Linear Quadratic Gausssian Control Design with Loop Transfer Recovery * Leonid Freidovich Department of Mathematics Michigan State University MI 48824, USA e-mail:[email protected]http://www.math.msu.edu/˜leonid/ December 4, 2003 * Presentation for ME-891, Fall 2003
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Linear Quadratic Gausssian Control Design with Loop Transfer Recovery [LQG…€¦ · · 2003-12-14Linear Quadratic Gausssian Control Design with ... LQR problem: Design state feedback
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Linear Quadratic Gausssian Control Design withLoop Transfer Recovery ∗
Leonid Freidovich
Department of MathematicsMichigan State University
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
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)
If C∗1D12 = 0 then Jlqr = ‖C1x‖22 + ‖D12u‖22.
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)
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 :
1. the closed-loop system (1), (2) is internally stable,
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
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,
2. A + B2Klqr is Hurwitz and limt→∞
x(t) = 0,
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,
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 :
1. the closed-loop system (6), (5) is internally stable,
2. the solution x(t) and the control signal u(t) satisfy certainspecifications.
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 :
1. the closed-loop system (6), (5) is internally stable,
2. the solution x(t) and the control signal u(t) satisfy certainspecifications.
Let us choose C1, D12, as before.
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 :
1. the closed-loop system (6), (5) is internally stable,
2. the solution x(t) and the control signal u(t) satisfy certainspecifications.
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).
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
2. observer cannot be too fast since transient with peaking is not ‘compatible’with small values of ‖C1x + D12u‖∞,
3. if w(t) ≡ 0 then limt→∞
x(t) = 0,
11
2. observer cannot be too fast since transient with peaking is not ‘compatible’with small values of ‖C1x + D12u‖∞,
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
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.
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.
• If the plant is nonminimum-phase then A + L(ε)C2 is not Hurwitz forsmall values of ε.
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.
• 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
LQR design of L(ε) ←Consider LQR problem for the system
˙x = Ax + B2u,x(0) = x0,
where A = A∗ and B2 = C∗2 .
Take C1(ε) = B∗2/ε (cheap control)
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 .
Take C1(ε) = B∗2/ε (cheap control) and D12 independent of ε so thatD∗
12D12 = R > 0 and C∗1D12 = 0.
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 .
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
Then [A + B2Klqr(ε)
]∗ = 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
Then [A + B2Klqr(ε)
]∗ = 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
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.