linear quadratic gausssian control design with loop transfer recovery [lqg…€¦ · ·...

Linear Quadratic Gausssian Control Design withLoop Transfer Recovery ∗

Leonid Freidovich

Department of MathematicsMichigan State University

MI 48824, USA

e-mail:[email protected]

http://www.math.msu.edu/˜leonid/

December 4, 2003

∗Presentation for ME-891, Fall 2003

http://www.math.msu.edu/~leonid/

1

Outline

1

Outline

(a) Linear quadratic regulation [LQR]

1

Outline


(b) H2 design [LQG]

1

Outline


(b) H2 design [LQG]

(c) Idea of loop transfer recovery [LTR]

1

Outline


(b) H2 design [LQG]


(d) Technical implementations of LQR/LTR

1

Outline


(b) H2 design [LQG]


(d) Technical implementations of LQR/LTR

(e) References

[LQR/LQG/LTR]

2

Linear Quadratic Regulation [LQR] ←x = Ax + B2u,x(0) = x0.

(1)

• (A,B2) is controllable.

2

Linear Quadratic Regulation [LQR] ←x = Ax + B2u,x(0) = x0.

(1)

• (A,B2) is controllable.

Goal: Design state feedback controller

u = K(s)x, [ K(s) is a proper real rational transfer function] (2)

depending on A and B2 such that for arbitrary x0 :

1. the closed-loop system (1), (2) is internally stable,

2. the solution x(t) and the control signal u(t) satisfy certainspecifications.

[LQR/LQG/LTR]

3

Let us choose C1, D12 and define performance index:

Jlqr = ‖z‖22 = ‖C1x + D12u‖22∆=

∞∫0

[C1x(t) + D12u(t)]∗[C1x(t) + D12u(t)] dt (3)

3


Jlqr = ‖z‖22 = ‖C1x + D12u‖22∆=

∞∫0

[C1x(t) + D12u(t)]∗[C1x(t) + D12u(t)] dt (3)

If C∗1D12 = 0 then Jlqr = ‖C1x‖22 + ‖D12u‖22.

3


Jlqr = ‖z‖22 = ‖C1x + D12u‖22∆=

∞∫0

[C1x(t) + D12u(t)]∗[C1x(t) + D12u(t)] dt (3)

If C∗1D12 = 0 then Jlqr = ‖C1x‖22 + ‖D12u‖22.

LQR problem: Design state feedback controller (2) depending on A, B2,C1, and D12 such that for arbitrary x0 :


2. the performance index (3) is the smallest.

[LQR/LQG/LTR]

4

The solution exists if

• D∗12D12 > 0,

• (C1, A) is detectable;

•[

A− jωI B2

C1 D12

]has full column rank for all ω.

[LQR/LQG/LTR]

5

Properties of the solution:

1. optimal controller K(s) = Klqr is the constant gain:

Klqr = −(D∗12D12)−1(B∗2X + D∗

12C1),

where X ≥ 0 is the stabilizing solution of the Riccati equation:

A∗X + XA−XB2(D∗12D12)−1B∗2X + C∗1 [I −D12(D∗

12D12)−1D∗12]C1 = 0,

with A = A−B2(D∗12D12)−1D∗

12C1,

5



Klqr = −(D∗12D12)−1(B∗2X + D∗

12C1),


A∗X + XA−XB2(D∗12D12)−1B∗2X + C∗1 [I −D12(D∗

12D12)−1D∗12]C1 = 0,

with A = A−B2(D∗12D12)−1D∗

12C1,

2. A + B2Klqr is Hurwitz and limt→∞

x(t) = 0,

5



Klqr = −(D∗12D12)−1(B∗2X + D∗

12C1),


A∗X + XA−XB2(D∗12D12)−1B∗2X + C∗1 [I −D12(D∗

12D12)−1D∗12]C1 = 0,

with A = A−B2(D∗12D12)−1D∗

12C1,

2. A + B2Klqr is Hurwitz and limt→∞

x(t) = 0,

3. guaranteed 6 dB (= 20 log 2) gain margin and 60o phase margin in bothdirections.

[LQR/LQG/LTR]

6

Linear Quadratic Gaussian [LQG or H2 ] problem ←Model:

x = Ax + B2u,y = C2x + D21u,x(0) = x0.

(4)

Assumptions:

• (A,B2) is controllable,

• (C2, A) is detectable.

[LQR/LQG/LTR]

7

Goal: Design output feedback controller

u = K(s)y, [ K(s) is a proper real rational transfer function] (5)

depending on A, B2, C2, and D21 such that for arbitrary x0 :



7






Let us choose C1, D12, as before.

7






Let us choose C1, D12, as before. It could be shown that if D12 6= 0,then the output controller minimizing Jlqr = ‖z‖2 = ‖C1x + D12u‖22 cannotbe proper (i.e. solution does not exist).

[LQR/LQG/LTR]

8

Considerx = Ax + B1w + B2u,y = C2x + D21u,x(0) = x0.

(6)

8


(6)

Define performance index

Jlqg = ‖Tw→z‖22 = max‖w‖2≤1

{‖z‖2∞‖w‖22

}= max‖w‖2≤1

{‖C1x + D12u‖2∞

‖w‖22

}. (7)

8


(6)

Define performance index

Jlqg = ‖Tw→z‖22 = max‖w‖2≤1

{‖z‖2∞‖w‖22

}= max‖w‖2≤1

{‖C1x + D12u‖2∞

‖w‖22

}. (7)

LQG problem: Design output feedback controller (5) depending on A, B1,B2, C1, C2, D12, and D21 such that for arbitrary x0 :


2. the performance index (7) is the smallest.[LQR/LQG/LTR]

9

The solution exists if

• D∗12D12 > 0,

•[

A− jωI B2

C1 D12

]has full column rank for all ω.

• D21D∗21 > 0 ,

•[

A− jωI B1

C2 D21

]has full row rank for all ω.

[LQR/LQG/LTR]

10


1. optimal K(s) is the observer based controller:

u = Klqgx,˙x = Ax + B2u + Llqg(C2x + D21u− y),x(0) = x0,

10


1. optimal K(s) is the observer based controller:

u = Klqgx,˙x = Ax + B2u + Llqg(C2x + D21u− y),x(0) = x0,

whereKlqg = Klqr

andLlqg = −(Y C∗2 + B1D

∗21)(D21D

∗21)−1,

where Y ≥ 0 is the stabilizing solution of the Riccati equation:

A∗Y + Y A− Y C∗2(D21D∗21)−1C2Y + B1[I −D∗

21(D21D∗21)−1D21]B∗1 = 0,

with A = A−B1D∗21(D21D

∗21)−1C2,

[LQR/LQG/LTR]

11

2. observer cannot be too fast since transient with peaking is not ‘compatible’with small values of ‖C1x + D12u‖∞,

11


3. if w(t) ≡ 0 then limt→∞

x(t) = 0,

11


3. if w(t) ≡ 0 then limt→∞

x(t) = 0,

4. in general, neither gain nor phase margins are guaranteed (have to checkrobustness for each particular design).

[LQR/LQG/LTR]

12

LQG with loop transfer recovery ←Consider the observer based controller for the system (4) with D21 = 0

x = Ax + B2u,y = C2x,x(0) = x0,

12

LQG with loop transfer recovery ←Consider the observer based controller for the system (4) with D21 = 0

x = Ax + B2u,y = C2x,x(0) = x0,

u = Klqrx,˙x = Ax + B2u + L(C2x− y),x(0) = x0.

(8)

Assumptions:

• (A,B2) is controllable,

• (C2, A) is detectable,

• Klqr is the optimal gain from the LQR problem.

[LQR/LQG/LTR]

13

Goal: Design the observer gain L so that the closed-loop system (8)recovers internal stability and some of the robustness properties (gain andphase margins) of the LQR design.

13

Goal: Design the observer gain L so that the closed-loop system (8)recovers internal stability and some of the robustness properties (gain andphase margins) of the LQR design.

Compare

-��

- x = Ax + B2u

�Klqr

6

u

u

w x

Figure 1: u = {Lt(s)}u

u = Klqr (sI −A)−1B2u︸︷︷︸x=Ax+B2u

def= {Lt(s)}u

[LQR/LQG/LTR]

14

and

-��

-

r�

x = Ax + B2u -

��Klqr

C2

observer

6

u

u x

w x y

Figure 2: u = {Lo(s)}u

u = Klqr [−(sI −A− LC2 −B2Klqr)−1LC2]︸︷︷︸observer: ˙x=Ax+B2[Klqrx]+LC2(x−x)

(sI −A)−1B2u︸︷︷︸x=Ax+B2u

def= {Lo(s)}u

14

and

-��

-

r�

x = Ax + B2u -

��Klqr

C2

observer

6

u

u x

w x y

Figure 2: u = {Lo(s)}u

u = Klqr [−(sI −A− LC2 −B2Klqr)−1LC2]︸︷︷︸observer: ˙x=Ax+B2[Klqrx]+LC2(x−x)

(sI −A)−1B2u︸︷︷︸x=Ax+B2u

def= {Lo(s)}u

We want Lo(s) ≡ Lt(s), but Lo(s) is biproper and Lt(s) is strictly proper.[LQR/LQG/LTR]

15

LQG/LTR problem: Design the observer gain L so that

1. the closed-loop system (8) is internally stable,

2. for all ω ∈ [0, ωmax] :Lo(jω) ≈ Lt(jω),

where ωmax � 1,target (under state feedback) open-loop transfer function is

Lt(s) = Klqr(sI −A)−1B2,

achieved (under output feedback) open-loop transfer function is

Lo(s) = KlqrTx→x(s)(sI −A)−1B2,

and ‘excited’ observer transfer function is

Tx→x(s) = [−(sI −A− LC2 −B2Klqr)−1LC2].[LQR/LQG/LTR]

16

Doyle – Stein formula ←It could be shown that if

−L[I − C2(jωI −A)−1L]−1 = B2[C2(jωI −A)−1B2]−1

thenLo(jω) = Lt(jω).

16




Supposelimε→0{εL(ε)} = −B2W,

where W is not singular and ε� 1.

16




Supposelimε→0{εL(ε)} = −B2W,

where W is not singular and ε� 1. Then

limε→0{B2W [εI + C2(jωI −A)−1B2W ]−1} = B2[C2(jωI −A)−1B2]−1

and hencelimε→0{Lo(jω)} = Lt(jω).

[LQR/LQG/LTR]

17

Remarks on(−L(ε)[I − C2(jωI −A)−1L(ε)]−1 ≈ B2[C2(jωI −A)−1B2]−1

)• In general, the choice L(ε) = −B2W/ε does not guarantee internal

stability of the closed-loop system.

17




• Since

−L(ε)[I −C2(jωI −A)−1L(ε)]−1 = −(jωI −A)(jωI − [A + L(ε)C2])−1L(ε),

17




• Since


if εL(ε) = −B2W + O(ε), then some of the eigenvalues of [A + L(ε)C2](i.e. poles of the observer error dynamics) approach the zeros of[C2(jωI −A)−1B2] (i.e. zeros of the plant) and the others approach infinityas ε→ 0.

17




• Since



• If the plant is nonminimum-phase then A + L(ε)C2 is not Hurwitz forsmall values of ε.

17




• Since



• If the plant is nonminimum-phase then A + L(ε)C2 is not Hurwitz forsmall values of ε.

• Some of the observer’s modes are fast. Therefore peaking phenomenonmust occur.

[LQR/LQG/LTR]

18

LQR design of L(ε) ←Consider LQR problem for the system

˙x = Ax + B2u,x(0) = x0,

where A = A∗ and B2 = C∗2 .

18


˙x = Ax + B2u,x(0) = x0,


Take C1(ε) = B∗2/ε (cheap control)

18


˙x = Ax + B2u,x(0) = x0,


Take C1(ε) = B∗2/ε (cheap control) and D12 independent of ε so thatD∗

12D12 = R > 0 and C∗1D12 = 0.

18


˙x = Ax + B2u,x(0) = x0,


Take C1(ε) = B∗2/ε (cheap control) and D12 independent of ε so thatD∗

12D12 = R > 0 and C∗1D12 = 0.

Suppose Klqr(ε) is the solution gain of the problem above for each ε > 0.

TakeL(ε) =

[Klqr(ε)

]∗.

[LQR/LQG/LTR]

19

Then [A + B2Klqr(ε)

]∗ = A + L(ε)C2

is Hurwitz

19


]∗ = A + L(ε)C2

is Hurwitz andL(ε) = −X(ε)C∗2R−1,

where X(ε) is the stabilizing solution of the Riccati equation

AX(ε) + X(ε)A∗ − X(ε)C∗2R−1C2X(ε) + B2B∗2/ε2 = 0.

19


]∗ = A + L(ε)C2

is Hurwitz andL(ε) = −X(ε)C∗2R−1,

where X(ε) is the stabilizing solution of the Riccati equation

AX(ε) + X(ε)A∗ − X(ε)C∗2R−1C2X(ε) + B2B∗2/ε2 = 0.

It could be shown that if the plant is of minimum phase (and left invertible)then lim

ε→0{ε2[AX(ε) + X(ε)A∗]} = 0 and

limε→0{εL(ε)} = lim

ε→0{−εX(ε)C∗2R−1} = −B2R

−1/2 = −B2W

[LQR/LQG/LTR]

20

Alternative designs of L(ε) ←

For the ‘broken’ closed-loop system

x = Ax + B2u,y = C2x + D21u,x(0) = x0,

20

Alternative designs of L(ε) ←

For the ‘broken’ closed-loop system

x = Ax + B2u,y = C2x + D21u,x(0) = x0,

u = Klqrx,˙x = Ax + B2u + L(C2x + D21u− y),x(0) = x0.

we have: Lt(s) = Klqr(sI −A)−1B2 and

Lo(s) = Klqr[−(sI−A−LC2−B2Klqr+LD21Klqr)−1]L[C2(sI−A)−1B2+D21].[LQR/LQG/LTR]

21

It could be shown that

Lt(s)− Lo(s) = M(s)[I + M(s)]−1(I −KlqrB2),

where M(s) = −Klqr([sI −A]−1 − LC2)−1(B2 + LD21).

Therefore it is reasonable to search for L that minimizes ‖M(s)‖2 or‖M(s)‖∞.

Also, a pole-placement technique can be used to design the observer gain aswell. However, it is not trivial and the system need to be transformed into thespecial coordinate basis.

[LQR/LQG/LTR]

22

References

[1] H. Kwakernaak and R. Sivan, Linear optimal control systems, N.Y.:Wiley-Interscience, 1972.

[2] J.C. Doyle, “Guaranteed margins for LQG regulators,” IEEE Trans. onAC, Vol. 23, pp. 756–757, 1978.

[3] J.C. Doyle and G. Stein, “Robustness with observers,” IEEE Trans. onAC, Vol. 24, pp. 607–611, 1979.

[4] J.C. Doyle and G. Stein, “Multivariable feedback design: concepts for aclassical/Modern synthesis,” IEEE Trans. on AC, Vol. 26, pp. 4–16, 1981.

[5] G. Stein and M. Athans, “The LQG/LTR procedure for multivariablecontrol design,” IEEE Trans. on AC, Vol. 32, pp. 105–114, 1987.

[LQR/LQG/LTR]

23

[6] J.C. Doyle, K. Glover, P.P. Khargonekar, and B.A. Francis, “State-spacesolution to standard H2 and H∞ control problems,” IEEE Trans. onAC, Vol. 34, pp. 831–847, 1989.

[7] A. Saberi, B.M. Chen, and P. Sanuti, Loop transfer recovery: analysisand design, London: Springer-Verlag, 1993.

[8] K. Zhou and J.C. Doyle, Essentials of robust control, N.J.: Prentice-Hall,1998.

[9] J.S. Freudenberg, Robustness with observers, Lecture notes, Universityof Michigan, 1999.

[10] C. Gokcek, ME-891: Robust and Optimal Control, Lecture notes,Michigan State University, 2003.

[11] B.M. Chen, Loop transfer recovery, Lecture notes, 2003.←

[LQR/LQG/LTR]

linear quadratic gausssian control design with loop transfer recovery [lqg…€¦ · ·...

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