978-0849390081 Part 6

subject Type Homework Help
subject Pages 9
subject Words 1318
subject Authors Dan Popa, Frank L. Lewis, Lihua Xie

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Answers to exercises of chapters 6-8
The authors would like to express their appreciation to Jun Xu, Shuai Liu, Nan Xiao and Keyou You for their help in
preparing the answers to the exercises in chapter 6-8.
ANSWERS TO EXERCISES OF CHAPTER 6
1. Since T
kkI,wehave
ε1
kIk¯
F˜
Σk¯
FTT
kk(ε1
kI¯
F˜
Σk¯
FT)T
k0(1)
Applying Schur complements, the following inequality holds
!¯
F˜
Σk¯
AT
k
¯
DT
k"!(ε1
k¯
F˜
Σk¯
FT)1T
k
kε1
kIk¯
F˜
Σk¯
FTT
k"!¯
F˜
Σk¯
AT
k
¯
DT
k"0(2)
Then we can write
APkAT+APkET#ε1
kIEPkET$1EPkAT
=!I
1"T!AYkATAYkHT+ε1
kD1DT
2
HYkAT+ε1
kD2DT
1Rεk+HYkHT"!I
1"
2=#Rεk+HYkHT$1FP
kAT
3=(ε1
kFP
kFT)1FP
kHT
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3. (i). Let the transformation matrix T=!0I
II"where Iis n×nidentity matrix. Then we consider the solution ˜
Σkin (6.17) in
the partitioned form ˜
Σk=!˜
Σ11,k ˜
Σ12,k
˜
Σ21,k ˜
Σ22,k ", where all blocks are n×nmatrices, and we detine ¯
Xk=˜
Σ11,k+˜
Σ12,k +˜
Σ21,k+˜
Σ22,k.
Multiplying both sides of RDE (6.17) from the left and the right by T, respectively, we have
Pk+1 =
AT
DT
1
Q1
2GT
T
Xk00
0ε1
kI0
00I
AT
DT
1
Q1
2GT
+AXkFT#ε1
kIFX
kFT$1FX
kAT(5)
Considering the assumption6.1.2, we know that Xk+1 >0if Xk>0. Since X0=Px0>0, then Xk>0(k>0).
˜zk=ε
1
2
kF˜xk
the following worst-case performance measure is satistied:
0"=(˜x0,˜w)Rn×l2[0,N]
˜xT
0Px0˜x0+'˜wk'2
2
Next, we consider the following linear system
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3
Obviously, (X1
kεkFTF)1(P1
kεkFTF)10when XkPk, which implies Xk+1 Pk+1 >0. Besides,
X0=P0=Px0>0, thus by induction, XkPk>0over [0,N].
4. For scalar system, minimizing trace(Pk)degenerates as minimizing Pk. By considering the monotonicity of the RDE
P
k{min
εk1
(Pk),P
k1{min
εk2
(Pk1),··· ,P
1{min
ε0
(P1),given P0}···}}
Then, (6.20) can be rewritten as
0 10 20 30 40 50 60 70 80 90 100
4
3
2
1
0
1
2
3
x
xr
xk

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