数字通信原理课后答案-第5-7章习题

5.3

解：信号带宽

B

(108

60)kHz

48kHzs

minf

2B(1

M

)

2

48

(1

0.25

)

108kHzN

248108

2.25，N

2，M

0.252

1-483142216-216-10848

60

1085.5解：第一fs1

2，3，4z子带为带通B

800Hz每个子带的最高频率为带宽的M

0因此:fs

2

fs3

f

s

4

2B

1600Hz5.8

解：被抽样信号s(t)

3.25

sin(

2ft)

3.25

sin(1600t)抽样周期：Ts

1/f

s

1/8000信号频率：f

800，每周期抽取10个样本值抽样时间样值量化器输出t=000.5t=1TS=0.2

π1.911.5t=2TS=0.4

π3.093.5t=3TS=0.6

π3.093.5t=4TS=0.8

π1.911.5t=5TS=π0-0.5t=6TS=1.2

π-

1.91-1.5t=7TS=1.4

π-3.09-3.5t=8TS=1.6

π-3.09-3.5t=9TS=1.8

π-1.91-1.5t=10TS=2

π0-0.55.9+1-1已知模拟信号抽样值得概率密度f(x)如果按4电平进行均匀量化，计算量化信噪比如果按8电平进行均匀量化，计算量化信噪比f(x)1x解：由图可知：模拟信号取值范围[-1,1]，a=-1

,

b=11) 4电平量化，M=4，量化间隔Δ=(b-a)/M=0.5

a06b

1

1S

x2

f

(x)dx

2

x2

(1

x)dx

量化噪声：N量化电平：mi=a+iΔ分层电平：qi=mi+mi-1信号功率：-1，-0.5，0，0.5，1-1，-0.5，0，0.5，110.50.5

02(x

0.25)

(1

x)dx

2(x

0.75)2

(1

x)dx

1/

48S

/

Nq

8(x

q

)2

f

(x)dxM

i1mimii1q量化信噪比：5.10

解：

6.02n

1.76

30

40

n

12

40

6.02n

3021

4.77

20

lg2V

2S

3A2

L2当输入信号为正弦波，线性编码，量化信噪比:min最小信噪比：SNR

m

SNR

A2V动态范围：

20

lg

Amin

40dBAmaxD

，Amax

Vq满载正弦波的信噪比：SNRdB

4.77

20

lg

D

6.02n

25.141）求输出编的码组，计算量化误差4096

40965

5y

0.9375

0.9375

4096

768

5

，量化单位：

输入幅度为-768，因为是负数，所以M1为“0”。查表5-3，768落在第5段，M2M3M4=“101”以512为起点，32为量化间隔,（768-512）/32=8M4M5M6M7=“1000”总的编码为“01011000”输出：（512+32×8+32/2）=-784量化误差：768-784=-16，

0.0195

1.95%40965误差值

16

16

2)

线性码输出“0,0011,0001,0000”6.1

解22a

S

k

S(k)

S

(k)

a

S

k

11Ed

(k)

EEd

2

(k)ES

2

(k)1

rS

k

G

Ni1e

i

reP量化器编码dq

(k

)器S(k

)S

(k)eSr

(k)I

(k

)d

(k

)++DPCM编1)2

2222222rrrr

rrR(1)

ES

(k

)Sr

(k

1)Ed

2

(k

)

R(0)

2R(1)

R(0)

2R(0)

2R(1)ES

(k

)

ES

(k

1)R(0)

S

(k

)S

(k

1)

ES

(kES

(k

)

2ES

(k

1)

ES

(k

)

2S

(k

)S

(k

1)

ES

(k

)

S

(k

1)

Ed

(k

)0.351

1

2.857

1

2 2

0.8252

R(0)R(1)R(0)

1

1ES

2

(k)GP

GP

Ed

2

(k)

2R(0)

2R(1)

2

6.2解：

3.131

0.825

0.825111

R(1)

R(1)R(0)

R(0)R(1)1

a1opt

R(0)1R

R(0)

R(1)R(1)

R(1)

a

R(i)

aP144

公式6-17R(i)N1

a1Ni1i1iopt

ES

2

(k

)G

P.optG

P.opta1opta1opt

R(0)1opt

ES

2

(k

)iopt

ES

2

(k

)SS

rss6.9

解：当s(k

)

0时，M

(

I

(k

)

)

1,

量化间隔不需要调整sr

(k

)

dq

(k

)

se

(k)

dq

(k

)

1sr

(k

1)

dq

(k

)

0.85sr

(k

1)I

(k

)为量化器输出码字，即dq

(k

)0.2s(0)

0,

sr

(0)

0所以k

1,2,10周期T

2

10,一个周期内(k)值及sr

(k)如下量化d

(k)

s(k)

0.85(k

)

(kdq

(krks(k)d(k)dq(k)/I(k)/M

/I(k)/Δ(k)sr(k)0000010.01010.58780.58780.62551.20.0120.62520.95110.41970.37530.90.01080.906330.95110.18070.12510.90.00970.894540.5878-0.1733-0.12510.90.00870.6361500001.00.008706-0.5878-0.5878-0.62551.20.0104-0.6257-0.9511-0.4197-0.37530.90.0094-0.90638-0.9511-0.1807-0.12510.90.0085-0.89549-0.58780.17330.12510.90.0077-0.636110000-10.00707.1

解2

fssssaTa

f

2f（2）（1）临界斜率过载时：

nX(n)=anTsc(n)Xr(n)=Xr(n-1)+1.5aTs000(-0.375aTs)11aTs11.875atS22aTs13.375aTs33aTs01.875aTs44aTs13.375aTs55aTs14.875aTs66aTs16.375aTs77aTs17.875aTs88aTs19.375aTs99aTs07.875aTs1010aTs19.375aTs7.10

解：已知：f

1kHz,fs

32kHz,fL

140Hz,fB

3400

160

3240Hz最大量化信噪比：

384.73(32

103

)38

2

(103

)2

3240

2

q

max

S

B

s

8

2

f

2

f3

f

3L

B6P

f

2

2

f

f1

b

s

q

max

2

2

S

S

n

MenLPCM1

4(L2

1)PL22SNR

S

/

3.142

160

34006(32103

)2

10384.71

S

1384.76(32103

)2

13.142

160

3400

S

1384.76(32103

)2

103.142

160

3400

n

S

ΔM调制PCM调制1

4LPCM0.1SNR

1

4LPCM0.1SNR

1

4LPCM0.001SNR

8

36W信息

输

率：6W

6

最小传输带宽：288W

/26

6W11码元长度：时隙长度：8.19.10

1

0

0

1

1

0

0

0

0

0

1

0

0

1

1

1

0

0

0

0

0A双极性归零码-A差分传号码CMI码数字双向码条件双向码9.40

1

1

0

1

0

0

0

0

1

0

0

0

1

1

0

0

0

0

0

0

0

1

0AMIHDB3B6ZSB3ZS9.511000

10101

11011

00001111000 010101

011011

1000019.7sff

s

fsfH

f

112(1

)fs(1

)fsfsfH1

f

(1

)

fsf(1

)

fsH2

f

(1

)2(1

)

2(1

)

2ss1

2h1

(t)

2(1

)ft(1

)

f

Sa

ssf

t122(1

)(1

)

f

Sa

2h

(t)

22sin(

212h(t)

h1

(t)

h2

(t)

sssssss

s

s

s2

2f

2t

2f

t)f

t)

sin(

22f

tf

t)

sin(

22ftf

t)

sin(

2ssf

t(1

)(1

)

f

Sa

ssf

t(1

)(1

)

f

Sa

f

t)

Sa(2f

t)

2

f

2

f

Sa(212s3）定时优于理想低通，不及升余弦滚降信号9.7bs2B

(1

)

f

1

f1

2

fb

2

f

s

1

1

ffs

b2

Tb

2Tb

2bsT

码元重复频率码元周期等效带宽传输带宽sff

s

fsfH

f

121s(1

)fs(1

)f9.174）7电平部分响应信号7电平部分响应信号传输为4进制信号查Q函数表得：3

S

4.754 63

N

S

183Nlog

2

4

2156( )

21015

N

3

S( )

Q3

SL2

1

N

8

4Q

42(L2

1)Pe

P

PeP

eb6( )

1.07

10

4

15

NL2Q

3

S9.174）7电平部分响应信号7电平部分响应信号传输为4进制信号查Q函数表得：3

S

4.754 63

N

S

183Nlog

2

4

2156( )

21015

N

3

S( )

Q3

SL2

1

N

8

4Q

42(L2

1)Pe

P

PeP

eb6( )

1.07

10

4

15

NL2Q

3

S9.24

解1：(51)8

(101001)2特征多项是：x5

x3

1a4a3a2a1a0a5脉冲序列a0a1a2a3a4a5=a0⊙a2100001100001020001003001000100110810110090110011011001111100111120011111111016111100脉冲序列a0a1a2a3a4a5=a4⊙a2110000001120000110210011012201101123110111241011102501110126111010271101012810101029010100301010003101000032100001输出序列：1000010010110011

111000110111010周期31；除全零00001~1111状态各出现一次；0出现15次，1出现16次；是m序列9.24

解2：(51)8

(101001)2特征多项是：x5

x3

1a0a1a2a3a4a5脉冲序列a4a3a2a1a0a5=a4⊙a2100001100001020001003001000100110810110090110011011001111100111120011111111016111100脉冲序列a0a1a2a3a4a5=a4⊙a21100000011200001102100110122011011231101112410111025011101261110102711010128101010290101003010100031010000321000019.24

解3：(51)8

(101001)2特征多项是：x5

x3

1a0a1a2a3a4a5脉冲序列a4a3a2a1a0a5=a4⊙a20000101000100200100010011071011008011000011111001111120111111110016111000脉冲序列a0a1a2a3a4a5=a4⊙a210001001102000110121011011221101112310111024011101251110102611010127101010280101002910100030010000311