ME 233, UC Berkeley, Spring Background Parseval s Theorem Frequency-shaped LQ cost function Transformation to a standard LQ

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1 ME 233, UC Berkeley, Spring 214 Xu Chen Lecture 1: LQ with Frequency Shaped Cost Function FSLQ Background Parseval s Theorem Frequency-shaped LQ cost function Transformation to a standard LQ Big picture why are we learning this: in standard LQ, Q and R are constant matrices in the cost function x T tqxt + ρu T trut dt 1 how can we introduce more design freedom for Q and R? Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

2 Connection between time and frequency domains Theorem Parseval s Theorem For a square integrable signal f t defined on [, f T tf tdt = 1 2π F T jωf jωdω 1D case: f t 2 dt = 1 F jω 2 dω 2π Intuition: energy in time-domain equals energy in frequency domain For the general case, f t can be acausal. We have f T tf tdt = 1 F T jωf jωdω 2π Discrete-time version: k= f T kf k = 1 2π F T e jω F e jω dω Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME History Marc-Antoine Parseval : French mathematician published just five but important mathematical publications in total source: Wikipedia.org Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

3 Frequency-domain LQ cost function From Parseval s Theorem, the LQ cost in frequency domain is = 1 2π x T tqxt + ρu T trut dt 2 X T jωqxjω + ρu T jωrujω dω 3 Frequency-shaped LQ expands Q and R to frequency-dependent functions: 1 X T jωq jωxjω + ρu T jωr jωujω dω 2π 4 Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME Frequency-domain LQ cost function Let Q jω = Q T f jωq f jω, X f jω = Q f jωx jω R jω = R T f jωr f jω, U f jω = R f jωu jω 4 becomes 1 2π Xf T jωx f jω + ρuf T jωu f jω dω which is equivalent to using Parseval s Theorem again xf T tx f t + ρuf T tu f t dt 5 Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

4 Frequency-domain LQ cost function Summarizing, we have: plant: { ẋ t y t = Ax t + Bu t = Cx t 6 new cost: xf T tx f t + ρuf T tu f t dt 7 filtered states and inputs: x t Q f s x f t, u t R f s u f t We just need to translate the problem to a standard one [which we know very well how to solve] Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME Frequency-domain weighting filters state filtering x t Q f s x f t a MIMO process in general: if x t R n and x f t R q, then Q f s is a q n transfer function matrix Q f s: state filter; designer s choice; can be selected to meet the desired control action and the performance requirements write Q f s = C 1 si A 1 1 B 1 + D 1 in the general state-space realization: { ż1 t = A 1 z 1 t + B 1 xt 8 x f t = C 1 z 1 t + D 1 xt Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

5 Frequency-domain weighting filters input filtering u t R f s u f t R f s: input filter; designer s choice; can be selected to meet the robustness requirements write R f s = C 2 si A 2 1 B 2 + D 2 in the general state-space realization: { ż 2 t = A 2 z 2 t + B 2 ut 9 u f t = C 2 z 2 t + D 2 ut Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME Back to time-domain design Combining 6, 8 and 9 gives the enlarged system xt A xt d z 1 t = B 1 A 1 z 1 t + dt z 2 t A 2 z 2 t }{{}}{{} x e t A e B B 2 }{{} B e ut and xt x f t = [D 1 C 1 ] z }{{} 1 t C e z 2 t u f t = [ C 2 ]x e t + D 2 ut Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

6 Summary of solution With the enlarged system, the cost xf T tx f t + ρuf T tu f t dt 1 translates to xe T tq e x e t + 2u T t [ ρd2 T C ] 2 x e t + u T tρd2 T D 2 }{{}}{{} N e R e Q e = D T 1 D 1 D T 1 C 1 C T 1 D 1 C T 1 C 1 ρc T 2 C 2 solution see appendix for more details: ut dt u t = Re 1 Be T P e + N e x e t = Kx t K 1 z 1 t K 2 z 2 t algebraic Riccati equation: A T e P e + P e A e B T e P e + N e T R 1 e B T e P e + N e + Q e = Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME Implementation structure of the FSLQ system: B si A 1 x u f D C z 2 2 si A 2 1 B 2 K 2 K 1 x f + C 1 si A z 1 B 1 D 1 K Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

7 Appendix: general LQ solution Consider LQ problems with cost x T tc}{{ T C} x t + 2u T tnx t + u T tru t dt 11 Q and system dynamics ẋ t = Ax t + Bu t assume A, B is controllable/stabilizable and A, C is observable/detectable the solution of the problem is u t = R 1 B T P + Nx t A T P + PA B T P + N T R 1 B T P + N + Q = Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME Appendix: general LQ solution Intuition: under the assumptions, we know we can stabilize the system and drive x t to zero. Consider V t = x T tpx t V V = = Adding 11 on both sides yields V V + x T t Q + PA + A T P V tdt x T t PA + A T P x t + 2x T tpbu t dt x t + 2x T t PB + N T u t + u T tru t dt 12 to minimize the cost, we are going to re-organize the terms in 12 into some squared terms Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME

8 Appendix: general LQ solution completing the squares : 2x T t PB + N T u t+u T tru t = R 1/2 u t + R 1/2 B T P + N x t x T t PB + N T R 1 B T P + N x t hence 12 is actually V V + J [ = x T t Q + PA + A T P PB + N T R 1 B T P + N x t + R 1/2 u t + R 1/2 B T P + N x t hence J min = V = x T Px is achieved when Q + PA + A T P PB + N T R 1 B T P + N = and u t = R 1 B T P + N x t 2 2 ] dt 2 2 Lecture 1: LQ with Frequency Shaped Cost Function FSLQ ME