北邮通信原理答案第九章
9.1
解:
① 由生成矩阵与检验矩阵的关系,易求:
⎛1101100⎫ ⎪H = 1011010⎪
1110001⎪⎝⎭ ② 由伴随式的公式,得
⎛1101100⎫ ⎪T
s =yH =(1101101) 1011010⎪=(001)
1110001⎪⎝⎭
T
9.2
解:
① 由H 阵求出G 阵:
⎛1000101⎫ ⎪ 0100111⎪G =
0010110⎪ ⎪ 0001011⎪⎝⎭
② 首先将m 分组,四位码一组,不足的用0补, 得m 1=1101,m 2=0110,m 3=1010,则C i =m i ⋅G
⎛1 0
∴C 1=(1101)
0 0⎝
C 2=m 2⋅G =(011C 3=m 3⋅G =(101
[***********]
101111011) 1)
1⎫⎪1⎪
=(1101001) ⎪0⎪1⎪⎭
9.3
解:利用G 的初等行变换来完成相应的变换
① 1行←→4行,2行←→3行,得:
⎛1011000⎫ ⎪ 0101100⎪G '=
0010110⎪ ⎪ 0001011⎪⎝⎭
② 2行+4行,1行+3行+4行,得:
⎛1000101⎫ ⎪ 0100111⎪G ''=
0010110⎪ ⎪ 0001011⎪⎝⎭
故G ''就是所求的系统码生成阵。
9.4
解:
① 由初等行变换得:
⎛1001110⎫ ⎪G ''= 0100111⎪
0011101⎪⎝⎭ ② 求相应的检验矩阵
⎛1011000⎫ ⎪ 1110100⎪H ''=
1100010⎪ ⎪ 0110001⎪⎝⎭
③ 由伴随式的定义,得到伴随式表
S =(S 0S 1S 2S 3) 陪集首
0000 0001 0010 0100 1000 1101 0111 1110 0011 0101 1001 1100 0110 1111 1010 1011
④ 由定义可知 d min =4
0000000 0000001 0000010 0000100 0001000 0010000 0100000 1000000 0000011 0000101 0001001 0010001 0100001 1000001 0001010 0001011
9.5
解:由题可知,该线性码为(7,3)码, d min =4,最多可纠一个错。而
⎛1101000⎫ ⎪ 0110100⎪H =
1110010⎪ ⎪ 1010001⎪⎝⎭
所以,其可纠差错图样及伴随式如下: S E 0001 0000001 0010 0000010 0100 0000100
1000 0111 1110 1011
0001000 0010000 0100000 1000000
9.6
解:
723
① h (x ) =(x +1) /g (x ) =1+x +x
⎛0010111⎫ ⎪
G (x ) = 0101110⎪
1011100⎪⎝⎭ ②
⎛10011
G ''(x ) = 01001
00111⎝其系统码
⎛101100
111010H ''=
110001
011000⎝而对应的
10⎫
⎪11⎪01⎪⎭
0⎫⎪0⎪0⎪⎪1⎪⎭
9.7
解:
已知(7,4)汉明码的生成矩阵和校验矩阵如下:
1⎫
⎪
⎛0111100⎫1⎪
⎪
⎪H =10110101 ⎪⎪ 1110001⎪0⎪⎝⎭, ⎭,
所以缩短码(6,3)只需要除去原(7,4)码生成矩阵的第一行,及对应校验矩阵的第一列就可以了,即
⎛0100101⎫⎛111100⎫ ⎪ ⎪G '= 0010111⎪H '= 011010⎪
0001110⎪ 110001⎪⎝⎭,⎝⎭
⎛1 0G =
0 0⎝
[***********]11
9.8
解:分别列出
23
① G (x ) =1+x +x
3
② G (x ) =1+x +x
234
③ G (x ) =1+x +x +x
34
④ G (x ) =1+x +x +x
234
H (x ) =1+x +x +x H (x ) =1+x +x 3+x 4 23
H (x ) =1+x +x H (x ) =1+x +x 3
23456
⑤ G (x ) =1+x +x +x +x +x +x ⑥ G (x ) =1+x
H (x ) =1+x
H (x ) =1+x +x 2+x 3+x 4+x 5+x 6
9.9
1524581035
1+x =(1+x +x +x +x +x +x )(1+x +x +x ) 证明:已知
2458103515
(1+x +x +x +x +x +x )(1+x +x +x ) =0mod (1+x ) , 即
所以 得证。
35
h (x ) =1+x +x +x 同时,有。
9.10
解:
151187532
① h (x ) =(1+x ) /g (x ) =x +x +x +x +x +x +x +1 ② 已知系统码的生成矩阵
⎛x n -1+r n -1(x ) ⎫ ⎪n -2
x +r n -2(x ) ⎪G = ⎪
⎪ x n -k +r (x ) ⎪
n -k ⎝⎭
⎛x 14+r 14(x ) ⎫
⎪13
x +r 13(x ) ⎪G = ⎪
⎪ x 4+r (x ) ⎪
4⎝⎭ 所以对于本题
由公式得
32
r 14=x 3+1,r 13=x +x +1,r 12=x 3+x 2+x +1,
32
r 11=x 3+x 2+x ,r 10=x +x +1,r 9=x +x ,
r 8=x 2+1,r 7=x 3+x +1,r 6=x 3+x 2, r 5=x 2+x ,r 4=x +1,
1⎛
1
1
1
G '''= I 11⨯11 1
1
1
0 0⎝所以
[1**********]
1001101101
1⎫⎪1⎪1⎪⎪0⎪1⎪⎪0⎪⎪1⎪1⎪⎪0⎪0⎪⎪1⎪⎭
③ 缩短为(8,4)后的生成矩阵和监督矩阵分别为
⎛0
0G '=
0 0⎝⎛1 0H '=
1 1⎝
[**************]0
[1**********]00⎫
⎪0⎪ I 4⨯4⎪1
⎪⎪1⎭
[***********][1**********]011
1⎫⎪0⎪0⎪⎪1⎪⎭
Y
1
7
9.11
解:
h (x ) =(1+x 15) /g (x ) =x 7+x 6+x 4+1
9.12
解:
只要满足G (x ) ⋅h (x ) =0mod (x +1) 即可,所以分别列出
2343442
(1) G (x ) =1+x +x +x +x ,H (x ) =(1+x +x )(1+x +x )(1+x +x )(1+x )
3423442
(2) G (x ) =1+x +x ,H (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x )
4234342
(3) G (x ) =1+x +x , H (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x ) 234344
(4) G (x ) =1+x +x , H (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x )
2343442
(5) G (x ) =1+x , H (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x +x ) 2343442
G (x ) =(1+x +x +x +x )(1+x +x ) H (x ) =(1+x +x )(1+x +x )(1+x ) (6) , 2344342G (x ) =(1+x +x +x +x )(1+x +x ) H (x ) =(1+x +x )(1+x +x )(1+x ) (7) , 2342344G (x ) =(1+x +x +x +x )(1+x +x ) H (x ) =(1+x +x )(1+x +x )(1+x ) (8) ,
2
15
(9)
G (x ) =(1+x +x 2+x 3+x 4)(1+x ) , H (x ) =(1+x 3+x 4)(1+x +x 4)(1+x +x 2)
3442342
G (x ) =(1+x +x )(1+x +x ) H (x ) =(1+x +x +x +x )(1+x +x )(1+x ) (10) , 3422344G (x ) =(1+x +x )(1+x +x ) H (x ) =(1+x +x +x +x )(1+x +x )(1+x ) (11) , 3423442G (x ) =(1+x +x )(1+x ) H (x ) =(1+x +x +x +x )(1+x +x )(1+x +x ) (12) , 4223434G (x ) =(1+x +x )(1+x +x ) H (x ) =(1+x +x +x +x )(1+x +x )(1+x ) (13) , 4G (x ) =(1+x +x )(1+x ) , H (x ) =(1+x +x 2+x 3+x 4)(1+x 3+x 4)(1+x +x 2) (14)
2G (x ) =(1+x +x )(1+x ) , H (x ) =(1+x +x 2+x 3+x 4)(1+x 3+x 4)(1+x +x 4) (15)
234344G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x ) , H (x ) =(1+x +x 2)(1+x ) (16)
234342G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x ) , H (x ) =(1+x +x 4)(1+x ) (17)
2343442
(18) G (x ) =(1+x +x +x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x )(1+x +x )
2344234
(19) G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x ) , H (x ) =(1+x +x )(1+x ) 2344342
(20) G (x ) =(1+x +x +x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x )(1+x +x )
2342344
(21) G (x ) =(1+x +x +x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x )(1+x +x ) 3442234
(22) G (x ) =(1+x +x )(1+x +x )(1+x +x ) , H (x ) =(1+x +x +x +x )(1+x )
3442342
(23) G (x ) =(1+x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x +x +x )(1+x +x ) 3422344
(24) G (x ) =(1+x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x +x +x )(1+x +x )
4223434
(25) (1+x +x )(1+x +x )(1+x ) , H (x ) =G (x ) =(1+x +x +x +x )(1+x +x ) 2343442
(26) G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x +x ) , H (x ) =(1+x )
2343442
(27) G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x )
2343424
(28) G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x ) 2344234
(29) G (x ) =(1+x +x +x +x )(1+x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x )
3442234
(30) G (x ) =(1+x +x )(1+x +x )(1+x +x )(1+x ) , H (x ) =(1+x +x +x +x )
9.13
解:
255
取本原多项式p (x ) =1+x +x ,周期P =2-1=31 此题中,令l =5,显然31不能整除2l -1=9, 故此Fire 码的生成多项式
g (x ) =(x 9+1)(1+x 2+x 5) =1+x 2+x 5+x 9+x 11+x 14 其码长n =LCM (9, 31) =279,
其监督位数r =n -k =m +2l -1=5+2⨯5-1=14 其信息位数k =n -r =279-14=265
, 265) 的循环码。能纠长度l =5或更短的任何突发差错。所以该Fire 码是一个(n , k ) =(279
9.14
解:
已知 t =3, n =15, m =4,由RS 码的性质 码距:d =2t +1=7个符号:7⨯4=28bit
监督段:n -k =2t =2⨯3=6个符号:6⨯4=24bit 码长:n =15个符号:15⨯4=60bit
因此该码是(n , k ) =(15, 9) 的RS 码,也可看作(n , k ) =(60, 36) 的二进制码。其生成多项式
g (x ) =(x +α)(x +α2)(x +α3)(x +α4)(x +α5)(x +α6)
=x 6+α10x 5+α14x 4+α4x 3+α6x 2+α9x +α6
9.15
解:
① 编码器结构
② 生成矩阵
⎛111001011⎫ ⎪ ⎪111001011G = ⎪
111001011 ⎪
⎪⎝⎭ ③ 状态转移图
④
a b c d
00100111
⑤由状态图作出转移函数,得d jm =6。
9.16
解:
① 初始状态为000,令输入为100……序列,则 ⎧g 1=a 1⊕a 3⎪
) ⎧g 1=g 2=(0101⎨g 2=a 1⊕a 3
⎨⎪g =a ⊕a ⊕a ) 123,所以
⎩g 3=(0111⎩31
②
000/0
9.17
解:
) ① g 1=a 1, g
2=a 1⊕a 2, ∴g 1=(010), g 2=(011
②
③
0000
0分支00
a(00)
a(00)
0011
b(10)
00
树根00
a(00)
110111
c(01)
1分支00
01
b(10)
10
d(11)
11
10
④
a b c d
00100111
9.18
解:
①
n =2, k =1, m =4
4
② g 1(x ) =1+x +x +x ;g 2(x ) =1+x +x ③
2
4
3
⎛1110100111⎫ ⎪ ⎪1110100111G = ⎪
1110100111 ⎪
⎪⎝⎭
) ⋅G 8行=([***********]100111) ④ c =(11010001
9.19
解:
① n =3, k =1, m =2
②
000000
0分支000
a(00)
a(00)
111110
b(10)
111
树根00
a(00)
001111110
c(01)
1分支00
000
b(10)
001
d(11)
001
110
a b c d
00100111
③ 利用状态图得出转移函数,由定义d f =6。
9.20
解:
22
① g 1(x ) =1+x ;g 2(x ) =x +x
⎛100111⎫ ⎪ ⎪100111G = ⎪
100111 ⎪
⎪⎝⎭ ②
0000
0分支00
a(00)
a(00)
1001
b(10)
10
树根00
a(00)
111101
c(01)
1分支10
01
b(10)
10
d(11)
11
00
a b c d
00100111
9.21
解: ①
②
(1)(4)g 1=(100) g 1(2)=(000) g 1(3)=(000) g 1=(111) (1)(2)(3)(4)g 2=(000) g 2=(100) g 2=(000) g 2=(101) (1)(2)(3)(4)g 3=(000) g 3=(000) g 3=(100) g 3=(110)
⎛1001
0101
0011
c =uG =(100101)
⎝③
[***********][***********][1**********]1
⎫⎪⎪⎪
⎪=([**************]1) ⎪0001⎪0001⎪0000⎪⎭
9.22
解:
① n =2, k =1, m =3
) ;g 2(x ) =(1101) ② g 1=(1000⎛110100
1101G =
11
⎝
③ c =uG =(11④
[1**********]1
⎪⎪⎪
⎪⎭
[1**********]101)
0⎫⎪
00/0