A p 1 ) )) yk) Robust version in the presence of plant uncertainties: C 1) = k r N+µ q, 1 )A p 1 )B p ) µ 1 q, 1 ) N )B + p 1 ) d b 4) where q, 1 ) : lowpass filter. e.g. erophase low pass α α 0 + α 1 α 0 + 2α 1 and µ is the order of B p ) Lecture 13: Internal Model Principle and Repetitive Control ME
13 Example dk) rk)+ + C 1) + Plant yk) disturbance period: N = 10; nominal plant: d B p 1 ) 1 A p 1 = ) ) ) C 1) = k r ) ) q, 1 ) q, 1 ) 10 ) Lecture 13: Internal Model Principle and Repetitive Control ME Additional reading ME233 course reader X. Chen and M. Tomiuka, An Enhanced Repetitive Control Algorithm using the Structure of Disturbance Observer, in Proceedings of 2012 IEEE/ASME International Conference on Advanced Intelligent Mechatronics, Taiwan, Jul , 2012, pp X. Chen and M. Tomiuka, New Repetitive Control with Improved Steadystate Performance and Accelerated Transient, IEEE Transactions on Control Systems Technology, vol. 22, no. 2, pp pages), Mar Lecture 13: Internal Model Principle and Repetitive Control ME
14 Summary 1. Big Picture review of integral control in PID design 2. Internal Model Principle theorems typical disturbance structures 3. Repetitive Control use of internal model principle design by pole placement design by stable poleero cancellation Lecture 13: Internal Model Principle and Repetitive Control ME
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