WIP, Description in progress
This part shows how to design dynamic outpur feedback control in mixed
H
2
{\displaystyle {\mathcal {H}}_{2}}
and
H
∞
{\displaystyle {\mathcal {H}}_{\infty }}
sense for the continuous time.
Consider the discrete-time generalized LTI plant
P
{\displaystyle {\mathcal {P}}}
with minimal state-space realization
x
˙
=
A
x
+
[
B
1
,
1
B
1
,
2
]
[
w
1
w
2
]
+
B
2
u
,
{\displaystyle {\dot {x}}=Ax+{\begin{bmatrix}B_{1,1}&B_{1,2}\end{bmatrix}}{\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix}}+B_{2}u,}
[
z
1
z
2
]
=
[
C
1
,
1
D
1
,
2
]
x
k
+
[
D
11
,
11
D
11
,
12
D
11
,
21
D
11
,
22
]
[
w
1
w
2
]
+
[
D
12
,
1
D
12
,
2
]
u
,
{\displaystyle {\begin{bmatrix}z_{1}\\z_{2}\end{bmatrix}}={\begin{bmatrix}C_{1,1}\\D_{1,2}\end{bmatrix}}x_{k}+{\begin{bmatrix}D_{11,11}&D_{11,12}\\D_{11,21}&D_{11,22}\end{bmatrix}}{\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix}}+{\begin{bmatrix}D_{12,1}\\D_{12,2}\end{bmatrix}}u,}
y
=
C
d
2
x
+
[
D
21
,
1
D
21
,
2
]
[
w
1
w
2
]
+
D
d
22
u
{\displaystyle y=C_{d2}x+{\begin{bmatrix}D_{21,1}&D_{21,2}\end{bmatrix}}{\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix}}+D_{d22}u}
A continuous-time dynamic output feedback LTI controllerwith state-space realization
(
A
c
,
B
c
,
C
c
,
D
c
)
{\displaystyle (A_{c},B_{c},C_{c},D_{c})}
is to be designed to minimize the
H
2
{\displaystyle {\mathcal {H}}_{2}}
norm of the closed-loop transfer matrix
T
11
(
s
)
{\displaystyle T_{11}(s)}
from the exogenous
input
w
1
{\displaystyle w_{1}}
to the performance output
z
1
{\displaystyle z_{1}}
while ensuring the H∞ norm of the closed-loop
transfer matrix
T
22
(
s
)
{\displaystyle T_{22}(s)}
from the exogenous input
w
2
{\displaystyle w_{2}}
to the performance output
z
2
{\displaystyle z_{2}}
is less than
γ
d
{\displaystyle \gamma _{d}}
,
where
T
11
(
s
)
=
C
C
L
1
,
1
(
s
I
−
A
C
L
)
−
1
B
C
L
1
,
1
,
{\displaystyle T_{11}(s)=C_{CL1,1}(sI-A_{CL})^{-1}B_{CL1,1},}
T
22
(
s
)
=
C
C
L
1
,
2
(
s
I
−
A
C
L
)
−
1
B
C
L
1
,
2
+
D
C
L
11
,
22
,
{\displaystyle T_{22}(s)=C_{CL1,2}(sI-A_{CL})^{-1}B_{CL1,2}+D_{CL11,22},}
A
d
C
L
=
[
A
+
B
2
D
c
D
~
−
1
C
2
B
2
(
I
+
D
c
D
~
−
1
D
22
)
C
c
B
c
D
~
−
1
C
2
A
c
+
B
c
D
~
−
1
D
22
C
c
]
{\displaystyle A_{d_{CL}}={\begin{bmatrix}A+B_{2}D_{c}{\tilde {D}}^{-1}C_{2}&B_{2}(I+D_{c}{\tilde {D}}^{-1}D_{22})C_{c}\\B_{c}{\tilde {D}}^{-1}C_{2}&A_{c}+B_{c}{\tilde {D}}^{-1}D_{22}C_{c}\end{bmatrix}}}
,
B
C
L
1
,
1
=
[
B
1
,
1
+
B
2
D
c
D
~
−
1
D
21
,
1
B
c
D
~
−
1
D
21
,
1
]
{\displaystyle B_{CL1,1}={\begin{bmatrix}B_{1,1}+B_{2}D_{c}{\tilde {D}}^{-1}D_{21,1}\\B_{c}{\tilde {D}}^{-1}D_{21,1}\end{bmatrix}}}
,
B
C
L
1
,
2
=
[
B
1
,
2
+
B
2
D
c
D
~
−
1
D
21
,
2
B
c
D
~
−
1
D
21
,
2
]
{\displaystyle B_{CL1,2}={\begin{bmatrix}B_{1,2}+B_{2}D_{c}{\tilde {D}}^{-1}D_{21,2}\\B_{c}{\tilde {D}}^{-1}D_{21,2}\end{bmatrix}}}
,
C
C
L
1
,
1
=
[
C
1
,
1
+
D
12
,
1
D
c
D
~
−
1
C
2
,
1
D
12
,
1
(
I
+
D
c
D
~
−
1
D
22
)
C
c
]
{\displaystyle C_{CL1,1}={\begin{bmatrix}C_{1,1}+D_{12,1}D_{c}{\tilde {D}}^{-1}C_{2,1}&D_{12,1}(I+D_{c}{\tilde {D}}^{-1}D_{22})C_{c}\end{bmatrix}}}
,
C
C
L
1
,
2
=
[
C
1
,
2
+
D
12
,
2
D
c
D
~
−
1
C
2
,
2
D
12
,
2
(
I
+
D
c
D
~
−
1
D
22
)
C
c
]
{\displaystyle C_{CL1,2}={\begin{bmatrix}C_{1,2}+D_{12,2}D_{c}{\tilde {D}}^{-1}C_{2,2}&D_{12,2}(I+D_{c}{\tilde {D}}^{-1}D_{22})C_{c}\end{bmatrix}}}
,
D
C
L
11
,
22
=
D
11
,
22
+
D
12
,
2
D
c
D
~
−
1
D
21
,
2
{\displaystyle D_{CL11,22}=D_{11,22}+D_{12,2}D_{c}{\tilde {D}}^{-1}D_{21,2}}
,
and
D
~
=
I
−
D
22
D
c
{\displaystyle {\tilde {D}}=I-D_{22}D_{c}}
.
Solve for
A
n
∈
R
n
x
×
n
x
,
B
n
∈
R
n
x
×
n
x
,
C
n
∈
R
n
u
×
n
x
,
D
n
∈
R
n
u
×
n
y
,
X
1
,
Y
1
∈
S
n
x
,
Z
∈
S
n
Z
1
,
{\displaystyle A_{n}\in \mathbb {R} ^{n_{x}\times n_{x}},B_{n}\in \mathbb {R} ^{n_{x}\times n_{x}},C_{n}\in \mathbb {R} ^{n_{u}\times n_{x}},D_{n}\in \mathbb {R} ^{n_{u}\times n_{y}},X_{1},Y_{1}\in \mathbb {S} ^{n_{x}},Z\in \mathbb {S} ^{n_{Z_{1}}},}
and
μ
∈
R
>
0
{\displaystyle \mu \in \mathbb {R} _{>0}}
that minimizes
J
(
μ
)
=
μ
{\displaystyle {\mathcal {J}}(\mu )=\mu }
subjects to
X
1
>
0
,
Y
1
>
0
Z
>
0
,
{\displaystyle X_{1}>0,\ Y_{1}>0\ Z>0,}
[
N
11
A
+
A
n
T
+
B
2
D
n
C
2
B
1
,
1
+
B
2
D
n
D
21
,
1
∗
X
1
A
+
A
T
X
1
+
B
n
C
2
+
C
2
T
B
n
T
X
1
B
1
,
1
+
B
n
D
21
,
1
∗
∗
−
I
]
<
0
{\displaystyle {\begin{bmatrix}N_{11}&A+A_{n}^{T}+B_{2}D_{n}C_{2}&B_{1,1}+B_{2}D_{n}D{21,1}\\*&X_{1}A+A^{T}X_{1}+B_{n}C_{2}+C_{2}^{T}B_{n}^{T}&X_{1}B_{1,1}+B_{n}D_{21,1}\\*&*&-I\end{bmatrix}}<0}
,
[
N
11
A
+
A
n
T
+
B
2
D
n
C
2
B
1
,
1
+
B
2
D
n
D
21
,
1
Y
1
T
C
1
,
2
T
+
C
n
T
D
12
,
2
T
∗
X
1
A
+
A
T
X
1
+
B
n
C
2
+
C
2
T
B
n
T
X
1
B
1
,
1
+
B
n
D
21
,
1
C
1
,
2
T
+
C
2
T
D
n
T
D
12
,
2
T
∗
∗
−
γ
d
I
D
11
,
22
T
+
D
21
,
2
T
D
n
T
D
12
,
2
T
∗
∗
∗
−
γ
d
I
]
<
0
{\displaystyle {\begin{bmatrix}N_{11}&A+A_{n}^{T}+B_{2}D_{n}C_{2}&B_{1,1}+B_{2}D_{n}D{21,1}&Y_{1}^{T}C_{1,2}^{T}+C_{n}^{T}D_{12,2}^{T}\\*&X_{1}A+A^{T}X_{1}+B_{n}C_{2}+C_{2}^{T}B_{n}^{T}&X_{1}B_{1,1}+B_{n}D_{21,1}&C_{1,2}^{T}+C_{2}^{T}D_{n}^{T}D_{12,2}^{T}\\*&*&-\gamma _{d}I&D_{11,22}^{T}+D_{21,2}^{T}D_{n}^{T}D_{12,2}^{T}\\*&*&*&-\gamma _{d}I\end{bmatrix}}<0}
,
[
Y
1
I
Y
1
C
1
,
1
T
+
C
n
T
D
12
,
1
T
∗
X
1
C
1
,
1
T
+
C
2
T
D
n
T
D
12
,
1
T
∗
∗
Z
]
>
0
{\displaystyle {\begin{bmatrix}Y_{1}IY_{1}C_{1,1}^{T}+C_{n}^{T}D_{12,1}^{T}\\*&X_{1}&C_{1,1}^{T}+C_{2}^{T}D_{n}^{T}D_{12,1}^{T}\\*&*&Z\end{bmatrix}}>0}
,
D
11
,
11
+
D
12
,
1
D
n
D
21
,
1
=
0
,
{\displaystyle D_{11,11}+D_{12,1}D_{n}D_{21,1}=0,}
[
X
1
I
∗
Y
1
]
>
0
,
{\displaystyle {\begin{bmatrix}X_{1}&I\\*&Y_{1}\end{bmatrix}}>0,}
tr
Z
<
μ
,
{\displaystyle Z<\mu ,}
where
N
11
=
A
Y
1
+
Y
1
A
T
+
B
2
C
n
+
C
n
T
B
2
T
{\displaystyle N_{11}=AY_{1}+Y_{1}A^{T}+B_{2}C_{n}+C_{n}^{T}B_{2}^{T}}
.
The controller is recovered by
A
c
=
A
K
−
B
c
(
I
−
D
22
D
c
)
−
1
D
22
C
c
,
{\displaystyle A_{c}=A_{K}-B{c}(I-D_{22}D_{c})^{-1}D_{22}C_{c},}
B
c
=
B
K
(
I
−
D
c
D
22
)
,
{\displaystyle B_{c}=B_{K}(I-D_{c}D_{22}),}
C
c
=
(
I
−
D
c
D
22
)
C
K
,
{\displaystyle C_{c}=(I-D_{c}D_{22})C_{K},}
D
c
=
(
I
+
D
K
D
22
)
−
1
D
K
,
w
h
e
r
e
<
m
a
t
h
>
[
A
K
B
K
C
K
D
K
]
=
[
X
2
X
1
B
2
0
I
]
−
1
(
[
A
n
B
n
C
n
D
n
]
−
[
X
1
A
d
Y
1
0
0
0
]
)
[
Y
2
T
0
C
2
Y
1
I
]
−
1
{\displaystyle D_{c}=(I+D_{K}D{22})^{-1}D_{K},where<math>{\begin{bmatrix}A_{K}&B_{K}\\C_{K}&D_{K}\end{bmatrix}}={\begin{bmatrix}X_{2}&X_{1}B_{2}\\0&I\end{bmatrix}}^{-1}({\begin{bmatrix}A_{n}&B_{n}\\C_{n}&D_{n}\end{bmatrix}}-{\begin{bmatrix}X_{1}A_{d}Y_{1}&0\\0&0\end{bmatrix}}){\begin{bmatrix}Y_{2}^{T}&0\\C_{2}Y_{1}&I\end{bmatrix}}^{-1}}
, and the matrices
X
2
{\displaystyle X_{2}}
and
Y
2
{\displaystyle Y_{2}}
satisfy
X
2
Y
2
T
=
I
−
X
1
Y
1
{\displaystyle X_{2}Y_{2}^{T}=I-X_{1}Y_{1}}
. If
D
22
=
0
{\displaystyle D_{22}=0}
, then
A
c
=
A
K
,
B
c
=
B
K
,
C
c
=
C
K
{\displaystyle A_{c}=A_{K},B_{c}=B{K},C_{c}=C_{K}}
and
D
c
=
D
K
{\displaystyle D_{c}=D_{K}}
.
Given
X
1
{\displaystyle X_{1}}
and
Y
1
{\displaystyle Y_{1}}
, the matrices
X
2
{\displaystyle X_{2}}
and
Y
2
{\displaystyle Y_{2}}
can be found using a matrix decomposition, such as
a LU decomposition or a Cholesky decomposition.
If
D
11
,
11
=
0
,
D
12
,
1
≠
0
,
and
D
21
,
1
≠
0
,
{\displaystyle D_{11,11}=0,D_{12,1}\neq 0,{\text{ and }}D_{21,1}\neq 0,}
then it is often simplest to choose
D
n
=
0
{\displaystyle D_{n}=0}
in order
to satisfy the equality constraint
D
11
,
11
+
D
12
,
1
D
n
D
21
,
1
=
0
,
{\displaystyle D_{11,11}+D_{12,1}D_{n}D_{21,1}=0,}
.
WIP, additional references to be added
A list of references documenting and validating the LMI.
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