Intro to Communication Systems Chap 3

8/13/2019 Intro to Communication Systems Chap 3

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Some Basic Definitions


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Busy Hour

The hour of the day when traffic is at its greatest.

It may not exactly an hour.

Here hour describes a specific amount of time.

Its different for different area, location or

environment.

It may be any occasion or festival for instance

Eid, Christmas or Boxing day etc.

It could be a duration of cricket or football match.

or simply it could be a lunch break.


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Traffic Unit Traffic is measure in Erlangsdenoted by E named after A K Erlang

(Danish). If Q calls of average duration T occur during time t, then the traffic

(A) is given by

A = Q . T / t Erlangs

t and T are measured in the same units. During a particular period of time (say, an hour) a user generates 1E

of traffic if they use the phone continuously.

If they only use the phone 10% of the time they generate 0.1E of

traffic.

If there are 100 users, on average using the phone 10% of the time

each, then the total traffic generated is 10E

Let T = average calling time in minutes.

Hence the offered traffic load is given by

A = Q.T / 60 Erlangs


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Grade of Service (GoS)

It is a measure of the probability ofNOTgetting a connection at all.

If the grade of service is adequateat thebusy hour, then it will be better during theremainder of the day i.e. smaller.

e.g. If the grade of service B= .005, thenone call in 200 would be lost because ofinsufficient equipment.


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Issues in planning a cellular system

Available spectrum is limited.

How many mobiles can be served at an

acceptable grade of service???Challenge- for a given bandwidthmaximise the number of customers with

an acceptable grade of service


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BLOCKING PROBABILITY

Is the probability of call being blocked or

the probability that user gets

NETWORK BUSYtone.

Blocking Probability , B, is given by

B = { A

N

/ N! } /{ A0

/ 0! + A

1

/ 1! + A

2

/2! + A

3/ 3! + A

4/ 4! + A

5/ 5! + . A

N/

N!}


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For example, if N = 5 and the offered traffic

A = 1.66 E, then the blocking probability is

given by

B = { 1.66.5 / 5! } / { 1.660 / 0! + 1.661 / 1! +

1.662 / 2! + 1.663 / 3! + 1.664 / 4! + 1.665 / 5! }

= 0.02 %

These values are available in tabular form e.g.

Erlang B Table

BLOCKING PROBABILITY Cntd....


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Let Q = maximum calls per hour in one cell= 3000

T = average calling time

= 1.76 minutesB = blocking probability

= 2%

Then the offered traffic, A, is given byA = Q T / 60

Now find No of Channels required ???/

Example 1


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Solution:

A= 3000 x 1.76 / 60 or 88 Erlangs.

B = 2%Hence from Erlang B Table, the maximum

number of channels, N , is given by N = 100.

Example 1 Cntd..........


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Q = maximum calls per hour in one cell

= 28000

T = average calling time

= 1.76 minutes

B = blocking probability

= 2%How many radio channels are needed?

Example 2


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Solution:

the offered traffic, A, is given by

A = Q T / 60 = 28000 x 1.76 / 60 = 821 Erlangs. B = 2%

From the Erlang B tables, this corresponds to

820 channels.

Example 2 Cntd..........


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How many customers can be served with 666voice channels if all the channels are used in

a) a single cell, and

b) Multiple cells (e.g 4, 7, )? Assume a blocking probability of 0.02. Let the

average duration of each call be 1.76 minutes.

Consider the advantages and disadvantages ofsingle and multiple cell operation.

How many cells per cluster are really needed ?

Example 3


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Single Cell

Call duration = 1.76 minutes Number of channels = 666

Blocking probability = 0.02

How many calls per hour can be handled?

Hence the traffic offered , an Erlangs, is given by A = Q (calls/hour) x 1.76 / 60

From Erlang B tables, for 2 % blocking

Channels offered traffic

600 587.2700 688.2

Hence for 666 channels, A = 66/100 x (688.2 - 587.2) + 587.2Erlangs

Hence Q = A x 60 / 1.76 = 22 291 calls / hour

Solution : 3


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Seven Cells per Cluster

Call duration = 1.76 minutes

Number of channels per cell ( 7 cells / cluster)

= 666 / 7 = 95

Blocking probability = 0.02

From Erlang B tables this corresponds to A = 83.1 E

Hence Q = 83.1 x 60 / 1.76 or 2833 calls per hour.

Hence the total number of calls per cluster is givenby 2833 x 7 or 19831 calls / hour.

This is smaller than the total number of calls

handled by a single cell with 666 channels.

Solution : 3 Cntd.....


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Four Cells per Cluster If we use 4 cells per cluster, the total number of

calls per cluster can be shown to be 20809 calls

/hour. i.e.number of cells/cluster calls / hour

1 22291

4 20809

7 19831



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So what is the advantage of dividing an areainto smaller cells?

The cell pattern can be repeated in order toincrease the total number of calls that can be

handled. For instance, if the single cell is sub-divided into a 7 cell pattern 4 times, then thenumber of calls per hour per cell is equal to2833 (N=666, B=0.02). Thus for the whole

area, the number of calls is given by

2833 x 7 x 4 = 79324 calls / hour.



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What are the disadvantages of this method ?

need more processing

more hand-offs

more infrastructure costs for cells sites etc.

need to address the issue of co-channel

interference from other cell clusters



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Subjective telephony tests

Subjects rated the quality of simulated and actualmobile telephone channels subjected to the rapid

Rayleigh fading encountered in UHF mobile

communications.

APMS test result show that for an RF C / N of 18 dB,

most listeners consider the channel to be good or

excellent. Hence C / N must be greater than or equal to

18 dB. Similarly listeners opinions on transmission quality in

the presence of co-channel interference showed that the

ratio of carrier to co-channel interference ( C / I ) must

be greater than or equal to 17dB.

How is the value of C / I specified ?


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Mathematical Relation of C/I &

Number of Channel

Now if C/I is given how to find number of

channels, mathematically ???Come on guyzzz now at least this should be done

by yourself !!!!!


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Here is the answer

Let C / I = 18 dB

Let a (propagation constant) be equal to 4.

Then

1018/10

= 1 / { 6 ( D / R )-a

}

Thus( D / R )

2= { 6 x 10 18/10 }

1/2

Hence D / R = 4.4.1

Now D / R is also given by

D / R = ( 3 N )1/2

Hence N = 6.5 or 7 i.e. 7 cells per cluster are needed to

satisfy the carrier to interference requirements.

Intro to Communication Systems Chap 3

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