7
which is used to rewritten (33) as
∗T
Q1 + Q3 Θ∗T
q1 ẋp = Q2 + Q3 Θq2 xp + Q3 up .
S UPPLEMENTARY M ATERIALS TO “A DAPTIVE
O UTPUT-F EEDBACK C ONTROL FOR A C LASS OF
M ULTI -I NPUT-M ULTI -O UTPUT P LANTS WITH
A PPLICATIONS TO V ERY F LEXIBLE A IRCRAFT ”1
Zheng Qu2 , Anuradha M. Annaswamy2 and Eugene Lavretsky3
are
Assume that Q1 , (Q1 + Q3 Θ∗T
q1 ) and (I +
invertible around the equilibrium. Taking inverse on both sides,
and noting
∗T −1
(Q1 + Q3 Θq )
1
A PPENDIX
−1
−1
∗T −1
−1 ∗T −1
Q1
− Q1 Q3 (I + Θq Q1 Q3 )
Θq Q1
1
1
|
{z
}
Θ∗T
q1
=
A. Proof of Lemma 1
|[·]0 as partial difProof: Define (·)/[·]0 = ∂(·)
∂[·]
ferential variables. Linearizing (3) around a trim point
[¨
0 , ˙0 , 0 , β̇0 , β0 , λ0 , u0 ]T yields

I
0
0

0
0
(MF F )
0
(MBF )
0
{z
(MF B )
0
(MBB )
0

0
0
0
+
}
0
0
∆MF F
∆MF B
∆MBF
∆MBB
{z
∆Q∗
0 ,˙ 0 ,β̇0 )
1 (¨
|
Q1 (0,0,0 ,0,β0 )








}

˙
¨
β̇
−1
−1
∗T −1
∗T
Q1
− Q1 Q3 Θq Q1
Q2 + Q3 Θq
xp
1
2
−1
−1
∗T −1
− Q1 Q3 Θq Q1
Q3 u p
+ Q1
1
h
i
−1
−1
∗T
∗T −1
∗T −1
∗T
= Q1 Q2 + Q1 Q3 Θq
− Θq Q1 Q2 − Θq Q1 Q3 Θq
xp
2
1
1
2
−1
∗T −1
+ Q1 Q3 I − Θq Q1 Q3 up
1


ẋp =









∗T
∗T
∗T 
∗T
∗T
= Ap + Bp Θq
− Θq Ap − Θq Bp Θq
 x + Bp I − Θq Bp up

1
1
2  p
1
2


{z
}
{z
}
|
|

Λ∗
Θ∗T
p
p







= 




|
0
T ) F load
−(KF F ) + (Jh
0 /
0
0
0
I
−Ce
0
0
T ) F load
(Jh
0 /β
0
T ) F load
−CRB + (Jhb
0 /β
0
{z
(38)
yields










|
(37)
−1
Θ∗T
q2 Q1 Q3 )
(39)
(40)
(41)

−1



−1
with Ap = Q1 Q2 , Bp = Q1 Q3 . Cp as in yp = Cp xp is the
selection matrix that picks out measurable states from xp .
}
Q2 (0,0,0 ,0,β0 )


+
0
∆KF F
∆KBF
0
0
∆CF F
∆CF B
∆CBF
∆CBB
{z
∆Q∗
0 ,˙ 0 ,β̇0 )
2 (¨
|









˙  + 

β

 | {z }
}
|
xp
0
BF /u
0
BB/u
0
{z
Q3
B. Proof of Lemma 3
Proof: The proof will be performed in a transformed
C2T C1T .
coordinate. Similar to B 2 , we part C T =
For a square plant model that has nonuniform input relative degree
two, there
exists an invertible transformation
(CB)−1 C
−1
B M , where CT =
Tin =
, Tin
=
N
T
C2 AT C2T C1 , B = B2 AB2 Bs1 , N and M
are chosen to satisfy NB = 0, CM = 0 and NM = I, that
transforms (10) into a new coordinate called “input normal
form” (See [21, Corollary 2.2.5] for proof). In this proof,
matrices in input normal form coordinate will be denoted
−1
with the subscript (·)in , as in xin = Tin x, Ain = Tin ATin
,
B 2,in = Tin B 2 (and therefore B2,in = Tin B2 and Bs1,in =
−1
Tin Bs1 ), B1,in = Tin B1 , Bz,in = Tin Bz , Cin = CTin
,
∗T
∗T −1
∗T
∗T −1
Ψ1,in = Ψ1 Tin and Ψ2,in = Ψ2 Tin . The input normal
form of the plant model (10) is


up
(33)
}
where, following the definition of terms in (4), the unknown
deviation terms are
T ) F load
∆MF B = −(Jh
0
/β̇0
T ) F load
∆MBB = −(Jhb
0
/β̇0
T ) F load
∆BB/λ = (Jhb
0 /λ
0
0
+ MF B/ β̇0
0
+ MBB/ β̇0
0
T ) F load
∆CF F = −(CF F )x − CF F /˙ ˙ 0 − CF B/˙ β0 + (Jh
0 /˙
0
0
0
0
T ) F load
∆CBF = −(CBF )x − CBF /˙ ˙ 0 − CBB/˙ β0 + (Jhb
0 /˙
0
0
0
0
∆CF B = −(CF B )x0 − CF F /β ˙ 0 − CF B/β β0
0
0
∆CBB = −(CBB )x0 − CBF /β ˙ 0 − CBB/β β0
0
0
T ) F load
∆MF F = −(Jh
0 /¨
0
T ) F load
∆MBF = −(Jhb
0 /¨
0
T ) F load
∆BF /λ = (Jh
0 /λ
0
0
∆KF F = MF F / ¨
0 0
∆KBF = MBF / ¨
0 0
(34)
T
load
T
load
BF/u0 = (Jh
)0 F/u
and BB/u0 = (Jhb
)0 F/u
. Without
0
0
load
loss of generality, we scale each input so that F/u
= I.
0
T
In realistic application, only [0 , β0 , u0 ] can be measured
accurately and therefore variables that depend on them can be
gain scheduled. [˙0 , ¨0 , β̇0 , λ0 ]T cannot be measured accurately
and therefore variables that depends on them are generally
unknown. As a result, Q1 , Q2 and Q3 are known but ∆Q∗1
and ∆Q∗2 are unknown. Examination on (34) using (4) shows
that


0
T
Jh
T
Jhb
0


∗
∆Q1 = 





0
|
0
F load
/¨
0
F load
/β̇0
{z
Θ∗T
q1
0
∗T
= Q3 Θq
1





2
ξ̇1
2
ξ̇2
1
ξ̇1
η̇
0
2
R2,1
2
R2,2
2
R1,1
0
U2

0




 = 



I

|
+ B2,in
(35)

}


+
|


∗
∆Q2 = 
0
T
Jh
T
Jhb
Me



0
|
∗T
= Q3 Θq
2
∂Jh
∂Jhb
∂ J˙h
∂ J˙hb
¨+
β
Me J˙h +
˙ +
β
∂
∂
∂ ˙
∂ ˙
!#
˙
˙
∂
J
∂
J
h ˙ +
hb β
Me J˙hb +
∂β
∂β
x0
{z
Θ∗T
q2
!
(36)
1 Accepted by 2016 American Control Conference, Boston, MA. This work
is supported by Boeing Strategic University Initiative.
2 Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, MA 02139, USA. IEEE fellow.
3 The Boeing Company, Huntington Beach, CA, 92647, USA. IEEE fellow.
U1
{z
Ain
2∗T
ψ20
y =
|
}
0

Im
1I
h
c m 
0

0
|
0
{z
}
B1,in
CAB2
{z
Cin
2
ξ1
2
ξ2
1
ξ1
η
{z
}|
xin





V1
Z
2∗T
ψ21
2∗T
ψ11
CBs1

V2
0
|
and
"
h
1
R2,1
1
R2,2
1
R1,1
1∗T
ψ21
{z
Ψ∗T
2,in
1∗
ψ11
{z
Ψ∗T
1,in
0


Im
 ∗
0
Λ u

0
0
{z
}
|
B2,in




+


}
ψ 2∗T
(n−rs )
ψ 1∗T
(n−rs )
i
xin
}
i
xin + Bz,in zcmd
(42)
}
xin .
}
P
Matrix Z ∈ R(n−rs )×(n−rs ) , where rs = i ri , is the zero
dynamics matrix whose eigenvalues are transmission zeros of
the plant model (see [21, Section 2.3]). It is noted that B1,in =
T
−1
0 × × × since
× × 0 0
and Ψ∗T
1 Tin =
Assumption 4 holds.
∗T
∗ −1
Define A∗in = Ain +B1,in Ψ∗T
1,in +B2,in Ψ2,in = Tin A Tin .
1
1∗
Examination of the elements of B 2,in and B 2,in , which are
8
1∗
1∗
1
1
defined as B 2,in = Tin B 2 , and B 2,in = Tin B 2 , respectively,
shows that

 0
a1 Im
0
1
 a11 Im
0 
1
,
Bs1,in = 
(43)
B 2,in = B2,in
 0
Ims 
0
0
By appealing to (27)(28)(29)(49)(51), the derivative of V has
the following bound
∗
∗ ∗
V̇ = eTmx A∗T
L∗ P + P AL∗ emx
e TΛ χ
− 2eTmx [P ∗ B21∗ − C T S2T ]Λ∗ Ψ
e Tm esy
− 2eTmx [P ∗ B21∗ − C T S2T ]Ψ
= −eTmx Q∗ emx ≤ 0.
and
2∗T
a01 Im + a11 ψ20
1

a1 Im
=

0
0

1∗
B 2,in
=
1∗
B2,in
Bs1,in
0
0
Ims
0


.

(44)
1
1
1∗
where B2,in
= Tin B2,in
, Bs1,in = Tin Bs1 and B2,in
=
1
1∗
1∗
T
B
.
It
is
noted
that
C
B
=
C
B
=
in 2,in
in 2,in
in 1 2
a1 CAB2 CBs1
has full rank by Assumption 3 and
1∗
1
Lemma 2. Examination on elements of B 2,in and B 2,in shows
that
1∗
B 2,in
1
= B 2,in + B2,in a11 Ψ∗T
in,m .
(45)
where
Ψ∗T
in,m =
2∗T
ψ20
0m×ms
∈ Rm×p
(46)
2∗T
where ψ20
is a subset of the elements in Ψ∗2,in as shown in
(42). It is noted that (44) also holds for (A∗in − Lin Cin ) for
∀Lin ∈ Rn×m . Transformation back to the original coordinate
proves the Lemma.
C. Proof of Lemma 4
∗
, B21∗ , C}
Proof: It has been proved that the Z{A
1∗
is exactly the eigenvalues of N 2,in A∗in Min with
T
1
0
Im 0 0
=
Min
=
and N 2,in
0
0
0
I
n−rs
h
i
1
1
T
T
(Min
Min )−1 Min
I − B 2,in (C in B 2,in )−1 C in (see [18]).
Some algebra shows that
"
#
a0
1∗
− a11
×
∗
(47)
N 2,in Ain Min =
1
0
Z
where Z is the zero dynamics matrix as in 42 whose eigenvalues are Z{A, B2 , C} (see [21, Section 2.3]).
D. Proof of Theorem 1
Proof: We propose a Lyapunov function candidate
V = eTmx P ∗ emx
h
i
h
i
e TΛ Γ−1 Λ∗ Ψ
eΛ + Tr Ψ
e Tm Γ−1 Ψ
em
+ Tr Ψ
ψλ
ψm
(48)
where P ∗ = P ∗T > 0 is the matrix that guarantees the SPR
1∗
properties of {A∗L∗ , B 2 , SC}, satisfying
P ∗ A∗L∗ + A∗L∗ P ∗ = −Q∗ < 0
1∗
P ∗B2
∗
∗T
for a Q = Q
= CT ST ,
(49)
(50)
> 0. Partition on both sides of (50) yields
P ∗ B21∗ B1 = C T S2T S1T .
(51)
(52)
e Λ (t) and Ψ
e m (t) are bounded as t → ∞, which
Then emx (t), Ψ
proves i). Applying Barbalat’s Lemma (using the fact that
ėmx (t) is bounded) shows that emx (t) → 0 as t → ∞, which
proves ii). From (28) and (16), the fact emx (t) → 0 implies
that ey (t) → 0, esy (t) → 0 and esy (t) → 0 as t → ∞, which
in turn implies that xm , as well as xm and ubl , is bounded.
Further, denote
epz (t) = z − zcmd , emz (t) = zm − zcmd .
(53)
´
From (9), it is noted ´that epz (t)dt is an element of x. From
(14), it is noted that (emz (t) − LI ey )dt is an element of xm
where LI are the rows of L corresponding to emz dynamics.
As a result, emx (t) → 0 as t → ∞, which, together with the
definition of emx as
h
i
∗T
e Tm (t)esy
emx = ex + B2 Λ∗ a11 Ψ1 xm + B2 a11 Ψ
e TΛ χ,
− B2 Λ∗ a11 Ψ
´
implies ex (t) → (B2 [·])dt and therefore
ˆ
ˆ
(epz − emz + LI ey )dt → (B2,I [·])dt = 0
(54)
(55)
as t → ∞ (since B2,I , the rows of B2 corresponding to wz
dynamics, is zero). Eq.(55), together with the fact that
ez (t) = z − zm = epz − emz
implies
ˆ
(56)
ˆ
ez (t)dt → −
(LI ey )dt
(57)
which has a bounded limit as t → ∞ (since ey (t) → 0).
Further, from (9) and (14), ėz (t) is bounded as t → ∞.
Applying Barbalat’s Lemma shows that ez (t) → 0 as t → ∞,
which proves iii).