solution of linear system theory and design 3ed for chi tsong chen
DESCRIPTION
Solution of Linear System Theory and DesignTRANSCRIPT
Linear System Theory and Design SA01010048 LING QING
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2.1 Consider the memoryless system with characteristics shown in Fig 2.19, in which u denotes the input and y the output. Which of them is a linear system? Is it possible to introduce a new output so that the system in Fig 2.19(b) is linear?
Figure 2.19 Translation: 考虑具有图 2.19 中表示的特性的无记忆系统。其中 u 表示输入，y 表示输出。 下面哪一个是线性系统？可以找到一个新的输出，使得图 2.19(b)中的系统是线性
的吗？ Answer: The input-output relation in Fig 2.1(a) can be described as:
uay *=
Here a is a constant. It is a memoryless system. Easy to testify that it is a linear system. The input-output relation in Fig 2.1(b) can be described as:
buay += *
Here a and b are all constants. Testify whether it has the property of additivity. Let:
buay += 11 *
buay += 22 *
then:
buuayy *2)(*)( 2121 ++=+
So it does not has the property of additivity, therefore, is not a linear system. But we can introduce a new output so that it is linear. Let:
byz −=
uaz *= z is the new output introduced. Easy to testify that it is a linear system. The input-output relation in Fig 2.1(c) can be described as:
uuay *)(=
a(u) is a function of input u. Choose two different input, get the outputs:
111 *uay =
Linear System Theory and Design SA01010048 LING QING
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222 *uay =
Assure:
21 aa ≠
then:
221121 **)( uauayy +=+
So it does not has the property of additivity, therefore, is not a linear system. 2.2 The impulse response of an ideal lowpass filter is given by
)(2
)(2sin2)(
0
0
tttt
tg−−
=ωω
ω
for all t, where w and to are constants. Is the ideal lowpass filter causal? Is is possible to built the filter in the real world? Translation: 理想低通滤波器的冲激响应如式所示。对于所有的 t，w 和 to，都是常数。理
想低通滤波器是因果的吗？现实世界中有可能构造这种滤波器吗？ Answer: Consider two different time: ts and tr, ts < tr, the value of g(ts-tr) denotes the output at
time ts, excited by the impulse input at time tr. It indicates that the system output at time ts is dependent on future input at time tr. In other words, the system is not causal. We know that all physical system should be causal, so it is impossible to built the filter in the real world.
2.3 Consider a system whose input u and output y are related by
>≤
==atfor
atfortutuPty a 0
)(:))(()(
where a is a fixed constant. The system is called a truncation operator, which chops off the input after time a. Is the system linear? Is it time-invariant? Is it causal? Translation: 考虑具有如式所示输入输出关系的系统，a 是一个确定的常数。这个系统称作 截断器。它截断时间 a 之后的输入。这个系统是线性的吗？它是定常的吗？是因果 的吗？ Answer: Consider the input-output relation at any time t, t<=a: uy =
Easy to testify that it is linear. Consider the input-output relation at any time t, t>a:
0=y
Easy to testify that it is linear. So for any time, the system is linear. Consider whether it is time-invariable. Define the initial time of input to, system input is u(t), t>=to. Let to<a, so It decides the system output y(t), t>=to:
Linear System Theory and Design SA01010048 LING QING
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≤≤
=totherforattfortu
ty0
)()( 0
Shift the initial time to to+T. Let to+T>a , then input is u(t-T), t>=to+T. System output:
0)(' =ty
Suppose that u(t) is not equal to 0, y’(t) is not equal to y(t-T). According to the definition, this system is not time-invariant.
For any time t, system output y(t) is decided by current input u(t) exclusively. So it is a causal system.
2.4 The input and output of an initially relaxed system can be denoted by y=Hu, where H is some mathematical operator. Show that if the system is causal, then
uHPPHuPyP aaaa ==
where Pa is the truncation operator defined in Problem 2.3. Is it true PaHu=HPau? Translation: 一个初始松弛系统的输入输出可以描述为：y=Hu，这里 H 是某种数学运算，
说明假如系统是因果性的，有如式所示的关系。这里 Pa 是题 2.3 中定义的截断函
数。PaHu=HPau 是正确的吗？ Answer: Notice y=Hu, so:
HuPyP aa =
Define the initial time 0, since the system is causal, output y begins in time 0. If a<=0,then u=Hu. Add operation PaH in both left and right of the equation:
uHPPHuP aaa =
If a>0, we can divide u to 2 parts:
≤≤
=totherforatfortu
tp0
0)()(
>
=totherfor
atfortutq
0)(
)(
u(t)=p(t)+q(t). Pay attention that the system is casual, so the output excited by q(t) can’t affect that of p(t). It is to say, system output from 0 to a is decided only by p(t). Since PaHu chops off Hu after time a, easy to conclude PaHu=PaHp(t). Notice that p(t)=Pau, also we have:
uHPPHuP aaa =
It means under any condition, the following equation is correct:
uHPPHuPyP aaaa ==
PaHu=HPau is false. Consider a delay operator H, Hu(t)=u(t-2), and a=1, u(t) is a step input begins at time 0, then PaHu covers from 1 to 2, but HPau covers from 1 to 3.
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2.5 Consider a system with input u and output y. Three experiments are performed on the system using the inputs u1(t), u2(t) and u3(t) for t>=0. In each case, the initial state x(0) at time t=0 is the same. The corresponding outputs are denoted by y1,y2 and y3. Which of the following statements are correct if x(0)<>0?
1. If u3=u1+u2, then y3=y1+y2. 2. If u3=0.5(u1+u2), then y3=0.5(y1+y2). 3. If u3=u1-u2, then y3=y1-y2.
Translation: 考虑具有输入 u 输出 y 的系统。在此系统上进行三次实验，输入分别为 u1(t), u2(t) 和 u3(t)，t>=0。每种情况下，零时刻的初态 x(0)都是相同的。相应的输出表
示为 y1,y2 和 y3。在 x(0)不等于零的情况下，下面哪种说法是正确的？ Answer: A linear system has the superposition property:
0221102211
022011 ),()(),()(
)()(tttyty
tttututxtx
≥+→
≥++
αααα
αα
In case 1:
11 =α 12 =α
)0()0(2)()( 022011 xxtxtx ≠=+αα
So y3<>y1+y2. In case 2:
5.01 =α 5.02 =α
)0()()( 022011 xtxtx =+αα
So y3=0.5(y1+y2). In case 3:
11 =α 12 −=α
)0(0)()( 022011 xtxtx ≠=+αα
So y3<>y1-y2. 2.6 Consider a system whose input and output are related by
=−≠−−
=0)1(0
0)1()1(/)()(
2
tuiftuiftutu
ty
for all t. Show that the system satisfies the homogeneity property but not the additivity property. Translation: 考虑输入输出关系如式的系统,证明系统满足齐次性,但是不满足可加性. Answer: Suppose the system is initially relaxed, system input:
)()( tutp α=
a is any real constant. Then system output q(t):
Linear System Theory and Design SA01010048 LING QING
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=−≠−−
=0)1(0
0)1()1(/)()(
2
tpiftpiftptp
tq
=−≠−−
=0)1(0
0)1()1(/)(2
tuiftuiftutuα
So it satisfies the homogeneity property. If the system satisfies the additivity property, consider system input m(t) and n(t),
m(0)=1, m(1)=2; n(0)=-1, n(1)=3. Then system outputs at time 1 are:
4)0(/)1()1( 2 == mmr
9)0(/)1()1( 2 −== nns
0)]0()0(/[)]1()1([)1( 2 =++= nmnmy
）1()1( sr +≠
So the system does not satisfy the additivity property. 2.7 Show that if the additivity property holds, then the homogeneity property holds for all rational numbers a . Thus if a system has “continuity” property, then additivity implies homogeneity. Translation: 说明系统如果具有可加性，那么对所有有理数 a 具有齐次性。因而对具有某种
连续性质的系统，可加性导致齐次性。 Answer: Any rational number a can be denoted by: nma /= Here m and n are both integer. Firstly, prove that if system input-output can be described
as following: yx →
then: mymx →
Easy to conclude it from additivity. Secondly, prove that if a system input-output can be described as following: yx →
then:
nynx // →
Suppose: unx →/ Using additivity:
nuxnxn →=)/(*
So: nuy =
nyu /=
Linear System Theory and Design SA01010048 LING QING
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It is to say that:
nynx // →
Then:
nmynmx /*/* →
ayax →
It is the property of homogeneity. 2.8 Let g(t,T)=g(t+a,T+a) for all t,T and a. Show that g(t,T) depends only on t-T. Translation: 设对于所有的 t,T 和 a，g(t,T)=g(t+a,T+a)。说明 g(t,T)仅依赖于 t-T。 Answer: Define:
Ttx += Tty −=
So:
2
yxt +=
2yxT −
=
Then:
)2
,2
(),( yxyxgTtg −+=
)2
,2
( ayxayxg +−
++
=
)22
,22
( yxyxyxyxg +−+
−+−+
+=
)0,(yg=
So:
0)0,(),(=
∂∂
=∂
∂xyg
xTtg
It proves that g(t,T) depends only on t-T. 2.9 Consider a system with impulse response as shown in Fig2.20(a). What is the zero-state response excited by the input u(t) shown in Fig2.20(b)?
Fig2.20
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Translation: 考虑冲激响应如图 2.20(a)所示的系统，由如图 2.20(b)所示输入 u(t)激励的零状 态响应是什么？ Answer: Write out the function of g(t) and u(t):
≤≤−≤≤
=212
10)(
tttt
tg
≤≤−≤≤
=211
101)(
tt
tu
then y(t) equals to the convolution integral:
∫ −=t
drrturgty0
)()()(
If 0=<t=<1, 0=<r=<1, 0<=t-r<=1:
∫=t
rdrty0
)(2
2t=
If 1<=t<=2:
)(ty ∫−
−=1
0
)()(t
drrturg ∫−
−+1
1
)()(t
drrturg ∫ −+t
drrturg1
)()(
)(1 ty= )(2 ty+ )(3 ty+
Calculate integral separately:
)(1 ty ∫−
−=1
0
)()(t
drrturg 10 ≤≤ r 21 ≤−≤ rt
∫−
−=1
0
t
rdr2
)1( 2−−=
t
)(2 ty ∫−
−=1
1
)()(t
drrturg 10 ≤≤ r 10 ≤−≤ rt
∫−
=1
1t
rdr2
)1(21 2−−=
t
)(3 ty ∫ −=t
drrturg1
)()( 21 ≤≤ r 10 ≤−≤ rt
∫ −=t
drr1
)2(2
1)1(22 −
−−=tt
)(ty )(1 ty= )(2 ty+ )(3 ty+ 2423 2 −+−= tt
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2.10 Consider a system described by
uuyyy −=−+••••
32
What are the transfer function and the impulse response of the system? Translation: 考虑如式所描述的系统，它的传递函数和冲激响应是什么？ Answer: Applying the Laplace transform to system input-output equation, supposing that the
System is initial relaxed:
)()()(3)(2)(2 sYssYsYssYsYs −=−+
System transfer function:
3
132
1)()()( 2 +
=−+
−==
ssss
sYsUsG
Impulse response:
tes
LsGLtg 311 ]3
1[)]([)( −−− =+
==
2.11 Let y(t) be the unit-step response of a linear time-invariant system. Show that the impulse
response of the system equals dy(t)/dt. Translation: y(t)是线性定常系统的单位阶跃响应。说明系统的冲激响应等于 dy(t)/dt. Answer: Let m(t) be the impulse response, and system transfer function is G(s):
ssGsY 1*)()( =
)()( sGsM =
ssYsM *)()( =
So:
dttdytm /)()( =
2.12 Consider a two-input and two-output system described by
)()()()()()()()( 212111212111 tupNtupNtypDtypD +=+
)()()()()()()()( 222121222121 tupNtupNtypDtypD +=+
where Nij and Dij are polynomials of p:=d/dt. What is the transfer matrix of the system? Translation: 考虑如式描述的两输入两输出系统， Nij 和 Dij 是 p:=d/dt 的多项式。系统的
传递矩阵是什么？ Answer: For any polynomial of p, N(p), its Laplace transform is N(s). Applying Laplace transform to the input-output equation:
)()()()()()()()( 212111212111 sUsNsUsNsYsDsYsD +=+
)()()()()()()()( 222121222121 sUsNsUsNsYsDsYsD +=+
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Write to the form of matrix:
)()(
)()()()(
2
1
2221
1211
sYsY
sDsDsDsD
=
)()(
)()()()(
2
1
2221
1211
sUsU
sNsNsNsN
)()(
2
1
sYsY
=1
2221
1211
)()()()( −
sDsDsDsD
)()(
)()()()(
2
1
2221
1211
sUsU
sNsNsNsN
So the transfer function matrix is:
)(sG =1
2221
1211
)()()()( −
sDsDsDsD
)()()()(
2221
1211
sNsNsNsN
By the premising that the matrix inverse:
1
2221
1211
)()()()( −
sDsDsDsD
exists. 2.11 Consider the feedback systems shows in Fig2.5. Show that the unit-step responses of the positive-feedback system are as shown in Fig2.21(a) for a=1 and in Fig2.21(b) for a=0.5. Show also that the unit-step responses of the negative-feedback system are as shown in Fig 2.21(c) and 2.21(d), respectively, for a=1 and a=0.5.
Fig 2.21 Translation: 考虑图 2.5 中所示反馈系统。说明正反馈系统的单位阶跃响应，当 a=1 时，如
图 2.21(a)所示。当 a=0.5 时，如图 2.21(b)所示。说明负反馈系统的单位阶跃响应
如图 2.21(c)和 2.21(b)所示，相应地，对 a=1 和 a=0.5。
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Answer: Firstly, consider the positive-feedback system. It’s impulse response is:
∑∞
=
−=1
)()(i
i itatg δ
Using convolution integral:
∑∞
=
−=1
)()(i
i itraty
When input is unit-step signal:
∑=
=n
i
iany1
)(
)()( nyty = 1+≤≤ ntn
Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(a) and Fig 2.21(b) shown.
Secondly, consider the negative-feedback system. It’s impulse response is:
∑∞
=
−−−=1
)()()(i
i itatg δ
Using convolution integral:
∑∞
=
−−−=1
)()()(i
i itraty
When input is unit-step signal:
∑=
−−=n
i
iany1
)()(
)()( nyty = 1+≤≤ ntn
Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(c) and Fig 2.21(d) shown.
2.14 Draw an op-amp circuit diagram for
uxx
−
+
−=
•
42
5042
[ ] uxy 2103 −=
2.15 Find state equations to describe the pendulum system in Fig 2.22. The systems are useful to
model one- or two-link robotic manipulators. If θ , 1θ and 2θ are very small, can you
consider the two systems as linear?
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Translation: 试找出图 2.22 所示单摆系统的状态方程。这个系统对研究一个或两个连接的机
器人操作臂很有用。假如角度都很小时，能否考虑系统为线性？ Answer: For Fig2.22(a), the application of Newton’s law to the linear movements yields:
)cossin()cos(cos2
2
2
θθθθθθ•••
−−==− mlldtdmmgf
)sincos()sin(sin2
2
2
θθθθθθ•••
−==− mlldtdmfu
Assuming θ and •
θ to be small, we can use the approximation θsin =θ , θcos =1.
By retaining only the linear terms in θ and •
θ , we obtain mgf = and:
umll
g 1+−=
••
θθ
Select state variables as θ=1x , •
= θ2x and output θ=y
uml
xlg
x
+
−
=•
/11
0/10
[ ]xy 01=
For Fig2.22(b), the application of Newton’s law to the linear movements yields:
)cos(coscos 112
2
112211 θθθ ldtdmgmff =−−
)cossin( 12
11111 θθθθ•••
−−= lm
)sin(sinsin 112
2
11122 θθθ ldtdmff =−
)sincos( 12
11111 θθθθ•••
−= lm
)coscos(cos 22112
2
2222 θθθ lldtdmgmf +=−
)cossin( 12
11112 θθθθ•••
−−= lm )cossin( 22
22222 θθθθ•••
−−+ lm
)sinsin(sin 22112
2
222 θθθ lldtdmfu +=−
)sincos( 12
11112 θθθθ•••
−= lm )sincos( 22
22222 θθθθ•••
−+ lm
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Assuming 1θ , 2θ and 1
•
θ , 2
•
θ to be small, we can use the approximation 1sinθ = 1θ ,
2sinθ = 2θ , 1cosθ =1, 2cosθ =1. By retaining only the linear terms in 1θ , 2θ and
1
•
θ , 2
•
θ , we obtain gmf 22 = , gmmf )( 211 += and:
211
21
11
211
)(θθθ
lmgm
lmgmm
++
−=••
ulmlm
gmmlm
gmm
222
21
211
21
212
1)()(+
+−
+=
••
θθθ
Select state variables as 11 θ=x , 12
•
= θx , 23 θ=x , 24
•
= θx and output
=
2
1
2
1
θθ
yy
:
+−+
+−=
•
•
•
•
4
3
2
1
22212121
1121121
4
3
2
1
0/)(0/)(10000/0/)(0010
xxxx
lmgmmlmgmm
lmgmlmgmm
xxxx
+ u
lm
22/1000
=
01000001
2
1
yy
4
3
2
1
xxxx
2.17 The soft landing phase of a lunar module descending on the moon can be modeled as shown in Fig2.24. The thrust generated is assumed to be proportional to the derivation of m, where m is the mass of the module. Then the system can be described by
mgmkym −−=•••
Where g is the gravity constant on the lunar surface. Define state variables of the system as:
yx =1 , •
= yx2 , mx =3 , •
= my
Find a state-space equation to describe the system. Translation: 登月舱降落在月球时，软着陆阶段的模型如图 2.24 所示。产生的冲激力与 m
的微分成正比。系统可以描述如式所示形式。g 是月球表面的重力加速度常数。定
义状态变量如式所示，试图找出系统的状态空间方程描述。 Answer: The system is not linear, so we can linearize it. Suppose:
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−
+−= ygty 2/2
−••
+−= ygty
−••••
+−= ygy
−
+= mmm 0
−••
= mm
So:
)( 0
−
+ mm−••
+− )( yg =
−•
− mk gmm )( 0
−
+−
+−−−
gmgm0
−••
=ym0
−•
− mk gmgm−
−− 0
−••
=ym0
−•
− mk
Define state variables as:
−−
= yx1 ,
−•−
= yx 2 , −−
= mx3 ,
−•−
= my Then:
•−
•−
•−
3
2
1
x
x
x
=
000000010
−
−
−
3
2
1
x
x
x
+
−
1/0
0mk−
u
−
y = [ ]001
−
−
−
3
2
1
x
x
x
2.19 Find a state equation to describe the network shown in Fig2.26.Find also its transfer function. Translation: 试写出描述图 2.26 所示网络的状态方程，以及它的传递函数。 Answer: Select state variables as:
1x : Voltage of left capacitor
2x : Voltage of right capacitor
3x : Current of inductor
Applying Kirchhoff’s current law:
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3x = RxLx /)( 32
•
−
RxxCu 11+=•
32 xxCu +=•
2xy =
From the upper equations, we get:
CuCRxx //11 +−=•
ux +−= 1
CuCxx //32 +−=•
ux +−= 3
LRxLxx // 323 −=•
32 xx −=
2xy =
They can be combined in matrix form as:
•
•
•
3
2
1
x
x
x
=
−−
−
110100
001
3
2
1
xxx
+ u
011
[ ]
=
3
2
1
010xxx
y
Use MATLAB to compute transfer function. We type: A=[-1,0,0;0,0,-1;0,1,-1]; B=[1;1;0]; C=[0,1,0]; D=[0]; [N1,D1]=ss2tf(A,B,C,D,1)
Which yields: N1 = 0 1.0000 2.0000 1.0000
D1 = 1.0000 2.0000 2.0000 1.0000 So the transfer function is:
122
12)( 23
2^
+++++
=sss
sssG1
12 +++
=ss
s
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2.20 Find a state equation to describe the network shown in Fig2.2. Compute also its transfer matrix.
Translation: 试写出描述图 2.2 所示网络的状态方程，计算它的传递函数矩阵。 Answer: Select state variables as Fig 2.2 shown. Applying Kirchhoff’s current law:
1u = 323121111 xRxLxxxCR ++++••
22211
••
=+ xCuxC
223
•
= xCx
3231212311 )( xRxLxxuxRu ++++−=•
3231 xRxLy +=•
From the upper equations, we get:
12131 // CuCxx −=•
232 / Cxx =•
12111131112113 ///)(// LuRLuLxRRLxLxx +++−−−=•
2113121 uRuxRxxy ++−−−=
They can be combined in matrix form as:
•
•
•
3
2
1
x
x
x
=
+−−− 12111
2
1
/)(/100/100
LRRLLCC
3
2
1
xxx
+
−
2
1
11
1
1 /0/1
/100
uu
LR
C
L
[ ]
−−−=
3
2
1
111xxx
Ry + [ ]
2
121
uu
R
Applying Laplace Transform to upper equations:
)(/1/1
)( 1
^
21111
21^
susCsCRRsL
RsLsy++++
+=
)(/1/1)/1)((
2
^
21111
2121 susCsCRRsL
sCRRsL++++
+++
++++
+=
sCsCRRsLRsLsG
21111
21^
/1/1)(
++++
++sCsCRRsL
sCRRsL
21111
2121
/1/1)/1)((
Linear System Theory and Design SA01010048 LING QING
16
2.18 Find the transfer functions from u to 1y and from 1y to y of the hydraulic tank system
shown in Fig2.25. Does the transfer function from u to y equal the product of the two transfer functions? Is this also true for the system shown in Fig2.14?
Translation: 试写出图 2.25 所示水箱系统从u 到 1y 的传递函数和从 1y 到 y 的传递函数。从
u 到 y 的传递函数等于两个传递函数的乘积吗？这对图 2.14 所示系统也是正确的吗？
Answer: Write out the equation about u , 1y and y :
111 / Rxy =
222 / Rxy =
dtyudxA /)( 111 −=
dtyydxA /)( 122 −=
Applying Laplace transform:
)1/(1/ 11
^
1
^sRAuy +=
)1/(1/ 221
^^sRAyy +=
)1)(1/(1/ 2211
^^sRAsRAuy ++=
So:
=^^
/ uy )/(^
1
^uy )/( 1
^^yy
But it is not true for Fig2.14, because of the loading problem in the two tanks.
3.1consider Fig3.1 ,what is the representation of the vector x with respect to the basis ( )21 iq ?
What is the representation of 1q with respect ro ( )22 qi ?图 3.1 中,向量 x 关于 ( )21 iq 的表
示是什么? 1q 关于 ( )22 qi 的表示有是什么?
If we draw from x two lines in parallel with 2i and 1q , they tutersect at 131 q and 23
8 i as
shown , thus the representation of x with respect to the basis ( )21 iq is ′
38
31
, this can
be verified from [ ]
=
=
=
38
31
1103
3831
31
22 iqx
To find the representation of 1q with respect to ( )22 qi ,we draw from 1q two lines in
parallel with 2q and 2i , they intersect at -2 2i and 223 q , thus the representation of 1q
with respect to ( )22 qi , is ′
−
232 , this can be verified from
[ ]
−
=
−=
=
23
22120
23
213
221 qiq
3.2 what are the 1-morm ,2-norm , and infinite-norm of the vectors [ ] [ ]′=′−= 111,132 21 xx , 问向量
[ ] [ ]′=′−= 111,132 21 xx 的 1-范数,2-范数和 −∞ 范数是什么?
1max3max
3)111(14)1)3(2(
31116132
2211
21222
222
12221
'121
12
3
1111
====
=++==+−+==
=++==++==
∞∞
=∑
iiii
ii
xxxxxx
xxxx
xxx
3.3 find two orthonormal vectors that span the same space as the two vectors in problem 3.2 ,求与题 3.2 中的两个向量
张成同一空间的两个标准正交向量. Schmidt orthonormalization praedure ,
[ ] [ ]
[ ]′===−=
′−==′−==
1113
1)(
132141132
2
22212
'122
1
1111
uu
qxqxqxu
uu
qxu
The two orthomormal vectors are
=
−=
111
31
13
2
141
21 qq
In fact , the vectors 1x and 2x are orthogonal because 01221 =′=′ xxxx so we can only
normalize the two vectors
==
−==
111
31
13
2
141
2
22
1
11 x
xq
xx
q
3.4 comsider an mn× matrix A with mn ≥ , if all colums of A are orthonormal , then
mIAA =′ , what can you say abort AA ′ ? 一个 mn× 阶矩阵A( mn ≥ ), 如果A的所有列都是
标准正交的,则 mIAA =′ 问 AA ′是怎么样?
Let [ ] [ ]mnijm aaaaA
×== 21 , if all colums of A are orthomormal , that is
∑=
=≠
==n
ililiii jiif
jiifaaaa
1
'
10
then
[ ]
[ ]
=≠≠≠
=
==
=′
=
=
=
=′
∑
∑∑
=
×==
jiifjiif
aageneralin
aaaa
a
a
a
aaaAA
Iaaaaaa
aaaaaaaaa
a
a
a
AA
jl
m
iil
nnjl
m
iil
m
iii
m
mi
m
mmmm
m
mi
m
10
)(
100010001
1
11
'
'
'2
'1
''2
'
'2
'1
'
'12
'11
'1
''2
'
'
'2
'1
if A is a symmetric square matrix , that is to say , n=m jlil aa for every nli 2,1, =
then mIAA =′
3.5 find the ranks and mullities of the following matrices 求下列矩阵的秩和化零度
−−=
−=
=
100022104321
011023114
100000010
321 AAA
134)(033)(123)(3)(3)(2)(
321
321
=−==−==−====
ANullityANullityANullityArankArankArank
3.6 Find bases of the range spaces of the matrices in problem 3.5 求题 3.5 中矩阵值域空间的基
the last two columns of 1A are linearly independent , so the set
100
,001
can be used as a
basis of the range space of 1A , all columns of 2A are linearly independent ,so the set
−
001
122
134
can be used as a basis of the range spare of 2A
let [ ]43213 aaaaA = ,where ia denotes of 3A 4321 aandaandaanda are
linearly independent , the third colums can be expressed as 213 2aaa +−= , so the set
{ }321 aaa can be used as a basis of the range space of 3A
3.7 consider the linear algebraic equation xx −
=
−−
−
101
213312
it has three equations and two
unknowns does a solution x exist in the equation ? is the solution unique ? does a solution exist if
[ ]′= 111y ?
线性代数方程 xx −
=
−−
−
101
213312
有三个方程两个未知数, 问方程是否有解?若有解是否
唯一?若 [ ]′= 111y 方程是否有解?
Let [ ],213312
21 aaA =
−−
−= clearly 1a and 2a are linearly independent , so rank(A)=2 ,
y is the sum of 1a and 2a ,that is rank([A y ])=2, rank(A)= rank([A y ]) so a solution x
exists in A x = y
Nullity(A)=2- rank(A)=0
The solution is unique [ ]′= 11x if [ ]′= 111y , then rank([A y ])=3≠ rank(A), that is
to say there doesn’t exist a solution in A x = y
3.8 find the general solution of
=
−−
123
100022104321
x how many parameters do you have ?
求方程
=
−−
123
100022104321
x 的同解,同解中用了几个参数?
Let
=
−−=
123
100022104321
yA we can readily obtain rank(A)= rank([A y ])=3 so
this y lies in the range space of A and [ ]′−= 101 ox p is a solution Nullity(A)=4-3=1 that
means the dimension of the null space of A is 1 , the number of parameters in the general solution
will be 1 , A basis of the null space of A is [ ]′−= on 121 thus the general solution of
A x = y can be expressed an
−
+
−
=+=
012
1
1001
ααnxx p for any real α
α is the only parameter 3.9 find the solution in example 3.3 that has the smallest Euclidean norm 求例 3 中具有最小欧氏
范数的解,
the general solution in example 3.3 is
−−
−+=
−
+
−+
−
=
2
1
21
1
21
42
1020
01
11
004
0
ααααα
ααx for
any real 1α and 2α
the Euclidean norm of x is
16168453
)()()42(
212122
21
22
21
221
21
+−−++=
−+−+−++=
αααααα
αααααx
−
−
−=⇒
=
=⇒
=−+⇒=∂
=−+⇒=∂
1116114
118
114
1116114
082502
042302
2
1
122
211 x
x
x
α
α
ααα
ααα has the smallest Euclidean
norm , 3.10 find the solution in problem 3.8 that has the smallest Euclidean norm 求题 3.8 中欧氏范数
最小的解,
−
+
−
=
012
1
1001
αx for any real α
the Euclidean norm of x is
−
−
=⇒=⇒=−⇒=∂
∂
+−=++−+−=
1616365
6102120
2261)2()1( 2222
xx
x
ααα
ααααα
has the
smallest Euclidean norm
3.11 consider the equation ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnubAubAubAxAnx nnn
where A is an nn× matrix and b is an 1×n column vector ,under what conditions on A and b will there
exist ]1[],1[],[ −nuuou to meet the equation for any ]0[],[ xandnx ?
令 A 是 nn× 的矩阵, b 是 1×n 的列向量,问在 A 和b 满足什么条件时,存在 ]1[],1[],[ −nuuou ,对所有的
]0[],[ xandnx ,它们都满足方程 ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnubAubAubAxAnx nnn ,
write the equation in this form
−−
== −
]0[
]2[]1[
],[]0[}{ 1
u
nunu
bAbAbxAnx nn
where ],[ 1bAbAb n− is an nn× matrix and ]0[}{ xAnx n− is an 1×n column vector , from the equation we
can see , ]1[],1[],[ −nuuou exist to meet the equation for any ]0[],[ xandnx ,if and only if
nbAbAb n =− ],[ 1ρ under this condition , there will exist ]1[],1[],[ −nuuou to meet the equation for any
]0[],[ xandnx .
3.12 given
=
=
=
1321
100
1000020001200012
bbA what are the representations of A with respect to
the basis { }bAbAbAb 32 and the basis { }bAbAbAb 32 , respectively? 给定
=
=
=
1321
100
1000020001200012
bbA 请问 A 关于{ }bAbAbAb 32 和基{ }bAbAbAb 32 的表
示分别是什么?
=⋅=
=⋅=
=
1163224
,
1441
,
1210
232 bAAbAbAAbAbA
we have [ ] [ ]
−
−
==∴
+−+−=
710018010
200018000
718208
3232
324
bAbAbAbbAbAbAbA
bAbAbAbbA
thus the representation of A
with respect to the basis { }bAbAbAb 32 is
−
−
=
710018010
200018000
A
bAbAbAbbA
bAbAbAbA
324
432
718208
148
128152
,
1245250
,
1122015
,
1674
+−+−=
=
=
=
=
[ ] [ ]
−
−
==∴
710018010
200018000
3232 bAbAbAbbAbAbAbA thus the representation of A with
respect to the basis { }bAbAbAb 32 is
−
−
=
710018010
200018000
A
3.13 find Jordan-form representations of the following matrices 写出下列矩阵的 jordan 型表示:
.20250
16200340
.200010101
.342
100010
,300020
1041
4321
−−=
−=
−−−=
= AAAA
the characteristic polynomial of 1A is )3)(2)(1()det( 11 −−−=−=∆ λλλλ AI thus the eigenvelues of 1A are
1 ,2 , 3 , they are all distinct . so the Jordan-form representation of 1A will be diagonal .
the eigenvectors associated with 3,2,1 === λλλ ,respectively can be any nonzero solution of
[ ]
[ ]
[ ]′=⇒=
′=⇒=
′=⇒=
1053
0142
001
3331
2221
1111
qqqA
qqqA
qqqA
thus the jordan-form representation of 1A with respect to
{ }321 qqq is
=
300020001
ˆ1A
the characteristic polynomial of 2A is
223
22 )1)(1)(1(243()det()( AiiAI −++++=+++=−=∆ λλλλλλλλ has
eigenvalues jandj −−+−− 11,1 the eigenvectors associated with
jandj −−+−− 11,1 are , respectively
[ ] [ ] [ ]′−−′−−′− jjandjj 211211,111 the we have
QAQj
jAandjj
jjQ 21
2
100010001
ˆ
220111111
−=
−−+−
−=
−−−+−−=
the characteristic polynomial of 3A is )2()1()det()( 233 −−=−=∆ λλλλ AI theus the
eigenvalues of 3A are 1 ,1 and 2 , the eigenvalue 1 has multiplicity 2 , and nullity
213)(3)( 33 =−=−−=− IArankIA the 3A has two tinearly independent eigenvectors
associated with 1 , [ ] [ ]
[ ]′−=⇒=−
′=′=⇒=−
1010)(
0100010)(
333
213
qqIA
qqqIA thus we have
QAQAandQ 31
3
200010001
ˆ
100011101
−=
=
−−=
the characteristic polynomial of 4A is 344 )det()( λλλ =−=∆ AI clearly 4A has lnly one
distinct eigenvalue 0 with multiplicity 3 , Nullity( 4A -0I)=3-2=1 , thus 4A has only one
independent eigenvector associated with 0 , we can compute the generalized , eigenvectors
of 4A from equations below
[ ]
[ ]
[ ]′−=⇒=
′−=⇒=
′=⇒=
430
540
0010
3231
2121
111
vvvA
vvvA
vvA
, then the representation of 4A
with respect to the basis { }321 vvv is
−−==
= −
450340
001
000100010
ˆ4
14 QwhereQAQA
3.14 consider the companion-form matrix
−−−−
=
010000100001
4321 αααα
A show that its
characterisic polynomial is given by 433
23
14)( αλαλαλαλλ ++++=∆
show also that if iλ is an eigenvalue of A or a solution of 0)( =∆ λ then
[ ]133 ′iii λλλ is an eigenvector of A associated with iλ
证明友矩阵 A 的特征多项式 433
23
14)( αλαλαλαλλ ++++=∆ 并且,如果 iλ 是 iλ 的一
个特征值, 0)( =∆ λ 的一个解, 那么向量 [ ]133 ′iii λλλ 是 A 关于 iλ 的一个特征向量
proof:
43223
14
432
31
4
432
1
4321
2
1det
10
det
1001det
100100
det)(
100010001
det)det()(
αλαλαλαλ
λαα
λλ
αλαλ
λλ
ααα
λλ
λαλ
λλ
λααααλ
λλ
++++=
−
+
−
++=
−−+
−−+=
−−
−+
=−=∆ AI
if iλ is an eigenvalue of A , that is to say , 0)( =∆ iλ then we have
=
=
−−−−
=
−−−−
111010000100001 2
3
2
3
4
243
22
31
2
34321
i
i
i
I
i
i
i
i
i
i
iii
i
i
i
λλλ
λ
λλλλ
λλ
αλαλαλα
λλλαααα
that is to say [ ]133 ′iii λλλ is an eigenvetor oa A associated with iλ
3.15 show that the vandermonde determinant
11114321
24
23
22
21
34
33
32
31
λλλλλλλλλλλλ
equals
)(41 ijji λλ −Π ≤<≤ , thus we conclude that the matrix is nonsingular or equivalently , the
eigenvectors are linearly independent if all eigenvalues are distinct ,证明 vandermonde 行列式
11114321
24
23
22
21
34
33
32
31
λλλλλλλλλλλλ
为 )(41 ijji λλ −Π ≤<≤ , 因此如果所有的特征值都互不
相同则该矩阵非奇异, 或者等价地说, 所有特征向量线性无关, proof:
41
241423132412
23132414424142313313
1224142313
2414423133
141312
24142313
2414423133
141312
144133122
142413
2312
22
141312
144133122
142413
2312
22
4321
24
23
22
21
34
33
32
31
))()()()()(()])()()(())()()(()[(
)())(())((
))(())((det
))(())((0))(())((0
det
)()()()()()(
det
11110
)()()(0)()()(0
det
1111
det
≤<≤Π=−−−−−−=
−−−−−−−−−−−=
−⋅
−−−−−−−−
−=
−−−−−−−−−−−
−=
−−−−−−−−−
=
−−−−−−−−−
=
ji
λλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλ
λλλλλλλλλλλλλλλλλλλλ
λλλλλλλλλλλλλλλλλλλλλλλλ
λλλλλλλλλλλλλλλλλλλλλλλλ
λλλλλλλλλλλλλλλλλλλλλλλλ
λλλλλλλλλλλλ
let a, b, c and d be the eigenvalues of a matrix A , and they are distinct Assuming the matrix is singular , that is abcd=0 , let a=0 , then we have
))()((111
detdet
1111000
det
222
222
333222
333
bcbdcdbcddcbdcb
bcddcb
dcbdcb
dcbdcbdcb
−−−−=
⋅−=
=
and from vandermonde determinant
))(())()()((
1111000
det
1111
det222
333
2222
3333
bdcdbcdabacbdcddcb
dcbdcb
dcbadcbadcba
−−=−−−−=
=
so we can see
−=
1111000
det
1111000
det222
333
222
333
dcbdcbdcb
dcbdcbdcb
,
that is to say dandcbabdcdbcd ,,,0))(( ⇒=−− are not distinet
this implies the assumption is not true , that is , the matrix is nonsingular let i
q be the
eigenvectors of A , 44332211
qaqAqaqAqaqAqaqA ====
( )
tindependenlinenrlyqqandq
so
dddddcbbbaaa
qqqq
qdqcqbqa
qdqcqbqa
qdqcqbqa
qqqq
iini
nii ⇒=≠=
=
⇒
=+++
=+++
=+++
=+++
×
×
×
000
0
1111
0
0
0
0
1
1
44
32
32
32
32
44332211
443
333
223
113
442
332
222
112
44332211
44332211
αα
αααα
αααα
αααα
αααα
αααα
3.16 show that the companion-form matrix in problem 3.14 is nonsingular if and only if
04 ≠α , under this assumption , show that its inverse equals
−−−−
=−
4
3
4
2
4
1
4
1
1100001000010
αα
αα
αα
α
A 证明题 3.14 中的友矩阵非奇异当且仅当
04 ≠α , 且矩阵的逆为
−−−−
=−
4
3
4
2
4
1
4
1
1100001000010
αα
αα
αα
α
A
proof: as we know , the chacteristic polynomial is
432
23
14)det()( αλαλαλαλλλ ++++=−=∆ AI so let 0=λ , we have
44 )det()det()1()det( α==−= AAA
A is nonsingular if and only if
04 ≠α
−−−−
=∴
=
⋅
−−−−
⋅
−−−−
=
⋅
−−−−
⋅
−−−−
−
4
3
4
2
4
1
4
1
4
4
3
4
2
4
1
4
4321
4
4321
4
3
4
2
4
1
4
1100001000010
1000010000100001
1100001000010
010000100001
1000010000100001
010000100001
1100001000010
αα
αα
αα
α
αα
αα
αα
α
αααα
αααα
αα
αα
αα
α
A
I
I
3.17 consider
=λλλ
λλλ
000
2/2
tTtTT
A with 0≠λ and T>0 show that [ ]′100 is an
generalized eigenvector of grade 3 and the three columns of
=0000002/222
tTTT
Q λλλ
constitute a chain of generalized eigenvectors of length 3 , venty
=−
λλ
λ
001001
1 tAQQ
矩阵 A 中 0≠λ ,T>0 , 证明 [ ]′100 是 3 级广义特征向量, 并且矩阵 Q 的 3 列组成长度
是 3 的广义特征向量链,验证
=−
λλ
λ
001001
1 tAQQ
Proof : clearly A has only one distinct eigenvalue λ with multiplicity 3
0100
000000000
100
00000
2/0
000000
00
100
)(
000
100
000000
00
100
)(
222
3
2222
2
=
=
=
=−
≠
=
=
=−
TTTT
IA
TTIA
λλλλ
λ
λλλ
these two equation imply that [ ]′100 is a generalized eigenvctor of grade 3 ,
and
=
=
=−
=
=
=−
00
000000
2/0
0)(
0100
00000
2/0
100
)(
2222222
222
TTT
TTT
TT
IA
TT
TTT
IA
λλλ
λλλ
λλ
λ
λλ
λλλ
λ
that is the three columns of Q consititute a chain of generalized eigenvectors of length 3
=
=
=
=
=
−
λλ
λ
λλλ
λλλ
λλ
λλ
λλ
λλ
λ
λλλ
λλλλ
λλ
λλλ
λλλ
001001
000
2/2/3
001001
1000002/
001001
000
2/2/3
1000002/
000
2/
1
2
22223222
222232222
AQQ
TTTTT
TTT
Q
TTTTT
TTT
TTT
AQ
3.18 Find the characteristic polynomials and the minimal polynomials of the following matrices 求下列矩阵的特征多项式和最小多项式,
)(
000000000000
)(
000000000001
)(
000000010001
)(
000000010001
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
dcba
λλ
λλ
λλ
λλ
λλ
λλ
λλ
λλ
)()()()()(
)()()()()(
)()()()()(
)()()()()()()(
144
14
213
413
312
412
23
1123
11
λλλλλλ
λλλλλλ
λλλλλλ
λλλλλλλλλλ
−=Ψ−=∆
−=Ψ−=∆
−=Ψ−=∆
−−=Ψ−−=∆
dcba
3.19 show that if λ is an eigenvalue of A with eigenvector x then )(λf is an eigenvalue
of )(Af with the same eigenvector x
证明如果λ是 A 的关于λ的特征向量,那么 )(λf 是 )(Af 的特征值, x 是 )(Af 关于
)(λf 的特征向量,
proof let A be an nn× matrix , use theorem 3.5 for any function )(xf we can define
1110)( −−+++= n
n xxxh βββ which equals )(xf on the spectrum of A
if λ is an eigenvalue of A , then we have )()( λλ hf = and
)()( AhAf =
xfxhxxx
xAAIxAhxAfxxAxxAxxAxAxxAf
nn
nn
kk
)()(
)()()(,,)(
1110
1110
3322
λλλβλββ
βββ
λλλλλλ
==+++=
+++==∴
====⇒
−−
−−
which implies that )(λf is an eigenvalue of )(Af with the same eigenvector x
3.20 show that an nn× matrix has the property 0=kA for mk ≥ if and only if A has
eigenvalues 0 with multiplicity n and index m of less , such a matrix is called a nilpotent matrix
证明 nn× 的矩阵在 0=kA 当且仅当A的 n 重 0特征值指数不大于m ,这样的矩阵被称为
归零矩阵, proof : if A has eigenvalues 0 with multiplicity n and index M or less then the Jordan-form
representation of A is
=
3
2
1
ˆ
JJ
JA where
mnma
nnJ
ii
l
ii
nn
i
ii
≤×
=
= ∑
=
×
1
010
10
from the nilpotent property , we have iki nkforJ ≥= 0 so if
,iinmamk ×≥≥ liallJ k
i ,,2,10 == and 0ˆ
3
2
1
=
=k
k
k
JJ
JA
then )0)ˆ(0)((0 === AfifonlyandifAfAk
If ,0 mkforAk ≥= then ,0ˆ mkforAk ≥= where
=
3
2
1
ˆ
JJ
JA , nnJ
l
ii
nni
i
i
i
ii
=
= ∑
=
×
111
λλ
λ
So we have
liformnand
linnk
kk
J
ii
ki
ki
nki
ii
ki
ki
ik
i
,2,1,0
,2,10)!1)(1(
! 11
=≤=∴
==
−+−
=
+−−
λ
λλ
λλλ
which implies that A has only one distinct eigenvalue o with multiplicity n and index m or less ,
3.21 given
=
100100011
A , find AteandAA 10310 , 求A的函数 AteandAA 10310 , ,
the characteristic polynomial of A is 22 )1()det()( −=−=∆ λλλλ AI
let 2210)( λβλββλ ++=h on the spectrum of A , we have
2110
21010
010
2110)1()1(
1)1()1(
0)0()0(
ββ
βββ
β
+=⋅′=′
++==
==
hfhfhf
the we have 98,0 220 =−== βββ
=
+
−=
+−=
100100911
100100101
9100100011
8
98 210 AAA
the compute
1021010
211031
0
2
1
0
21103
210103
0103103
=−==
⇒+=⋅++=
=
βββ
βββββ
βA
=
+
−=
+−=
100100
10211
100100011
102100100011
101
102101 2103 AAA
to compute Ate :
122
1
2 2
1
0
21
210
00
+−=−−=
=⇒
+=++=
=
tt
tt
t
t
etetee
tee
e
ββ
β
βββββ
β
−+−−
=
+−+
−−+
=
++=
t
t
tttt
tttt
At
ee
eteee
eeee
AAIAe
00110
11
100100111
)12(100100011
)22(100010001
2210 βββ
3.22 use two different methods to compute Ate for A1 and A4 in problem 3.13
用两种方法计算题 3.13 中A1和A4 的函数tAAt ee 2,
method 1 : the Jordan-form representation of A1 with respect to the basis
[ ] [ ] [ ]{ }′′′ 105014001 is { }3,2,1ˆ1 diagA =
−−=
=
=
==∴
−
−
t
t
ttttt
t
t
ttt
t
t
t
tAtA
ee
eeeee
ee
eeeCd
Cc
ee
eCb
QwhereQQeeCa
3
2
33
3
2
32
1
3
2
1ˆ
0000
)(5)(4
100010541
0000
54
100010541
000000
100010541
100010541
,11
+−−+
++=
−=
=∴
−−==
=
−
120250161200
232/541
10010
2/1
450340001
450340
001
000100010
ˆ
22
2
1
4
44
4
tttttttt
ttt
QQee
QwhereQQeA
tAtA
tA
method 2: the characteristic polynomial of 1A is ).3)(2)(1()( −−−=∆ λλλλ let
2210)( λβλββλ ++=h on the spectrum of 1A , we have
)2(21
234
25
33
93)3()3(32)2()2(
)1()1(
323
322
330
22103
2102
10
ttt
ttt
ttt
t
t
t
eee
eee
eee
ehfehf
ehf
+−=
−++−=
+−=
⇒++==′++==++==
β
β
β
ββββββ
βββ
−−=
+−+
−+−+
+−+−
+−=
++=
t
t
ttttt
ttt
ttt
ttt
ttt
ttt
tA
ee
eeeee
eee
eeeeee
eeeeee
AAIe
3
2
32
32
32
32
32
32
212210
0000
)(5)(4
90004040121
)2(21
300020
1041)
234
25(
330003300033
1 βββ
the characteristic polynomial of 4A is 3)( λλ =∆ , let 2210)( λβλββλ ++=h on the
spectrum of 4A ,we have
22
1
0
2:)0()0(
:)0()0(1:)0()0(
β
ββ
=′′=′′
=′=′==
thfthf
hf
thus
+−−+
++=
+
−−
++=
++=
120250161200
232/541
000000450
2/20250
162002340
22
2
2
241410
4
tttttttt
ttt
tI
AAIe tA βββ
3.23 Show that functions of the same matrix ; that is )()()()( AfAgAgAf = consequently
we have AeAe AtAt = 证明同一矩阵的函数具有可交换性，即 )()()()( AfAgAgAf = 因
此有 AeAe AtAt = 成立
proof: let 1
110
1110
)(
)(−
−
−−
+++=
+++=n
n
nn
AAIAfAAIAg
ααα
βββ
( n is the order of A)
kik
k
nkii
n
nk
kik
k
ii
n
k
nnn
nin
n
ii
nin
n
ii
AA
AAAAIAgAf
)()(
)()()(
1
22
0
1
0
2211
1
11
1
0011000
−+−=
−
=−
=
−
=
−−−−
=
−−−
−
=
∑∑∑∑
∑∑
+=
+++++++=
βαβα
βαβαβαβαβαβα
)()()()()()(1
22
0
1
0AgAfAAAgAf k
ik
k
nkii
n
nk
kik
k
ii
n
k=+= −
+−=
−
=−
=
−
=∑∑∑∑ βαβα
let AteAgAAf == )(,)( then we have AeAe AtAt =
3.24 let
=
3
2
1
000000
λλ
λC , find a B such that CeB = show that of 0=iλ for some
I ,then B does not exist
let
=
3
2
1
000000
λλ
λC , find a B such that CeB = Is it true that ,for any nonsingular c ,there
exists a matrix B such that CeB = ?
令
=
3
2
1
000000
λλ
λC 证明若 0111 =⋅⋅ λλλ ,则不存在 B 使 CeB =
若
=
3
2
1
000000
λλ
λC ,是否对任意非奇异 C 都存在 B 使, CeB = ?
Let λλ λ == ef ln)( so BeBf B == ln)(
==
3
2
1
ln000ln000ln
lnλ
λλ
CB where 3,2,1,01 => iλ , if 0=λ for some i ,
1lnλ does not exist
for
=
λλ
λ
000001
C we have
=
′
==λ
λλλ
λλλλ
ln000ln001ln
ln000ln00nlln
lnCB ,
where 0,0 ≤> λλ if then λln does mot exist , so B does mot exist , we can conclude
that , it is mot true that , for any nonsingular C THERE EXISTS a B such that cek =
3.25 let )()(
1)( AsIAdjs
AsI −∆
=− and let m(s) be the monic greatest common divisor of
all entries of Adj(Si-A) , Verify for the matrix 2A in problem 3.13 that the minimal polynomial
of A equals )()( sms∆
令, )()(
1)( AsIAdjs
AsI −∆
=− , 并且令 m(s)是 Adj(Si-A)的所有元素的第一最大公因子,
利用题 3.13 中 2A 验证 A 的最小多项式为 )()( sms∆
verification :
1)()1(00
0)2)(1(0)1(0)2)(1(
)(,200
010101
)2(\)1()()2()1()(
200010001
ˆ
200010101
2
2
33
−=
−−−
−−−−=−
−−
−=−
−−=−−=∆
=
−=
ssmsos
sssss
AsIAdjs
SS
AsI
ssssss
AA
j
we can easily obtain that )()()( smss ∆=j
3.26 Define [ ]122
11
01
)(1)( −−
−−− ++++∆
=− nnnn RsRsRsR
sAsI where
innn RandsssAsIs ααα ++++=−=∆ −− 2
21
12:)det()( are constant matrices theis
definition is valid because the degree in s of the adjoint of (sI-A) is at most n-1 , verify
IARnARtr
IAAAIARRn
ARtr
IAAIARRARtr
IAIARRARtr
IRARtr
nnn
nnnn
nnnn
αα
ααααα
αααα
ααα
α
+=−=
++++=+=−
−=
++=+=−=
+=+=−=
=−=
−
−−−−
−−−−
11
122
11
1211
1
212
2121
3
11011
2
00
1
0)(1
)(
2)(
2)(
1)(
where tr stands for the trase of a matrix and is defined as the sum of all its diagonal entries this
process of computing ii Randα is called the leverrier algorithm
定义 ][)(
1:)( 122
11
01
−−−−− ++++
∆=− nn
nn RSRSRSRs
ASI 其中 )(s∆ 是 A 的特征多项
式 innnn RandsssAsIs ααα ++++=−=∆ −− 2
21
1:)det()( 是常数矩阵,这样定义
是有效的, 因为 SI-A 的伴随矩阵中 S 的阶次不超过 N-1 验证
IARnARtr
IAAAIARRn
ARtr
IAAIARRARtr
IAIARRARtr
IRARtr
nnn
nnnn
nnnn
αα
ααααα
αααα
ααα
α
+=−=
++++=+=−
−=
++=+=−=
+=+=−=
=−=
−
−−−−
−−−−
11
122
11
1211
1
212
2121
3
11011
2
00
1
0)(1
)(
2)(
2)(
1)(
其中矩阵的迹 tr 定义为其对角元素之和, 这种计算 iα 和 iR 的程式被称为 leverrier 算法.
verification:
implieswhichSI
SSSSIARSARRSARRSARRSR
RSRSRSRASI
nnnnn
nnnnnn
nnnn
)()(
)()()(
])[(
12
21
1
1212
121
010
122
11
0
∆=+++++=
+−+−+−+=
++++−
−−−
−−−−−
−−−−
αααα
][)(
1:)( 122
11
01
−−−−− ++++
∆=− nn
nn RSRSRSRs
ASI ’
where nnnn ssss ααα ++++=∆ −− 2
21
1)(
3.27 use problem 3.26 to prove the cayley-hamilton theorem 利用题 3.26 证明 cayley-hamilton 定理 proof:
IARIARR
IARRIARR
IR
nn
nnn
αα
αα
=−=−
=−=−
=
−
−−−
1
121
212
101
0
multiplying ith equation by 1+−inA yields ( ni 2,1= )
IARARAAR
ARARAARARA
ARA
nn
nnn
nnn
nnn
nn
αα
α
α
=−=−
=−
=−
=
−
−−−
−−−
−−
1
122
1
221
12
2
1101
10
then we can see 012
2101
10
12
21
1
=−−++−++=
+++++
−−−−
−−−
nnnnnn
nnnnn
ARRAARRARARAIAAAA
ααααthat is
0)( =∆ A
3.28 use problem 3.26 to show
[ ]IssAssAsAs
AsI
nnnnnn )()()(
)(1
)(
12
113
2122
11
1
−−−−−−
−
++++++++++∆
=
−
ααααα
利用题 3.26 证明上式, Proof:
[ ]IssAssAsAs
IAAASAAA
SIAASIASs
RSRSRSRs
ASI
nnnnnn
nnnn
nnnn
nnn
nnnn
)()()()(
1])(
)()([)(
1
][)(
1:)(
12
113
2122
11
122
11
433
12
321
221
1
122
11
01
−−−−−−
−−−−
−−−−
−−−
−−−−−
++++++++++∆
=
+++++++++
++++++∆
=
++++∆
=−
ααααα
αααααα
ααα
another : let 10 =α
[ ]IssAssAsAs
IAAASAAASIAASIAS
ASASs
ASs
SRs
RSRSRSRs
ASI
nnnnnn
nnnn
nnnn
nnn
n
i
Ini
N
i
Ini
n
ii
n
i
Lli
n
i
inini
nnnn
)()()()(
1])(
)()()(
1)(
1)(
1
][)(
1:)(
12
113
2122
11
122
11
433
12
321
221
1
1
0
11
0
01
1 1
0
1
0
11
122
11
01
−−−−−−
−−−−
−−−−
−−−
−
=
−−−
=
−−
−
=
−
=−
−
=
−−−−
−−−−−
++++++++++∆
=
+++++++++
++++++
+++∆
=
∆=
∆=
++++∆
=−
∑∑
∑ ∑∑
ααααα
αααααα
ααα
αα
α
3.29 let all eigenvalues of A be distinct and let iq be a right eigenvector of A associated with
iλ that is iIi qqA λ= define [ ]nqqqQ 21= and define
== −
np
pp
QP
2
1
1 :; ,
where ip is the ith row of P , show that ip is a left eigenvector of A associated with iλ , that
is iii pAp λ= 如果 A 的所有特征值互不相同 , iq 是关于 iλ 的一个右特征向量 ,即
iIi qqA λ= ,定义 [ ]nqqqQ 21= 并且
== −
np
pp
QP
2
1
1 :;
其中 ip 是 P 的第 I 行, 证明 ip 是 A 的关于 iλ 的一个左特征向量,即 iii pAp λ=
Proof: all eigenvalues of A are distinct , and iq is a right eigenvector of A associated with iλ ,
and [ ]nqqqQ 21= so we know that 11
3
2
1
ˆ −− ==
= PAPAQQA
λλ
λ
PAPA ˆ=∴ That is
=
⇒
=
nnnnn
n p
pp
Ap
ApAp
p
pp
A
p
pp
λ
λλ
λλ
λ
22
11
2
1
2
1
2
12
1
: so iii pAp λ= , that is , ip is a
left eigenvector of A associated with iλ
3.30 show that if all eigenvalues of A are distinct , then 1)( −− ASI can be expressed as
∑ −=− −
iii
pqs
ASIλ
1)( 1 where iq and ip are right and left eigenvectors of A associated
with iλ 证 明 若 A 的 所 有 特 征 值 互 不 相 同 , 则 1)( −− ASI 可 以 表 示 为
∑ −=− −
iii
pqs
ASIλ
1)( 1其中 iq 和 ip 是 A 的关于 iλ 的右特征值和左特征值,
Proof: if all eigenvalues of A are distinct , let iq be a right eigenvector of A associated with iλ ,
then [ ]nqqqQ 21= is nonsingular , and
=−
np
pp
Q
2
1
1: where is a left eigenvector of
A associated with iλ ,
∑ ∑
∑∑
∑
=−−
=
−−
=−−
=
−−
iiiiii
iiiiii
iiiii
iii
pqpqss
pqpqss
Apqpqss
ASIpqs
))((1
)(1)(1
)(1
λλ
λλλ
λ
That is ∑ −=− −
iii
pqs
ASIλ
1)( 1
3.31 find the M to meet the lyapunov equation in (3.59) with
==
−−
=33
322
10CBA what are the eigenvalues of the lyapunov equation ? is the
lyapunov equation singular ? is the solution unique ? 已知A,B,C,求M 使之满足(3.59) 的 lyapunov 方程的特征值, 该方程是否奇异>解是否唯一?
=⇒=
−
∴
=+
30
1213
MCM
CMBAM
3)det()()1)(1()det()( +=−=∆++−+=−=∆ λλλλλλλ BIjjAI BA
The eigenvalues of the Lyapunov equation are
jjjj −=+−−=+=++−= 231)231 21 ηη
The lyapunov equation is nonsingular M satisfying the equation
3.32 repeat problem 3.31 for
−
=
==
−−
=3
333
121
1021 CCBA with two
different C ,用本题给出的 A, B, C, 重复题 3.31 的问题,
αα
αanyforMCM
solutionNoCM
CMBAM
−
=⇒=
−−
⇒=
−−
∴
=+
31111
1111
2
1
1)()1()( 2 −=∆+=∆ λλλλ BA
the eigenvalues of the lyapunov equation are 01121 =+−==ηη the lyapunov equation is
singular because it has zero eigenvalue if C lies in the range space of the lyapunov equation , then solution exist and are not unique , 3.33 check to see if the following matrices are positive definite or senidefinite 确定下列矩阵
是否正定或者正半定,
−
−
332313
322212
312111
201000100
202013232
aaaaaaaaaaaaaaaaaa
1.
∴<−=−=
202013232
07921332
det is not positive definite , nor is pesitive
semidefinite
2. )21)(21(201
0010
det −−+−=
−
−λλλ
λλ
λ it has a negative eigenvalue 21− ,
so the second matrix is not positive definite , neither is positive demidefinte ,,
3 the third matrix ‘s prindipal minors ,0,0,0 332211 ≥≥≥ aaaaaa `
0det0det0det0det
332313
322212
312111
3323
3222
3313
3111
2212
2111 =
=
=
=
aaaaaaaaaaaaaaaaaa
aaaaaaaa
aaaaaaaa
aaaaaaaa
that is all the principal minors of the thire matrix are zero or positive , so the matrix is positive semidefinite , 3.34 compute the singular values of the following matrices 计算下列矩阵的奇值,
−
−=
−
−
−
−
−
−
−
5222
0110
21
012101
4221
012101
)6)(1(674)5)(2()( 2 −−=+−=−−−=∆ λλλλλλλ
the eigenvalues of
−
−
−
−
0110
21
012101
are 6 and 1 , thus the singular values of
−
−012101
are 6 and 1 ,
=
−
−20665
4221
4221
)4123
225)(41
23
225(36)20)(5()( −−+−=−−−=∆ λλλλλ the eigenvalues of
−
−4221
4221
are 4123
22541
23
225
−+ and , thus the singular values of
−4221
are 70.1)4123
225(70.4)41
23
225( 2
12
1=−=+ and
3.35 if A is symmetric , what is the ralatimship between its eigenvalues and singular values ? 对称矩阵 A 的特征值与奇异值之间有什么关系?
If A is symmetric , then 2AAAAA =′=′ LET iq be an eigenvector of A associated with
eigenvaue iλ that is , ),3,1(, niqqA iii == λ thus we have
),3,1(22 niqqAqAqA iiiiiii ==== λλλ
Which implies 2iλ is the eigenvalue of niforA ,2,12 = , (n is the order of A )
So the singular values of A are || iλ nifor ,2,1= where iλ is the eigenvalue of A
3.36 show [ ] m
n
mmn
n
n babbb
a
aa
I ∑=
+=
+1
212
1
1det
证明上式成立
let [ ]n
n
bbbB
a
aa
A 212
1
=
= A is 1×n and B is n×1
use (3.64) we can readily obtain
[ ] [ ]
∑≤
+=
+=
+
n
mmm
n
nn
n
n
ba
a
aa
bbbIbbb
a
aa
I
1
2
1
211212
1
1
detdet
3.37 show (3.65) 证明(3.65) proof: let
−−
=
=
=
n
m
n
m
n
m
IsBAIsP
IsBIsQ
IsAIsN 0
0
then we have
−−
=
−=
BAsIAssIQP
sIBsABsI
NPn
m
n
m
00
because
PsPQQPPsPNNP
detdetdet)det(detdetdet)det(
====
we have det(NP)=det(QP)
And )det()det()det()det(
)det()det()det()det(
BAsISBAsIsIQPABsISsIABsINP
nm
nm
mn
nm
−=−=
−=−=
3.38 Consider yxA = , where A is nm× and has rank m is yAAA ′′ −1)( a solution ? if not ,
under what condition will it be a solution? Is yAAA 1)( −′′ a solution ?
nm× 阶矩阵 A 秩为 m , yAAA ′′ −1)( 是不是方程 yxA = 的解? 如果不是, 那么在
什么条件下, 他才会成为该方程的解? yAAA 1)( −′′ 是不是方程的解?
A is nm× and has rank m so we know that nm ≤ , and AA′ is a square matrix of mm×
rankA=m , rank( nmrankAAA ≤=≤′ ) ,
So if nm ≠ ,then rank( AA′ )<n , 1)( −′AA does mot exist m and yAAA ′′ −1)( isn’t a
solution if yxA =
If m=n , and rankA=m , so A is nonsingular , then we have rank( AA′ )=rank(A)=m , and
A yAAA ′′ −1)( =A yyAAA =′′ −− 11 )( that is yAAA ′′ −1)( is a solution ,
RankA=M
∴ Rank( AA′ )=m AA′ is monsingular and 1)( −′AA exists , so we have
yyAAAA =′′ −1)( , that is , yAAA 1)( −′′ is a solution of yxA = ,
PROBLEMS OF CHAPTER 4 4.1 An oscillation can be generated by 一个振荡器可由下式描述：
XX
−
=0110
试证其解为：
Show that its solution is )0(cossinsincos
)( Xtttt
tX
−
=
Proof: )0()0()( 0110
XeXetXt
At
−== ,the eigenvalues of A are j,-j;
Let λββλ 10)( +=h .If teh λλ =)( ,then on the spectrum of A,then
tjtejjh
tjtejjhjt
jt
sincos)(
sincos)(
10
10
−==−=−
+==+=−ββ
ββ then
tt
sincos
1
0
==
ββ
so
−
=
−
+
=+=
tttt
ttAIAhcossinsincos
0110
sin1001
cos)( 10 ββ
)0(cossinsincos
)0()0()( 0110
Xtttt
XeXetXt
At
−
===
−
4.2 Use two different methods to find the unit-step response of 用两种方法求下面系统的单位阶
跃响应：
UXX
+
−−
=11
2210
[ ]XY 32=
Answer: assuming the initial state is zero state. method1:we use (3.20) to compute
−+
++=
+−
=−−
−
ss
ssss
AsI2
1222
122
1)( 2
11
then tAt ettt
tttAsILe −−−
−−
+=−=
sincossin2sinsincos
))(( 11 and
)22(5
)22(5)()()( 22
1
++=
++=−= −
sssssssBUAsICsY
then tety t sin5)( −= for t>=0
method2:
t
tt
tt
Att tA
t tA
te
tetetete
C
BeeCABdeC
dBueCty
−
−−
−−
−−
−
=
−−−+
−−=
−==
=
∫∫
sin5
1sin3cos1sin2cos
015.01
)(
)()(
01
0
)(
0
)(
t
tt
t
t
for t>=0 4.3 Discretize the state equation in Problem 4.2 for T=1 and T=π .离散化习题 4.3 中的状态方
程，T 分别取 1 和π Answer:
][][][
][][]1[0
kDUkCXkY
kBUdekXekXT AAT
+=
+=+ ∫ αα
For T=1,use matlab: [ab,bd]=c2d(a,b,1) ab =0.5083 0.3096 -0.6191 -0.1108 bd =1.0471 -0.1821
[ ] ][32][
][1821.0
0471.1][
1108.06191.03096.05083.0
]1[
kXkY
kUkXkX
=
−
+
−−
=+
for T=π ,use matlab: [ab,bd]=c2d(a,b,3.1415926) ab =-0.0432 0.0000 -0.0000 -0.0432 bd =1.5648 -1.0432
[ ] ][32][
][0432.1
5648.1][
0432.0000432.0
]1[
kXkY
kUkXkX
=
−
+
−
−=+
4.4 Find the companion-form and modal-form equivalent equations of 求系统的等价友形和模式
规范形。
UXX
+
−−
−=
101
220101002
[ ]XY 011 −=
Answer: use [ab ,bb,cb,db,p]=canon(a,b,c,d,’companion) We get the companion form ab = 0 0 -4 1 0 -6 0 1 -4 bb = 1 0 0 cb = 1 -4 8 db =0 p =1.0000 1.0000 0 0.5000 0.5000 -0.5000
0.2500 0 -0.2500
UXX
+
−−−
=001
410601400
[ ]XY 841 −=
use use [ab ,bb,cb,db,p]=canon(a,b,c,d) we get the modal form ab = -1 1 0 -1 -1 0 0 0 -2 bb = -3.4641 0 1.4142 cb = 0 -0.5774 0.7071 db = 0 p = -1.7321 -1.7321 -1.7321 0 1.7321 0 1.4142 0 0
UXX
−+
−−−
−=
4142.104641.3
200011011
[ ]XY 7071.05774.00 −=
4.5 Find an equivalent state equation of the equation in Problem 4.4 so that all state variables have their larest magnitudes roughly equal to the largest magnitude of the output.If all signals are required to lie inside 10± volts and if the input is a step function with magnitude a,what is the permissible largest a?找出习题 4.4 中方程的等价状态方程使所有状态变量的最大量几乎等于
输出的最大值。如果所有的信号需要在 10± 伏以内，输入为大小为 a 的的阶跃函数。求所
允许的最大的 a 值。 Answer: first we use matlab to find its unit-step response .we type
a=[-2 0 0;1 0 1;0 -2 -2]; b=[1 0 1]';c=[1 -1 0]; d=0; [y,x,t]=step(a,b,c,d) plot(t,y,'.',t,x(:,1),'*',t,x(:,2),'-.',t,x(:,3),'--')
0 1 2 3 4 5 6-0.6-0.5-0.4-0.3-0.2-0.1
00.10.20.30.40.50.60.70.80.9
11.11.2
x2
x1
x3
y
so from the fig above.we know max(|y|)=0.55, max(|x1|=0.5;max(|x2|)=1.05,and max(|x3|)=0.52
for unit step input.Define ,,5.0, 332211 xxxxxx === then
UXX
+
−−
−=
101
2405.005.0
002
[ ]XY 021 −=
the largest permissible a is 10/0.55=18.2
4.6 Consider Ubb
XX
+
=
2
1
00λ
λ [ ]XccY 11=
where the overbar denotes complex conjugate.Verify that the equation can be transformed into
XCyuBXAX 1=+=
with [ ])Re(2)Re(21010
11111 cbcbCBA λλλλλ
−=
=
+−
= ,by using the
transformation XQX = with
−−
=11
11
bbbb
Qλλ
。试验证以上变换成立。
Answer: let XQX = with
−−
=11
11
bbbb
Qλλ
,we get
Ubb
XQXQ
+
=
2
1
00λ
λ [ ] XQccY 11=
−−
−=−
11
11
11
1
)(1
bbbb
bbQ
λλλλ
so
UBXAUX
Ubbbb
Xbbbb
bbbb
bb
Ubb
bbbb
bbX
bbbb
bbbb
bb
Ubb
QXQQX
+=
+
+−
=
−−
+
−−
−−
−=
−−
−+
−−
−−
−=
+
= −−
1010
)(0
)(1
)(1
)(1
00
)(1
00
111111
11
12
12
11
11
2
1
11
11
1111
11
11
11
11
2
111
λλλλ
λλλλλλ
λλλλ
λλ
λλλλλλ
λλ
λλλλ
λλ
[ ] [ ] [ ] XCXbcbccbcbXbbbb
ccXQccY 11111111111
111111 =+−−=
−−
== λλλλ
4.7 Verify that the Jordan-form equation
[ ]Xccccccy
U
bbbbbb
XX
321321
3
2
1
3
2
1
11
01
1
=
+
=
λλ
λλ
λλ
can be transformed into
[ ]XCCCyuBBB
XAIA
IAX 3212
2
000
0=
+
=
where iCandBA ,, are defined in Problem 4.6 and 2I is the unit matrix of order 2.
试验证变换成立，其中 iCandBA ,, 是习题 4.6 中定义的形式。 2I 为二阶单位阵。
PROOF: Change the order of the state variables from [x1 x2 x3 x4 x5 x6]’ to [x1 x4 x2 x5 x3 x6]’ And then we get
[ ] [ ]XCCCXccccccy
uBBB
XAIA
IA
bbbbbb
xxxxxx
xxxxxx
X
321332211
3
2
1
2
2
3
3
2
2
1
1
6
3
5
2
4
1
6
3
5
2
4
1
000
0
11
11
==
+
=
+
=
=
λλ
λλ
λλ
4.8 Are the two sets of state equations
[ ]XyUXX 011011
100220212
−=
+
=
and
[ ]XyUXX 011011
100120112
−=
+
−=
equivalent?Are they zero-state equivalent?上面两组状态方程等价吗？它们的零状态等价吗？ Answer:
[ ]
[ ]
2
22
1
11
)2(1
011
)2(00)2(2)1)(2(0)1(2)1()1)(2(
)1()2(011
011
100220212
011)()(ˆ
−=
−−−−−−−−
−−−
=
−−−−−−
−=−=
−
−
s
ssssssss
ss
ss
sBAsICsG
[ ]
[ ]
2
22
1
12
)2(1
011
)2(00)2()1)(2(0)1()1()1)(2(
)1()2(011
011
100120112
011)()(ˆ
−=
−−+−−++−
+−−
=
+−−−−−
−=−=
−
−
s
ssssssss
ss
ss
sBAsICsG
obviously, )(ˆ)(ˆ21 sGsG = ,so they are zero-state equivalent but not equivalent.
4.9 Verify that the transfer matrix in (4.33)has the following realization:
U
NN
NN
X
III
IIII
X
r
r
qr
qqr
+
−−
−−
=
−− 1
2
1
1
2
1
00000
0000
αα
αα
[ ]XIY q 000 =
This is called the observable canonical form realization and has dimension rq.It is dual to (4.34). 验证式(4.33)具有如上的实现。这叫做能观标准型实现，维数为 rq。它与式(4.34)对偶。 Answer:
)()(
1)(
)(1,,
)(,
)(
2,2,
)]([
][)(
111
2
2
1
1
11
1
11
1
21
211
rr
qrq
r
q
r
r
ii
iq
r
iiiq
iiii
r
r
NNssd
BAsIC
then
Isd
ZIsd
sZIsd
sZ
thens
ZIZIsZ
ris
ZZriAZsZ
getweAsIZZZC
thenZZZAsIC
define
++=−
===
−=−=
==⇒==
−=
=−
−−
−−
=−
=
−
−
∑∑
αα
this satisfies (4.33). 4.10 Consider the 21× proper rational matrix 考虑如下有理正则矩阵
[ ]
[ ]42322
223
1241312
213
11
432
23
1421
1)(ˆ
ββββββββ
αααα
++++++×
+++++=
ssssssssss
ddsG
Show that its observable canonical form realization can be reduced from Problem 4.9 as 试证习题 4.9 种系统的能观标准形实现能降低维数如下表示：
+
−−−−
=
4241
3231
2221
1211
4
3
2
1
000100010001
ββββββββ
αααα
XX
[ ] [ ]UddXy 210001 +=
Answer: In this case,r=4,q=1,so
][],[],[],[,1 42414323132221212111 ββββββββ ===== NNNNI q
its observable canonical form realization can be reduced from 4.9 to 4.10 4.11 Find a realization for the proper rational matrix求下面正则有理传递函数矩阵的一种实现。
+
−−−
+
−−++
=
+
+−
+−
++−
+=
++−
++−
+=
1100
2634
2322
231
1100
22
13
)2)(1(32
12
212
)2)(1(32
12
)(ˆ
2 sss
ss
sss
s
ss
ss
sss
ssG
so the realization is 4.12 Find a realization for each column of
)(ˆ sG in Problem
4.11,and then
connect them.as shown in fig4.4(a).to obtain a realization of )(ˆ sG .What is the dimension of this
realization ?Compare this dimension with the one in Problem 4.11.求习题 4.11 中 )(ˆ sG 每一列的
实现，在如图 4.4(a)所示把它们连结得到 )(ˆ sG 的一种实现。比较其与习题 4.11 的维数。
Answer:
222
222
2
111
111
1
10
2232
01
0123
10
23
22
)2)(1(1
10
2232
)2)(1(1)(ˆ
10
32
10
32
11
22
11)(ˆ
UXY
UXX
ssss
sss
sG
UXY
UXX
ssssG
C
C
+
−−−
=
+
−−=
+
−−
+
−++
=
+
−−−
++=
+
−
=
+−=
+
−+
=
−+
=
These two realizations can be combined as
UXY
UXX
+
−−−−−
=
+
−−
−−
=
1100
26233422
00001001
001000012030
0203
UXY
UXX
+
−−−−
=
+
−−
−=
1100
223322
001001
010230
001
the dimension of the realization of 4.12 is 3,and of 4.11 is 4.
4.13 Find a realization for each row of )(ˆ sG in Problem 4.11 and then connect them,as shown in
Fig.4.4(b),to obtain a realization of )(ˆ sG .What is the dimension of this realization of this
realization?Compare this dimension with the ones in Problems 4.11 and 4.12. 求习题 4.11 中
)(ˆ sG 每一行的实现，在如图 4.4(b)所示把它们连结得到 )(ˆ sG 的一种实现。比较其与习题
4.11、4.12 的维数。 Answer:
[ ] [ ] [ ][ ]
[ ] [ ]
[ ] [ ] [ ] [ ][ ] [ ]
[ ] [ ] 222
222
2
111
111
21
11012623
0213
112623)2)(1(
111)1(2)2(3)2)(1(
1)(ˆ
000134
220213
342223
13242)2)(1(
1)(ˆ
UXY
UXX
sss
ssss
sG
UXY
UXX
sss
ssss
sG
C
C
+=
−−−−
+
−−
=
+−−+−−++
=++−+−++
=
+=
−
+
−−
=
−+++
=−+++
=
These two realizations can be combined as
UXY
UXX
+
=
−−−−−
+
−−
−−
=
1100
01000001
262334
22
0200130000020013
the dimension of this realization is 4,equal to that of Problem 4.11.so the smallest dimension is of Problem 4.12.
4.14 Find a realization for
++
++−
=3432322
343)612()(ˆ
ss
sssG
求
++
++−
=3432322
343)612()(ˆ
ss
sssG 的一种实现
Answer:
[ ] [ ]
[ ][ ]UXY
UXX
sss
sssG
3/224
9/6793/1303
34
9/6793/1303/34
13/2243432322
343)612()(ˆ
−+=
−+−=
−+
+−=
++
++−
=
4.16 Find fundamental matrices and state transition matrices for 求下列状态方程的基本矩阵
和状态转移矩阵：
Xe
XandXt
Xt
−−
=
=
101
010 2
Answer:
for the first case: 25.0
22222
10 2121
)0()()(ln
)0()()(t
t
extxtdttxdtxx
xdttxtxxx
=⇒=⇒=
+=⇒= ∫
we have
=⇒
=
=⇒
= ∫
2
2
5.00
5.0
)(10
)0(01
)(01
)0(t
t d
e
etXXandtXXtt
=
=Φ
=
−
−−
∫∫∫
∫
)(5.0
5.05.01
5.00
5.0
5.00
5.0
0
5.00
5.0
20
20
220
20
0 2
2
2
2
2
0
101
01
),(
01
)(
t.thusindependenlinearly are states initial twoThe
tt
t
t
ddt
t
t d
t
t d
t
t d
e
ee
e
e
e
ett
ande
etX
ttttttt
tt
for the second case:
t
tttt dttdt
tt
extxxx
exeexcdteexetx
xextxexxt
−
−−−
=⇒−=
+−=+∫∫=
⇒+−=+−=
∫)0()(
)0())(0(5.0])0([)(
)0()(
2222
120 21
2122
11
0
we have
−=⇒
=
=⇒
=
−
−−
t
ttt
eee
tXXande
tXX)(5.0
)(10
)0(0
)(01
)0(
the two initial states are linearly independent;thus
−+−=
−
−=Φ
−=
−
−−−−
−
−
−−
−
−−
−
−−
tt
tttttttt
t
ttt
t
ttt
t
ttt
eeeeeeee
eeee
eeee
tt
ande
eeetX
0
0000
0
000
0)(5.0)(5.0
)(5.00
)(5.00
),(
)(5.00
)(
3
1
0
4.17 Show ).(),(/),( 00 tAttttt Φ−=∂Φ∂
试证 ).(),(/),( 00 tAttttt Φ−=∂Φ∂
Proof: first we know
)(),(/),(0
),(),()(
),(),(),(),()(
),(),(),(
),()),(),((),(
),()(),(
00
00
100
100
000
000
00
tAttttt
ttt
tttAt
tttttttttA
ttt
ttttt
ttt
ttttt
ttthen
tttAt
tt
Φ−=∂Φ∂⇒=
∂Φ∂
Φ+=∂
Φ∂Φ+ΦΦ=
∂Φ∂
Φ+Φ∂
Φ∂=
∂ΦΦ∂
=∂
Φ∂
Φ=∂
Φ∂
−−
4.18 Given
=
)()()()(
)(2221
1211
tatatata
tA ,show
+=Φ ∫ ttt daatt
t
t))()((exp),(det 22110
0
已知 A(t),试证
+=Φ ∫ ttt daatt
t
t))()((exp),(det 22110
0
Proof:
ttt
ttt
φφφφφφφφφφφφφφφφ
φφφφφφφφφφφφ
φφφφ
φφφφ
daa
daa
t
t
t
t
ettthen
cgetwe
Ittwithcett
soaaaa
aaaaaaaatt
so
aaaa
∫=Φ
=
=Φ∫=Φ
Φ+=−+=+−+++−+=
−+−=−∂∂
=Φ∂∂
=
+
+
02211
02211
))()((
0
00
))()((
0
2211122122112211
2122112112222212211121221212112221121111
122122111221221112212211
2221
1211
2221
1211
2221
1211
),(det
1
),(),(det
det)())(()()()()(
)()(det
4.20 Find the state transition matrix of 求状态转移矩阵
xt
tx
−
−=
cos00sin.
Answer:
t
t
extxtxxextxxtx
sin2222
1cos1111
)0()(cos
)0()(sin−
−
=⇒−=
=⇒⋅−=
then we have
=⇒
=
=⇒
= −
+−
t
t
etXXand
etXX sin
cos1 0)(
10
)0(0
)(01
)0(
the two initial states are linearly independent;thus
=
=Φ
=
+−
−−
−
+−
−
+−
−
+−
0
0
0
0
sinsin
coscos1
sin
cos1
sin
cos1
0
sin
cos1
00
00
00
),(
00
)(
tt
tt
t
t
t
t
t
t
ee
ee
ee
tt
ande
etX
4.21 Verify that BtAtCeetX =)( is the solution of 试验证 BtAtCeetX =)( 为下面方程的解：
CxXBAXX =+= )0(
Verify:first we know AeAeet
AtAtAt ==∂∂
,then
DEQCCeeX
XBAXBeCeCeAeCeet
X BtAtBtAtBtAt
..)0(
)()()(
00 ==
+=+=∂∂
=
4.23 Find an equivalent time-invariant state equation of the equation in Problem 4.20. 求习题 4.20 中方程的等价时不变状态方程。
Answer: let )()()(0
0)()( sin
cos11 txtPtxand
ee
tXtP t
t
=
==
−−
Then 0)(0)( == txtA
4.24 Transform a time-invariant ),,( CBA into ))(),(,0( tCtB is by a time-varying equation
transformation.求把(A,B,C)转换成 ))(),(,0( tCtB 的时变状态转移方程。
Answer: since (A,B,C)is time-invariant,then
AtetX =)(
)(0
10
1
1
0)()(),(
)()()()()()(
ttA
At
At
etXtXttCetCXtC
BeBtXtBtXtP
−−
−−
−
==Φ
==
==
=
4.25 Find a time-varying realization and a time-invariant realization of the inpulse response
tettg λ2)( = .求冲击响应的时变实现及时不变实现。
Answer:
[ ]
−=
+−=−=−=
−
−
−
−−
λt
λt
λt
λλλ
λtλtλ
tt
ttttt
eee
eteet
ettettgtg
ttt
tt
2
2
22)(2
2
)2()()(),(
the time-varying realization is
[ ] )(2)(
)()(000000000
)(
2
2
txeteetty
tuet
tee
txtx
ttt
t
t
t
λλλ
λ
λ
λ
−=
+
=
−
−
−
[ ]xy
tuxx
nrealizatioinvianttimethegetwegusss
etLsg t
200
)(001
010001
33
:),41.4(sin33
2][)(ˆ
32
3232
=
+
−=
−−+−
==
λλλ
λλλλ
4.26 Find a realization of tt t cos)(sin),( )( −−= tettg . Is it possible to find a time-invariant
state equation realization?求 tt t cos)(sin),( )( −−= tettg 的一种实现，问能否找到一种时不变
状态方程实现。 Answer: clearly we can get the time-varying realization of
)()()cos)((sincos)(sin),( )( tttt tt NtMeteettg tt === −−− using Theorem 4.5
we get the realization.
)(sin)()(cos)()()(
txtetyttuetutNtx
t
t
−=
==
but we can’t get g(t) from it,because )(),( tt −≠ tgtg ,so it’s impossible to find a
time-invariant state eqution realization.
5.1 Is the network shown in Fing5.2 BIBO stable ? If not , find a bounded input that will excite an unbound output 图 5.2 的网络 BIBO 是否稳定?如果不是.请举出 一般激励无界输出的有界输入. From Fig5.2 .we can obtain that
xyyux
=−=
Assuming zero initial and applying Laplace transform then we have
1)(ˆ)(ˆ
)(ˆ2 +
==s
ssusysy
ts
sLsgLtg cos]1
[)]([)( 211 =
+== −−
because
∫ ∫ ∫ ∑∫∞ ∞ ∞
=
π+
π+
+−+==0 0
20
0
232
212
1 cos)1(cos|cos||)(|x
k
k
kk tdttdtdttdttg
=1+∑∞
=
+ π+π−π+π−0
21
231 )sin()2[sin()1(
k
k k
=1+∑∞
=
+ π−π−−0
21
231 ]sin[sin)1()1(
k
kk
=1+∑∞
=
++ −⋅−−0
1212 2)1()1(k
kk ）（
=1+2∑∞
=01
k
=∞
Which is not bounded .thus the network shown in Fin5.2 is not BIBO stable ,If u(t)=sint ,then we have
y(t)= ∫ ∫ τ−τ=τττ−t t
tdt0 0
)sin(cossin)cos( τd = ∫ =τt
tttd0
sin21sin
21
we can see u(t)is bounded ,and the output excited by this input is not bounded
5.2 consider a system with an irrational function )(ˆ sy .show that a necessary condition for the system to be BIBO stable is that |g(s)| is finite for all Res≥ 0 一个系统的传递函数是 S 的非有理式。证明该系统 BIBO 稳定的一个必要条件是对所有 Res≥ 0. |g(s)|有界。 Proof let s= ,ζ+λ j then
∫∞ ζ−λ− ==ζ+λ=
0)()(ˆ)(ˆ dteetgjgsg tjt ∫ ∫
∞ ∞ λ−λ− ζ−0 0
sin)(cos)( tdtetgjtdtetg tt
If the system is BIBO stable ,we have
+∞<≤∫∞
0|)(| Mdttg . for some constant M
which implies shat ∫∞
0)( dttg is convergent .
For all Res= ,0≥λ We have
|g(t) |cos te t ζλ− |)(| tg≤
|g(t0 |)(||sin tgte t ≤ζλ−
so ∫∞ λ−
0cos)( tdtetg t and ∫
∞ λ− ζ0
sin)( tdtetg t are also convergent , that is .
∫∞ λ− +∞<=ζ
0cos)( Ntetg t
∫∞ λ− +∞<=ζ
0sin)( Ltdtetg t
where N and L are finite constant. Then
| 21
2221
22 )())((|)(ˆ LNLNsg +=−+= is finite ,for all Res 0≥ .
Another Proof, If the system is BIBO stable ,we have .|)(|0
+∞<≤∫∞
Mdttg for some constant M .And
.
For all Res 0≥ . We obtain | =|)(ˆ sg ∫ ∫∞ ∞− +∞<≤≤
0 0
)(Re |0(|)(| Mdttgdtetg ts .
That is . | |)(ˆ sg is finite for all Res 0≥ 5.3
Is a system with impulse g(t)=1
1+t
BIBO stable ? how about g(t)=t te− for t ?0≥
脉冲响应为个 g(t)=1
1+t
的系统是否 BIBO 稳定? 0,)( ≥= − ttetg t 的系统又如何?
Because ∞=+=+
=+∫ ∫
∞ ∞)1ln(
11|
11|
0 0ttd
tdt
t
∫ ∫∞ ∞ ∞−−−− +∞<=−−==
0 0 0 1|)(|| tttt etedttedtte
the system with impulse response g(t)=1
1+t
is not BIBO stable ,while the system with impulse response g(t)= tte−
for 0≥t is BIBO stable.
5．4 Is a system with transfer function )1(
)(ˆ2
+=
−
sesg
s
BIBO stable
一个系统的传递函数为 )1(
)(ˆ2
+=
−
sesg
s
，问该系统是否 BIBO 稳定。
Laplace transform has the property . If ).()( atftg −= then )(ˆ)(ˆ sfesg sα−=
We know 0,]1
1[1 ≥=+
−− tes
L t . Thus 2,]1
1[ )2(1 ≥=+
−−α−− tes
eL ts
So the impulse response of the system is )2()( −−= tetg . for 2≥t
∫∫∞ −−∞
+∞<===2
)2(
01|)(| dtedttg t
the system is BIBO stable
5.5 Show that negative-feedback shown in Fig . 2.5(b) is BIBO stable if and only if and the gain a has a magnitude less than 1.For a=1 ,find a bounded input )(tr that will excite an unbounded output. 证明图 2.5(b) 中的负反馈系统 BIBO 稳定当且增到 a 的模小于 1。对于 a=1 情况，找出一有界输入 r(t), 她所产
生的输出是无界的。 Poof ,If r(t)= )(tδ .then the output is the impulse response of the system and equal
+−δ−−δ−−δ= )4()3()1()( 43 taratatg =∑∞
=
+ −δ−1
1 )()1(t
ii ita
the impulse is definded as the limit of the pulse in Fig.3.2 and can be considered to be positive .thus we have
∑∞
=
−δ=1
)(|||)(|t
i itatg
and ∫ ∑ ∫ ∑∞ ∞
=
∞ ∞
=
∞
∞+<−
==−δ=0
10
1 |)|1(||||)(|||)(|
t t aa
ii adtitadttg 1||0||
<≥
aifaif
which implies that the negative-feedback system shown in Fig.2.5(b) is BIBO stable if and only if the gain a has magitude less than 1. For 1=a .if we choose ).sin()( π= ttr clearly it is bounded the output excited by )sin()( π= ttr is
∑
∫ ∑ ∑ ∑ ∑∞
=
∞
+
∞
=
∞
=
∞
=
++++
π−=
π−−=π−π−=π−π−=π−π−=τττ−=
1
01 1 1 1
1111
1)sin(
)sin()1()1()sin()1()sin()1()sin()1()()()(
i
t
i i i i
iiiii
t
titititdrtgty
And y(t) is not bounded
5.6 consider a system with transfer function )1()2()(ˆ
+−
=sssg 。 what are the steady-state responses by
u ( t )=3.for 0≥t .and by u ( t )=sin2t. for ?0≥t
一个系统的传递函数是)1()2()(ˆ
+−
=sssg ，分别由 0.3)( ≥= ttu 和 0.2sin)( ≥= tttu
所产生的稳态响应是什么？
The impulse response of the system is 0.3)(]1
3[`]12[)( 11 ≥−δ=
+−=
+−
= −−− tets
LssLtg t
。
And we have
∫ ∫∫ ∫ ∫∞ ∞ −∞ ∞ ∞ −− +∞<=+=+δ=+δ≤−δ=
0 00 0 04313)()3||)((||3)(||)(| dtedttdtetdtetdttg ttt
so the system is BIBO stable using Theorem 5.2 ,we can readily obtain the steady-state responses if u ( t )=3for 0≥t .,then as )(, tyt ∞→ approaches 6323)0(ˆ −=⋅−=⋅g if u ( t )=sin2t for ,0≥t then ,as )(. tyt ∞→ approaches
)25.1sin(26.1)3arctansin(5102)(ˆsin(|)(ˆ| +τ=+τ=τ∠+ττ ttjgtjg ）
5.7 consider =x ux
−+
−02
10101
y=[ -2 3] ux 2− is it BIBO stable ? 问上述状态方程描述的系统是否 BIBO 稳定？ The transfer function of the system is
[ ] [ ] 202
110
)1)(1(10
11
32202
10101
32)(ˆ1
−
−
−
−++−=−
−
−≤
−+−=
−
s
sssssg
=1
2221
4++−
=−+ s
ss
The pole of is –1.which lies inside the left-half s-plane, so the system is BIBO stable 5.8 consider a discrete-time system with impulse response
kkkg )8.0(][ = for ,0≥k is the system BIBO stable ?
一个离散时间系统的脉冲响应序列是 kkkg )8.0(][ = ， ,0≥k 。问该系统是否 BIBO 稳定？ Because
∑ ∑ ∑∞
=
∞
=
∞
=
⋅−==0 0 0
)8.0()8.01(2.0
1)8.0(|][|k k k
kk kkkg =
∑ ∑∑∞
=
∞
=
∞
=
+ +∞<=−
⋅==⋅−⋅0 00
1 208.01
8.02.0
18.02.0
1])8.0()8.0([2.0
1k k
k
k
kk kk
G[k]is absolutely sumable in ∞,0[ ]. The discrete-time system is BIBO stable . 5.9 Is the state equation in problem 5.7 marginally stable? Asymptotically stable ? 题 5.7 中的状态方程
描 述的系统是否稳定？是否渐进稳定？ The characteristic polynomial
)1)(1(10
101det)det()( −λ+λ=
−λ
−+λ=−λ=λ∆ AI
the eigenralue Ihas positive real part ,thus the equation is not marginally stable neither is Asymptotically stable.
5.10 Is the homogeneous stable equation xx
−=
000000101
.marginally stable ? Asymptotically stable?
齐次状态方程 xx
−=
000000101
，是否限界稳定？
The characteristic polynomial is )1(00
00101
det)det()( 2 +λλ=
λλ
−+λ=−λ=λ∆ AI
And the minimal polynomial is )1()( +λλ=λψ . The matrix has eigenvalues 0 , 0 . and –1 . the eigenvalue 0 is a simple root of the minimal .thus the equation is marginally
stable The system is not asymptotically stable because the matrix has zero eigenvalues
5 。11 Is the homogeneous stable equation xx
−=
000000101
marginally stable ? asymptotically stable ?
齐次状态方程 xx
−=
000000101
是否限界稳定？是否渐进稳定？
The characteristic polynomial ）（ 100
10101
det)( 2 +λλ=
λ−λ−+λ
=λ∆ .
And the minimal polynomial is )1()( 2 +λλ=λψ ..the matrix has eigenvalues 0. 0. and –1 . the eigenvalue 0 is a repeated root of the minimal polymial ).(λΨ so the equation is not marginally stable ,neither is asymptotically stable . 5.12 Is the discrete-time homogeneous state equation
][100010109.0
]1[ kxkx
−−
=+
marginally stable ?Asymptotically stable ?
离散时间齐次状态方程 ][100010109.0
]1[ kxkx
−−
=+ 是否限界稳定？是否渐进稳定？
The characteristic of the system matrix is ),9.01()( 2 −λ−λ=λ∆ （） .and the minimal polynomial is ),9.01()( −λ−λ=λψ （） the equation is not asymptotically stable because the matrix has eigenvalues 1. whose magnitudes equal to 1 . the equation is marginally stable because the matrix has all its eigenvalues with magnitudes less than or to 1 and the eigenvalue 1 is simple root of the minimal polynomial . )(λψ .
5.13 Is the discrete-time homogeneous state equation
][100010109.0
]1[ kxkx
−−
=+ marginally stable ?
asymptotically stable ?
离散时间齐次状态方程 ][100010109.0
]1[ kxkx
−−
=+ 是否稳定？是否渐进稳定？
Its characteristic polynomial is ),9.01()( 2 −λ−λ=λ∆ （） and its minimal polynomial is
),9.01()( 2 −λ−λ=λψ （） the matrix has eigenvalues 1 .1 .and 0.9 . the eigenvalue 1 with magnitude equal to 1 is a repeated root of the minimal polynimial )(λψ . .thus the equation is not marginally stable and is not asymptotically stable .
5.14 Use theorem 5.5 to show that all eigenvalues of
−−
=15.0
10A have negative real parts .
用定理 5.5 证明 A 的特征值都具有负实部.
Poof :For any given definite symmetric matrix N where
=
cbba
N with a>0 .and 02 >− bac
The lyapunov equation NMAMA −=+′ can be written as
−−−−
=
−−
+
−−
cbba
mmmm
mmmm
15.010
115.00
2221
1211
2221
1211
that we have
−−−−
=
−+−−−−+−
cbba
mmmmmmmmmmm
222112222111
2212112112
25.05.0)(5.0
thus
camamm
bcam
5.0
25.05.1
22
2112
11
+===
−+=
=
2221
1211
mmmm
M is the unique solution of the
lyapunor equation and M is symmetric and 0
02 >−
>
baca
⇒00
>>
ca
025.05.1
0,0)25.05.1(
0]32)2[(161)436(
161]16)6[(
161)25.05.1(
11
22
22222
>−+=⇒
>>>≥+
≥+−=−+=−+=−+∴
bcamca
bacca
acaaccaaccaacca
221122211
2221
1211 )5.)(25.05.1(det acabcammmmmmmm
−−+−+=−=
0)](8)2()22[(
81
)4884(81
222
22
>−+−+−=
−−++=
bacbcba
bcabcaca
we see is positive definite because for any given positive definite symmetric matrix N, The lyapunov equation NMAMA −=+′ has a unique symmetric solution M and M is positive definite , all eigenvalues of A have negative real parts , 5.15 Use theorem 5.D5show that all eigencalues of the A in Problem 5.14 magnitudes less than 1. Poof : For any given positive definite symmetric matrix N , we try to find the solution of discrete Lyapunov equation M
NMAAM =′− Let .
=
cbba
N where a>0 and 02 >− bac
and M is assumed as
=
2221
1211
mmmm
M ,then
=
++−−+−+−
=
−−
−−
−
=′−
cbba
mmmmmmmmmmm
mmmm
mmmm
MAAM
211211221221
2221122211
2221
1211
2221
1211
)50)(5.025.0
15.010
115.00
（。
this bcambcammbcam5
165
125
12,52
54
54,
54
53
58
22211211 −+=−+==−+=
=
2221
1211
mmmm
M is the unique symmetric solution of the discrete Lyapunov equation
,NMAAM =′− And we have
>>
⇒
−>
++=−+
000
3296416)38(
2
222
caa
baca
accaacca
bacca 1616)38( 2 >>+⇒ 0438 >−+⇒ bca 054
53
58
11 >−+=⇒ bcam
0)](5)2()22[(54
)48534(54
])244()161212)(4(38[(251det
222
222
221122211
2221
1211
>−+−+−=
−−+++=
−+−−+−+=−=
bacbcba
bcabacbca
bcabcabammmmmmmm
we see M is positive definite .use theorem 5.D5,we can conclude that all eigenvalues of A have magnitudes less than 1. 5.16 For any distinct negative real iλ and any nonzero real ia ,show that the matrix
λ−
λ+λ−
λ+λ−
λ+λ−
λ−
λ+λ−
λ+λ−
λ+λ−
λ−
=
3
23
23
23
13
13
32
32
2
22
12
12
31
31
21
21
1
21
2
2
2
aaaaa
aaaaa
aaaaa
M is positive definite .
正明上定义的矩阵 M 是正宗的 （ iλ 为相异的负实数， ia 为非零实数）。
Poof : .3,2,1, =λ ii are distinct negatire real numbers and ia are nonzero real numbers , so we have
012
),1(21 >λ
−a
(2) . ）22121
22
212
21
22
21
21
22
21
2
22
12
12
21
21
1
21
)(1
41(
)(42
2detλ+λ
−λλ
=λ+λ
−λλ
=
λ−
λ+λ−
λ+λ−
λ−
aaaaaaaaa
aaa
22121
212
21
2212121
22
2121
221
)(1
414
424
λ+λ>
λλλλ>λ+λ∴
λ−λ=λλ−λλ+λ+λ=λλ−λ+λ
and）（
〉）（）（
λ−
λ+λ−
λ+λ−
λ−
2
22
12
12
21
21
1
21
2
2detaaa
aaa
= 22121
22
21 )(
14
1(λ+λ
−λλ
aa ) >0
(3)
λ+λλ
−λλ+λλ+λ
λ−
λ+λ
λ+λλ+λλ
−λ+λλ+λ
λ−
λ
λ+λλ+λλ
−=
λ−
λ+λ−
λ+λ−
λ+λ−
λ−
λ+λ−
λ+λ−
λ+λ−
λ−
−=
221
1
33121
1
32
3121
1
322
12
3
2
31211
23
22
21
32313
32212
31211
23
22
21
22
1))(210
))(212
210
1121
det
2111
12
11
1121
det
）（（
（）（
）
）
aaa
aaaM
0det
.0))(
4
,0)(
,0.04
1,021
))(4
)(1
)(41det
21
))(212
212
21det
21
3213121
1
312
1
213
1
213
1
32
23
22
21
1
3213121
12
32312
1
213
1
32
23
22
21
1
2
3121
1
322
13
1
32
12
1
2
23
22
21
1
<∴
>λ+λλ+λλ+λ
λ
>λ+λλ
λ−>
λ+λλλ
−>λ+λλ
λ−
λλ>
λ−
λ+λλ+λλ+λ
λ+
λ+λ−
λ+λλλ
−λ+λλ
λ−
λλλ−=
λ+λλ+λ
λ−
λ+λ−
λ+λλ
−λλ+λ
λ−
λλ−=
=
=
M
aaa
aaa
aaa
）（）（
）（）（
）（）（）（
））（
（））（
）（）（
（
From (1), (2) and (3) , We can conclude that M is positive definite 5.17 A real matrix M (not necessarily symmetric ) is defined to be positive definite if 0>′ XMX for any non zero X .Is it true that the matrix M is positive definite if all eigenvalues of Mare real and positive or you check its positive definiteness ? 一个实矩阵 M 被定义为正定的，如果对任意非零 X 都有 0>′ XMX 。如果 M 的所有特征值都
是正实数，是不是就有 M 正定/ 或者如果 M 的所有主子式都是正，M 是不是正定呢？如果不
是，要如何确定它的正定性呢？ If all eigenvalues of M are real and positive or if all its leading pricipal minors are positive ,the matrix M may not be positive definite .the exampal following can show it .
Let
−=
1081
M , Its eigenvalues 1,1 are real and positive and all its leading pricipal minors are
positive , but for non zero [ ] ′= 21X we can see 011<−=′MXX . So M is not positive
definite .
Because MXX ′ is a real number . we have XMXXMXXMX ′=′′=′′ )( , and
XMMXXMXXMXXMX )]21[
21
21 ′+′=′′+′=′
（
This ifonlyandifXMX ,0>′ XMMX )]21[ ′+′（ >0, that is , M is positive and the matrix
)21 MM ′+（ is symmetric , so we can use theorem 3.7 to cheek its positive definiteness .
5.18 show that all eigenvalues of A have real parts less than 0<µ− if and only if ,for any given positive definite symmetric matrix N , the equation NMMAMA −=µ++′ 2 has a unique symmetric solution M and M is positive definite . 证明 A 的所有特征值的实部小于 0<µ− 当且对任意给定的正定对称阵 N。方程
NMMAMA −=µ++′ 2 有唯一解 M，且 M 是正定的。 Proof : the equation can be written as NIAMIA −=µ++′µ+ )()( Let IAB µ+= , then the equation becomes NMBMB −=+′ . So all eigenvalues of B have megative parts if and only if ,for any given positive definite symmetric matrix N, the equation
NMBMB −=+′ has a unique symmetric solution m and is positive definitive . And we know IAB µ+= , so ( )AIIAIBI −µ−λ=µ−−λ=−λ )(det)det()det( , that is , all eigenvalues of A are the eigenvalues of B subtracted µ− . So all eigenvalues of A are real parts less than 0<µ− .if and only all eigenvalues of B have negative real parts , eguivalontly if and only if , for any given positive symmetric matrix N the equation NMMAMA −=µ++′ 2 has a unique symmetric solution M and M is positive definite . 5.19 show that all eigenvalues of A have magnitudes less than ρ if and only if , for any given
positive definite symmetric matrix N ,the equation NMAAM 22 ρ=′−ρ . Has a unique symmetric solution M and M is positive definite . 证明 A 的所有特征值的模都小于 ρ 当且仅当对任意给定的对称正定 N ， 方程
NMAAM 22 ρ=′−ρ 有唯一解 M ，且 M 是对称正定阵。
Proof : the equation can be written as NAMcM =ρ′− − )()( 1 let AB 1−ρ= , then
).det()det()det( 11 AIAIBI −ρλρ=ρ−λ=−λ −− that is ,all eigenvalues of A are the eigenvalues B multiplied by ρ ,so all eigenvalue of B have msgritudes less than 1 . equivalently if and only if ,
for any given positive definite symmetric matrix N the equation NMAAM 22 ρ=′−ρ has a unique symmetric solution M and M is positive definite . 5．20 Is a system with impulse response ||||2),( τ−−=τ tetg , for τ≥t , BIBO stable ? how about
?cos)sin(),( )( tetg t τ−−=τ
脉冲响应为 ||||2),( τ−−=τ tetg ， τ≥t ， 的系统是否 BIBO 稳定？ ?cos)sin(),( )( tetg t τ−−=τ 的系统是否 BIBO 稳定？
τ=τ=τ ∫∫∫ −−−−− dededet
t
tt
t
ttt
t
tt
000
||||||2||||2 || 、
<<−−><−−≥−
=−
−−
−−−
00)2(00)2(0)(
02
02
02
0
0
0
tandtifeeetandtifeee
tifeee
ttt
ttt
ttt
∞+<≤ 2
)1(|sin||cos|sin||cos)sin(| 0)()( tttt ettette −τ−−τ−− −⋅≤⋅⋅=
∫ ∫ −−τ−τ−− −⋅=τ⋅=τ⋅t
t
ttt
t
tt etdeetdet0
0
0
)1(|sin||sin|)(|sin| )()(
]1,0(, 00 ∈∴≤ −ttett
∫ +∞<≤τ⋅ τ−−t
t
t sodet0
0)(|sin| )(
∫ ∫ +∞<≤⋅⋅≤τ⋅ −−−τ−−t
t
ttt
t
tt etetet0
0
0
1|sin||sin||cos)(sin| )()(
Both systems are BIBO stable .
5.21 Consider the time-varying equation xteyutx2
2 −=+τ= is the equation stable ? marginally stable ? asymptotically stable ?
时变方程 xteyutx2
2 −=+τ= 是否 BIBO 稳定 限界稳定？ 渐进稳定？ 22 )()0()()(20( tt etxextxttxtx =⇒=⇒= is a fundamental matrix
⇒ its stable transition matrix is 02
),( 0ttett −=φ
⇒ Its impulse response is 2222
),( τ−τ−− =⋅=τ eeetg tt
+∞<π=π⋅=τ=τ=⋅τ≤τ=ττ ∫∫ ∫∫
∞+ τ−τ−∞+
∞−
τ−τ−
222),(
0
22
0
22
0
dededededtgt
t
t
t
so the equation is BIBO stable .
00 ,|||),(| 02
02
tteett tttt ≥==φ −− is not bounded ,that is . there does not exist a finite
constant M such that .),( 0 +∞<≤φ Mtt so the equation is not marginally stable and is not asymptotically stable . 5.22 show that the equation in problem 5.21 can be transformed by using
,)( xtpx = with 2
)( tetp −= , into xyuexx t =+= − 2
.0 is the equation BIBO stable ?marginally stable ? asymptotically ? is the transformation a lyapunov transformation ?
证 明 题 5 。 21 方 程 是 可 通 过 ,)( xtpx = with 2
)( tetp −= 变 换 成
xyuexx t =+= − 2
.0 该方程是否 BIBO 稳定？界限稳定？渐进稳定？该变换是不是
lyapunov 变换 ?
proof : we have found a fundamental matrix of the equation in problem 5.21 .is .)(2tetx = so
from theorem 4.3 we know , let .)()21 tetxtp −−
==（ and ,)( xtpx = , then we have
0)( =tA
0)().1))().))()2221
======= −−−tDtDeetBtCtCetBtxtB ttt （（（（（
and the equation can be transformed into xyuexx t =+= − 2
.0
the impulse response is invariant under equivalence transformation , so is the BIBO stability . that is marginally stable . 1)()0()(.0 =⇒=⇒→= txxtxxx is a fundamental matrix ⇒ the state transition matrix is +∞<≤=φ=φ 11|),(|1),( 00 tttt so the equation is
marginally stable . 0|),(|, 0 →φ∞→ ttt (不趋于零)so the equation is not asymptotically stable . from theorem 5.7 we know the transformation is mot a lyapunov transformation
5.23 is the homogeneous equation 0001
03 ≥
−−
= − tforxe
x t
marginally stable ? asymptotically stable?
齐次方程 0,001
03 ≥
−−
= − txe
x t 是否限界稳定？渐进稳定？
=⇒
−+=
=⇒
−=−=
⇒
−−
=
−
−
−
−
−−
141
0
)0(41)0()0(
41)(
)0()(
001
4
124
12
11
13
2
113
t
t
t
t
tt
ee
x
xxextx
extx
xexxx
xe
x
is a fundamental matrix of the equation .
==Φ⇒ +−
+−−
141
0)()()(
0
0
40
1tt
tt
ee
tXtXt is the state
transition of the equation.
−+=φ −+−+−
∞000 34
0 41
411,max||),(|| ttttt eeett for all 00 ≥t
and 0tt ≥ 41
410;10 00 4 ≤≤≤≤ +−+− tttt ee soee ttt ,0
41
41;10 00 3 ≤≤−≤≤ −+−
45
41
411
43
00 34 ≤−+≤ −+− ttt ee . +∞<≤φ ∞ 45||),(|| 0tt the equation is marginally
stable .if 0tt − is fixed to some constant , then 0||),(|| 0 →φ tt (不趋于零)。 ,∞→tas so the equation is not asymptotically stable .
for all t and 0t with 0tt ≥ . we have 41
410;10 00 4 ≤≤≤≤ +−+− tttt ee . every entry of
),( 0ttφ is bounded , so the equation is marginally stable . because the (2,2)th entry of ),( 0ttφ does not approach zero as ,∞→t the equation is not
asymptotically .
6.1 is the stable equation
xy
xx
]121[001
331100010
=
µ
+
−−−=
controllable ? observable ?题
述状态方程是否可控？ 是否可观？
1121121
021(
001
3331
100010
(])([ 2
=
−−−ρ=
ρ
=
−−−ρ=ρ
））（
）BAABB
thus the state equation is controllable ,
but it is not observable .
6.2 is the state equation
xy
uxx
]101[000110
121100010
=
+
−−=
controllable ? observable?
述状态方程是否可控？ 是否可观？
,1232.2)( =−=−=ρ nB using corollary , we have
3)020000010110
(])([ =
ρ=ρ ABB ,Thus the state equation is controllable
3)420130
101()(
2
=
−−ρ=
ρ
CACAC
. Thus the state equation is observable
6.3 Is it true that the rank of [B AB ]1BAn− equals the rank of ][ 2 BABAAB n ? If
not ,under what condition well it be true ?
ρ ( [B A ][ 2 BABAAB n B ]1BAn− ]=ρ ()是否成立？若否，问在什么条件下该成
立？
It is not true that the rank of [B AB ]1BAn− equals the rank of ][ 2 BABAAB n .
If A is nonsingular {( ρ ( ][ 2 BABAAB n )= ρ (A [B AB ]1BAn− ]= ρ ( [B
AB ]1BAn− (see equation (3.62) ) then it will be true
6.4 show that the state equation uB
xAAAA
x
+
=
01
2221
1211 is controllable if and only if the pair
)( 2122 AA IS controllable 证明题述状态方程可控当且仅当 )( 2122 AA 可控？
Proof : 6.5 find a state equation to describe the network shown in fig.6.1 and then check its controllability
and observability 求图 6.1 中网络的一个状态方程,是否可控?可观?
The state variables 1x and 2x are choosen as shown ,then we have
uxyxx
uxx
20
2
22
11
+−==+=+
thus
a state equation describing the network can be es pressed as
uxx
y
Uxx
xx
2]10[
01
1001
2
1
2
1
2
1
+
−−
+
−
−=
checking the controllability and observability :
1)10
01()(
1)10
01(]([
=
−
−ρ=
ρ
=
−
−ρ=ρ
CAC
ABB The equation is neither controllable nor obserbable
6.6 find the controllability index and obserbability index of the state equation in problems 6,1 and 6.2 求题 6.1 和 6.2 中状态方程的可控性指数和可观性指数.
The state equation in problem 6.1 is
uxx
y
Uxx
xx
2]10[
01
1001
2
1
2
1
2
1
+
−−
+
−
−=
the equation is controllable ,so
we have 1)(,3.1)1,min( ===+−≤+−≤µ≤ brankpnwherepnpnnpn
333.1 =µ≤µ≤+−≤µ≤∴ iepnpn
[ ]( ) 1,12
=∴=
ρ
ρ=ρ V
ACACC
ACC
c
The controllability index and cbservability index of the state eqyation in problem 6.1 are 3 and 1 ,
respectively .the state equation in problem 6.2 is
[ ]xy
uxx
101000110
120100010
−
+
−=
The state equation in is both controllable and observable ,so we have
1.1 +−≤ν≤+−≤µ≤∴ pnpnandpnpn
1.2)(,3 ===== crankqBrankpnwhere
3,233.223 =ν=µ≤µ≤≤µ≤∴
The controllability index and cbservabolity index of the state equation in problem 6.2 are 2 and 3 ,respectively rank(c)=n .the state equation is controllable so
1)(.1 =µ⇒==+−≤µ≤ nBrankpwherepnpn
6.7 what is the controllability index of the state equation IUXAx += WHERE I is the unit matrix ?
状态方程 IUXAx += 的可控性指数是什么？
Solution . ][][ 1212 −− == nn AAAIBABAABBC
C is an 2nn× matrix. So rank(C) n≤ . And it is clearly that the first n columns of C are
linearly independent. This rank(C) n≤ .and niforui ,1,01 ==
The controllability index of the state equation IUXAx += is 1),,,max( 21 == nuuuu
6.8 reduce the state equation xyuxx ]11[11
1341
=
+
−
−= . To a controllable one . is the
reduced equation observable ? 将题述状态方程可控的，降维后方程是否可观？
Solution :because [ ]( ) 21)3331
()( <=
ρ=ρ=ρ ABBC
The state equation is not controllable . choosing
== −
01111PQ and let xpx = then we
have
−
=
−
−
−
== −
5043
0111
1441
11101PAPA
=
−
==01
11
1110
PBB . [ ] [ ]120111
111 =
== −CPC
thus the state equation can be reduced to a controllable subequation
ccc xyuxx 23 =+= and this reduced equation is observable
6.9 reduce the state equation in problem 6.5 to a controllable and observable equation . 将题 6.5 中的状
态方程降维为可控且可观方程. Solution : the state equation in problem 6.5 is
uxy
Uxx
2]10[01
1001
+−−
+
−
−=
it is neither controllable nor observable
from the form of the state equation ,we can reaelily obtain
uxx
y
Uxx
xx
2]10[
01
1001
2
1
2
1
2
1
+
−−
+
−
−=
thus the reduced controllable state equation can be independent of cx . so the equation can be further
reduced to y=2u .
6.10 reduce the state equation
[ ]xy
uxx
1011010010
00001000000000100001
2
2
1
1
1
=
+
λλ
λλ
λ
= to a controllable and
observable equation 将状态方程降维为可控可观方程。
Solution the state equation is in Jordan-form , and the state variables associated with 1λ are
independent of the state variables associated with 2λ thus we can decompose the state equation into
two independent state equations .and reduce the two state equation controllable and observable equation ,
respectively and then combine the two controllable and observable equations associated with 1λ and
2λ . Into one equation ,that is we wanted .
Decompose the equation .
[ ]
[ ] 222
22
11
1
1
1
1
1010
01
110010
001001
xyuxx
xyuxx
=
+
λ
λ=
=
+
λλ
λ=
Where 212
1 , yyyxx
x +=
=
Deal with the first sub equation .c
Choosing 1111
10001010
: xpxletandPQ =
λ== − then we have
λλλ−
=
λ
λλ
λ
−== −
1
1
12
1
1
1
11
11
0002110
10001010
001001
10000101C
PPAA
=
λ−==
001
010
1000000111
11 PBB [ ] [ ]1110001010
110 11
11 λ=
== − CPCC
Thus the sub equation can be reduced to a controllable one
[ ]
211
)(
21
01
10
211
1
111
11
21
1
<=
λλλ
ρ=ρ
λ=
+
λλ−
=
O
xy
uxx
c
cc
the reduced controllable equation is not observable choosing
λ=
101 1P .and let ,
cc xpx 11 = then we have
[ ] [ ]0110
11
01
01
101
10
101
210
101
111
11
1
11
1
211
1
=
λ−λ=
=
λ=
λ
λ=
λ−
λλ−
λ=
C
B
A
thus we obtain that reduced controllable and observable equation of the sub equation associated with
011
01111 :
c
cco
xyuxx
=+λ=λ
deal with the second sub equation :
2)1
10()(
22 =
λ
ρ=ρ C it is controllable
21)0
10()(
22 <=
λ
ρ=ρ O It is not observable choosing
=
0110
p and let
22 xpx = then we have
[ ] [ ]010110
10
01
10
0110
10
0110
01
0110
2
2
2
2
2
22
=
=
=
=
λ
λ=
λ
λ
=
C
B
A
thus the sub equation can be reduced to a controllable and observable
ico
icoico
xy
uxx
=
+λ=
2
2
combining the two controllable and observable state equations . we obtain the reduced controllable and observable state equation of the original equation and it is
[ ] co
coco
xy
uxx
11
11
00
2
1
=
+
λ
λ=
6.11 consider the n-dimensional state equation uoxcyuBxAx
+=+=
the rank of its controllability matrix is
assumed to be nn <1 . Let 1Q be an 1nn× matrix whose columns are any 1n linearly independent
columns of the controllability matrix let 1p be an nn ×1 matrix such that 111 InQp = ,where 1In
is the unit matrix of order 1n .show that the following 1n -dimensional state equation
uDxCQyuBPxAQPx
+=+=
11
11111
is controllable and has the same transfer matrix as the original state equation
n 维状态方程uoxcyuBxAx
+=+=
。假设共可控性矩阵的秩为 nn <1 ，令 1Q 为可控性矩阵中任意 1n 个线
性无关组成的 1nn× 矩阵，而 1p 是 P 发矩阵且 111 InQp = ，其中 1In 是 1n 阶单位阵。证明下列 1n
维状态方程uDxCQy
uBPxAQPx+=
+=
11
11111
可控且与原状态方程具有相同的传函。
Let
== −
2
11 :PPQP , where 1P is an nn ×1 matrix and 2P is an nnn ×− )( 1 one ,and we
have
1
1
1
1
22
11
2212
21112211 0
0
nn
nn
nn
nn IQP
IQPisthatI
II
QPQPQPQP
PQIPQPQQP−− =
==
=
==+=
Then the equivalence trans formation xpx = will transform the original state equation into
[ ]
uB
xx
AAA
uBpBp
xx
AQPAQPAQPAQP
uBpp
xx
QQApp
xx
c
c
c
C
C
c
c
c
c
c
c
+
=
+
=
+
=
0012
2
1
2212
2111
2
121
2
1
[ ]
[ ]
[ ] uDxx
CC
uDxx
CQCQ
uDxx
QQCy
c
ccc
c
c
c
c
+
=
+
=
+
=
21
21
where 111111 ,.. nnisAandCPCBPBAQPA cCcc ×=== the reduced 1n -dimensional
state equation uDxCQy
uBPxAQPx
+=
+=
11
11111
is controllable and has the sane transfer matrix as the original
state equation
6.12 In problem 6.11 the reduction procedure reduces to solving for 1P in .11 IQP = how do you
solve 1P ?
题 6.11 中的降维过程为求解 .11 IQP = 中的 1P , 该如何求 1P ?
Solution : from the proof of problem 6.11 we can solve 1P in this way
ρ (C)= ρ ][ 2 BABAAB n = nn <1
we fprm the nn < matrix ]1 1([ nn qqqQ = , where the first 1n columns are any 1n linearly
independent columns of C , and the remaining columns can aribitrarily be chosen as Q is nonsingular .
Let
== −
2
11 :PPQP , where 1P is an nn ×1 matrix and 111 InQp =
6.13 Develop a similar statement as in problem 6.11 for an unobservable state equation .对一不可观状
态方程，如题 6.11 般低哦降维为可观方程.
Solution : consider the n-dimensional state equation uDxCyuBxAx
+=+=
the rank of its observability matrix
is assumed to be nn <1 .let 1P be an nn ×1 matrix whose rows are any 1n linearly independent
rows of the observability matrix . let 1Q be an 1nn× matrix such that 111 InQp = .where 1In is
the unit matrix of order 1n , then the following 1n -dimensional state equation
uDxCQyuBPxAQPx
+=+=
11
11111
is observable and has the same transfer matrix as the original state equation 6.14 is the Jordan-form state equation controllable and observable ?
下 Jordan-form 状态方程是不是可控，可观？
uxx
−+
=
001101101123111112012
1000000010000001100000002000000020000000200000012
xy
−=
011011000021111113122
solution : 2)001101
(3)123111112
( =
ρ=
ρ and
so the state equation is controllable ,however ,it is not observable because
32)110211312
( ≠=
ρ
6.15 is it possible to find a set of ijb and a set of ijc such that the state equation is
controllable ? observable ?是否可能找到一组 ijb 和一组 ijc 使如下状态方程可控？可观？
uccccccccccccccc
yu
bbbbbbbbbb
xx
=
+
=
3534333231
2524232221
1514131211
5251
4241
3231
2221
1211
1000001000011000001000011
solution : it is impossible to find a set of ijb such that the state equation is observable ,because
32(
5251
4241
2221
<≤
ρ ）
bbbbbb
it is possible to find a set of ijc such that the state equation is observable .because
3(
353331
252321
151311
≤
ρ ）
CCCCCCCCC
The equality may hold for some set of ijc such as ICCCCCCCCC
=
353331
252321
151311
the state equation is
observable if and only if 3(
353331
252321
151311
=
ρ ）
CCCCCCCCC
6
6.16 consider the state equation
[ ]xCCCCCy
u
bbbbb
xx
222112111
22
21
12
11
1
12
22
11
11
1
000000
0000000000
=
+
αbbα
αb−bα
λ
=
it is the modal form discussed in (4.28) . it has one real eigenvalue and two pairs of complex conjugate eigenvalues . it is assumed that they are distinct , show that the state equation is
controllable if and only if 2,100;0 211 =≠≠≠ iforborbb ii it is observable if
and only if 2,100;0 211 =≠≠≠ iforcorcc ii
题中的状态方程是（4.82）中讨论的规范型。它有一个实特征值和两对复特征值，并假定它
们都是互不相同的。证明该状态方程可控当且仅当
2,100;0 211 =≠≠≠ iforborbb ii 可观当且仅当
2,100;0 211 =≠≠≠ iforcorcc ii
proof:；controllability and observability are invariant under any equivalence transformation . so we introduce a nonsingular matrix is transformed into Jordan form.
−
−=
−
−= −
ji
iip
jj
jj
p
000110000000011000001
,
5.05.00005.05.0000
005.05.00005.05.0000001
1
+−+−
==
b−αb+α
b−αb+α
λ
== −
)(5.0)(5.0)(5.0)(5.0
00000000000000000000
2221
22211
1211
1211
1
22
22
11
11
1
1
jbbjbbjbbjbb
b
PBB
JJ
JJ
PAPA
[ ]222222211211121111 jCCjCCjCCjCCCCPC −+−+== −
Using theorem 6.8 the state equation is controllable if and only if
;0)(5.0;0)(5.0;0 222112111 ≠±≠±≠ jbbjbbb equivalently if and only if
2,1;0;0;0 211 =≠≠≠ iforborbb ii
The state equation is observable if and only if
;0;0;0 2221121111 ≠±≠±≠ jccjccc equivalently , if and only if
2,100;0 211 =≠≠≠ iforcorcc ii
6.17 find two- and three-dimensional state equations to describe the network shown in fig 6.12 .
discuss their controllability and observability 找到图 6.12 所示网络的两维和三维状态方程描述。讨论它们的可控和可观性， solution: the network can be described by the following equation :
321332121 ;;3;2
2xyxxxxxxxxu
=−−==−=+
they can be arranged as
213
13
12
11
221
221
223
223
112
112
xxxy
uxx
uxx
uxx
−−==
+=
+=
−−=
so we can find a two-dimensional state equation to describe the network as
following : 2132
1
2
1
223112
0223
0112
xxxyuxx
xx
−−==
−+
−=
and it is not controllable because 21)
223
112
223
112
112
()(
2
<=
−−
−
ρ=ρ c
however it is observable because 2)0221
11()( =
−−ρ=ρ o
we can also find a three-dimensional state equation to describe
it : [ ]xyuxxx
xxx
100
221223112
00221
00223
00112
3
2
1
3
2
1
=
−
+
−
=
and it is neither controllable nor
observable because [ ] 32)(31)(2
2 <=
ρ<=ρ
CACAC
BAABB
6．18 check controllability and observability of the state equation obtained in problem 2.19 . can uou give a physical interpretation directly from the network 检验题 2.19 中所得状态方程的可控性和可观性。能否直接从该网络给出一个物理解释？ Solution : the state equation obtained in problem 2.19 is
[ ]xyuxxx
xxx
010011
110100
001
3
2
1
3
2
1
=
+
−=
It is controllable but not observable because
[ ] 2110100
010)(3
110101
111)(
2
2 ≤
−−−ρ=
ρ=
−ρ=ρ
CACAC
BAABB
A physical interpretation I can give is that from the network , we can that if 0)0(0 1 ≠== axu
and 0)0(2 =x then 0≡y that is any [ ]Tax *0)0( = and 0)( ≡tu yield the same output
0)( ≡tY thus there is no way to determine the initial state uniquely and the state equation describing
the network is not observable and the input is has effect on all of the three variables . so the state equation describing the network is controllable 6.19 continuous-time state equation in problem 4.2 and its discretized equations in problem 4.3 with sampling period T=1 and π . Discuss controllability and observability of the discretized equations . 考虑题 4.2 中的连续时间状态以及题 4.3 中它的离散化状态方程（T=1 and π）. 讨论离散状态
方程的可控性和可观性
solution : the continuous-time state equation in problem 4.2 is [ ]xyuxx 3211
2210
=
+
−−
=
the discretized equation :
[ ] [ ] [ ]kueconTT
eTconTkxeTTTe
TeeTTkx
T
T
TT
TT
++
+−+
−−+
=+−
−
−−
−−
)sin2(1
)sin21
23(
23
)sin(cossin2sin)sin(cos
1
[ ] [ ] [ ]kxky 32=
T=1
−
=
−−
=1821.0
0471.11108.06191.0
3096.05083.0dd BA
T= π :
−
=
=
0432.15648.1
0432.0000432.0
dd BA
The xontinuous-time state equation is controllable and observable , and the sustem it describes is single-input so the drscretized equation is controllable and observable if and only if
[ ] 2,12/2|| =π≠λ−λ mformjI im whenever [ ] 0=λ−λ jR ie the eigenvalues of a are
–1+j and –1-j . so [ ] 2|| =λ−λ jI im for T=1 . 2,122 =≠π manyform .so the
discretized equation is neither controllable nor observable . for T= π 2m=1 .so the descretized equation is controllable and observable .
6.20 check controllability and observabrlity of [ ]xyuxt
x 1010
010
=
+
= 检验可控性和
观性。
Solution :
−
−=t
MtAM1
)( 01 the determinant of the matrix [ ]
−−
=t
MM1
1010 is 1. which
is nonzero for all t . thus the state equation is controllable at every t。
Its state transition matrix is
τ−=Φ
−
τ∫)(5.0
5.05.0
0 20
20
22
0
1),(
tt
t
t
t
e
deett and
[ ] [ ])20
2
)20
20
22
(5.0
(5.0
5.05.0
0 00
110),()( t
t
t
t ee
deettC −τ
−τ
ττ
=
ϑ−=Φτ ∫
[ ]
τ
=
τ⋅
=
∫
∫
−τ
−τ−τ
de
dee
ttW
t
t t
tt
t t
1
020
2
20
21
020
2
000
00
),( )(5.0)(5.0100
We see . ),( 100 ttW is singular for all 0t and t , thus the state equation is not observable at any 0t .
6.21 check controllability and observability of [ ]xeyue
xx tt
−− =
+
−
= 01
1000
solution :
textxxtx −== )0()()0()( 2211 its state transition matrix is
=
==Φ +−−
−
00 001
001
001
)(),( 01
0 tttt eeetXtXtt ）（ and
[ ] [ ]0
0
20 0
001
0),()(
110
01)(),(
tt
t
tt
ee
etC
eeeBt
+τ−+τ−
−
−τ−τ+−
=
=τΦτ
=
=ττΦ
WE compute [ ]
−−−−
=τ
=τ
=
−−
−
−−
−−
− ∫∫ )()()(1
11
),(0
20
0020
00 ttettettett
deee
dee
ttW tt
tt
t tt
ttt
t tc
Its determinant is identically zero for all to and 0t , so the state equation is not controllable at any 0t
[ ] τ
=τ
= ∫∫ +τ−
+τ−+τ− d
ede
ettW
t
t ttt
t tc0 0
0
0 0 242
20 000
11
),( its determinant is identically zero for all
0t and 1t , so the state equation is not observable at any 0t .
6.22 show shat ( ))(),(( tBtA is controllable at 0t if and only if ))(),(( tBtA ′′− is observable at 0t
证明 ))(),(( tBtA 在 0t 可控当且仅当 ))(),(( tBtA ′′− 在 0t 可观。
Proof: Itttt =ΦΦ ),(),( 00 where ),( 0ttΦ is the transition matrix of ))()()( txtAtx =
0),(),()(),(),(),(),()(
),(),(),(
),(),(),(
000000
000
000
=Φ∂∂
Φ+=Φ∂∂
Φ+ΦΦ=
∂Φ∂
Φ+Φ∂
Φ∂=ΦΦ
∂∂
∴
ttt
tttAttt
tttttttA
ttt
ttttt
tttttt
t
There we have )(),(),(
00 tAttt
tt+Φ−=
∂Φ∂
and. ),()(),(
00 tttAt
ttΦ′+′−=
∂Φ∂
So ),( 0 ttΦ′ is the transition matrix of ))()()( txtAtx ′= ))(),(( tBtA is controllable at 0t if and only
if there exists a finite 01 tt > such that ττΦ′τ′ττΦ= ∫ dtBBtttWt
tc ),()()(),(),( 1101
0
is nonsingular
))(),(( tBtA ′′− is observable at 0t if and only if there exists a finite 01 tt > such that
ττΦ′τ′ττΦ= ∫ dtBBtttWt
tc ),()()(),(),( 0001
0
is nonsingular.
For time-invariant systems , show that （A,B）is controllable if and only if (-A,B) is controllable , Is true foe time-varying systems ? 证明对时不变系统，（A,B）可控当且仅当（-A,B）可控。 对时变系统是否有相同结论？ Proof: for time-invariant system (A,B) is controllable if and only if
[ ] nBAABBC n =ρ=ρ −11 )( (assuming a is nn×
(-A,B) is controllable if and only if [ ] nBAABBC n =ρ=ρ −12 )(
we know that any column of a matrix multiplied by nonzero constant does not change the rank of the
matrix . so =ρ )( 1C )( 2Cρ the conditions are identically the same , thus we conclude that (A,B)is
controllable if and only if (-A,B)is controllable .for time-systems, this is not true .
7.1 Given ）（ 2)(1
1)( 2 +−−
=ss
ssg . Find a three-dimensional controllable realization check its
observalility
找）（ 2)(1
1)( 2 +−−
=ss
ssg 的一三维可控性突现，判断其可观性。
Solution : ）（）（ 22
12)(1
1)( 222 −−+−
+−−
=sss
sss
ssg using (7.9) we can find a
three-dimensional controllable realization as following
[ ]xyuxx 110001
010001212
−=
+
−= this realization is not observable because
(s-1)and ）（ 2)(12 +− ss are not coprime
7.2 find a three-dimensional observable realization for the transfer function in problem 7.1 check its controllability
找题 7.1 中 ）（ 2)(1
1)( 2 +−−
=ss
ssg 的一三维可控性突现，判断其可观性。
Solution : using (7.14) ，we can find a 3-dimensional observable realization for the transfer
function : [ ]xyuxx 0011
10
002101012
=
−+
−= and this tealization is not
controllable because (s-1) and ）（ 2)(12 +− ss are not coprime
7.3 Find an uncontrollable and unobservable realization for the transfer function in problem 7.1 find also a minimal realization 求题 7.1 中传函的一个不可控且不可观突现以及一个最小突现。
Solution : 23
1)2)(1(
12)(1
1)( 22 ++=
++=
+−−
=ssssss
ssg）（
a minimal realization is ,
[ ]xyuxx 1001
0123
=
+
−−= an uncontrollable and unobservable realization
is [ ]xyuxx 010001
100001023
=
+
−−=
7．4 use the Sylvester resultant to find the degree of the transfer function in problem 7.1 用 Sylvester resultant 求题 7.1 中传函的阶 solution :
−−−−α
−−−−−
=
01000002010011020112110
001211000012
s
rank(s)=5. because all three D-columns of s are linearly independent . we conclude that s has only
two linearly independent N-columns . thus deg 2)(ˆ =sy
7.5 use the Sylvester resultant to reduce 14
122 −−
ss
to a coprime fraction 用 Sylvester resultant
将14
122 −−
ss
化简为既约分式。
Solution : ssusranks ===
−−
−−
= 11,3)(
040020041120
0011
nullT
s
−= 10
21
21)( 1 thus we have ssDssN +=⋅+=
21)(0
21)( and
121
21
21
1412
2 +=
+=
−−
ssss
7.6 form the Sylvester resultant of ）（ ss
ssg22)(ˆ
2 ++
= by arranging the coefficients of
)(sN and )(sD in descending powers of s and then search linearly independent columns in
order from left to right . is it true that all D-columns are linearly independent of their LHS
columns ?is it true that the degree of )(ˆ sg equals the number of linearly independent N-columns?
）（ ssssg
22)( 2 +
+= 将 )(sN 和 )(sD 按 s 的降幂排列形成 Sylvester resultant ，然后从左
至右找出线性无关列。所有的 D-列是否都与其 LHS 线性无关？ )(ˆ sg 的阶数是否等于线性
无关 N-列的数回？
Solution ； )()(:
22)(ˆ
2 sNsD
ssssg =++
=）（
1)(ˆdeg =sg
012
2
012
2
)(
)(
NSNSNsNDSDSDsD
++=
++=
from the Sylvester resultant :
=
2000122001120001
s 23)( == usrank (the number of linearly independent
N-columns
7.7 consider )()()(ˆ
212
21
sNsD
ssssg =
∂+α+β+β
=）（
and its realization
[ ]xyuxx 2121
01
01ββ=
+
α−α−=
show that the state equation is observable if and only if the Sylvester resultant of
)()( sNandsD is nonsingular . 考 虑)()()(ˆ
212
21
sNsD
ssssg =
∂+α+β+β
=）（
及 其 实 现
[ ]xyuxx 2121
01
01ββ=
+
α−α−= 证 明 该 状 态 方 程 可 观 及 其 当 且 仅 当
)()( sNandsD 的 Sylvester resultant 非奇异
proof: [ ]xyuxx 2121
01
01ββ=
+
α−α−= is a controllable canonical form realization of
)(ˆ sg from theorem 7.1 , we know the state equation is observable if and only id )()( sNandsD
are coprime
from the formation of Sylvester resultant we can conclude that )()( sNandsD are coprime if
and only if the Sylvester resultant is nonsingular thus the state equation is observable if and only if the
Sylvester resultant of )()( sNandsD is nonsingular
7.8 repeat problem 7.7 for a transfer function of degree 3 and its controllable-form realization , 对一
个 3 阶传函及其可控型实现重复题 7.7
solution : a transfer function of degree 3 can be expressed as)()()(ˆ
322
13
322
13
1
sDsN
ssssss
sg =α+∂+α+β+β+β+β
=）（
where )()( sNandsD are coprime
writing )(ˆ sg as ）
（
322
13
0320222
010 )())(ˆ
α+α+α+βα+β+βα−β+βα−β
=sss
sssg , we can obtain a controllable
canonical form realization of )(ˆ sg
[ ] uxy
uxx
0033022011
321
001
010001
β+βα−ββα−ββα−β=
+
α−α−α−=
the Sylvester resultant of )()( sNandsD is nonsingular because )()( sNandsD are
coprime , thus the state equation is observable .
7.9 verify theorem 7.7 fir 2)1(1)(ˆ+
=s
sg .对 2)1(1)(ˆ+
=s
sg 验定定理 7.7
verification : 2)1(1)(ˆ+
=s
sg is a strictly proper rational function with degree 2, and it can be
expanded into an infinite power series as
++−+−+⋅= −−−−−− 654321 54320)(ˆ sssssssg 1−s
[ ]( )( ) 2,1,.22,2)2,2(
1)1,2()2,1(00)1,1(==++=
====lkveyerforLKTT
TTTrr
rrrr
7.10 use the Markov parameters of 2)1(1)(ˆ+
=s
sg to find on irreducible companion-form realization
用马尔科夫参数求 2)1(1)(ˆ+
=s
sg 的一个不可简约的有型实现
solution 2
2110
)2,2(
5)4(3)2(0)(ˆ 654321
=
−
=
++−++−++⋅= −−−−−−
rrT
sssssssg
deg )(ˆ sg =2
[ ]0110
2110
0112
3221
2110
3221
)2,2()2,2(~1
1
=
=
−−
=
−
−=
−
−
−==∴
−−
cb
TTA
the triplet ),,( cbA is an irreducible com-panion-form realization
7.11 use the Markov parameters of 2)1(1)(ˆ+
=s
sg to find an irreducible balanced-form realization
用 )(ˆ sg 的马尔科夫参数求它的一个不可简约的均衡型实现.
Solution :
−
=21
10)2,2(T
Using Matlab I type
[ ] [ ][ ]
)1,1(),1,(;)(**)(
;*;*);(),(,,
32;2121;10
OccbcinvoinvA
lslcslkossqrtsltsvdlsk
ttt
===
′====−−=−=
ϑ
This yields the following balanced realization
[ ]xy
xx
5946.05946.05946.0
5946.02929.07071.0
7071.07071.1
−−=
−
+
−−
−=
7.12
Show that the two state equation [ ]xyuxx 2201
1012
=
+
= and
[ ]xyuxx 0221
1102
=
+
−−
= are realizations of )2()22(
2 −−+
sss , are they
minimal realizations ? are they algebraically equivalent?
证明以上两状态方程都是 )2()22(
2 −−+
sss
的突现,他们是否最小突现?是否代数等价?
Proof : 2
2)2)(1(
)1(2)2(
)22(2 −
=−+
+=
−−+
ssss
sss
[ ] [ ]
[ ] [ ]2
221
11
)2)(1(1
02
1
0221
1102
02
22
01
110
)2)(1(1
21
2201
1012
22
1
1
−=
+−+−
−=
+
−
−=
−
−−−=
−−−
−
−
ssss
ss
s
ss
ssss
s
thus the two state equations are realizations of )2()22(
2 −−+
sss the degree of the transfer function
is 1 , and the two state equations are both two-dimensional .so they are nor minimal realizations . they are not algebraically equivalent because there does not exist a nonsingular matrix p such that
[ ] [ ]0222
01
01
1102
1012
1
1
=
=
−−
=
−
−
P
P
PP
7.13 find the characteristic polynomials and degrees of the following proper rational matrices
++
++
=
131
131
ˆ1
ss
s
ss
sG
+++
+++=
)2)(1(1
21
)2)(1(1
)1(1
ˆ2
1
sss
sssG
+
++
+
++
+=
s
s
ss
s
ss
sG 15
1
41
21
23
)1(1
ˆ 2
3
note that each entry of )(ˆ sGs has different poles from other entries
求正则有理矩阵 )(ˆ1 sG , )(ˆ
2 sG 和 )(ˆ3 sG 的特征多项式和阶数.
Solution : the matrix )(ˆ1 sG has
1,
31,
13,1
++++
ss
sss
s and det )(ˆ
1 sG =0 as its minors
thus the characteristic polynomial of )(ˆ1 sG is s(s+1)(s+3) and )(ˆ
1 sG =3 the matrix )(ˆ2 sG has
2)1(1+s
,)2)(1(
1++ ss
, 2
1+s
, )2)(1(
1++ ss
. det 23
2
2 )2()1(1)(ˆ
+++−−
=sssssG as
its minors , thus the characteristic polynomial of 5)(ˆ)2()1()(ˆ2
232 =++= sGandsssG d
Every entry of )(ˆ3 sG has poles that differ from those of all other entries , so the characteristic
polynomial of )(ˆ3 sG is )5)(4()3)(2()1( 22 +++++ ssssss and 8)(ˆ
3 =sGd
7.14 use the left fraction
−
−
=−
111
)(ˆ1
sss
sG to form a generalized resultant as in (7.83), and
then search its linearly independent columns in order from left to right ,what is the number of linearly
independent N-columns ?what is the degree of )(ˆ sG ? Find a right coprime fraction lf )(ˆ sG ,is the
given left fraction coprime?用 )(ˆ sG 的左分式构成(7.83)中所示的广义终结阵,然后从左到右找出其
线性无关列.线性无关的 N-列的数目是多少> )(ˆ sG 的阶是多少?求出 )(ˆ sG 的一个既约右分式.题
给的左分式是否既约?
Solution : )()(:1
11)(ˆ 1
1
sNsDss
ssG −
−
=
−
−
=
Thus we have
−
=
−
+
=
−
=
11
)(
1101
00101
)(
sN
sss
ssD
And the generalized resultant
0 1 1 0 0 00 0 -1 0 0 01 0 0 0 1 1-1 1 0 0 0 -10 0 0 1 0 00 0 0 -1 1 0
S
=
rank s=5 ,
the number of linearly independent N-colums is l. that is u=1,
[ ]1100
100001)(DNDN
snull−−
−=
So we have
1)(ˆdeg
01
)(ˆ
01
00
01
)(
10)(
1
===
=
+=
+
+=
=⋅+=
−
usG
ssG
sSN
SSSD
the given left fraction is not coprime . 7.15 are all D-columns in the generalized resultant in problem 7.14 linearly independent .pf their LHS
columns ?Now in forming the generalized resultant ,the coefficient matrices of )(SD and )(SN are
arranged in descending powers of s , instead of ascending powers of s as in problem 7.14 .is it true that
all D-columns are linearly independent of their LHS columns? Doe’s he degree of )(ˆ sG equal the
number of linearly independent N-columns ?does theorem 7.M4hold ?题 7.14 中的广义终结阵中是否
所有的 D-列都与其 LHS 列线性无关?现将 )(SD 和 )(SN 的系数矩阵按 S得降幂.排列构成另一种
形式的广义终结阵.在这样的终结阵中是否所有的 D-列也与其 LHS 列线性无关? )(ˆ sG 的阶数是
否等于线性无关 N-列的数目?定理 7.M4 是否成立?
Solution : because ,01 ≠D ALL THE D-columns in the generalized resultant in problem7.14 are
linearly independent of their LHS columns
Now forming the generalized resultant by arranging the coefficient matrices of )(SD and )(SN in
descending powers of s :
5
100000110000011100001110000011000001
00
00
00
1100
11
=
−
−−
−
=
= Srank
NDNDND
NDS
we see the 0D -column in the second colums block is linearly dependent of its LHS columns .so it is
not true that all D-columns are linearly independent of their LHS columns .
the number of linearly independent N -columns is 2 and the degree of )(ˆ sG is I as known in problem
7.14 , so the degree of )(ˆ sG does not equal the number of linearly independent N -columns , and the
theorem 7.M4 does not hold .
7.16, use the right coprime fraction of )(ˆ sG obtained in problem 7.14 to form a generalized tesultant as
in (7.89). search its linearly independent rows in order from top to bottom , and then find a left coprime
fraction of )(ˆ sG 用题 7.14 中得到的 )(ˆ sG 的既约右分式,如式(7.89)构造一个广义终结阵, 从上至
下找出其线性无关行, 求出 )(ˆ sG 的一个既约左分式.
Solution :
==
= −
01
)()(01
)(ˆ 1 sNssDssG
The generalized resultant
0,13
000010100000001010
21 ===
= γγTrankT
[ ]
[ ]′−=
′=
1001)
010100001010
(
100)000001010
(
mull
mull
[ ]
+=
=
+
=
−=−−∴
01
)(
100
0001
1000
)(
000100010001
0000
sN
sssD
DNDN
thus a left coprime fraction of
−−+
5.05.05.05.02
ssss
is
=
−
01
100
)(ˆ1s
sG
7.17 find a right coprime fraction of
+
++
=
sss
ss
ss
sG22
121
)(ˆ
2
23
2
snd then a minimal realization
求 )(ˆ sG 的一个既约右分式及一个最小突现]
solution : )()(:22
121
)(ˆ 1
2
23
2
SNSD
sss
ss
ss
sG −=
+
++
=
where 2
2
322
3
0021
2110
0201
22)12(1
)(
0001
1000
0000
0000
00
)(
ssss
ssssN
ssss
ssD
+
+
=
+++
=
+
+
+
=
=
the generalized resultant is
=
00000000000000010000000000100000000021000001000021000010000010002100000102002100001001001000210000000200210000001001000000000000200000000000100
S
rank 1,2,9 21 === µµs
the monic null vectors of the submatrices that consist of the primary dependent 2N -columns and
1N -columns are , respectively
[ ][ ]
′
−−−−−−−−
=
−
−
∴
−−−−−=
′−−−=
000015.000010005.011
5.005.25.25.005.25.0
1005.0115.005.25.0115.0005.005.25.22
2
2
1
1
0
0
DNDN
DN
ZZ
220 0 0.5 0.5 1 0 0.5 0.5
( )0.5 0.5 0 1 0 0 0.5 0.5
0.5 2.5 1 0 0.5 2.5( )
2.5 2.5 1 0 2.5 2.5
s s sD s s s
s
sN s s
s
+ = + + = − − − −
+ = + = +
thus a right coprime fraction of )(ˆ sG is
−−+
++
=5.05.0
5.05.05.25.25.25.0
)(ˆ2
ssss
ss
sG
we define
=
=
10010
)(0
0)(
2 ssL
ss
sH
then we have
11
1
1 0.5 0.5 0 0( ) ( ) ( )
0 1 0 0.5 0.5
1 0.5 2.5( ) ( )
1 2.5 2.5
1 0.5 1 0.50 1 0 1
1 0.5 0.5 0 0 0.5 0.25 0.250 1 0 0.5 0.5 0 0.5 0.5
hc
hc ic
D s H s L s
N s L s
D
D D
−
−
−
= + − −
=
− = =
− = = − − − −
thus a minimal realization of )(ˆ sG is
0.5 0.25 0.25 1 0.51 0 0 0 00 0.5 0.5 0 1
1 0.5 2.51 2.5 2.5
x x u
y x
− − − − = +
=
8.1 given [ ]xyuxx 1121
1112
=
+
−
=
find the state feedback gain k so that the feedback system has –1 and –2 as its eigenvalues .
compute k directly with using any equivalence transformation 对题给状态方程, 不用等价交换,
直接求状态反馈增益 k 使状态反馈系统的特征值为-1,-2.
Solution : introducing state feedback [ ]xkkru 21−= , we can obtain
[ ] rxkkxx
+
−
−
=21
21
1112
21
the new A-matrix has characteristic polynomial
)35()32(
)1)(21()21)(2()(
21212
2121
+−+−++=
−−−−+−+−=∆
kkskks
kkkskssf
the desired characteristic polynomial is 23)2)(1( 2 ++=++ ssss ,equating
332 21 =−+ kk and 235 21 =+− kk , yields
14 21 == kandk so the desired state feedback gain k is [ ]14=k
8.2 repeat problem 8.1 by using (8.13), 利用式(8.13)重解题 8.1.
solution :
[ ] [ ]
[ ] trollablecanisbAsogularnonisCCbAbD
CC
k
ssssssssss
f
),(,sin7
17
27
47
1
1241
1031
1031
1632)3(3
23)2)(1()(131)1)(2()(
1
1
2
2
−
−=
==
=
−=
−=−−−=
++=++=∆
+−=+−−=∆
−
−
using (8.13)
[ ] [ ]147
17
27
47
1
1031
161 =
−
−
−== −CCkk
8.3 Repeat problem 8.1 by solving a lyapunov equation 用求解 lyapunov 方程的方法重解题 8.1.
solution : ),( bA is controllable
selecting [ ]1120
01=
−
−= kandF then
[ ] [ ]1401319
11
01319
1391
1310
1
1
=
−==
−=
=⇒=−
−
−
Tkk
TTkbTFAT
8.4 find the state foodback gain for the state equation uxx
+
−=
101
100110211
so that the resulting system has eigenvalues –2 and –1+j1 . use the method you think is the simplest by hand to carry out the design 求状态反馈增益. 使题给状态方程的状态反馈系统具
有-2 和-1+j1 特征值/
solution : ),( bA is controllable
[ ] [ ]
−−−=
−−=
=
−
−=
=−−−−−=
+++=−++++=∆
−+−=−=∆
−
−
121232011
111210211
100310631
100310
331537)1(436)3(4
464)11)(11)(2()(133)1()(
1
1
23
233
CC
CC
k
sssjsjssssssss
f
Using(8.13) [ ] [ ]84715121
232011
100310631
5371 −=
−−−
= −CCkk
8.4 Consider a system with transfer function )3)(2)(1(
)2)(1()(ˆ+−+
+−=
ssssssg is it possible to
change the transfer function to )3)(2(
)1()(ˆ++
−=
ssssg f by state feedback? Is the resulting
system BIBO stable ?asymptotically stable?
是否能够通过状态反馈将传函 )(ˆ sg 变为 )(ˆ sg f ?所得系统是否 BIBO 稳定?是否逐渐稳定?
Solution : )3()2)2)(1(
)3)(2()1()(ˆ
2 +++−
=++
−=
ssss
ssssg f
We can easily see that it is possible to change )(ˆ sg to )(ˆ sg f by state feedback and the resulting
system is asymptotically stable and BIBO stable .
8.6 consider a system with transfer function )3)(2)(1(
)2)(1()(ˆ+−+
+−=
ssssssg is it possible to
change the transfer function to 3
1)(ˆ+
=s
sg f by state feedback ? is the resulting system BIBO
stable? Asymptotically stable ?
能否通过状态反馈将传函 )(ˆ sg 变为 )(ˆ sg f ?所得系统是否 BIBO 稳定?是否逐渐稳定?
Solution ; )3)(2)(1(
)2)(1(3
1)(ˆ++−
+−=
+=
sssss
ssg f
It is possible to change )(ˆ sg to )(ˆ sg f by state feedback and the resulting system is
asymptotically stable and BIBO stable .. however it is not asymptotically stable . 8.7 consider the continuous-time state equation
[ ]xyuxx 002101
100110211
=
+
−=
let xkpru −= , find the feedforward gain p and state feedback gain k so that the resulting
system has eigenvalues –2 and 11 j±− and will track asymptotically any step reference input
对题给连续时间状态方程, 令 xkpru −= 求前馈增益p 和状态反馈增益 k 使所得系统特
征值为-2 和 11 j±− , 并且能够渐进跟踪任意阶跃参考输入.
Solution :
[ ]
[ ]84715
5.084
464)(
133882
101
1100
)1(1
110
)1(1
)1(1
)1(1
11
002)()(ˆ
3
3
23
23
2
2
232
−=
===∴
+−++=∆
−+−+−
=
−
−−
−−
−−−
=−=
k
P
ssss
sssss
s
ss
ssss
bAsIcsg
f
bα
(obtained at problem 8.4) 8.8 Consider the discrete-time state equation
[ ] ][002][
][101
100110211
]1[
kxky
kuxkx
=
+
−=+
find the state feedback gain so that the resulting system has all eigenvalues at x=0 show that for any initial state the zero-input response of the feedback system becomes identically zero for
3≥k 对题给离散时间状态方程,求状态反馈增益使所得系统所有特征值都在 Z=0, 证明对任意
初始状态,反馈系统的零输入响应均为零( 3≥k )
solution ; ),( bA is controllable
[ ]
[ ] [ ]251121
232011
100310631
133
133
)(133)(
1
3
3
=
−−−
−==
−=
=∆
−+−=∆
− kCCk
k
zzzszz
f
The state feedback equation becomes
[ ]
[ ] ][002][
][101
][151
110440
][101
][251101
100110211
]1[
kxky
krkxkrkxkx
=
+
−−−
−−=
+
−
−=+
denoting the mew-A matrix as A , we can present the zero-mput response of the feedback system
as following ]0[][ xACkyk
zi =
compute k
A
1
1
000100010
000100010
−
−
=
=
QQA
QQA
K
k
using the nilpotent property
30000100010
≥=
kfor
K
so we can readily obtain 0]0[0][ == xCkyzi for 3≥k
consider the discret-time state equation in problem 8.8 let for ][][][ kxkkprku −= where p
is a feedforward gain for the k in problem 8.8 find a gain p so that the output will track any step
reference input , show also that y[k]=r[k] for 3≥k ,thus exact tracking is achieved in a finite number of sampling periods instead of asymptotically . this is possible if all poles of the resulting system are placed at z=0 ,this is called the dead-beat design
对题8.8中的离散时间状态方程,令 ][][][ kxkkprku −= ,其中 p 是前馈增益,对题8.8中所得
k ,求一增益 p 使输出能够跟踪任意阶跃参考输入,并证明 3≥k 时 y[k]=r[k].实际的跟踪是在
无穷多采样周期中得到的,而不是渐进的. 当反馈系统的所有标点都被配置在 Z=0 时,这样的
跟踪是可以做到的,这就是所谓的最小[拍设计. Solution ;
3
2
323
2
88)(ˆ
][][][)(133
882)(ˆ
zzzpzg
kxkkprkuzzzsz
szzg
f+−
=
−==∆−+−
+−=
),( bA is controllable all poles of )(ˆ zg f can be assigned to lie inside the section in fig 8.3(b)
under this condition , if the reference input is a step function with magnitude a , then the output
y[k] will approach the constant +∞→⋅ kasag f )1(ˆ
thus in order for y[k] to track any step reference input we need 1)1(ˆ =fg
5.012)1(ˆ =⇒== ppg f
the resulting system can be described as
[ ] ][002][][][
][5.0
05.0
][151
110440
][
kxkykrbkxA
krkxkx
=+=
+
−−−
−−=
the response excited by r[k] is
][]0[][1
0
1mrbACxACky
K
M
mKK∑−
=
−−+=
sokforAknowweasK
,30, ≥=
3]3[4]2[4]1[]3[]2[]1[][
2
≥−+−−−=−+−+−=
kforkrkrkrkrbAckrbAckrbcky
for any step reference input r[k]=a the response is y[k]=(1-4+4) a =a =r[k] for 3≥k
8.10 consider the uncontrollable state equation uxx
+
−−
=
1110
1000010000200012
is it possible
to find a gain k so that the equation with state feedback xkru −= has eigenvalues –2, -2 ,
-1 ,-1 ? is it possible to have eigenvalues –2, -2 ,–2 ,-1 ? how about –2 ,-2, –2 ,-2 ? is the equation
stabilizable?对题给不可控状态方程,是否可以求出一增益 k 使具有反馈 xkru −= 的方程具
有特征值-2. –2, -1, -1, ?特征值-2, -2, -2,-1?特征值-2,-2,-2,-2?该方程是否镇定? Solution: the uncontrollable state equation can be transformed into
ub
xx
AA
uxx
xx
c
c
c
c
c
c
c
c
c
+
=
+
−
−
=
000
0001
1000031000010400
Where the transformation matrix is
−−
=−
1111011104210410
1P
),( cC bA is controllable and )( CA is stable , so the equation is stabilizable . the eigenvalue of
)( CA ,say –1 , is not affected by the state feedback ,while the eigenvalues of the controllable
sub-state equation can be arbitrarily assigned in pairs . so by using state feedback ,it is possible for the resulting system to have eigenvalues –2 ,-2 , -1, -1, also eigenvalues –2, -2, -2, -1. It is impossible to have –2, -2, -2, -2, as it eigenvalues because state feedback can not affect the eigenvalue –1. 8.11 design a full-dimensional and a reduced-dimensional state estimator for the state equation in
problem 8.1 select the eigenbalues of the estimators from { }223 j±−−
对题 8.1 中的状态方程, 设计一全维状态估计器和一将维估计器.估计器的特征值从
{ }223 j±−− 中选取.
Solution : the eigenvalues of the full-dimensional state estimator must be selected as 22 j±− ,
the A-, −b and −c matxices in problem 8.1 are
[ ]1121
1112
=
=
−
= cbA
the full-dimensional state estimator can be written as ylubxclAx ++=−= ˆ)(
let
=
2
1
ll
l then the characteristic polynomial of )(ˆ clAx −= is
1912844)2(
23)3(
)1)(1()1)(2()(
21
222121
21221
=−=⇒++=++=
−−+−++=
−+++−+−=∆′
llsss
llsllslllslss
thus a full-dimensional state estimor with eigenvalues 22 j±− has been designed as following
yuxx
−+
+
−−
=1912
21
ˆ1820
1314
designing a reduced-dimensional state estimator with eigenvalue –3 :
(1) select |x| stable matrix F=-3 (2) select |x| vector L=1, (F,L) IS controllable
(3) solve T=: TA-FT= cl
let T :
[ ]
[ ] [ ] [ ]
=⇒
=+
−
=
214
213
1131112
2121
21
T
tttt
ttT
(4) THE l-dimensional state estimutor with eigenvalue –3 is
−
−=
=
++−=
−
zy
zy
x
yuxz
215214
214
215
11ˆ
21133
1
8.12 consider the state equation in problem 8.1 .compute the transfer function from r to y of the state feedback system compute the transfer function from r to y if the feedback gain is applied to the estimated state of the full-dimensional estimator designed in problem 8.11 compute the transfer function from r to y if the feedback gain is applied to the estimated state of the reduced-dimensional state estimator also designed in problem 8.11 are the three overall transfer functions the same ? 对题 8.1 中的状态方程,计算状态反馈系统从 r 到 y 的传送函数.若将反馈增益应用到题 8.11中所设计的全维状态估计器所估计的估计状态,计算从 r 到 y 的传送函数,若将反馈增益应用
到题 8.11 中所设计的降维状态估计器的估计状态, 计算从 r 到 y的传送函数, 这三个传送函
数是否相同? Solution
(1)
[ ])2)(1(
4321
11
)2)(1(9
02
1
11
)(ˆ 1
++−
=
+++
+=
+−=∴
−==+=−
→
sss
sss
s
bkbAsIcyxkruxcyubxAx
yx
(2)
[ ]
[ ] [ ]
[ ]xy
rxx
xxrx
yuxx
rxx
xrxuxx
2121
19191212
ˆ2028
1210
111912
ˆ1421
21
ˆ1820
1314
1912
21
ˆ1820
1314ˆ
21
ˆ2814
1112
ˆ1421
21
1112
21
1112
=
+
−−+
−−
=
−+
−
+
−−
=
−+
+
−−
=
+
−
−
=
−
+
−
=
+
−
=
[ ]
=
+
−−−−
−−−−−
=
∴
xx
y
rxx
xx
ˆ0011
2121
ˆ20281919
1210121228111412
[ ]
)2)(1(43
)84)(2)(1()84)(43(
163222732883
2121
202819191210121228111412
0011ˆ
2
2
234
23
1
++−
=++++
++−=
++++−++
=
+−−−
−−−−−
=∴
−
→
sss
sssssss
sssssss
s
ss
g yr
(3)
[ ]
[ ] [ ]
[ ]xy
rzx
xzy
rzz
rzx
rzxx
rzy
xx
rrxxuxx
11211342
21164
21164
11)215
21414(
21133
21
12663
23211213
21
12663
112211
1112
21
215214
ˆ2814
1112
21
21
ˆ2814
1112
21
1112
=
+−
=
+
−
−−+−=
+
+
=
+
−
−−
−
−
=
+
−
−
−
−
=
−
+
−
−
=
+
−
=
[ ]
[ ]
)2)(1(43
)3)(2)(1()3)(43(
61161253
211321
4221
16421
1641262321632113
011)(ˆ
011
211321
4221
16421
1641262321631213
23
2
1
++−
=+++
+−=
+++−+
=
+
+−−
−−−
=∴
=
+
−
−−
=
∴
−
→
sss
sssss
sssss
ss
ssg
zx
y
rzx
zx
yr
these three overall transfer functions are the same , which verifies the result discussed in section 8.5.
8.13 let
=
−=
20210000
0012321301000010
BA find two different constant matrices k such that
(A-BK) has eigenvalues j34 ±− and j45 ±− 求两个不同的常数阵使(A-BK)具有特征值
j34 ±− 和 j45 ±−
solution : (1) select a 44× matrix
−−−
−−−
=
54004500
00430034
F
the eigenvalues of F are j34 ±− and j45 ±−
(2)if ),(,01000001
11 kFk
= is observable
−−−−==
−−−−
−−−−−
=⇒=−
−
2253.28747.141758.1190670.3710000.22000.140000.1682000.606
1982.02451.00043.00006.00185.01731.00714.01322.0
0191.00193.00273.00126.00042.00005.00059.00013.0
1111
1111
TKK
TKBFTAT
(3)IF
=
01001111
2K ),( 2kF is observable
−−−−==
−
−−−−−
=⇒=−
−
5853.24593.78815.595527.1852509.30893.02145.559824.252
2007.02473.00037.00048.00245.03440.00607.02036.00306.00443.00147.00399.00081.00024.00071.00046.0
1222
2222
TKK
TKBFTAT