Wireless Communications 3

8/9/2019 Wireless Communications 3

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Radio Capacity Assume that the propagation constant n = 4

and the minimum carrier-to-interference ratio

is 18 dB. Substituting the values of n and C/I in the

last equation, N = 7 cells/cluster.

Therefore, the total allocated bandwidth B t is

divided into N groups of channels.

With Bc denoting the channel bandwidth, theradio capacity of the system is defined as the

number of channels per cell or

radio capacity k = Bt/(NBc) channels/cell


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Other Capacity Measures:

1. For a given blocking probabil ity, kchannel per cell, the traffic A in Erlangs per

cell can be determined from Erlang B tables. 2. If the cell area is a square km, the resulting

traffic density is A/a Erlang/km2

3. Knowing the average number of calls peruser in the busy hour and the averageholding time, the cell capacity defined as thetotal number of users per cell can be

calculated as well as the number of users persquare km.

4. The system capacity can be calculated by

multiplying the cell capacity by the total areaof the system.


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Erlang B Table


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How many users can be supported for 0.5%

blocking probability for the following numberof trunked channels in a blocked calls clearedsystem? (a) 1, (b) 5,(c) 10, (d) 20, (e) 100.

Assume each user generates 0.1 Erlangs oftraffic.

a) one user could be supported on onechannel

b) Given C = 5 , GOS = 0.005, A = 1 .I 3

U =A/Au = 1.13/0.1 = 11 users

(c) Given C = 10,A = 3.96 U =A/Au =3.96/0.1=39 users

d) Given C=20, A=11.1, U=111 users

e) Given C=100, A=80.0, U=809 users


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Example

Assume that the radio capacity k = 45channel/cell, blocking probability = 0.002,cell area =12.5 square km, average holdingtime =100 s, average arrival rate is 0.8calls/user in a busy hour, and the total areais 7000 square km.

From Erlang table, traffic/cell = 36.5 Erlang

traffic per user =100*0.8/3600 Erlang

traffic density = 35.6/12.5 = 2.86 Erlang/km2and number of users/km2 =128.75 users/km2

total number of users in the system =

128.75*7000= 901,250 users.


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Example 3.5

SystemA has 394 cells with 19 channels each.

system B has 98 cells with 57 channels each,and system C has 49 cells. each with 100

channels. Find the number of users that can be

supported at 2% blocking if each user averages2 calls per hour at an average call duration of 3

minutes.

Traffic intensity per user,Au = H = 2 x (3/60) =0.1 Erlangs


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For system A, GOS = 0.02 and C = 19, A=12Erlangs.and U=120 users

total number of subscribers=120 x 391 = 17280. For system A, GOS = 0.02 and C = 57, A=45

Erlangs.and U=450 users

total number of subscribers=450x 98 = 44 100. For system A, GOS = 0.02 and C =100 , A=88

Erlangs.and U=880 users

total number of subscribers =880x 49 = 43 120

total number of subscribers in the 3 systems=47280 + 44100 + 43 120 = 134500 users.


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Adjacent channel interference

Adjacent channel interference can be

minimized through careful filtering andchannel assignments.

By keeping the frequency separation

between each channel in a given cell as

large as possible, the adjacent channel

interference may be reduced considerably. separate adjacent channels in a cell by as

many as N channel bandwidths


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d f

dn

the near-far effect

20

1/ =fn dd

For a path loss exponent n = 4, this is equal to -52 dB


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Adjacent channel interference can be

minimized through careful filtering andchannel assignments.

a channel separation greater than six isneeded to bring the adjacent channel

interference to an acceptable level


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Improving Capacity in Cellular


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Improving Capacity in Cellular

Systems

Cell Splitting


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Sectoring


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only 2 of them interfere with the center cell.


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Wireless Communications 3