Sponsored Links
This presentation is the property of its rightful owner.
1 / 71

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


  • 97 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

第二章 蜂窝移动通信系统

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript



2.1

2.2

2.3

2.4


2.1





BSS

BSS

MSC2

SS7

SS

MSC

MSC1

PSTN

SS

OMS

2-1

  • PSTNPSDN

  • SS (OMS)BSSMS


  • BSSSS

    • BSSSS

    • SSSS

  • OMS

  • MS

    • SIM



  • 1-10



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


--


--

  • ()



--

  • ---


--


--

120


(b)

(c)

(a)

--


  • AMFM


( )

  • N: N=i2+ij+j2ijN4712




  • Sk(k<S)SN

    S=kN

  • NMC

    C=MkN=MS

  • CM:N,M.


  • N.,N,N,.


  • j

  • 60i

i=3

j=2

N=19


2.2


2.2


2.2


2.2

    • -90dBm~-100dBm

  • (MSC)

  • TDMA

  • MAHO


2.2


2.2


2.3


2.3.1


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


N

N

D

D

()

  • Q

  • QQ


C---

I---

K

N


D

4

N


  • TACS

N6.49N =7 7

  • GSM

GSM


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

1(C/I)s=18dB

2(C/I)s=12dB


k


2.3.2


2.3.3


  • .

  • 4

  • .



  • (a) (b)


120


N=7

  • 6

  • 12023

  • 601


  • RD/R

  • N


  • 73()(Tx/Rx)


  • MSC


  • RzD

  • Dz/Rz=4.6 (N=7)

  • D/R=3 (N=3)

  • 7/3

N=7


2.4

  • BB


2.4

    • :


2.4

    • (1)()SA

      (/)S(/)A(Erlang)


2.4

  • 1 (): A() = S(/)(/)

  • A1 11

  • 20 = 20(/)3


2.4

0(0)

AA0B


2.4

  • B B ()


2.4 B

B()ABnBAn


2.1


2.1


  • 24 1kk=10%-15%

  • C(/) T(/)k


  • 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


mn mn

a=0.01(/)(a)


1

  • 200.01Erl,,1%,?

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

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


2

  • 10,300,0.03Erl,?5%

  • n=10, m.n=300

  • B

  • B=20%

  • AA=9Erl.AB=5%,B,n=14,14-10=4



  • t

  • C

s=10s =1.2/ min

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



  • 1101.220


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%


  • Login