ME 233 Advanced Control II Lecture 13 Frequency-Shaped Linear Quadratic Regulator (ME233 Class Notes pp.FSLQ1-FSLQ5)
•  Parseval's theorem •   Frequency-shaped LQR – Implementation •   Frequency-shaped LQR  with reference input Infinite-Horizon LQR (review) nth order  LTI system:
Find the optimal control:
which minimizes the cost functional:
3
•  Let f(k) be a map from the integers to R n •   Its (symmetric) Fourier transform is defined by
and
4
where
5
6 Frequency Cost Function By Parseval's theorem, the cost function:
is equivalent to the cost function
7 Frequency-Shaped Cost Function Key idea: Make matrices  Q  and R functions of frequency
where
8 Frequency-Shaped Cost Function Define the state and input filters 9 Frequency-Shaped Cost Function
can be written
10 Realizing the filters using LTI’s Let
be realized by
so that
is causal or strictly causal. 11 Realizing the filters using LTI’s Let be realized by zշ(k + 1) = Aշzշ(k) + Bշu(k) uf(k) = C2z2(k) + D2u(k) (with D T D 2 > 0) so that Rf(z) = C2(zI — A2)−1 B2 + D2 is causal (but not strictly causal) 12
Consider a simplified model of a voice coil motor and suspension (from control input u(k) to read/write head position y(k)) uncertainty nominal model actual plant x(k + 1) = Ax(k) + Bu(k)
(ヴ)
we want zero steady state (i.e. dc) error under a step input set position cost function weight to go to ∞ as ω → 0 Example Set weight on |Y (ejw)|²  to
|2
Example
X f (ejw)
we want zero steady state (i.e. dc) error under a step input set position cost function weight to go to ∞ as ω → 0 15
X(ejw) X f (ejw state space realization z1(k + 1) = A1z1(k) + B1x(k) xf(k) = C1z1(k) + D1x(k) Example
state space realization
16
Apply control design to nominal model
FS-LQR is a dynamic state feedback Sufficient condition for robustness
17
Go(z)
sufficient condition for robustness (by small gain theorem) :
potential lack of robustness
r T {X*(e'w')Q(e'w')X(e'jw')+ pU*(e'w')R(e'w')U(e'w')} dw - T increase control penalty at high-frequencies Example: Frequency Control Weight R(jω) {X*(ejw)Q(ejw)X(ejw) + pU*(ejw)R(ejw)U(ejw)} dw π U*(ejw)Uf(ejw) Example U(ejw) U f (ejw) Rf(ejw Rf(s): α 2- s² + 2Çwns + wñ s² + 2ζ(αωn)s + (αωn)² X  — : 2.5 Discretize using ZOH ( = 1.5 Wn = 7500 ζ = 0.6 Rf(z) 20
C 71 {X*(ejw)Q(ejw)X(ejw) + pU*(ejw)R(ejw)U(ejw)} dw T Example O( e)
ρ = 106 Robustness condition is satisfied U(z) X(z) K(z) Φ(z) 1 + Δ(z) B f Go(z)
11
22
state space realization z1(k + 1) = A1z1(k) + B1x(k) state space realization zշ(k + 1) = Aշzշ(k) + Bշu(k) uf(k) = C2z2(k) + D2u(k)
24
We can combine state equations and output as follows:
25
xe(k + 1) = Aexe(k) + Beu(k) Extended System Cost
Using
xe(k) De the cost can be expressed
Minimize
Subject to
This is a standard LQR problem! 28 FSLQR solution The optimal control law is uº(k) =  —Kexe(k) Ke = [BT PBe + DT De]−1[BT PAe + DT Ce] where P is the solution of the DARE P = AT PAe + CT Ce −[Aξ PBe + Cξ De][Bξ PBe + Dξ De]−1[Bξ PAe + Dξ Ce] for which Ae — BeKe  is Schur
The optimal control law exists if • (Ae, Be) stabilizable • The state-space realization C e (zl-A e ) -1B e + D e has no transmission zeros on the unit circle
The optimal control law exists if the following hold: 1. (A,B) is stabilizable 2.
(You will be asked to show this for homework)
Condition 3 from the existence conditions:
is equivalent to the condition that The state space realization for R f has no transmission zeros on the unit circle
Condition 4 from the existence conditions
is a stronger version of the condition that None of the unit circle eigenvalues of A are transmission zeros of the state space realization for Q f (The latter is not enough for FSLQR existence) Implementation • Control
= −K xx(k) − K 1z1(k) − K 2z2(k) 34
35
K(z) = [I + K2Φ2(z)B2]−1 [Kx + K1Φ1(z)B1] 36 State-space realization for K(z) u(k) = −Kxx(k) − K1z1(k) − K2z2(k) z1(k + 1) = A1z1(k) + B1x(k) zշ(k + 1) = Aշzշ(k) + Bշu(k) = A2z2(k) + B2(-Kxx(k) - K1z1(k) - K2z2(k)) 0 z1(k) z1(k + 1)| _ |    A1 + zշ(k + 1)| ￣ |−B₂K₁　A₂ − B₂K₂| zz(k) B1 x(k) — BշK x −u(k) = [K1 K2] [ z1(k) ] | + Kxx(k) | zշ(k)
•  For simplicity, we will assume a scalar output
•  Assume that we want to design a FSLQR that will achieve asymptotic output convergence to a reference input
•   Assume that the   reference input R  satisfies R
0 where Âr(z) has its zeros on the unit circle Reference input examples a)   Constant disturbance:
Then, Âr(z) = z − 1 b) Sinusoidal reference of  known frequency: r(k) = D sin(ω k + φ) Then, Âr(z) = z² — 2 cos(ω) z + 1 40 Reference input examples c)   Periodic reference of  known period  N
Then, In all of these three examples, the polynomial  Âr(z) has its zeros on the unit circle. FSLQR with reference input •   Define the reference frequency weight
where
R
0 We can choose this This is determined by the reference we are trying to follow 42
• with
used for achieving   lim   e(k) = 0 (we will show why later) 43 Frequency-Shaped Cost Function Define the state, input, and output filters 44
J =  J_  { X*(ejw)CTQ*(ejw)Qr(ejw)CX(ejw) + X*(eiw)Q*(eiw)Q f(eiw)X(eiw) + U*(ejw)R*f(ejw)Rf(ejw)U(ejw) { dw can be written J =  f_π {Y*(ejw)Yr(ejw) + X*(ejw)X f(ejw) + U*(ejw)U f(ejw)} dω 45 Realizing the filters using LTI’s Let
be realized by
so that
is causal or strictly causal. 46 Realizing the filters using LTI’s Let be realized by zշ(k + 1) = Aշzշ(k) + Bշu(k) uf(k) = C2z2(k) + D2u(k) (with D T D 2 > 0) so that
is causal (but not strictly causal)
Let
be realized by
so that
is causal or strictly causal. 48 Cost Function Realization Using Parseval's theorem, 8 J =  ∑  {yf (k)yr(k) + xf (k)xf (k) + uf (k)uf (k) {
where, x(k + 1)] A 0 0 0 x(k) zr(k + 1) BrC Ar 0 0 zr(k) u(k z1(k十1) B1 0 A1 0 z2(k) 0 Aっ0 yr(k)] | DrC Cr 0 0 x(k) zr(k) z1(k) 0 u(k) D2' u f ( k ) 0 0 0 C2 z2(k) z2(k + 1) | 0 X f ( k ) D1 0 0 C1
xe(k + 1) = Aexe(k) + Beu(k)
Using
Ve xe(k) De the cost can be expressed
0 x(k)
Minimize
Subject to
This is a standard LQR problem! 52 Solution The optimal control law is uº(k) =  —Kexe(k) Ke = [BT PBe + DT De]−1[BT PAe + DT Ce] where P is the solution of the DARE P = AT PAe + CT Ce −[Aξ PBe + Cξ De][Bξ PBe + Dξ De]−1[Bξ PAe + Dξ Ce] for which Ae — BeKe  is Schur 53
The optimal control law exists if • (Ae, Be) stabilizable • The state-space realization C e (zl-A e ) -1B e + D e has no transmission zeros on the unit circle
The optimal control law exists if the following hold: 1. (A,B) is stabilizable 2.
The optimal control law exists if the following hold: (A r , B r ) is stabilizable 5. 6. (C n A ) has no unobservable modes on the unit circle
. Conditions  1-4 are the same as for the FSLQR without a reference input . Conditions 5-6 are met if the realization of Q, is minimal • Condition 7 is a stronger version of the condition that none of the unit circle or unstable eigenvalues of A, are transmission zeros of C(zl-A) -1 B, the open- loop relationship between u and y – The condition here is not enough to guarantee FSLQR existence for reference tracking
• Control
= −K xx(k) − K rzr(k) − K 1z1(k) − K 2z2(k) 58 59
where
60
stable 61
•  Unit 0: Probability •  Unit 1: State-space control,estimation Finished •   Unit 2: Input/output control •   Unit 3: Adaptive control
Finite-horizon results • Kalman filter . Optimal LQR •   Optimal LQG –  state feedback –  output feedback Infinite-horizon results • Optimal LQR •  Kalman filter •   Optimal LQG –  output feedback •   Frequency-shaped  LQR
•  Continuous-time versions of: – Kalman filter – Optimal LQG – Frequency-shaped LQR •   Loop transfer recovery Slides will be posted for reference
A collection of SISO input/output control design techniques •   Disturbance observer •   Pole placement, disturbance rejection, and tracking control •   Repetitive control and the internal model principle •   Minimum variance regulators