1
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
k∆k≤I,wehave
ε−1
kI−∆k¯
F˜
Σk¯
FT∆T
k−∆k(ε−1
kI−¯
F˜
Σk¯
FT)∆T
k≥0(1)
Applying Schur complements, the following inequality holds
!¯
F˜
Σk¯
AT
k
¯
DT
k“!(ε−1
k−¯
F˜
Σk¯
FT)−1−∆T
k
−∆kε−1
kI−∆k¯
F˜
Σk¯
FT∆T
k“!¯
F˜
Σk¯
AT
k
¯
DT
k“≥0(2)
Then we can write
APkAT+APkET#ε−1
kI−EPkET$−1EPkAT
=!I
Ω1“T!AYkATAYkHT+ε−1
kD1DT
2
HYkAT+ε−1
kD2DT
1Rεk+HYkHT“!I
Ω1“
Ω2=#Rεk+HYkHT$−1FP
kAT
Ω3=−(ε−1
k−FP
kFT)−1FP
kHT
2
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
0ε−1
kI0
00I
AT
DT
1
Q1
2GT
+AXkFT#ε−1
kI−FX
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
3
Obviously, (X−1
k−εkFTF)−1−(P−1
k−εkFTF)−1≥0when Xk≥Pk, which implies Xk+1 ≥Pk+1 >0. Besides,
X0=P0=Px0>0, thus by induction, Xk≥Pk>0over [0,N].
4. For scalar system, minimizing trace(Pk)degenerates as minimizing Pk. By considering the monotonicity of the RDE
P∗
k←{min
εk−1
(Pk),P∗
k−1←{min
εk−2
(Pk−1),··· ,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