Solution-of-Linear-System-Theory-and-Design-3ed-for-Chi-Tsong-Chen

Prévia do material em texto

Linear System Theory and Design SA01010048 LING QING 
 1 
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 
 2 
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
)(2sin
2)(
0
0
tt
tttg
−
−
=
ω
ω
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
atfortu
tuPty 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 
 3 
 


 ≤≤
=
totherfor
attfortu
ty
0
)(
)( 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: 
 


 ≤≤
=
totherfor
atfortu
tp
0
0)(
)( 
 


 >
=
totherfor
atfortu
tq
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. 
Linear System Theory and Design SA01010048 LING QING 
 4 
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: 
 02211
02211
022011 ),()(
),()(
)()(
tttyty
tttutu
txtx
≥+→



≥+
+
αα
αα
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
tuif
tuiftutu
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 
 5


=−
≠−−
=
0)1(0
0)1()1(/)(
)(
2
tpif
tpiftptp
tq 
 



=−
≠−−
=
0)1(0
0)1()1(/)(2
tuif
tuiftutuα
 
 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 
 6 
 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 += 
2
yxT −= 
 Then: 
 )
2
,
2
(),( yxyxgTtg −+= 
 )
2
,
2
( ayxayxg +−++= 
 )
22
,
22
( yxyxyxyxg +−+−+−++= 
 )0,(yg= 
 So: 
 0)0,(),( =
∂
∂
=
∂
∂
x
yg
x
Ttg
 
 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 
Linear System Theory and Design SA01010048 LING QING 
 7 
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
)(
tt
tt
tg 
 



≤≤−
≤≤
=
211
101
)(
t
t
tu 
 then y(t) equals to the convolution integral: 
 ∫ −=
t
drrturgty
0
)()()( 
 If 0=<t=<1, 0=<r=<1, 0<=t-r<=1: 
 ∫=
t
rdrty
0
)(
2
2t
= 
 If 1<=t<=2: 
 )(ty ∫
−
−=
1
0
)()(
t
drrturg ∫
−
−+
1
1
)()(
t
drrturg ∫ −+
t
drrturg
1
)()( 
 )(1 ty= )(2 ty+ )(3 ty+ 
 Calculate integral separately: 
 )(1 ty ∫
−
−=
1
0
)()(
t
drrturg 10 ≤≤ r 21 ≤−≤ rt 
 ∫
−
−=
1
0
t
rdr
2
)1( 2−−
=
t
 
 )(2 ty ∫
−
−=
1
1
)()(
t
drrturg 10 ≤≤ r 10 ≤−≤ rt 
 ∫
−
=
1
1t
rdr
2
)1(
2
1 2−
−=
t
 
 )(3 ty ∫ −=
t
drrturg
1
)()( 21 ≤≤ r 10 ≤−≤ rt 
 ∫ −=
t
drr
1
)2(
2
1)1(2
2 −
−−=
tt 
 )(ty )(1 ty= )(2 ty+ )(3 ty+ 242
3 2 −+−= tt 
Linear System Theory and Design SA01010048 LING QING 
 8 
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
1
32
1
)(
)()( 2 +
=
−+
−
==
sss
s
sY
sUsG 
 Impulse response: 
 te
s
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): 
s
sGsY 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 +=+ 
Linear System Theory and Design SA01010048 LING QING 
 9 
 Write to the form of matrix: 
 











)(
)(
)()(
)()(
2
1
2221
1211
sY
sY
sDsD
sDsD
= 











)(
)(
)()(
)()(
2
1
2221
1211
sU
sU
sNsN
sNsN
 
 





)(
)(
2
1
sY
sY
=
1
2221
1211
)()(
)()( −






sDsD
sDsD












)(
)(
)()(
)()(
2
1
2221
1211
sU
sU
sNsN
sNsN
 
 So the transfer function matrix is: 
 )(sG =
1
2221
1211
)()(
)()( −






sDsD
sDsD






)()(
)()(
2221
1211
sNsN
sNsN
 
 By the premising that the matrix inverse: 
 
1
2221
1211
)()(
)()( −






sDsD
sDsD
 
 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。 
Linear System Theory and Design SA01010048 LING QING10 
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
iany
1
)( 
 )()( 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
iany
1
)()( 
 )()( 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 





−
+




−
=
•
4
2
50
42
 
 [ ] 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? 
Linear System Theory and Design SA01010048 LING QING 
 11 
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(cos
2
2
2
θθθθθθ
•••
−−==− mll
dt
dmmgf 
 )sincos()sin(sin
2
2
2
θθθθθθ
•••
−==− mll
dt
dmfu 
 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: 
 u
mll
g 1
+−=
••
θθ 
 Select state variables as θ=1x , 
•
= θ2x and output θ=y 
 u
ml
x
lg
x 





+





−
=
•
/1
1
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 ldt
dmgmff =−− 
)cossin( 1
2
11111 θθθθ
•••
−−= lm 
 )sin(sinsin 112
2
11122 θθθ ldt
dmff =− 
)sincos( 1
2
11111 θθθθ
•••
−= lm 
 )coscos(cos 22112
2
2222 θθθ lldt
dmgmf +=− 
)cossin( 1
2
11112 θθθθ
•••
−−= lm )cossin( 2
2
22222 θθθθ
•••
−−+ lm 
 )sinsin(sin 22112
2
222 θθθ lldt
dmfu +=− 
)sincos( 1
2
11112 θθθθ
•••
−= lm )sincos( 2
2
22222 θθθθ
•••
−+ lm 
 
Linear System Theory and Design SA01010048 LING QING 
 12 
 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: 
2
11
2
1
11
21
1
)(
θθθ
lm
gm
lm
gmm
+
+
−=
••
 
u
lmlm
gmm
lm
gmm
22
2
21
21
1
21
21
2
1)()(
+
+
−
+
=
••
θθθ 
 Select state variables as 11 θ=x , 12
•
= θx , 23 θ=x , 24
•
= θx and output 






=





2
1
2
1
θ
θ
y
y
: 
 
























+−+
+−
=














•
•
•
•
4
3
2
1
22212121
1121121
4
3
2
1
0/)(0/)(
1000
0/0/)(
0010
x
x
x
x
lmgmmlmgmm
lmgmlmgmm
x
x
x
x
+ u
lm 











22/1
0
0
0
 






=





0100
0001
2
1
y
y












4
3
2
1
x
x
x
x
 
 
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: 
Linear System Theory and Design SA01010048 LING QING 
 13 
 
−
+−= 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
=










000
000
010


















−
−
−
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: 
Linear System Theory and Design SA01010048 LING QING 
 14 
 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
=










−
−
−
110
100
001










3
2
1
x
x
x
+ u










0
1
1
 
 [ ]










=
3
2
1
010
x
x
x
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
sssG
1
1
2 ++
+
=
ss
s
 
 
Linear System Theory and Design SA01010048 LING QING 
 15 
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
LRRLL
C
C










3
2
1
x
x
x
+ 














 −
2
1
11
1
1 /
0
/1
/1
0
0
u
u
LR
C
L
 
 [ ]










−−−=
3
2
1
111
x
x
x
Ry + [ ] 





2
1
21 u
u
R 
Applying Laplace Transform to upper equations: 
)(
/1/1
)( 1
^
21111
21
^
su
sCsCRRsL
RsLsy
++++
+
= 
)(
/1/1
)/1)((
2
^
21111
2121 su
sCsCRRsL
sCRRsL
++++
++
+ 



++++
+
=
sCsCRRsL
RsLsG
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 13
1 q and 23
8 i as 
shown , thus the representation of x with respect to the basis ( )21 iq is 
′




3
8
3
1
, this can 
be verified from [ ]














=










=


=
3
8
3
1
11
03
3
8
3
1
3
1
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 22
3 q , thus the representation of 1q 
with respect to ( )22 qi , is 
′



−
2
32 , this can be verified from 
[ ]







−






=







−
=





=
2
3
2
21
20
2
3
2
1
3
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
2
1222
22
2
1222
1
'
121
12
3
1
111
====
=++==+−+==
=++==++==
∞∞
=
∑
iiii
i
i
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 , 
 
 
[ ] [ ]
[ ]′===−=
′−==′−==
111
3
1)(
132
14
1132
2
2
2212
'
122
1
1
111
u
u
qxqxqxu
u
u
qxu
 
The two orthomormal vectors are 










=










−=
1
1
1
3
1
1
3
2
14
1
21 qq 
In fact , the vectors 1x and 2x are orthogonal because 01221 =′=′ xxxx so we can only 
normalize the two vectors 










==










−==
1
1
1
3
1
1
3
2
14
1
2
2
2
1
1
1 x
x
q
x
x
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
i
liliii jiif
jiif
aaaa
1
'
1
0
 
 then 
[ ]
[ ]



=≠
≠≠
=
==














=′
=










=










=














=′
∑
∑∑
=
×
==
jiif
jiif
aageneralin
aaaa
a
a
a
aaaAA
I
aaaaaa
aaaaaa
aaa
a
a
a
AA
jl
m
i
il
nnjl
m
i
il
m
i
ii
m
mi
m
mmmm
m
mi
m
1
0
)(
100
010
001
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 










−−=









 −
=










=
1000
2210
4321
011
023
114
100
000
010
321 AAA 
134)(033)(123)(
3)(3)(2)(
321
321
=−==−==−=
===
ANullityANullityANullity
ArankArankArank
 
 
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 






























1
0
0
,
0
0
1
can be used as a 
basis of the range space of 1A , all columns of 2A are linearly independent ,so the set 



















−




















0
0
1
1
2
2
1
3
4
 can be used as a basis of the range spare of 2A 
let [ ]43213 aaaaA = ,where ia denotes of 3A 4321 aandaandaandaare 
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 −










=










−
−
−
1
0
1
21
33
12
 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 −










=










−
−
−
1
0
1
21
33
12
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 [ ],
21
33
12
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 










=










−−
1
2
3
1000
2210
4321
x how many parameters do you have ?
求方程










=










−−
1
2
3
1000
2210
4321
x 的同解,同解中用了几个参数? 
Let 










=










−−=
1
2
3
1000
2210
4321
yA we can readily obtain rank(A)= rank([A y ])=3 so 
this y lies in the range space of A and [ ]′−= 101 oxp 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 












−
+











−
=+=
0
1
2
1
1
0
0
1
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
1
0
2
0
0
1
1
1
0
0
4
0
α
α
αα
α
ααx for 
any real 1α and 2α 
the Euclidean norm of x is 
16168453
)()()42(
2121
2
2
2
1
2
2
2
1
2
21
2
1
+−−++=
−+−+−++=
αααααα
αααααx
 




















−
−
−
=⇒
=
=
⇒
=−+⇒=
∂
=−+⇒=
∂
11
16
11
4
11
8
11
4
11
16
11
4
08250
2
04230
2
2
1
12
2
21
1 x
x
x
α
α
αα
α
αα
α 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, 












−
+











−
=
0
1
2
1
1
0
0
1
αx for any real α 
the Euclidean norm of x is 


















−
−
=⇒=⇒=−⇒=
∂
∂
+−=++−+−=
1
6
1
6
3
6
5
6
102120
2261)2()1( 2222
x
x
x
αα
α
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
nu
nu
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 =− ],[ 1ρ under this condition , there will exist ]1[],1[],[ −nuuou  to meet the equation for any 
]0[],[ xandnx . 
 
3.12 given 












=












=












=
1
3
2
1
1
0
0
1000
0200
0120
0012
bbA what are the representations of A with respect to 
the basis { }bAbAbAb 32 and the basis { }bAbAbAb 32 , respectively? 给定












=












=












=
1
3
2
1
1
0
0
1000
0200
0120
0012
bbA 请问 A 关于{ }bAbAbAb 32 和基{ }bAbAbAb 32 的表
示分别是什么? 












=⋅=












=⋅=












=
1
16
32
24
,
1
4
4
1
,
1
2
1
0
232 bAAbAbAAbAbA 
we have [ ] [ ]












−
−
==∴
+−+−=
7100
18010
20001
8000
718208
3232
324
bAbAbAbbAbAbAbA
bAbAbAbbA
 thus the representation of A 
 
 
with respect to the basis { }bAbAbAb 32 is 












−
−
=
7100
18010
20001
8000
A 
bAbAbAbbA
bAbAbAbA
324
432
718208
1
48
128
152
,
1
24
52
50
,
1
12
20
15
,
1
6
7
4
+−+−=












=












=












=












=
 
[ ] [ ]












−
−
==∴
7100
18010
20001
8000
3232 bAbAbAbbAbAbAbA thus the representation of A with 
respect to the basis { }bAbAbAb 32 is 












−
−
=
7100
18010
20001
8000
A 
 
3.13 find Jordan-form representations of the following matrices 
写出下列矩阵的 jordan 型表示: 
.
20250
16200
340
.
200
010
101
.
342
100
010
,
300
020
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 










=
300
020
001
ˆ
1A 
the characteristic polynomial of 2A is 
 
 
2
23
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 
QAQ
j
jAand
jj
jjQ 2
1
2
100
010
001
ˆ
220
111
111
−=










−−
+−
−
=










−
−−+−−=the characteristic polynomial of 3A is )2()1()det()(
2
33 −−=−=∆ 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 3
1
3
200
010
001
ˆ
100
011
101
−=










=










−
−= 
the characteristic polynomial of 4A is 
3
44 )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 










−
−==










= −
450
340
001
000
100
010
ˆ
4
1
4 QwhereQAQA 
 
 
 
3.14 consider the companion-form matrix 











 −−−−
=
0100
0010
0001
4321 αααα
A show that its 
characterisic polynomial is given by 43
3
2
3
1
4)( 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 43
3
2
3
1
4)( 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: 
4322
3
1
4
43
2
3
1
4
432
1
4321
2
1
det
1
0
det
10
01det
10
01
00
det)(
100
010
001
det)det()(
αλαλαλαλ
λ
αα
λ
λ
αλαλ
λ
λ
ααα
λ
λ
λ
αλ
λ
λ
λ
ααααλ
λλ
++++=






−
+





−
++=










−
−+










−
−+=












−
−
−
+
=−=∆ AI
 
if iλ is an eigenvalue of A , that is to say , 0)( =∆ iλ then we have 














=














=













 −−−−
=

























 −−−−
1110100
0010
0001 2
3
2
3
4
2
43
2
2
3
1
2
3
4321
i
i
i
I
i
i
i
i
i
i
iii
i
i
i
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
αλαλαλα
λ
λ
λαααα
 
that is to say [ ]133 ′iii λλλ is an eigenvetor oa A associated with iλ 
 
 
 
3.15 show that the vandermonde determinant 














1111
4321
2
4
2
3
2
2
2
1
3
4
3
3
3
2
3
1
λλλλ
λλλλ
λλλλ
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














1111
4321
2
4
2
3
2
2
2
1
3
4
3
3
3
2
3
1
λλλλ
λλλλ
λλλλ
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
12
24142313
2414423133
141312
24142313
2414423133
141312
144133122
14
2
413
2
312
2
2
141312
144133122
14
2
413
2
312
2
2
4321
2
4
2
3
2
2
2
1
3
4
3
3
3
2
3
1
))()()()()((
)])()()(())()()(()[(
)(
))(())((
))(())((
det
))(())((0
))(())((0
det
)()()(
)()()(
det
1111
0
)()()(0
)()()(0
det
1111
det
≤<≤Π=
−−−−−−=
−−−−−−−−−−−=
−⋅





−−−−
−−−−
−=










−−−
−−−−
−−−−
−=










−−−
−−−
−−−
=














−−−
−−−
−−−
=














ji
λλλλλλλλλλλλ
λλλλλλλλλλλλλλλλλλλλ
λλ
λλλλλλλλ
λλλλλλλλλλ
λλλλλλ
λλλλλλλλ
λλλλλλλλλλ
λλλλλλ
λλλλλλλλλ
λλλλλλλλλ
λλλλλλ
λλλλλλλλλ
λλλλλλλλλ
λλλλ
λλλλ
λλλλ
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
1111
0
0
0
det
222
222
333
222
333
bcbdcdbcddcb
dcb
bcd
dcb
dcb
dcb
dcb
dcb
dcb
−−−−=










⋅−=










=














and from vandermonde determinant 
 
 
))(())()()((
1111
0
0
0
det
1111
det
222
333
2222
3333
bdcdbcdabacbdcd
dcb
dcb
dcb
dcba
dcba
dcba
−−=−−−−=














=














 
so we can see 














−=














1111
0
0
0
det
1111
0
0
0
det
222
333
222
333
dcb
dcb
dcb
dcb
dcb
dcb
 , 
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 ==== 
( )
tindependenlinenrlyqqand
q
so
ddd
ddc
bbb
aaa
qqqq
qdqcqbqa
qdqcqbqa
qdqcqbqa
qqqq
ii
ni
nii ⇒=
≠
=
=














⇒








=+++
=+++
=+++
=+++
×
×
×
00
0
0
1
1
1
1
0
0
0
0
1
1
44
32
32
32
32
44332211
44
3
33
3
22
3
11
3
44
2
33
2
22
2
11
2
44332211
44332211
α
α
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
1
1000
0100
0010
α
α
α
α
α
α
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
1
1000
0100
0010
α
α
α
α
α
α
α
A 
proof: as we know , the chacteristic polynomial is 
43
2
2
3
1
4)det()( αλαλαλαλλλ ++++=−=∆ AI so let 0=λ , we have 
 
 
4
4 )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
1
1000
0100
0010
1000
0100
0010
0001
1
1000
0100
0010
0100
0010
0001
1000
0100
0010
0001
0100
0010
0001
1
1000
0100
0010
α
α
α
α
α
α
α
α
α
α
α
α
α
α
αααα
αααα
α
α
α
α
α
α
α
A
I
I
 
3.17 consider 









=
λ
λλ
λλλ
00
0
2/2
tT
tTT
A with 0≠λ and T>0 show that [ ]′100 is an 
generalized eigenvector of grade 3 and the three columns of 










=
000
00
02/222
tT
TT
Q λ
λλ
 
constitute a chain of generalized eigenvectors of length 3 , venty 










=−
λ
λ
λ
00
10
01
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










=−
λ
λ
λ
00
10
01
1 tAQQ 
Proof : clearly A has only one distinct eigenvalue λ with multiplicity 3 
 
 
0
1
0
0
000
000
000
1
0
0
000
00
2/0
000
000
00
1
0
0
)(
0
0
0
1
0
0
000
000
00
1
0
0
)(
222
3
2222
2
=




















=






























=










=−
≠










=




















=










=−
T
TTT
IA
TT
IA
λ
λλλ
λ
λλ
λ
 
these two equation imply that [ ]′100 is a generalized eigenvctor of grade 3 , 
and 










=




















=










=−










=




















=










=−
0
0
0000
00
2/0
0
)(
01
0
0
000
00
2/0
1
0
0
)(
2222222
222
T
T
T
T
TT
T
T
IA
T
T
T
TT
IA
λ
λ
λ
λ
λλ
λ
λ
λ
λ
λ
λ
λλ
λ
 
that is the three columns of Q consititute a chain of generalized eigenvectors of length 3 










=










=




















=




















=




















=
−
λ
λ
λ
λ
λλ
λλλ
λ
λ
λ
λ
λλ
λ
λ
λ
λ
λλ
λλλ
λ
λλ
λ
λλ
λλλ
00
10
01
00
0
2/2/3
00
10
01
100
00
02/
00
10
01
00
0
2/2/3
100
00
02/
00
0
2/
1
2
22223222
222232222
AQQ
TT
TTT
T
TT
Q
TT
TTT
T
TT
T
TT
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, 
)(
000
000
000
000
)(
000
000
000
001
)(
000
000
010
001
)(
000
000
010
001
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
dcba
















































λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
 
 
 
)()()()()(
)()()()()(
)()()()()(
)()()()()()()(
14
4
14
2
13
4
13
3
12
4
12
2
3
112
3
11
λλλλλλ
λλλλλλ
λλλλλλ
λλλλλλλλλλ
−=Ψ−=∆
−=Ψ−=∆
−=Ψ−=∆
−−=Ψ−−=∆
d
c
b
a
 
 
 
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 
1
110)(
−
−+++=
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 =
xfxh
xxx
xAAIxAhxAf
xxAxxAxxAxAxxAf
n
n
n
n
kk
)()(
)()()(
,,)(
1
110
1
110
3322
λλ
λβλββ
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
J
J
J
A  where 
mnma
nn
J
ii
l
i
i
nn
i
ii
≤×
=










= ∑=
×
1
0
10
10


 
from the nilpotent property , we have i
k
i nkforJ ≥= 0 so if 
 
 
,ii nmamk ×≥≥ liallJ
k
i ,,2,10 == and 0ˆ
3
2
1
=










=
k
k
k
J
J
J
A  
then )0)ˆ(0)((0 === AfifonlyandifAfAk 
If ,0 mkforAk ≥= then ,0ˆ mkforAk ≥= where 










=
3
2
1
ˆ
J
J
J
A  , nnJ
l
i
i
nni
i
i
i
ii
=










= ∑
=
×
11
1
λ
λ
λ


 
So we have 
 
liformnand
li
nnk
kk
J
ii
k
i
k
i
nk
i
ii
k
i
k
i
i
k
i



,2,1,0
,2,10
)!1)(1(
! 11
=≤=∴
==
















−+−
=
+−−
λ
λ
λ
λλλ
 
which implies that A has only one distinct eigenvalue o with multiplicity n and index m or less , 
 
3.21 given 










=
100
100
011
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 
21
10
210
10
0
10
2110)1()1(
1)1()1(
0)0()0(
ββ
βββ
β
+=⋅′=′
++==
==
hf
hf
hf
 
the we have 98,0 220 =−== βββ 










=










+










−=
+−=
100
100
911
100
100
101
9
100
100
011
8
98 210 AAA
 
the compute 
 
 
102
101
0
21103
1
0
2
1
0
21
103
210
103
0
103103
=
−=
=
⇒
+=⋅
++=
=
β
β
β
ββ
βββ
βA










=










+










−=
+−=
100
100
10211
100
100
011
102
100
100
011
101
102101 2103 AAA
 
 to compute Ate :
1
22
1
2 2
1
0
21
210
0
0
+−=
−−=
=
⇒
+=
++=
=
tt
tt
t
t
ete
tee
te
e
e
β
β
β
ββ
βββ
β
 










−
+−−
=










+−+










−−+










=
++=
t
t
tttt
tttt
At
e
e
eteee
eeee
AAIAe
00
110
11
100
100
111
)12(
100
100
011
)22(
100
010
001
2
210 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
e
e
eeeee
e
e
eee
Cd
Cc
e
e
e
Cb
QwhereQQeeCa
3
2
33
3
2
32
1
3
2
1ˆ
00
00
)(5)(4
100
010
541
00
00
54
100
010
541
00
00
00
100
010
541
100
010
541
,11
 
 
 










+−−
+
++
=




















−
=
=∴










−
−==










=
−
120250
161200
232/541
100
10
2/1
450
340
001
450
340
001
000
100
010
ˆ
22
2
1
4
44
4
tt
tt
tttt
t
tt
QQee
QwhereQQeA
tAtA
tA
 
method 2: the characteristic polynomial of 1A is ).3)(2)(1()( −−−=∆ λλλλ let 
2
210)( λβλββλ ++=h on the spectrum of 1A , we have 
)2(
2
1
2
34
2
5
33
93)3()3(
32)2()2(
)1()1(
32
3
32
2
33
0
2210
3
210
2
10
ttt
ttt
ttt
t
t
t
eee
eee
eee
ehf
ehf
ehf
+−=
−++−=
+−=
⇒
++==′
++==
++==
β
β
β
βββ
βββ
βββ
 









 −−
=










+−+










−+−+










+−
+−
+−
=
++=
t
t
ttttt
ttt
ttt
ttt
ttt
ttt
tA
e
e
eeeee
eee
eee
eee
eee
eee
AAIe
3
2
32
32
32
32
32
32
2
12210
00
00
)(5)(4
900
040
40121
)2(
2
1
300
020
1041
)
2
34
2
5(
3300
0330
0033
1 βββ
the characteristic polynomial of 4A is 
3)( λλ =∆ , let 2210)( λβλββλ ++=h on the 
spectrum of 4A ,we have 
2
2
1
0
2:)0()0(
:)0()0(
1:)0()0(
β
β
β
=′′=′′
=′=′
==
thf
thf
hf
 
 
 
thus 










+−−
+
++
=










+










−−
+
+=
++=
120250
161200
232/541
000
000
450
2/
20250
16200
2340
22
2
2
2
41410
4
tt
tt
tttt
t
tt
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
1
110
)(
)(
−
−
−
−
+++=
+++=
n
n
n
n
AAIAf
AAIAg
ααα
βββ


 ( n is the order of A) 
 
k
ik
k
nki
i
n
nk
k
ik
k
i
i
n
k
n
nn
n
in
n
i
i
n
in
n
i
i
AA
AAAAIAgAf
)()(
)()()(
1
22
0
1
0
22
11
1
1
1
1
0
011000
−
+−=
−
=
−
=
−
=
−
−−−
=
−
−−
−
=
∑∑∑∑
∑∑
+=
+++++++=
βαβα
βαβαβαβαβαβα 
)()()()()()(
1
22
0
1
0
AgAfAAAgAf kik
k
nki
i
n
nk
k
ik
k
i
i
n
k
=+= −
+−=
−
=
−
=
−
=
∑∑∑∑ βαβα 
let AteAgAAf == )(,)( then we have AeAe AtAt = 
 
3.24 let 










=
3
2
1
00
00
00
λ
λ
λ
C , find a B such that CeB = show that of 0=iλ for some 
I ,then B does not exist 
let 










=
3
2
1
00
00
00
λ
λ
λ
C , 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
00
00
00
λ
λ
λ
C 证明若 0111 =⋅⋅ λλλ ,则不存在 B 使 Ce
B = 
若










=
3
2
1
00
00
00
λ
λ
λ
C ,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
ln00
0ln0
00ln
ln
λ
λ
λ
CB where 3,2,1,01 => iλ , if 0=λ for some i , 
1lnλ does not exist 
for 










=
λ
λ
λ
00
00
01
C we have 










=









 ′
==
λ
λ
λλ
λ
λ
λλ
ln00
0ln0
01ln
ln00
0ln0
0nlln
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)( AsIAdj
s
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)( AsIAdj
s
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
010
101
)2(\)1()()2()1()(
200
010
001
ˆ
200
010
101
2
2
33
−=










−
−−
−−−−
=−










−
−
−
=−
−−=−−=∆










=









 −
=
ssmso
s
ss
sss
AsIAdj
s
S
S
AsI
ssssss
AA
j
we can easily obtain that )()()( smss ∆=j 
 
3.26 Define [ ]1221101 )(
1)( −−
−−− ++++
∆
=− nn
nn RsRsRsR
s
AsI  where 
in
nn RandsssAsIs ααα ++++=−=∆ −− 22
1
1
2:)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 
IAR
n
ARtr
IAAAIARR
n
ARtr
IAAIARRARtr
IAIARRARtr
IRARtr
nnn
nn
nn
nnnn
αα
ααααα
αααα
ααα
α
+=−=
++++=+=
−
−=
++=+=−=
+=+=−=
=−=
−
−−
−−
−−−−
1
1
12
2
1
1
121
1
1
21
2
212
1
3
1101
1
2
0
0
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:)( 12
2
1
1
0
1
−−
−−− ++++
∆
=− nn
nn RSRSRSR
s
ASI  其中 )(s∆ 是 A 的特征多项
式 in
nnn RandsssAsIs ααα ++++=−=∆ −− 22
1
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
IAR
n
ARtr
IAAAIARR
n
ARtr
IAAIARRARtr
IAIARRARtr
IRARtr
nnn
nn
nn
nnnn
αα
ααααα
αααα
ααα
α
+=−=
++++=+=
−
−=
++=+=−=
+=+=−=
=−=
−
−−
−−
−−−−
1
1
12
2
1
1
121
1
1
21
2
212
1
3
1101
1
2
0
0
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: 
implieswhich
SI
SSSSI
ARSARRSARRSARRSR
RSRSRSRASI
nn
nnn
nnn
nnn
nn
nn
)(
)(
)()()(
])[(
1
2
2
1
1
121
2
12
1
010
12
2
1
1
0
∆=
+++++=
+−+−+−+=
++++−
−
−−
−−−
−−
−−
−−
αααα 


 
 
 
][
)(
1:)( 12
2
1
1
0
1
−−
−−− ++++
∆
=− nn
nn RSRSRSR
s
ASI  ’ 
where n
nnn ssss ααα ++++=∆ −− 22
1
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: 
IAR
IARR
IARR
IARR
IR
nn
nnn
α
α
α
α
=−
=−
=−
=−
=
−
−−−
1
121
212
101
0

 
multiplying ith equation by 1+−inA yields ( ni 2,1= ) 
IAR
ARAAR
ARARA
ARARA
ARA
nn
nnn
nnn
nnn
nn
α
α
α
α
=−
=−
=−
=−
=
−
−−−
−−−
−−
1
12
2
1
2
21
12
2
1
101
1
0

 
then we can see 
012
2
101
1
0
1
2
2
1
1
=−−++−++=
+++++
−−−
−
−
−−
nnn
nnn
nn
nnn
ARRAARRARARA
IAAAA

 αααα
that is 
0)( =∆ A 
 
 
3.28 use problem 3.26 to show 
[ ]IssAssAsA
s
AsI
n
nnnnn )()()(
)(
1
)(
1
2
1
13
21
22
1
1
1
−
−−−−−
−
++++++++++
∆
=
−
ααααα 
 
利用题 3.26 证明上式, 
Proof: 
 
 
[ ]IssAssAsA
s
IAAASAAA
SIAASIAS
s
RSRSRSR
s
ASI
n
nnnnn
nn
nn
nn
nn
nnn
nn
nn
)()()(
)(
1
])(
)()([
)(
1
][
)(
1:)(
1
2
1
13
21
22
1
1
12
2
1
1
43
3
1
2
3
21
22
1
1
12
2
1
1
0
1
−
−−−−−
−−
−−
−−
−−
−−−
−−
−−−
++++++++++
∆
=
+++++++++
++++++
∆
=
++++
∆
=−
ααααα
αααααα
ααα




another : let 10 =α 
[ ]IssAssAsA
s
IAAASAAA
SIAASIAS
ASAS
s
AS
s
SR
s
RSRSRSR
s
ASI
n
nnnnn
nn
nn
nn
nn
nnn
n
i
In
i
N
i
In
i
n
ii
n
i
L
li
n
i
inin
i
nn
nn
)()()(
)(
1
])(
)()(
)(
1
)(
1
)(
1
][
)(
1:)(
1
2
1
13
21
22
1
1
12
2
1
1
43
3
1
2
3
21
22
1
1
1
0
1
1
0
01
1 1
0
1
0
11
12
2
1
1
0
1
−
−−−−−
−−
−−
−−
−−
−−−
−
=
−−
−
=
−−
−
=
−
=
−
−
=
−−−−
−−
−−−
++++++++++
∆
=
+++++++++
++++++
+++
∆
=
∆
=
∆
=
++++
∆
=−
∑∑
∑ ∑∑
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
p
p
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
p
p
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
λ
λ
λ
 
PAPA ˆ=∴ That is 














=














⇒
























=














nnnn
n
n p
p
p
Ap
Ap
Ap
p
p
p
A
p
p
p
λ
λ
λ
λ
λ
λ



22
11
2
1
2
1
2
1
2
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 
∑ −=−
−
ii
i
pq
s
ASI
λ
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
∑ −=−
−
ii
i
pq
s
ASI
λ
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
p
p
Q

2
1
1: where is a left eigenvector of 
A associated with iλ , 
∑ ∑
∑∑
∑
=−
−
=
−
−
=−
−
=
−
−
iiiii
i
iiiii
i
iiii
i
ii
i
pqpqs
s
pqpqs
s
Apqpqs
s
ASIpq
s
))((1
)(1)(1
)(1
λ
λ
λ
λλ
λ
 
That is ∑ −=−
−
ii
i
pq
s
ASI
λ
1)( 1 
 
 
 
 
3.31 find the M to meet the lyapunov equation in (3.59) with 






==





−−
=
3
3
3
22
10
CBA 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? 






=⇒=





−
∴
=+
3
0
12
13
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
3
3
3
1
21
10
21 CCBA with two 
different C ,用本题给出的 A, B, C, 重复题 3.31 的问题, 
α
α
α
anyforMCM
solutionNoCM
CMBAM






−
=⇒=





−−
⇒=





−−
∴
=+
311
11
11
11
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
201
000
100
202
013
232
aaaaaa
aaaaaa
aaaaaa
 
 
 
1. 










∴<−=−=





202
013
232
0792
13
32
det is not positive definite , nor is pesitive 
semidefinite 
2. )21)(21(
201
00
10
det −−+−=










−
−
λλλ
λ
λ
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 =










=





=





=





aaaaaa
aaaaaa
aaaaaa
aaaa
aaaa
aaaa
aaaa
aaaa
aaaa
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, 






−
−
=










−
−






−
−





−






−
−
52
22
01
10
21
012
101
42
21
012
101
 
)6)(1(674)5)(2()( 2 −−=+−=−−−=∆ λλλλλλλ 
the eigenvalues of 










−
−






−
−
01
10
21
012
101
are 6 and 1 , thus the singular values of 






−
−
012
101
are 6 and 1 , 






=




−





−
206
65
42
21
42
21
 
)41
2
3
2
25)(41
2
3
2
25(36)20)(5()( −−+−=−−−=∆ λλλλλ the eigenvalues of 





−





−
42
21
42
21
 are 41
2
3
2
2541
2
3
2
25
−+ and , thus the singular values of 
 
 





−
42
21
are 70.1)41
2
3
2
25(70.4)41
2
3
2
25( 2
1
2
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(22niqqAqAqA iiiiiii ==== λλλ 
Which implies 2iλ is the eigenvalue of niforA ,2,1
2 = , (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
m
mn
n
n babbb
a
a
a
I ∑
=
+=


























+
1
21
2
1
1det 

 
证明上式成立 
let [ ]n
n
bbbB
a
a
a
A 
 21
2
1
=












= A is 1×n and B is n×1 
use (3.64) we can readily obtain 
[ ] [ ]
∑
≤
+=


























+=


























+
n
m
mm
n
nn
n
n
ba
a
a
a
bbbIbbb
a
a
a
I
1
2
1
21121
2
1
1
detdet



 
 
 
3.37 show (3.65) 证明(3.65) 
proof: let 
 
 






−
−
=





=





=
n
m
n
m
n
m
IsB
AIsP
IsB
IsQ
Is
AIsN 0
0
 
 then we have 






−
−
=




 −
=
BAsI
AssIQP
sIBs
ABsI
NP
n
m
n
m
0
0
 
because 
PsPQQP
PsPNNP
detdetdet)det(
detdetdet)det(
==
==
 
 we have det(NP)=det(QP) 
And 
)det()det()det()det(
)det()det()det()det(
BAsISBAsIsIQP
ABsISsIABsINP
n
m
nm
m
n
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 





−
=
01
10 
试证其解为： 
Show that its solution is )0(
cossin
sincos
)( X
tt
tt
tX 





−
= 
Proof: )0()0()( 01
10
XeXetX
t
At 




−== ,the eigenvalues of A are j,-j; 
Let λββλ 10)( +=h .If 
teh λλ =)( ,then on the spectrum of A,then 
 
tjtejjh
tjtejjh
jt
jt
sincos)(
sincos)(
10
10
−==−=−
+==+=
−ββ
ββ
 then 
t
t
sin
cos
1
0
=
=
β
β
 
so 





−
=





−
+





=+=
tt
tt
ttAIAh
cossin
sincos
01
10
sin
10
01
cos)( 10 ββ 
)0(
cossin
sincos
)0()0()( 01
10
X
tt
tt
XeXetX
t
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 





+





−−
=
1
1
22
10 
[ ]XY 32= 
Answer: assuming the initial state is zero state. 
method1:we use (3.20) to compute 






−
+
++
=





+
−
=−
−
−
s
s
sss
s
AsI
2
12
22
1
22
1
)( 2
1
1 
then tAt e
ttt
ttt
AsILe −−− 





−−
+
=−=
sincossin2
sinsincos
))(( 11 and 
)22(
5
)22(
5)()()( 22
1
++
=
++
=−= −
sssss
ssBUAsICsY 
then tety t sin5)( −= for t>=0 
method2: 
t
tt
tt
Att tA
t tA
te
tete
tete
C
BeeCABdeC
dBueCty
−
−−
−−
−−
−
=






−−
−+





 −−
=
−==
=
∫
∫
sin5
1sin3cos
1sin2cos
01
5.01
)(
)()(
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
kBUdekXekX
T 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
][
1108.06191.0
3096.05083.0
]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
][
0432.00
00432.0
]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










+










−−
−
=
1
0
1
220
101
002
 
[ ]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










+










−
−
−
=
0
0
1
410
601
400
 
[ ]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









−
+










−
−−
−
=
4142.1
0
4641.3
200
011
011
 
[ ]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
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
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










+










−−
−
=
1
0
1
240
5.005.0
002
 
[ ]XY 021 −= 
the largest permissible a is 10/0.55=18.2 
4.6 Consider U
b
b
XX 





+





=
2
1
0
0
λ
λ [ ]XccY 11= 
where the overbar denotes complex conjugate.Verify that the equation can be transformed into 
XCyuBXAX 1=+=
 
with [ ])Re(2)Re(2
1
010
11111 cbcbCBA λλλλλ
−=





=





+−
= ,by using the 
transformation XQX = with 





−
−
=
11
11
bb
bb
Q
λ
λ
。试验证以上变换成立。 
Answer: let XQX = with 





−
−
=