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第二章 蜂窝移动通信系统 PowerPoint PPT Presentation


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第二章 蜂窝移动通信系统. 内容. 2.1 蜂窝小区的概念和特点 2.2 信道切换策略 2.3 干扰和信道容量 2.4 电信业务流量与服务等级. 2.1 蜂窝小区的概念和特点. 实现系统在其覆盖区内良好的语音和数据通信,这样的通信网就是 移动通信网。 移动通信 网络结构 蜂窝式 组网理论 移动通信网的 基本组成 蜂窝移动通信系统图示 各子系统功能. 多址接入. 空中网络. 切换和位置更新. 频率复用和蜂窝小区. 服务区内各基站的相互连接. 地面网络. 基站与固定网络. 移动通信网络结构. 蜂窝式组网理论. 无线蜂窝式小区覆盖

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第二章 蜂窝移动通信系统

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2.1

2.2

2.3

2.4


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2.1


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BSS

BSS

MSC2

SS7

SS

MSC

MSC1

PSTN

SS

OMS

2-1

  • PSTNPSDN

  • SS (OMS)BSSMS


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  • BSSSS

    • BSSSS

    • SSSS

  • OMS

  • MS

    • SIM


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  • 1-10


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f

f

f

f

f

f

f

1

2

1

2

1

2

1

f

f

f

f

f

f

f

1

2

3

1

2

3

1

--

BS


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--


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--

  • ()


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--

  • ---


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--


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--

120


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(b)

(c)

(a)

--


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  • AMFM


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( )

  • N: N=i2+ij+j2ijN4712


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  • Sk(k<S)SN

    S=kN

  • NMC

    C=MkN=MS

  • CM:N,M.


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  • N.,N,N,.


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  • j

  • 60i

i=3

j=2

N=19


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2.2


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2.2


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2.2


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2.2

    • -90dBm~-100dBm

  • (MSC)

  • TDMA

  • MAHO


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2.2


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2.2


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2.3


2 3 1

2.3.1


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i

1

6

4

D

2

5

j

7

3

1

7

6

4

1

5

6

D

  • D :D120ijD

    D:

    N:

    R:

    :

N=7


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N

N

D

D

()

  • Q

  • QQ


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C---

I---

K

N


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D

4

N


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  • TACS

N6.49N =7 7

  • GSM

GSM


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MSo(C/I)sNW=25MHZB=25KHZ

1(C/I)s=18dB

2(C/I)s=12dB


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k


2 3 2

2.3.2


2 3 3

2.3.3


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  • .

  • 4

  • .


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  • (a) (b)


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120


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N=7

  • 6

  • 12023

  • 601


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  • RD/R

  • N


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  • 73()(Tx/Rx)


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  • MSC


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  • RzD

  • Dz/Rz=4.6 (N=7)

  • D/R=3 (N=3)

  • 7/3

N=7


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2.4

  • BB


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2.4

    • :


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2.4

    • (1)()SA

      (/)S(/)A(Erlang)


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2.4

  • 1 (): A() = S(/)(/)

  • A1 11

  • 20 = 20(/)3


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2.4

0(0)

AA0B


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2.4

  • B B ()


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2.4 B

B()ABnBAn


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2.1


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2.1


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  • 24 1kk=10%-15%

  • C(/) T(/)k


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  • 25100min4min ,9%0.36minA=s 0.36/60=0.06Erl

  • 10010,A=6ErlB=4.3%

  • a=0.01a=0.05a=0.02-0.03a=0.08


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mn mn

a=0.01(/)(a)


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1

  • 200.01Erl,,1%,?

    :n=20,B=1%,2.1 A=12Erl

    m.n=A/ =12/0.01=1200


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2

  • 10,300,0.03Erl,?5%

  • n=10, m.n=300

  • B

  • B=20%

  • AA=9Erl.AB=5%,B,n=14,14-10=4


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  • t

  • C

s=10s =1.2/ min

20 sP(r>20)=p(r>0)e-0.1(1-0.2)20=0.04038


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  • 1101.220


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B=1%395,1/,2,N=71200?

:

  • n=57 B=1%B A=44.2Erl

    m.n= A/ =44.2/(1/30)=1326

    =A(1-B)/n=44.299%/57=77%

  • 1200

    n=57/3=19 B=1% A=11.23Erl

    m.n= A/ =11.23/(1/30)=336

    =A(1-B)/n=11.2399%/19=58.5%


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