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

Title Solution of Linear System Theory and Design 3ed for Chi-Tsong Chen Mathematical Analysis Algebra Eigenvalues And Eigenvectors Numerical Analysis 949.2 KB 106
```                            Ch 2: Mathematical Descriptions of Systems
Ch 3: Linear Algebra
Ch 4: State-Space Solutions and Realizations
Ch 5: Stability
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
5.12
5.13
5.14
5.15
5.16
5.17
5.18
5.19
5.20
5.21
5.22
5.23
Ch 6: Controllability and Observability
Ch 7: Minimal Realizations and Coprime Fractions
Ch 8: State Feedback and State Estimators
```
##### Document Text Contents
Page 1

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。

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 =

Page 2

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
tt

tg

=
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

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

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:

mohammed
Highlight

mohammed
Highlight

Page 53

UXY

UXX

+

−−−

=

+

−−

=

11
00

223
322

00
10
01

010
230

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。

[ ] [ ] [ ][ ]

[ ] [ ]

[ ] [ ] [ ] [ ][ ] [ ]

[ ] [ ] 222

222

2

111

111

21

1101
26
23

02
13

112623
)2)(1(

1
11)1(2)2(3

)2)(1(
1

)(ˆ

0001
34

22
02
13

3422
23

1
3242

)2)(1(
1

)(ˆ

UXY

UXX

s
ss

ss
ss

sG

UXY

UXX

s
ss

ss
ss

sG

C

C

+=

−−
−−

+

=

+−−+−−
++

=++−+−
++

=

+=

+

=

−+
++

=−+
++

=

These two realizations can be combined as

UXY

UXX

+

=

−−
−−

+

=

11
00

0100
0001

26
23
34

22

0200
1300
0002
0013

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 



+
+

+
+−

=
343
2322

343
)612(

)(ˆ
s
s

s
s

sG

Page 54



+
+

+
+−

=
343
2322

343
)612(

)(ˆ
s
s

s
s

sG 的一种实现

[ ] [ ]

[ ]
[ ]UXY

UXX

ss
s

s
s

sG

3/224

9/6793/130
3

34

9/6793/130
3/34

1
3/224

343
2322

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

X
e

XandX
t

X
t

=

=

10
1

0
10 2



for the first case:
25.0

22222

10 2121

)0()()(ln

)0()()(
t

t

extxtdttxdtxx

xdttxtxxx

=⇒=⇒=

+=⇒= ∫

we have

=⇒

=

=⇒

= ∫

2

2

5.0
0

5.0

)(
1
0

)0(
0
1

)(
0
1

)0(
t

t d

e

e
tXXandtXX

tt

=

=

∫∫∫

)(5.0

5.05.0
1

5.0
0

5.0

5.0
0

5.0

0

5.0
0

5.0

2
0

2
0

22
0

2
0

0 2

2

2

2

2

0

1
0
1

0
1

),(

0
1

)(

t.thusindependenlinearly are states initial twoThe

tt

t

t

ddt

t

t d

t

t d

t

t d

e

ee

e

e

e

e
tt

and
e

e
tX

ttttttt

tt

for the second case:

t

tttt dttdt

tt

extxxx

exeexcdteexetx

xextxexx
t

−−−

=⇒−=

+−=+∫∫=

⇒+−=+−=

)0()(

)0())(0(5.0])0([)(

)0()(

2222

120 21

212
2

11

0

we have

 −
=⇒

=

=⇒

=

−−

t

ttt

e
ee

tXXand
e

tXX
)(5.0

)(
1
0

)0(
0

)(
0
1

)0(

the two initial states are linearly independent;thus

mohammed
Highlight

Page 105

(3)

[ ]

[ ] [ ]

[ ]xy

rzx

x
z
y

rzz

rzx

rzxx

r
z
y

xx

rrxxuxx

11
21
13

42
21

164
21

164

11)
215

214
14(

21
13

3

2
1

126
63

2321
1213

2
1

126
63

11
22
11

11
12

2
1

215
214

ˆ
28
14

11
12

2
1

2
1

ˆ
28
14

11
12

2
1

11
12

=

+−



=

+

−+−=

+

+

=

+

−

−

=

+

−

=

−

+

−

=

+

=

[ ]

[ ]

)2)(1(
43

)3)(2)(1(
)3)(43(

6116
1253

21
13
2
1

42
21

164
21

164
1262321
632113

011)(ˆ

011

21
13
2
1

42
21

164
21

164
1262321
631213

23

2

1

++

=
+++

+−
=

+++
−+

=

+

+−−

−−

=∴

=

+

=

ss
s

sss
ss

sss
ss

s
s

s
sg

z
x

y

r
z
x

z
x

yr

these three overall transfer functions are the same , which verifies the result discussed in section
8.5.

8.13 let

=

=

20
21
00
00

0012
3213
0100
0010

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 ±−

Page 106

solution : (1) select a 44× matrix

−−

−−

=

5400
4500

0043
0034

F

the eigenvalues of F are j34 ±− and j45 ±−

(2)if ),(,
0100
0001

11 kFk 

= is observable

 −−−−
==

−−
−−

−−
−−−

=⇒=−

2253.28747.141758.1190670.371
0000.22000.140000.1682000.606

1982.02451.00043.00006.0
0185.01731.00714.01322.0

0191.00193.00273.00126.0
0042.00005.00059.00013.0

1
111

1111

TKK

TKBFTAT

(3)IF 

=

0100
1111

2K ),( 2kF is observable

 −−−−
==

−−
−−−

=⇒=−

5853.24593.78815.595527.185
2509.30893.02145.559824.252

2007.02473.00037.00048.0
0245.03440.00607.02036.0
0306.00443.00147.00399.0
0081.00024.00071.00046.0

1
222

2222

TKK

TKBFTAT