H2 -Optimal Control
• The H2 -norm: Computation and Interpretation
• Response of systems to white noise and coloring filters
• Spectral factorization and the Positiv Real Lemma
• Optimal H2 -synthesis by static state-feedback
• Optimal H2 -estimation (Kalman filtering)
• Optimal H2 -synthesis by output feedback (Separation Principle)
Related Reading
[KK]: 8, 9, 11 (stationary version)
Doyle, Glover, Khargonekar and Francis, IEEE TAC 34 (1989) 831-847.
1/82
The H2 -Norm
Consider the LTI system with state-space description
ẋ = Ax + Bw, z = Cx
and with transfer matrix
T (s) = C(sI − A)−1 B of dimension p × q.
Here w is a disturbance input and z is an output of interest that is
desired to be small. A quantification of the effect of the input w onto
the output z is the so-called H2 -norm of the transfer matrix.
Definition 1 Let T have all its poles in the open left half-plane.
The H2 -norm of T is defined as
s Z
∞
1
kT k2 :=
kT (iω)k2F dω
2π −∞
with k.kF denoting the Frobenius matrix norm.
2/82
Relation to Hardy-Spaces
The Hardy-space H2p×q consists of all matrices S of dimension p × q
whose elements are analytic functions in the open right half-plane s.th.
Z ∞
1
2
kS(r + iω)k2F dω is finite.
kSk2 := sup
r>0 2π −∞
For all such functions one can show that the limit
T̂ (iω) := lim S(r + iω)
r&0
exists for almost all ω ∈ R, that ω → T̂ (iω) is square integrable over
R, and that kSk2 actually equals kT̂ k2 as defined on the previous slide.
One can as well show that H2p×q is a Hilbert-space.
The subspace of all strictly proper and stable real rational matrices
is denoted as RH2p×q . The subspace RH2p×q is dense in H2p×q .
This links our discussion to the fascinating area of Hardy-spaces in Math.
3/82
Parseval and Payley-Wiener
The space Lp×q
2 [0, ∞) denotes all real-matrix-valued functions on the
positive half-line that are square integrable. For any F ∈ Lp×q
2 [0, ∞)
the Fourier-transform is defined, in the sense of a limit in the mean, as
Z ∞
F̂ (iω) =
e−iωt F (t) dt.
0
Parseval’s theorem states that
Z ∞
Z ∞
1
2
kF (t)kF dt =
kF̂ (iω)k2F dω.
2π −∞
0
One can show that F̂ is in H2p×q . Parseval’s theorem just means that the
p×q
Fourier transform is a linear isometry Lp×q
. A version of
2 [0, ∞) → H2
the Payley-Wiener theorem establishes that this map is even surjective.
p×q
Therefore Lp×q
are actually isometrically isomorphic.
2 [0, ∞) and H2
See: B.A. Francis, A course in H∞ -Control, Springer LNCIS 88, 1987.
4/82
Computation
It is a beautiful fact that the H2 -norm of a stable transfer matrix can
be computed algebraically on the basis of a state-space realization.
Theorem 2 Let A be Hurwitz and T (s) = C(sI − A)−1 B. Then:
1. kT k22 = tr CPc C T where APc + Pc AT + BB T = 0.
2. kT k22 = tr B T Po B where AT Po + Po A + C T C = 0.
R∞ T
Proof of 2.: Recall that Po = 0 eA t C T CeAt dt. Using Parseval’s theorem we infer
Z ∞
1
2
tr
T (iω)∗ T (iω) dω =
kT k2 =
2π
−∞
Z ∞
= tr
[CeAt B]T [CeAt B] dt = tr B T Po B .
0
The proof of 1. proceeds in an analogous fashion.
5/82
Inequality Characterization
The following characterizations of a bound on the H2 -norm will be useful
for controller synthesis proofs.
Lemma 3 A is Hurwitz and kT k22 < γ iff there exists X 0 with
AT X + XA + C T C ≺ 0 and tr(B T XB) ≺ γ.
Proof of if. Since AT X + XA ≺ 0 and X 0, A is Hurwitz. By
subtracting the Lyapunov equation for Po we infer AT (X − Po ) +
(X − Po )A ≺ 0 and thus Po ≺ X since A is Hurwitz. This implies
tr(B T Po B) ≤ tr(B T XB) < γ and thus kT k22 < γ.
Proof of only if. Consider the solution P of the perturbed Lyapunov
equation AT P + P A + C T C + I = 0 with > 0. Clearly P → Po
for → 0. Since tr(B T Po B) ≺ γ, we can hence fix some 0 with
tr(B T P0 B) ≺ γ. Then X = P0 does the job.
6/82
Deterministic Interpretation
Let ek be the standard unit vector of dimension q and let zk (.) denote
the response of
ẋ = Ax, z = Cx, x(0) = Bek
Recall that this is just the response to an impulse in the k-th input.
Since zk (t) = CeAt Bek we infer
Z ∞
Z ∞
AT t T
At
T
T T
e C Ce dt Bek = eTk B T Po Bek .
zk (t) zk (t) dt = ek B
0
0
After summing over k the right-hand side is tr(B T Po B) = kT k22 .
The squared H2 -norm is the sum of the energies of the transients of
“the” impulse responses:
q Z ∞
X
kzk (t)k2 dt = kT k22 .
k=1
0
7/82
Random Vectors
Uncertain outcomes of experiments are modeled by random vectors
x = (x1 · · · xn )T . Here x is a vector of n random variables x1 , . . . , xn
and is characterized by its distribution function Fx : Rn → R which
admits the following interpretation: If (ξ1 · · · ξn )T ∈ Rn , the probability
for the event x1 ≤ ξ1 , ..., xn ≤ ξn to happen is given by Fx (ξ1 , ..., ξn ).
Fx (ξ1 , ..., ξn ) has the density fx (τ1 , ..., τn ) in case that
Z ξ1 Z ξn
Fx (ξ1 , ..., ξn ) =
···
fx (τ1 , ..., τn ) dτn · · · dτ1 for all ξ ∈ Rn .
−∞
−∞
The rv x is Gaussian if its distribution function has the density
1
1
T −1
exp
(τ − m) R (τ − m)
fx (τ ) = p
2
(2π)n det(R)
where m ∈ Rn and R ∈ Rn×n is symmetric and positive definite.
8/82
Expectation and Covariance
Suppose g : Rn → Rk×l is Borel measurable. If the random vector
x = (x1 · · · xn )T has the density fx (τ1 , ..., τn ), the expectation of
g(x1 , ..., xn ) is a matrix in Rk×l and defined as
Z +∞ Z +∞
E[g(x1 , ..., xn )] =
···
g(τ1 , ..., τn )fx (τ1 , ..., τn ) dτn · · · dτ1 .
−∞
−∞
Examples. With g(τ ) = τ we obtain the expectation E[x] of x itself.
With g(τ, σ) = (τ −E[x])(σ−E[y])T we obtain the covariance matrix
cov(x, y) := E[(x − E[x])(y − E[y])T ]
of the two random vectors x and y. Then
cov(x, x) = E[(x − E[x])(x − E[x])T ] = E[xxT ] − E[x]E[x]T < 0
is the auto-covariance matrix x. Its trace is called variance of x.
Moreover, E[xxT ] is the second order moment matrix of x and its
trace E[xT x] is the second order moment of x.
9/82
Wiener-Processes
In control the disturbance w on slide 2 is often considered as white
noise, which is associated with irregular signals having a flat spectrum.
Loosely speaking, this boils down to viewing w as the derivative of the
(normalized) Wiener-process or Brownian motion. It would take us too
far astray and it is not required for our purposes to develop the whole
theory of stochastic differential equations (based on Îto calculus).
Instead let us just collect some basic facts that are required in the sequel.
1) There exists a Wiener-process W (t) for t ≥ 0 with intensity 1:
• Initialized at zero: W (0) = 0 with probability one.
• Independent increments: For all 0 ≤ t1 ≤ t2 ≤ t3 ≤ t4 , the random
variables W (t2 ) − W (t1 ) and W (t4 ) − W (t3 ) are independent.
• Gaussian increments: For all 0 ≤ t1 ≤ t2 , the increment W (t2 ) −
W (t1 ) is Gaussian with expectation 0 and variance t2 −t1 = 1|t2 −t1 |.
10/82
Simulation of Standard Wiener Process
Realizations of Wiener Process
8
6
4
2
0
−2
−4
0
1
2
3
4
5
6
7
8
9
10
• Sample paths are continuous with probability one.
• W (t) for t > 0 is Gaussian with density fW (t) (τ ) =
τ2
√ 1 e− 2t
2πt
.
• The process W itself is Gaussian: For all k ∈ N and all pairwise
different time instances t1 , . . . , tk > 0 the random vector
T
W (t1 ) · · · W (tk )
is Gaussian.
11/82
Wiener-Processes
2) For a square integrable real-valued function f on [a, b] (0 ≤ a ≤ b),
Z b
the random variable
f (t) dW (t)
a
is defined analogously to standard Lebesgue-Stieltjes integrals as follows:
• For a step-function s(.) which takes the values s1 , . . . , sN on the
partition [tk , tk+1 ], k = 1, . . . , N (a = t1 < · · · < tN +1 = b), define
Z b
N
X
s(t) dW (t) :=
sk [W (tk+1 ) − W (tk )].
a
k=1
• With a sequence sν of step-functions that converges to f in L2 [a, b],
Z b
Z b
sν (t) dW (t),
f (t) dW (t) := lim
ν→∞ a
|
{z
}
{z
}
|a
I
Iν
2
in the sense of E[(I − Iν ) ] → 0 for ν → ∞.
12/82
Wiener-Processes
3) The integral is a Gaussian random variable. Just by direct calculation
for step-functions and taking limits, one proves for x, y ∈ L2 [a, b] that
Z b
E
x(t) dW (t) = 0,
a
Z
E
b
Z
x(t) dW (t)
a
b
Z b
y(τ ) dW (τ ) =
x(t)y(t) dt.
a
a
If Ŵ is a Wiener-process that is independent from W then
Z b
Z b
E
x(t) dŴ (t)
y(τ ) dW (τ ) = 0.
a
a
All these properties about expectation and correlation of integrals
of deterministic functions with respect to Wiener-processes are the
only facts from stochastics that remain unproven here.
L. Arnold, Stochastic Differential Equations: Theory and Applications, Wiley, 1974
13/82
Wiener-Processes
The q-dimensional Wiener-process W = col(W1 , . . . , Wq ) is a vector
of q Wiener processes W1 , . . . , Wq (with intensity 1) that are pairwise
independent.
If X and Y are matrix-valued functions of dimension p × q with square
integrable elements on [a, b] (0 ≤ a ≤ b), then the random vectors
Z b
Z b
x=
X(t) dW (t), y =
Y (t) dW (t)
a
a
of dimension p are defined elementwise. They both have mean zero and
and their correlation matrix is given by
Z b
T
E[xy ] =
X(t)Y (t)T dt.
a
Proof. Verify these properties elementwise.
14/82
White Noise and System Response
Let us come back to slide 2 with w being interpreted as the “derivative
of a Wiener process”; we (formally) denote it by Ẇ . In this sense “W
can be obtained by integrating white noise”:
Z t
Z t
W (t) = “
Ẇ (τ ) dτ “ =
dW (τ ) for t ≥ 0.
0
0
The middle expression is NOT sensible mathematically. But we can now
just define precisely what we mean by the state-response of a linear
system to a white noise input and a random initial condition ξ. Tacitly,
we assume that ξ is Gaussian and independent from W (t) for all t ≥ 0.
Definition 4 The response of the linear system ẋ = Ax + B Ẇ
with x(0) = ξ (and ξ independent from W ) is defined as
Z t
At
x(t) := e ξ +
eA(t−τ ) B dW (τ ) for t ≥ 0.
0
15/82
White Noise and System Response
According to our preparatory remarks x(.) is a Gaussian process.
If just applying the rules given above, the expectation of x(t) and the
covariance matrix of x(t1 ) and x(t2 ) are easily determined.
Theorem 5 Let x(.) be the response of ẋ = Ax + B Ẇ , x(0) = ξ.
Then
E[x(t)] = eAt E[ξ] for t ≥ 0
and
cov(x(t1 ), x(t2 )) = e
At1
cov(ξ, ξ)e
AT t2
Z
+
t1
eA(t1 −τ ) BB T eA
T (t −τ )
2
dτ
0
for 0 ≤ t1 ≤ t2 .
The specialization to A = 0 and ξ = 0 implies x(t) = BW (t) and thus
E[BW (t)] = 0 as well as E[BW (t)W (t)T B T ] = tBB T .
16/82
Stochastic Interpretation of H2 -Norm
Corollary 6 Let eig(A) ⊂ C− and x(.), z(.) be the state- and
output-responses of ẋ = Ax + B Ẇ , z = Cx, x(0) = ξ. Then
E[x(t)] → 0 and E[z(t)] → 0 for t → ∞. Moreover:
• The state’s auto-covariance satisfies lim cov(x(t), x(t)) = Pc .
t→∞
• The output-variance satisfies lim tr cov(z(t), z(t)) = kT k22 .
t→∞
kT k22 is the asymptotic variance limt→∞ E[z(t)T z(t)] − E[z(t)]T E[z(t)]
of the output of a stable linear system driven by white noise.
Proof. E[x(t)] = eAt E[ξ] → 0 and E[z(t)] → 0 for t → 0 are clear.
T
Since lim eAt cov(ξ, ξ)eA t = 0, we infer that lim cov(x(t), x(t)) is
t→∞
t→∞
Z t
Z t
T
T
lim
eA(t−τ ) BB T eA (t−τ ) dτ = lim
eAσ BB T eA σ dσ = Pc .
t→∞
t→∞
0
0
This also implies
lim tr cov(z(t), z(t)) = lim tr C[cov(x(t), x(t))]C T = tr(CPc C T ).
t→∞
t→∞
17/82
Example
√
Consider ẋ = −px + 2p Ẇ , z = x for p > 0. Simulate the system and
plot the output process and a histogram thereof for p = 10, 100:
Squared H2−norm: 1
Estimated variance 1.0747
4
300
2
200
0
100
−2
−4
0
5
10
0
−5
Squared H2−norm: 1
400
2
300
0
200
−2
100
5
5
Estimated variance 1.0272
4
−4
0
0
10
0
−5
0
5
18/82
Colored Noise
Definition 7 We say that w̃ is colored noise if there exists
(Ã, B̃, C̃) with eig(Ã) ⊂ C− such that w̃ is the output of
x̃˙ = Ãx̃ + B̃ Ẇ , w̃ = C̃ x̃, x̃(0) = 0.
This is also expressed as w̃ emerging through filtering white noise with
the coloring filter T̃ (s) = C̃(sI − Ã)−1 B̃.
The response of the linear system
ẋ = Ax + B w̃, z = Cx + Dw̃, x(0) = ξ
driven by colored noise w̃ is defined as the output of
0
ẋ
A B C̃
x
x(0)
ξ
=
+
Ẇ ,
=
˙x̃
x̃
x̃(0)
0
B̃
0 Ã
x
z = C DC̃
.
x̃
19/82
Remarks
Rt
• Since w̃(t) = 0 C̃eÃ(t−τ ) B̃ dW (τ ) for t ≥ 0, we can assume w.l.o.g.
that (Ã, B̃, C̃) is minimal, and that the coloring filter T̃ (and not
its realization) captures all properties of w̃. This raises the question
about how to determine such coloring filters in practice.
• The response of a linear system to colored noise is reduced to that
for white noise and for the series interconnection of the system and
the coloring filter.
• In the theory of stochastic differential equations it is shown that our
definitions are consistent. More precisely, the response of
ẋ = Ax + B w̃, z = Cx + Dw̃, x(0) = ξ
can indeed be represented as it should, namely as
Z t
At
x(t) = Ce ξ +
CeA(t−τ ) B w̃(τ ) dτ + Dw̃(t) for t ≥ 0,
0
if defining the integral appropriately.
20/82
Coloring Filters
Clearly E[w̃(t)] = 0 for all t ≥ 0. For a fixed τ ∈ R let us consider the
asymptotic covariance matrix of w̃(t) and w̃(t + τ ):
R(τ ) := lim E[w̃(t + τ )w̃(t)T ].
t→∞
It turns out that R(τ ) can be easily obtained algebraically as follows:
Theorem 8 If à is Hurwitz and P̃ is
Lyapunov equation ÃP̃ + P̃ ÃT + B̃ B̃ T
(
C̃eÃτ P̃ C̃ T
R(τ ) =
T
C̃ P̃ e−Ã τ C̃ T
the unique solution of the
= 0 then
for τ ≥ 0,
for τ < 0.
Note that R(−τ )T = R(τ ). Since eig(Ã) ⊂ C− , R(τ ) decays exponentially for τ → ±∞ and has a well-defined Fourier transform.
Definition 9 The Fourier transform R̂ of R is called the spectral
density of the process w̃.
21/82
Proof
For t ≥ 0 and τ > 0 the filter state satisfies
Z t
T
T
E[x̃(t + τ )x̃(t) ] =
eÃ(t+τ −σ) B̃ B̃ T eà (t−σ) dσ =
0
=e
Ãτ
Z
t
eÃσ B̃ B̃ T eÃ
Tσ
dσ → eÃτ P̃ for t → ∞.
0
Similarly for t ≥ 0 and τ ∈ [−t, 0] we get
Z t+τ
T
T
E[x̃(t + τ )x̃(t) ] =
eÃ(t+τ −σ) B̃ B̃ T eà (t−σ) dσ =
0
Z
=
t+τ
eÃσ B̃ B̃ T eÃ
T (σ−τ )
dσ → P̃ e−Ã
Tτ
for t → ∞.
0
The proof is finished with the observation
R(τ ) = lim C̃E[x̃(t + τ )x̃(t)T ]C̃ T .
t→∞
22/82
Coloring Filters
Theorem 10 The spectral density of w̃ is given by
R̂(iω) = T̃ (iω)T̃ (iω)∗ .
Hence R̂(iω) is Hermitian and positive semi-definite for all ω ∈ R.
Proof. By definition R̂(iω) equals
Z ∞
Z 0
−ÃT τ T
−iωτ
C̃ dτ +
C̃e−iωτ eÃτ P̃ C̃ T dτ =
C̃ P̃ e
e
0
−∞
Z ∞
iωτ ÃT τ
−1
e e
= C̃ P̃
dτ + (iωI − Ã) P̃ C̃ T =
0
= C̃ P̃ (iωI − Ã)−∗ + (iωI − Ã)−1 P̃ C̃ T .
Now note that ÃP̃ + P̃ ÃT + B̃ B̃ T = 0 also satisfies the equation
(iωI − Ã)P̃ + P̃ (iωI − Ã)∗ = B̃ B̃ T and thus
P̃ (iωI − Ã)−∗ + (iωI − Ã)−1 P̃ = (iωI − Ã)−1 B̃ B̃ T (iωI − Ã)−∗ .
Plugging this into the above relation proves the formula.
23/82
Example
Consider the colored noise on slide 18. The filter has the realization
√ −p − 2p
T̃ =
1
0
and hence P̃ = 1. Therefore
−p|τ |
R(τ ) = e
√
√
2p
2p
2p
and R̂(iω) =
.
= 2
p + iω p − iω
ω + p2
The graphs for p = 10 (blue) and p = 100 (red) look as follows:
R(τ)
|T~(iω)|
0
1
10
0.8
0.6
−2
10
0.4
0.2
0
−0.1
−4
−0.05
0
0.05
0.1
10
−2
10
0
10
2
10
4
10
24/82
Coloring Filters: Construction
We are now prepared for discussing how to determine coloring filters
in practice. By measurements one (statistically) estimates the spectral
density R̂(iω) of the process under scrutiny. One then approximates the
experimentally determined spectral density by G(iω) where G(s) is a
strictly proper real rational function without poles in C0 and with
G(iω) = G(iω)∗ and G(iω) < 0 for all ω ∈ R.
(∗)
Finally, the coloring filter is obtained by spectral factorization.
Theorem 11 Suppose that the strictly proper real-rational function
G(s) without poles in C0 satisfies (∗). Then there exists a strictly
proper and stable transfer matrix T with G(s) = T (s)T (−s)T .
This implies G(iω) = T (iω)T (iω)∗ such that T is a coloring filter for
modeling noise with the spectral density G, just as we desired. Often T
is also called a spectral factor of G.
25/82
Example
Consider the transfer function G(s) =
1−s2
.
s4 −13s2 +36
We have
1 + ω2
> 0 for ω ∈ R.
ω 4 + 13ω 2 + 36
Hence, by Theorem 11, G has a spectral factorization. This can be seen
directly as follows. The numerator and denominator of G(s) can be
factorized as
G(iω) =
(1 + s)(1 − s) and (3 + s)(2 + s)(2 − s)(3 − s)
respectively. The symmetric location of the poles and zeros with respect
to the imaginary axis is a consequence of G(iω) being real and positive
for ω ∈ R. Obviously, both transfer functions
1+s
1−s
and T+ (s) =
T− (s) =
(3 + s)(2 + s)
(3 + s)(2 + s)
are spectral factors of G. They distinguish themselves in that T− shares
it stable zero(s) with G, while T+ shares its anti-stable zero(s) with G.
26/82
System Description for Design
Let us now consider the following generalized plant
ẋ = Ax + Bw w + Bu
z = Cz x + Dzw w + Dz u
y = Cx + Dw w + Du
w
z
y
P
u
with an external disturbance input w (that cannot be influenced), a
control input u, a performance output z (to-be-rendered small) and
a measurement output y.
Recall how various concrete configurations of disturbance rejection as
discussed in the lecture on regulation can be subsumed to this paradigm.
The goal is to find a feedback controller which stabilizes this system
and which minimizes the H2 -norm of the closed-loop transfer matrix.
For simplicity of the exposition let us assume Dzw = 0 and D = 0.
27/82
The H2 -Control Problem
Open-loop system P :
ẋ = Ax + Bw w + Bu
z = Cz x
+ Dz u
y = Cx + Dw w
Controller K:
ẋK = AK xK + BK y
u = C K xK
w
z
P
y
u
K
Controlled system described as ξ˙ = Aξ + Bw, z = Cξ with


A
BCK
Bw
A B
=  BK C AK BK Dw  .
C D
Cz Dz CK
0
Problem: Minimize the H2 -norm kC(sI −A)−1 Bk2 of the controlled
closed-loop system over all controllers which render A Hurwitz.
28/82
LQG-Control
One particular scenario is worth mentioning. Consider the system
ẋ = Ax + B1 Ẇ1 + Bu
with control input u and process noise B1 Ẇ1 . Suppose the measurements Cx are corrupted by white noise Ẇ2 where W1 and W2 are independent Wiener processes. The measured output hence is
y = Cx + D2 Ẇ2 .
As in LQ-control, we are interested in keeping the linear combinations
C1 x and D1 u of the states and the controls small. Hence we choose
C1 x
z=
D1 u
as the performance output. The LQG-control goal is to find a stabilizing
controller which minimizes lim tr cov(z(t), z(t)), which equals
t→∞
lim E[z(t)T z(t)] = lim E x(t)T C1T C1 x(t) + u(t)T D1T D1 u(t) .
t→∞
t→∞
29/82
LQG-Control
Let us define the following generalized plant
ẋ = Ax + B1 0 w + Bu
C1
0
z =
x+
u
0
D1
y = Cx + 0 D2 w.
Finding a stabilizing output-feedback controller which minimizes the
asymptotic variance of z for a while noise w is the classical so-called
Linear-Quadratic-Gaussian (LQG) optimal control problem.
In view of our preparation this is merely a special case of H2 -control!
Note that the process and measurement noises are defined via the
Wiener-processes B1 W1 and D2 W2 with auto-covariances
E B1 W1 (t)W1 (t)T B1T = B1 B1T t, E D2 W2 (t)W2 (t)T D2T = D2 D2T t.
Hence B1 B1T and D2 D2T are the intensities of these processes.
30/82
LQG-Control
If Ẇ1 is not white but colored noise w̃1 , one absorbs the related coloring
filter T (s) = C̃(sI − Ã)−1 B̃ into the generalized plant and solves the
H2 -problem for the weighted generalized plant
0 0
ẋ
A B1 C̃
x
B
=
+
w+
u
x̃
x̃
0
B̃ 0
0 Ã
C1 0
x
0
z =
+
u
0 0
x̃
D1
x
y = C 0
+ 0 D2 w.
x̃
The block-diagram of the weighted generalized plant is more instructive:
w̃1
z
P
y
w2
T
w1
u
31/82
Static State-Feedback Synthesis
Let us first consider the case that the whole state is available for control.
For this purpose we consider the open-loop system
ẋ = Ax + Bw w + Bu
z = Cz x + Dz u
under the following hypotheses:
• (A, B) is stabilizable.
• (A, Cz ) has no unobservable modes in C0 .
• DzT Cz Dz = 0 I .
Controlling the system by static state-feedback as
u = −F x
leads to the closed-loop system
ẋ = (A − BF )x + Bw w
z = (Cz − Dz F )x.
32/82
Optimal H2 -Control by Static State-Feedback
The goal is to infimize
k(Cz − Dz F )(sI − A + BF )−1 Bw k2
over all F such that A − BF is Hurwitz. It turns out that the infimum
is actually attained. Moreover, both the optimal value and an optimal
state-feedback gain can be computed by solving an LQ-Riccati equation.
Theorem 12 Let P denote the stabilizing solution of
AT P + P A − P BB T P + CzT Cz = 0.
Then the optimization problem
γopt :=
min
F , A−BF Hurwitz
k(Cz − Dz F )(sI − A + BF )−1 Bw k22
has the optimal value
γopt = tr(BwT P Bw )
and F = B T P is an optimal solution.
33/82
Proof
Let γ > γopt . Then there exists F for which A − BF is Hurwitz and
such that k(Cz − Dz F )(sI − A + BF )−1 Bw k22 < γ. Due to Lemma 3
we can choose X 0 with
(A − BF )T X + X(A − BF ) + (Cz − Dz F )T (Cz − Dz F ) ≺ 0
and tr(BwT XBw ) < γ. Exploiting DzT Cz = 0 and DzT Dz = I allows to
rearrange to
AT X + XA + CzT Cz − XBB T X + (B T X − F )T (B T X − F ) ≺ 0.
Therefore AT X + XA + CzT Cz − XBB T X ≺ 0 and thus P ≺ X by
the result on slide 68. This implies that
tr(BwT P Bw ) 4 tr(BwT XBw ) < γ.
Since γ > γopt was arbitrary we conclude
tr(BwT P Bw ) ≤ γopt .
34/82
Proof
Let us now choose F = B T P . Note that A − BF = A − BB T P is
Hurwitz, just because of the choice of P as the stabilizing solution of
the ARE. Moreover with the ARE we trivially have
AT P + P A + CzT Cz − P BB T P + (B T P − F )T (B T P − F ) = 0.
As above this can be re-arranged to
(A − BF )T P + P (A − BF ) + (Cz − Dz F )T (Cz − Dz F ) = 0
Viewed as a Lyapunov equation, this shows that
k(Cz + Dz F )(sI − A − BF )−1 Bw k22 = tr(BwT P Bw ).
This proves that tr(BwT P Bw ) is attained by the stabilizing state-feedback
gain F = B T P which indeed confirms
tr(BwT P Bw ) = γopt
and optimality of F .
35/82
Example
For the following model of a DC-motor ([F] p.20)
0 1
0
φ
ẋ =
x+
u, x =
.
0 −1
1
ω
design a full information state-feedback controller which tracks φ. This
results in the following response:
Control action
200
0
−200
0
2
4
6
Reference (red) and output
8
10
8
10
2
0
−2
0
2
4
6
36/82
Example
Assume that the control input is affected by colored noise with filter
134.2/(s + 10) and having H2 -norm 30 in order to clearly display its
effect. We get the following substantially deteriorated response:
Control action
200
0
−200
0
2
4
6
Reference (red) and output
8
10
8
10
2
0
−2
0
2
4
6
37/82
Example
With the coloring filter included in the system description, design an
optimal H2 -state-feedback gain for the cost function
2000|φ|2 + 0.3|u|2 .
The feed-forward gain is adjusted appropriately. The cost has been tuned
so that the noise-free response resembles the one we obtained earlier:
Control action
200
0
−200
0
2
4
6
Reference (red) and output
8
10
8
10
2
0
−2
0
2
4
6
38/82
Example
The noisy closed-loop response shows that the effect of the noise onto
the to-be-tracked output is visibly reduced:
Control action
200
0
−200
0
2
4
6
Reference (red) and output
8
10
8
10
2
0
−2
0
2
4
6
39/82
Kalman Filtering
Consider again the full generalized plant
ẋ = Ax + Bw w + Bu, z = Cz x + Dz u, y = Cx + Dw w.
If w = 0, an observer for this system is defined as
x̂˙ = Ax̂ + Bu + L(y − ŷ), ẑ = Cz x̂ + Dz u, ŷ = C x̂
where L is taken with eig(A − LC) ⊂ C− ; then the observer state
asymptotically reconstructs the system’s state.
If w does not vanish and is a white-noise disturbance, the quantity
lim E (z(t) − ẑ(t))T (z(t) − ẑ(t))
(err)
t→∞
can serve as a measure for how well ẑ(t) approximates z(t) for t → ∞.
An observer for which the asymptotic variance (err) of z − ẑ is
minimized is called a Kalman Filter for the generalized plant.
40/82
Optimal H2 -Observer Synthesis
By considering the dynamics of the state-error ξ = x − x̂, one easily
checks that the transfer matrix from w to z − ẑ admits the description
ξ˙ = (A − LC)ξ + (Bw − LDw )w, z − ẑ = Cz ξ.
H2 -optimal observer synthesis problem
Find L which renders A − LC Hurwitz and minimizes the H2 -norm
of the transfer matrix from w to z − ẑ.
We stress again that this formulation admits various interpretations,
according to what can be subsumed to H2 -norm minimization.
The problem is solved under the following simplifying assumptions:
• (A, C) is detectable.
• (A, Bw ) has no uncontrollable modes in C0 .
• Dw BwT DwT = 0 I .
41/82
Optimal H2 -Observer Synthesis
We hence need to determine L which renders A − LC Hurwitz and
which minimizes
kCz (sI − A + LC)−1 (Bw − LDw )k2 .
The solution to this problem is dual to Theorem 12.
Theorem 13 Let Q denote the stabilizing solution of
AQ + QAT − QC T CQ + Bw BwT = 0.
Then the optimization problem
γopt :=
min
L, A−LC Hurwitz
kCz (sI − A + LC)−1 (Bw − LDw )k22
has the optimal value
γopt = tr(Cz QCzT )
and L = QC T is an optimal solution.
42/82
Proof by Duality
Note that we have
kCz (sI − A + LC)−1 (Bw − LDw )k2 =
= k(BwT − DwT LT )(sI − AT + C T LT )−1 CzT k2 .
This observation reduces the problem to one of static state-feedback
control for the system
x̌˙ = AT x̌ + CzT w̌ + C T ǔ, ž = BwT x̌ + DwT w̌.
An application of Theorem 12 finishes the proof. (Provide the details.)
43/82
Remarks
• Note that the optimal observer does not depend on Cz or Dz ! In
particular it is as well an optimal H2 -observer for the full state x
(with the choices Cz = I and Dz = 0) with optimal value tr(Q).
• An optimal H2 -estimator is an unstructured LTI system with inputs
u and y, which generates an asymptotic state-estimate x̂ such that
Cz x̂+Dz u is an optimal estimate of z in the H2 -sense. One can prove
that general estimators do not offer any benefit over observers!
• In view of the stochastic interpretation of the H2 -norm, optimal H2 observers minimize the asymptotic variance of z − ẑ if w is white
noise. Then the optimal observer is the celebrated Kalman-Filter.
R.E. Kalman, A new approach to linear filtering and prediction problems, Journal of basic Engineering, 82 (1960) 35-45. (13560 citations
in Google Scholar as of February 7, 2013!)
44/82
Example
Consider again the two-compartment model ([AM] pp.85):
ċ =
y =
−k0 − k1 k1
k2
−k2
1 0 c + λ2 w 2
c+
0
λ1
w1 +
b0
0
u
k0 > 0, k1 > 0, k2 > 0
in which the second state and the output are corrupted by (independent)
white noises w1 and w2 and λ1 and λ2 scale their intensities. For k0 = 1,
k1 = 1, k2 = 2 and b0 = 1 we design a pole-placing observer with pole
locations −1.5, −1.6 and compare it with the Kalman filter. Note that
the poles have been chosen such that the responses resemble those for
the Kalman filter if the intensities λ1 and λ2 are small (noise-free case).
The simulations on the next slides illustrate a substantial reduction in
noise sensitivity if using the Kalman filter (in the lower plots).
45/82
Example
System and observer responses as well as errors for λ1 = λ2 = 0.01:
2
2
1
1
0
0
−1
−1
−2
0
5
10
−2
0
2
2
1
1
0
0
−1
−1
−2
0
5
10
−2
0
1
0
−1
5
10
−2
0
5
10
5
10
1
0
−1
5
10
−2
0
46/82
Example
System and observer responses as well as errors for λ1 = λ2 = 0.1:
4
2
0
−2
0
5
10
2
4
1
2
0
0
−1
−2
−2
−4
0
5
10
2
1
5
10
5
10
1
0
0
0
−1
−2
−1
−2
0
−3
0
5
10
−4
0
−2
5
10
−3
0
47/82
Example
System and observer responses as well as errors for λ1 = λ2 = 1:
4
4
4
2
2
2
0
0
0
−2
−2
−2
−4
0
5
10
4
2
0
−2
0
5
10
−4
0
5
10
−4
0
4
4
2
2
0
0
−2
−2
−4
0
5
10
−4
0
5
10
5
10
48/82
The Output-Feedback H2 -Control Problem
Open-loop system P :
ẋ = Ax + Bw w + Bu
z = Cz x
+ Dz u
y = Cx + Dw w
Controller K:
ẋK = AK xK + BK y
u = C K xK
w
z
P
y
u
K
Controlled system described as ξ˙ = Aξ + Bw, z = Cξ with


A
BCK
Bw
A B
=  BK C AK BK Dw  .
C D
Cz Dz CK
0
Problem: Minimize the H2 -norm kC(sI −A)−1 Bk2 of the controlled
closed-loop system over all controllers which render A Hurwitz.
49/82
Hypotheses
We derive a solution to this optimal synthesis problem in terms of AREs
under the following assumptions:
1. (A, B) is stabilizable and (A, C) is detectable.
Are required for the existence of a stabilizing controller.
Bw
0
T
T
Dw =
.
2. Dz Cz Dz = 0 I and
Dw
I
It’s essential that Dz and Dw have full column and row rank. The
other properties are introduced to simplify the formulas.
3. (A, Cz ) has no unobservable and (A, Bw ) no uncontrollable modes
on the imaginary axis.
Are required for the existence of stabilizing solutions of AREs.
A fully general solution without hypotheses can be obtained with LMIs.
50/82
Output-Feedback Control: Riccati Equation Solution
Under the given hypotheses the H2 -problem admits an optimal solution.
This main result of this lecture is formulated as follows.
Theorem 14 With the stabilizing solutions P and Q of the ARE’s
AT P + P A − P BB T P + CzT Cz = 0,
AQ + QAT − QC T CQ + Bw BwT = 0,
an H2 -optimal controller is given as
ẋK = (A − BB T P − QC T C)xK + QC T y, u = −B T P xK
and the corresponding optimal closed-loop H2 -norm is
p
tr(BwT P Bw ) + tr(B T P QP B).
The proof is given in the appendix. The following remarks on the interpretation and on generalizations of this result are essential.
51/82
Output-Feedback Control: The Separation Principle
The H2 -optimal state-feedback gain is F = B T P while L = QC T is the
H2 -optimal observer gain. With these, the optimal H2 -controller can be
written as ẋK = (A − BF − LC)xK + Ly, u = −F xK or as
ẋK = AxK + Bu + L(y − ŷ), ŷ = CxK , u = −F xK .
• Hence the controller is an optimal H2 -observer for the system’s state.
• If the system state is available for control then u = −F x is optimal.
Otherwise, one just has to replace x by an optimal estimate xK of
the state and optimally control the system with u = −F xK . In short:
Optimal state-feedback + Optimal estimation = Optimal output-feedback.
• The controller can as well be seen as to provide an H2 -optimal estimate −F xK of the unavailable but to-be-implemented signal −F x.
The “extra cost” for this estimation is tr(F QF T ) if compared to the
cost for optimal state-feedback controller.
52/82
Output-Feedback Control: Comments
In Matlab’s robust control toolbox the command h2syn allows to solve
the H2 -control problem. The algorithm is applicable for systems P
ẋ = Ax + Bw w + Bu
z = Cz x + Dzw w + Dz u
y = Cx + Dw w + Du
w
z
y
P
u
whose descriptions satisfy the following hypotheses:
1. (A, B) is stabilizable and (A, C) is detectable.
There exists some N with Dzw + Dz N Dw = 0.
A − iωI B
2. Dz and
have full column rank for all ω ∈ R.
Cz
Dz
A − iωI Bw
3. Dw and
have full row rank for all ω ∈ R.
C
Dw
These properties are natural generalizations of those on slide 50.
53/82
How to Enforce Satisfaction of Hypotheses?
Condition 1 is necessary for the existence of a stabilizing controller which
renders the closed-loop transfer matrix strictly proper.
If condition 2 fails choose Cze , Dze such that


A − iωI B
Dz
Dz  have full column rank ∀ ω ∈ R.
,  Cz
Dze
Cze
Cze
If condition 3 fails choose Bwe , Dwe such that
A − iωI Bw Bwe
Dw Dwe ,
have full row column ∀ ω ∈ R.
C
Dw Dwe
These properties can be always achieved, as the following simple but
rough choices demonstrate:
Bwe
I 0
I 0
Cze Dze =
and
=
.
0 I
Dwe
0 I
54/82
How to Enforce Satisfaction of Hypotheses?
Then define, with small > 0, the perturbed open-loop system P :
ẋ
z
ze
y
=
=
=
=
Ax + Bw w + Bwe we + Bu
Cz x + Dzw w + Dz u
Cze x + Dze u
Cx + Dw w + Dwe we + Du
z
ze
y
P
w
we
u
The perturbed system satisfies the required hypothesis and one can find
an optimal H2 -controller K with optimal value γ . Now interconnect
K with the the original system P with closed-loop transfer matrix T .
One can show that K also stabilizes P and achieves an H2 -norm
level of at least γ :
kT k2 ≤ γ .
Moreover, γ converges monotonically to the optimal achievable H2 norm level for P (although an optimal controller might not exist).
55/82
Example
In the lectures we provide a demo on designing an output-feedback
tracking controller for the motor on slide 36. We obtain the following
responses for a pole-placement and LQG-synthesis with low noise-levels:
Control action
20
0
−20
0
2
4
6
Reference (red) and output
8
10
8
10
8
10
8
10
2
0
−2
0
2
4
6
Control action
20
0
−20
0
2
4
6
Reference (red) and output
2
0
−2
0
2
4
6
56/82
Example
For increased noise-levels we see the benefit of LQG-control:
Control action
20
0
−20
0
2
4
6
Reference (red) and output
8
10
8
10
8
10
8
10
2
0
−2
0
2
4
6
Control action
20
0
−20
0
2
4
6
Reference (red) and output
2
0
−2
0
2
4
6
57/82
Example
However, comparisons of this sort can be misleading:
• Obviously the response of the pole-placement observer to non-zero initial conditions is faster than that of the LQG controller. This explains
its higher sensitivity to noise. Slowing down the observer poles does
not alter the tracking behavior (a lot), but it reduces the sensitivity
to noise.
• For this simple example, one can obtain similar designs by poleplacement and LQG-synthesis after tuning. In practice, modern synthesis tools rather serve to reduce the time required for tuning a
controller, while the optimality properties are not that crucial.
• LQG output-feedback controllers suffer from an essential deficiency:
There are no guarantees for robustness! This lead to the development
of dedicated tools for robust controller synthesis. You are now wellprepared to enter this exciting field of control.
58/82
Appendix
59/82
Proof of Theorem 11: Step 1
Choose a minimal realization G(s) = CG (sI −AG )−1 BG . Since G has no
poles in C0 , the matrix AG has only eigenvalues in C− or C+ . In suitable
coordinates we can assume that the realization has the structure


A1 0 B1
 0 A2 B2  , eig(A1 ) ⊂ C− , eig(A2 ) ⊂ C+ .
C1 C2 0
Therefore
G(s) = C1 (sI − A1 )−1 B1 + C2 (sI − A2 )−1 B2 .
Observe that
G(−s)T = B2T (−sI − AT2 )−1 C2T + B1T (−sI − AT1 )−1 C1T .
Since G(iω)∗ = G(−iω)T , we have G(s) = G(−s)T for s ∈ C0 and
hence for all s ∈ C (different from poles of G(s) and G(−s)T ). Thus
the stable and anti-stable parts in the additive decomposition of G(s)
and G(−s)T coincide.
59/82
Proof of Theorem 11: Step 1
Therefore
C1 (sI −A1 )−1 B1 = B2T (−sI −AT2 )−1 C2T = (−B2T )(sI −(−AT2 ))−1 C2T .
By realization minimality, there exists a non-singular T such that
A1 = −T AT2 T −1 , B1 = T C2T , C1 = −B2T T −1 .
Performing yet another state-coordinate change proves the following
intermediate fact.
G admits a minimal realization


A 0
B
 0 −AT −C T  with eig(A) ⊂ C− .
C BT
0
Observe that minimality of the realization is equivalent to (A, B) being
controllable and (A, C) being observable.
60/82
Proof of Theorem 11: Step 2
If we define J(s) = C(sI − A)−1 B we infer
G(iω) = J(iω) + J(iω)∗ < 0 for all ω ∈ R.
Hence J is said to be positive real. This has the following consequence,
the result of which is the celebrated Positive Real Lemma.
There exists some X = X T such that
AX + XAT < 0 and B + XC T = 0.
If we then just factorize B̃ B̃ T = AX + XAT , we infer as on slide 23:
C(iωI − A)−1 B̃ B̃ T (iωI − A)−∗ C T =
= −CX(iωI − A)−∗ C T − C(iωI − A)−1 XC T for ω ∈ R.
Due to XC T = −B, the right-hand side is J(iω)∗ + J(iω) = G(iω).
This proves that T (s) = C(sI − A)−1 B̃ is the filter to-be-constructed.
61/82
Proof of the Positive Real Lemma
For ν > 0 let us consider the perturbed algebraic Riccati equation
AX + XAT − ν(XC T + B)(XC T + B)T = 0.
It suffices to show that this ARE has a solution Xν which does converges
to some X for ν → ∞. Indeed, we can then conclude AXν +Xν AT < 0
which implies AX + XAT < 0; moreover from
1
(AXν + Xν AT ) = (Xν C T + B)(Xν C T + B)T
ν
we infer Xν C T + B → 0 and thus XC T + B = 0.
In order to show existence of Xν , we observe that the ARE can be
written as
(A − νBC)X + X(A − νBC)T − νXC T CX − νBB T = 0.
(∗)
Clearly ((A − νBC)T , C T ) is controllable.
62/82
Proof of the Positive Real Lemma
According to slide 66, we need to show for any ω ∈ R that
(A − νBC)T − iωI
−νC T C
det
6= 0.
νBB T
−(A − νBC) − iωI
(∗ ∗)
For this purpose note that G(−iω) + ν1 I 0 and hence
1
J(−iω)∗ + J(−iω) + I 0.
ν
In particular the left-hand side is non-singular; it is also clearly (check!)
the Schur complement of

 T
A − iωI
0
CT

0
−A − iωI −B 
(∗ ∗ ∗)
1
T
I
B
C
ν
with respect to the left-upper block. Since A has no eigenvalues in C0
we infer that (∗ ∗ ∗) is non-singular. Hence the Schur-complement with
respect to the right-lower block is also non-singular. This is (∗ ∗).
63/82
Proof of the Positive Real Lemma
Let Xν− and Xν+ denote the stabilizing and anti-stabilizing solution of
(ARE). Now observe for ν ≥ µ that
AXν− + Xν− AT − µ(Xν− C T + B)(Xν− C T + B)T =
= (ν − µ)(Xν− C T + B)(Xν− C T + B)T < 0.
Again by slide 66 we infer Xν− 4 Xµ− . Combined with a similar argument
for Xν+ we conclude
Xµ+ 4 Xν+ 4 Xν− 4 Xµ− for ν ≥ µ.
This implies that Xν− is non-increasing and bounded from below for
increasing ν. Hence it converges for ν → ∞ as we desired to show.
Remark: The same argument applies to Xν+ . We get two limits which
lead to different spectral factors that are distinct in the properties of
their zeros. A detailed exposition goes beyond this course.
64/82
Example for Theorem 11
2
1−s
Consider G(s) = s4 −13s
2 +36 . Determine a minimal realization of G and
block-diagonalize the state-matrix. One then gets

 −0.06
−2 1.73
.
0.13
0 −3 J =
−0.87 0.5
0
By solving the perturbed ARE we obtain the two approximate solutions
as in the Positive Real Lemma:
−1.02 0.16
−1.12 0.94
, X+ =
.
X− =
0.16 −0.19
0.94 −6.12
These lead to the two spectral factors
−s − 1
−s + 1
T− (s) = 2
, T+ (s) = 2
.
s + 5s + 6
s + 5s + 6
The first shares the stable zero with G, while the second one shares the
anti-stable zero with G. This is not a coincidence but a general property!
65/82
Riccati Theory: Addendum I
With controllable (A, B), positive definite R and just a symmetric Q,
let us consider the algebraic Riccati equation
AT P + P A − P BR−1 B T P + Q = 0.
Under these hypotheses the following statements are equivalent:
A −BR−1 B T
• H=
does not have an eigenvalue in C0 .
−Q
−AT
• The ARE has a unique solution P− for which A − BR−1 B T P− is
Hurwitz. P− is called the stabilizing solution.
• The ARE has a unique solution P+ for which A − BR−1 B T P+ is
anti-Hurwitz. P+ is called the anti-stabilizing solution.
If P satisfies AT P + P A − P BR−1 B T P + Q < 0 (or 0) then
P+ 4 P 4 P− (or P+ ≺ P ≺ P− ).
66/82
Riccati Theory: Addendum I
Equivalence of the first two items is Theorem V-7. Equivalence to the
third item is a direct consequence as discussed in the exercises.
The relations among the various solutions is proved as in Theorem V-10,
by exploiting
AT P2 +P2 A−P2 BR−1 B T P2 +Q−(AT P1 +P1 A−P1 BR−1 B T P1 +Q) =
= (A − BR−1 B T P1 )T ∆ + ∆(A − BR−1 B T P1 ) − ∆BR−1 B T ∆
for ∆ = P2 − P1 .
For example for P2 = P and P1 = P+ we infer
(A − BR−1 B T P+ )T ∆ + ∆(A − BR−1 B T P+ ) < ∆BR−1 B T ∆ < 0.
Since A − BR−1 B T P+ is anti-stable, we infer ∆ < 0 and thus P < P+ .
(If P satisfies the strict Riccati inequality, all derived inequalities are
strict as well.)
67/82
Riccati Theory: Addendum II
Suppose that (A, B) is stabilizable and (A, C) has no unobservable
modes on the imaginary axis. Then the ARE
AT P + P A − P BB T P + C T C = 0
(ARE)
has a unique stabilizing solution P which also satisfies P < 0.
Let us reveal a useful relation to the set of all solutions X 0 of the
so-called strict algebraic Riccati inequality (ARI)
AT X + XA − XBB T X + C T C ≺ 0.
(ARI)
If X 0 satisfies (ARI) then P ≺ X.
Remark. Note that all solutions of (ARI) must be non-singular. Hence
any positive semi-definite solution is actually positive definite.
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Proof: Step 1
The kernel N (P ) of P is A-invariant and contained in N (C).
Indeed, with x satisfying P x = 0, we infer from xT (ARE)x = 0 that
kCxk2 = 0 and thus Cx = 0. Then (ARE)x = 0 implies P Ax = 0.
This proves the claim.
Let the columns of T2 form a basis of N (P ) and expand to a non-singular
matrix T = (T1 T2 ). Then
P1 0
T
T PT =
and P1 0.
0 0
Due to AN (P ) ⊂ N (P ) and N (P ) ⊂ N (C) we also infer
A1 0
B1
−1
−1
T AT =
, T B=
, CT = C1 0 .
A21 A2
B2
W.l.o.g. (coordinate-change - check!) we can assume that the matrices
are already given in this form while X has no particular structure.
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Proof: Step 2
By inspection, the ARE reads as AT1 P1 +P1 A1 −P1 B1 B1T P1 +C1T C1 = 0.
This implies for Q1 = P1−1 (recall invertibility!) that
(A1 + Q1 C1T C1 )T = −P1 (A1 − B1 B1T P1 )P1−1 .
Therefore A + Q1 C1T C1 is anti-stable. (All eigenvalues are in the open
right half-plane.) We also have
A1 Q1 + Q1 AT1 − B1 B1T + Q1 C1T C1 Q1 = 0.
Similarly, Y = X −1 satisfies AY + Y AT − BB T + Y C T CY ≺ 0. Just
by considering the left-upper block, we infer for
Y1 Y12
Y =
Y21 Y2
that Y1 satisfies the strict algebraic Riccati inequality
A1 Y1 + Y1 AT1 − B1 B1T + Y1 C1T C1 Y1 ≺ 0.
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Proof: Step 3 - The Coup de Grâce
By last property on slide 66 we conclude Q1 Y1 . (This requires some
work; please check!) Due to
−1
Y1 Y12
X1 X12
=
Y21 Y2
X21 X2
and the block-inversion formula we have
−1
= X1 − X12 X2−1 X21 .
P1 = Q−1
1 ≺ Y1
With X2 0 we infer
P1 0
X1 − X12 X2−1 X21 0
≺
.
0 0
0
X2
I X12 X2−1
A congruence transformation with
finally leads to
0
I
P1 0
X1 X12
≺
0 0
X21 X2
which finishes the proof.
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Proof of Theorem 14
If γopt denotes the optimal value of the H2 -problem, the first step is
2
to show tr(BwT P Bw + B T P QP B) ≤ γopt
. Choose γ > γopt . We can
then find AK , BK , CK which render A Hurwitz and such that kC(sI −
A)−1 Bk22 < γ 2 . By simply expanding the controller realization with
stable uncontrollable or unobservable modes, we can assume that AK
has at least the same dimension as A. Moreover, by slide 6 there exists
some X 0 with AT X + X A + C T C ≺ 0 and tr(B T X B) < γ 2 . With
Z := B T X B + I and suitably small > 0 we then infer
X XB
T
T
A X + X A + C C ≺ 0,
0, tr(Z) < γ 2 . (1)
BT X Z
A key step of the proof is to block-factorize X in the partition of A.
More precisely, let (X U ) denote the upper block row of X . We can
assume that U has full row rank and that the right-lower block of X is
non-singular (after perturbation if necessary). This assures:
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Proof of Theorem 14
There exist X, U and Y , V such that
Y 0
I VT
X =
and S has full row rank.
X U
I 0
| {z }
| {z }
S
(2)
R
Let us first prepare some relations for the proof. In view of (1) we need
SX S T SX B
T
T
T
T
T
SA X S + SX AS + SC CS ,
BT X S T Z
which are equal to
T
T
T
T
T
SA R + RAS + SC CS ,
RS T RB
B T RT Z
.
(3)
First note that
T
RS =
Y 0
X U
I I
V 0
=
Y Y
Y X
where we exploited the fact that this matrix is symmetric.
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Proof of Theorem 14
Moreover, all other blocks in (3) can be explicitly written as
Y 0
Bw
Y Bw
RB =
=
,
X U
BK Dw
XBw + U BK Dw
I I
T
CS = Cz Dz CK
= Cz + Dz CK V Cz
V 0
and
T
RAS =
=
Y 0
A BCK
I I
=
X U
V 0
BK C AK
Y 0
A + BCK V
A
=
=
X U
BK C + AK V BK C
Y (A + BCK V )
YA
X(A + BCK V ) + U BK C + U AK V XA + U BK C
.
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Proof of Theorem 14
Therefore the matrices in (3) read as


Y
Y
Y Bw
T
R1 R21
X XBw + U BK Dw 
,  Y
R21 R2
T
Bw Y (∗)T
Z
(4)
with blocks
R1 = (A + BCK V )T Y +Y (A + BCK V )+(Cz + Dz CK V )T (Cz + Dz CK V ),
R2 = AT X + XA + U BK C + (U BK C)T + CzT Cz ,
R21 = AT Y + X(A + BCK V ) + U BK C + U AK V + CzT Cz .
With the equation for P and with ∆ = X − P let us note that R2 and
R21 can be written as
R2 = AT ∆ + ∆A + U BK C + (U BK C)T + P BB T P,
R21 = AT (Y − P ) + ∆A + (P + ∆)BCK V + U BK C + U AK V + P BB T P.
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Proof of Theorem 14
After these preparations we continue the proof. Since S has full row
rank, (1) clearly implies that R1 ≺ 0, R2 ≺ 0 and that the second
matrix in (4) is positive definite. By slide 34, R1 ≺ 0 implies P ≺ Y .
On the other hand, by taking the Schur-complement we infer
X
XBw + U BK Dw
I
−
Y I Bw 0.
T
T
(XBw + U BK Dw )
Z
Bw
Combining the last two inequalities implies (recalling X − P = ∆) that
∆
∆Bw + U BK Dw
0.
(∆Bw + U BK Dw )T
Z − BwT P Bw
With L = −∆−1 U BK we have ∆Bw + U BK Dw = ∆(Bw − LDw ) and
therefore Z − BwT P Bw − (Bw − LDw )T ∆(Bw − LDw ) 0. This shows
tr(Bw − LDw )T ∆(Bw − LDw ) < γ 2 − tr(BwT P Bw )
(∗)
due to (1). Moreover R2 ≺ 0 as at the bottom of slide 75 reveals
(A − LC)T ∆ + ∆(A − LC) + P BB T P ≺ 0.
(∗∗)
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Proof of Theorem 14
By slide 6 the inequalities (∗), (∗∗) show that A − LC is Hurwitz and
A − LC Bw − LDw 2
< γ 2 − tr(BwT P Bw ).
BT P
0
2
This allows to apply our result on H2 -estimation on slide 41. Due the
choice of Q we can hence conclude
tr(B T P QP B) + tr(BwT P Bw ) < γ 2 .
Since γ > γopt was arbitrary we finally get
2
.
tr(B T P QP B) + tr(BwT P Bw ) ≤ γopt
This concludes the first part of the proof.
Equality is shown by constructing an optimal controller. For this purpose
we follow the above steps as much as possible in the reverse order,
which includes a motivation for the structure of the to-be-constructed
controller.
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Proof of Theorem 14: Controller Construction
Due to our result on H2 -estimation on slide 41 (applied for Cz replaced
with B T P ) we know that A − QC T C is Hurwitz and leads to
A − QC T C Bw − QC T Dw 2
= tr(B T P QP B).
BT P
0
2
Therefore the solution ∆ of
(A − QC T C)T ∆ + ∆(A − QC T C) + P BB T P = 0
(5)
tr(Bw − QC T Dw )T ∆(Bw − QC T Dw ) = tr(B T P QP B).
(6)
satisfies
Next we try to achieve R1 = 0, R2 = 0 and R21 = 0 on slide 75. If
recalling the state-feedback problem, we clearly have R1 = 0 for Y = P ,
V = I and CK = −B T P . (We stress that we could take ANY nonsingular V at this point - the constructed controller would then just
have a different state-space realization!)
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Proof of Theorem 14: Controller Construction
Next enforce R2 = 0 and R21 = 0 if represented as at the bottom of
slide 75. Indeed, (5) implies R2 = 0 with U = −∆ and BK = QC T .
Moreover, with all the choices so far we have
R21 = ∆A − (P + ∆)BB T P − ∆QC T C − ∆AK + P BB T P =
∆(A − BB T P − QC T C − AK ) = 0
with AK = A − BB T P − QC T C. Further take X = ∆ + P such that
the two representations of R2 and R21 on slide 75 are indeed correct.
Let us finally define R and S as in (2). Note that S is non-singular and
P + ∆ −∆
X =
−∆
∆
is easily seen to be the unique solution of (2). This renders all equations
on slides 73-75 satisfied.
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Proof of Theorem 14: Controller Construction
Now we can conclude the proof. The constructed controller is exactly
the one on slides 51 and 52. Since A − BB T P and A − QC T C are
Hurwitz and due to the observer structure, it is certainly stabilizing.
Moreover, by construction we assured R1 = 0, R2 = 0, R21 = 0 which
implies SAT RT + RAS T + SC T CS T and thus AT X + X A + C T C = 0
by using (2) and exploiting that S is now square and non-singular.
With the formula for X on the previous slide we finally infer
tr(B T X B) = tr(BwT P Bw + BwT ∆Bw + DwT CQ∆QC T Dw ) =
= tr(BwT P Bw + (Bw − QC T Dw )T ∆(Bw − C T QDw )) =
= tr(BwT P Bw ) + tr(B T P QP B)
due to (6). This finishes the proof.
80/82
References
• [KK] H.W. Knobloch, H. Kwakernaak, Lineare Kontrolltheorie, Springer-Verlag
Berlin 1985
• [AM] K.J. Aström, R.M. Murray, Feedback Systems: An Introduction for Scientists
and Engineers, Princeton University Press, Princeton and Oxford, 2009 (available
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Dover Publications, 2005
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• R. Brockett, Finite dimensional linear systems, Wiley, 1970
• W.M. Wonham, Linear multivariable control, a geometric approach, SpringerVerlag, 3 1985
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References
• J. Zabczyk, Mathematical Control Theory: An Introduction, Birkhäuser, 2007
• B.A. Francis, A course in H∞ -control theory, Springer, 1987
• H.K. Khalil, Nonlinear Systems, Prentice Hall, 3 2002
• L. Arnold, Stochastic Differential Equations: Theory and Applications, Wiley, 1974
• A. Packard, State-space Youla Notes, Course in Multivariable Control Systems,
UC Berkeley, 2008
• J.M. Coron, Control and nonlinearity, Mathematical Surveys and Monographs,
2007
• Y. Yuan, G.-B. Stan, S. Warnick, and J.M. Goncalves, Minimal realization of the
dynamical structure function and its application to network reconstruction, IEEE
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