Lecture Cellular Concept

8/12/2019 Lecture Cellular Concept

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Wireless Communications

Systems (EEE464)

Dr. Guftaar Ahmad

Office: Room 412Email: [email protected]



• In 1946, the first public mobile telephone is put intoservice, known as improved mobile telephone service(IMTS)

– A single, high powered transmitter mounted on a tall towernear the center of a metropolitan area, covering distance of50km, with a fixed number of available channels

• Good coverage

• Impossible to reuse the frequency.• Service poor due to call blocking.

– When metropolitan area grows, higher power is needed butmore subscribers are not able to connect.



Cellular Concept

• Replacing a single, high power transmitter with many low powertransmitters, each providing coverage to only a small area

• Each base station is allocated a portion of the total number offrequency channels available

to the entire system, andnearby base stations areassigned different groupsof channels.

• Neighboring cells are

assigned different groupsof channels in order tominimize interference

• The same set of channels isthen reused at differentgeographical locations.



• Without increasing thenumber of the channels – Coverage expand easily – Cope with increasing

demand for service easily

Cellular Concept



Cellular Concept

• In a wireless cellular system the service area is dividedinto a collection of geographical areas called cells

• Each cell has a Base Station (BS) at the center of the

cell (usually)• Wireless (mobile) telephones are connected to the Base

Station via a wireless link

• A number of base stations are connected via wire-line

to a Mobile Switching Center (MSC)• Finally, the MSCs are connected via wire-line to the

Public Switched Telephone Network (PSTN)



Cell and Cluster

• Frequency Planning:

– The design process of selecting and allocating channel groups for all ofthe BS within a system .

• Footprint:

– is the actual radio coverage of a cell and is determined from fieldmeasurements or propagation prediction models. For systematic cellplanning, a regular shape is assumed for the footprint.

– Coverage contour should be circular. To avoid overlap or leaving gaps,only three regular shapes allowed: equilateral triangle, square, hexagon.

– The hexagon has been chosen due to its maximum area coverage. The

fewest number of cells can cover a region.





Cluster Size=7



• Capacity : If a cluster is replicated times within thesystem, then

– If cell size does not change and we reduce , then

the capacity increases. How? – If cell size decreases and we do not change , then

the capacity increases. Why?

• Frequency reuse factor:• In order to connect without gap, the allowable must

satisfy , for non-negative .

C

C = MkN = MS

N

1=N

N = i2 + ij + j2

N

i; j

M

N



• To find the nearestco-channel neighbors:

– move cells along

any chain ofhexagons

– Then turn 60° counter-clock wise

and move cells

ii

j j



Some possible value of N N





• Example: If a total of 33.6 MHz of bandwidth isallocated to a particular FDD cellular system which usestwo 25 kHz simplex channels to provide full duplex

voice and control channels, compute the number ofchannels available per cell if a system uses (a) four-cellreuse; (b) seven cell reuse. (c) 12 cell reuse. – Total bandwidth = 33.6 MHz

– Channel bandwidth = 25 KHz×

2 = 50KHz/Duplex channel – Total available channel 33600/50 =672 channels

– For N=4, No of channels available per cell = 672/4 = 168





• Honework#1-1: continued on the previous examples

– If 1.5 MHz of the allocated spectrum is dedicated to controlchannels, determine an equitable distribution of the control

channels and voice channels in each cell for each of the threesystems



Channel Assignment

• Objectives: increasing capacity and minimizing interference

• Fixed channel assignment: – Each cell is assigned a pre-determined set of frequency channels.

– If all the channels in the same cell are used, the new calls areblocked.

– Variations: borrowing strategy. MSC, interference consideration

• Dynamic channel assignment:

– No pre-determined assignment of frequency channels is made. – When a call arrives, the base station ask the mobile switchingcentre (MSC) to allocate a free channel.

– MSC must take into account the co-channel interference inchannel.

– Reduce blocking probability and increase capacity.







• False triggering: handoff due to momentary fadingrather than moving away from the base station – Solution: The base station monitors the received signal

strength for a certain period of time before triggering the

handoff procedure.

• Dwell Time: The time over which a user remains withinone cell is called the dwell time. – The statistics of the dwell time are important for the practical

design of handover algorithms. – The statistics of the dwell time vary greatly, depending on the

speed of the user and the type of radio coverage.



• In 1G, each BS constantly monitors the signal strengthsof all of its reverse voice channels to determine therelative location of each mobile user with respect to theBS. This information is forwarded to the MSC who

makes decisions regarding handover. (1G)• In 2G, Mobile assisted handover (MAHO) : The

mobile station measures the received power fromsurrounding BSs and continually reports the results ofthese measurements to the serving BS.

• Intersystem handoff: Required if a mobile stationmoves from one cellular system to a different cellularsystem. Change from MSC to MSC



• Prioritizing Handover: Guard channel concept

– Dropped call is considered a more serious event than callblocking. Channel assignment schemes therefore must givepriority to handover requests.

– A fraction of the total available channels in a cell is reservedonly for handover requests. However, this reduces the totalcarried traffic. Dynamic allocation can improve this.

– Queuing of handover requests is another method to decreasethe probability of forced termination of a call due to a lack ofavailable channel. The time span over which a handover isusually required leaves room for queuing handover request.



• Practical handoff consideration – High speed users and low speed users have vastly different

dwell times which might cause a high number of handoverrequests for high speed users. MSC burned!!!

• The Umbrella Cell approach willhelp to solve this problems. – Different antenna height and power

– High speed users are serviced

by large (macro) cells, while low speed users arehandled by small(micro) cells.



`

• A hard handover does “break before make”, i.e., The oldchannel connection is broken before the new allocated channelconnection is setup. This obviously can cause call dropping.

• In soft handover, we do “make before break”, i.e., The new

channel connection is establishedbefore the old channel connectionis released.



Co-Channel Interference

• Co-channel cell : The set of cells using the same set offrequencies as the target cell.

• Interference tier :

A set of co-channelcells at the samedistance from thereference cell.

• Interference depends on radius and distance



• Co-channel reuse ratio:

• Signal to noise ratio (SNR): – improves by increasing the signal power

• Signal to interference ratio (SIR): – cannot be improved by changing the signal power

• Power law: – and are the power and the distance at the reference

point, is the path loss exponent.

– SIR is

P r=I = P r=Pi0

i=1 I i

P r=N 0

P r = P 0( dd0

)¡n

P 0 d0

n

P r=I = R¡n=Pi0

i=1(Di)¡n

Q = D

R =

p 3N Q =

D

R =

p 3N





• More accurate modeling

SIR = R¡n

2(D¡R)¡n+2(D+R)¡n+2D¡n



Adjacent Channel Interference

• Reasons

– For signals adjacent in frequency

– Resulting from imperfect receiver filters, leakage.

• Near-far effect

– Power control suchthat powers appear atBS equally.

• Solution

– Accurate filtering

– channel assignmentadjacent channelsassigned to differentcells



Trunking & Grade of Service

• Trunking – The idea is to accommodate a large number of users in a limited radio

spectrum – The concept is to allow the users to share the small amount of available

channels.

• Trunking Theory: – exploits the statistical behavior of users so that a fixed number of channel

may accommodate a large number of users, also in the sense oflikelihood.

• Grade of Service (GOS)

– is a measure of the ability of the user to access a trunked system duringthe busiest hour. – GOS is typically given as the likelihood that a call is blocked, or the

likelihood of a call experiencing a delay greater than a certain queuingtime.



• Erlang [Danish mathematician]

– The measure of traffic intensity nowadays bears his name.

– 1 Erlang: 1call-hour per hour or 1 call-minute per minute

– Example: A radio channel that is occupied for 30 minutes during an hour

carries 0.5 Erlang of traffic.

• Set-up Time: The time required to allocate a trunked radiochannel to a requesting user.

• Blocked Call: Call which cannot be completed at time of request,

due to congestion.• Holding Time: Average duration of a typical call. Denoted by

(in seconds).H H



• Traffic Intensity: Measure of channel time utilization, which isthe average channel occupancy measured in Erlangs. This is adimensionless quantity and may be used to measure the timeutilization of single or multiple channels. Denoted by .

• Load: Traffic intensity across the entire trunked radio system,measured in Erlangs.

• GOS: A measure of congestion which is specified as theprobability of a call being blocked (for Erlang B), or the

probability of a call being delayed beyond a certain amount oftime (for Erlang C).

• Request Rate: The average number of call requests per unit timefrom ONE user. Denoted by seconds-1.

AA

¸̧



• The traffic intensity from one user is

• For a system containing user, the total trafficintensity is (regardless of the number of channels):

• For a channel trunked system, the traffic intensityper channel is given by

• For given and , GOS can be computed fromtrunking theory and queuing theory. (omitted….).

Au = ¸H Au = ¸H

A = UAuA = UAu

Ac = A=C = U Au=C Ac = A=C = U Au=C

AA C C

U U

C C



• Two types of trunked system:

• First type offers no queuing for call requests. If nochannels are available, the request is blocked and theusers need to try later. – This type of trunking is called “blocked calls cleared”.

– Called Erlang B system and the probability that a call isblocked (GOS) is

• Example: AMPS system

is designed for GOS of

2% blocking.

P r[blocking] =

AC

C !P

C

k=0

Ak

k!P r[blocking] =

AC

C !P

C

k=0

Ak

k!



Erlang B



Erlang B



• Example 1: How many users can be supported for0.5% blocking probability for the following number oftrunked channels in a blocked calls cleared system? (a) 5

(b) 10 (c) 20 (d) 100. Assume each user generates 0.1Erlangs of traffic.

– (a) , from figure we know

. The total number of users is

– (b) . The total number of users is – (c) 120

– (d) 900

C = 5; Au = 0:1; GOS = 0:005C = 5; Au = 0:1; GOS = 0:005

A = 4A = 4

U = A=Au = 11U = A=Au = 11A = 1:1A = 1:1

U = A=Au = 40U = A=Au = 40

Interestingly, increases when increases, this means…. U=C U=C C C



`

• Example 2: An urban area has a population of 2 millionresidents. Two competing trunked mobile networks(systems A and B) provide cellular service in this area.System A has 394 cells with 19 channels each, system B

has 98 cells with 57 channels each. Find the number ofusers that can be supported at 2% blocking if each useraverages 2 calls per hour at an average call duration of 3minutes. Assuming that all two trunked systems areoperated at maximum capacity, compute the percentage

market penetration of each cellular provider.Given – Probability of blocking=0.02 – Traffic intensity per user: Erlangs.Au = ¸H = 2£ 3=60 = 0:1Au = ¸H = 2£ 3=60 = 0:1



– System A: C=19

From Erlang B figure, the total traffic is 12 Erlangs.

The number of users per cell is then

The total number of subscribers is then 120×

394=47280 The market penetration is 47280/2000000=2.36%

– System B: C=57

From Erlang B figure, the total traffic is

The number of users per cell is The total number of subscribers is

The market penetration is

U = A=Au = 120U = A=Au = 120

450×98=44100.

?

U = A=Au = 450U = A=Au = 450??

45 Erlangs.

? 44100/200000=2.205%.



• Example 3: A certain city has an area of 1300 square miles and iscovered by a cellular system using a 7-cell reuse pattern. Eachcell has a radius of 4 miles and the city is allocated 40 MHz ofspectrum with a full duplex channel bandwidth of 60 kHz.

Assume a GOS of 2% for an Erlang B system. If the offeredtraffic per user is 0.03 Erlangs, compute (a) the number of cellsin the service area, (b) the number of channels per cell, (c) trafficintensity of each cell, (d) the maximum carried traffic, (e) the

total number of users that can be served for 2% GOS, (f) thenumber of mobiles per unique channel, (g) the theoreticalmaximum number of users that could be served at one time bythe system.



– The area of the cell is . The number of the cellis then

– The number of channel per cell

– From Erlang B figure, at 2%,

– Maximum carried traffic=the total intensity of the system=

Erlangs

– The total user supported is

– The number of unique channel is . So thenumber of users per unique channel is

– The theoretical maximum number of users at one time is

, which is 3.4% of the supportable users.

2:6R2 = 41:62:6R2 = 41:6

1300=41:6 ¼ 311300=41:6 ¼ 31

40000=60=7 ¼ 9540000=60=7 ¼ 95

A = 84A = 84

2604=0:03 = 868002604=0:03 = 86800

40000=60 = 66640000=60 = 66686800=666 ¼ 13086800=666 ¼ 130

31£95 = 294531£95 = 2945

31£ 84 = 260431£ 84 = 2604



• The second type provides queue to hold calls whenblocked. If there is no channel available, the call requestmay be delayed until a channel becomes free. – This type is called “blocked calls delayed”.

– The GOS of this type is defined as the probability that thecall will not have immediate access to a channel. Erlang Cformula:

P r[delay > 0] =

AC

AC + C !³

1¡ AC

PC ¡1

k=0Ak

k!

´P r[delay > 0] =

AC

AC + C !³

1¡ AC

PC ¡1

k=0Ak

k!

´



– The probability that a call is delayed more than seconds isgiven by (can also be used as GOS):

– The average delay for all calls in a queued system is givenby

and the average delay for those calls which are already inqueue is

DD

D = P r[delay > 0] H

C ¡

A

D = P r[delay > 0] H

C ¡

A

H=(C ¡A)H=(C ¡A)

P r[delay > t] =P r[delay > 0]P r[delay > tjdelay > 0]

=P r[delay > 0] expµ¡

(C ¡A)t

H ¶

P r[delay > t] =P r[delay > 0]P r[delay > tjdelay > 0]

=P r[delay > 0] expµ¡

(C ¡A)t

H ¶

tt



Erlang C (delay>0)



• Example 4: A hexagonal cell within a 4-cell system hasa radius of 1.387km. A total of 60 channels are used

within the entire system. If the load per user is 0.029

Erlangs, and λ=1 call/hour, compute the following foran Erlang C system that has a 5% probability of a delayed

call

– (a) How many users per square km will this system support?

– (b) What is the probability that a delayed call will have to waitfor 10s?

– (c) What is the probability that a call will be delayed for morethan 10s?



– The area covered per cell is

– The number of channels per cell is 60/4=15.

– From Erlang C chart, at GOS=5%, C=15, we got A=9Erlangs.

– The number of users supported per cell is 9/0.029=310.

– The number of users per square kilometer is 310/5=62.

– The average holding time is

– The probability for a delayed call to wait for 10 s is

– The probability that a call will be delayed for more than 10 s is

2:6£ 1:3872 = 5 km22:6£ 1:3872 = 5 km2

exp(¡(C ¡A)t=H ) = exp(¡(15¡ 9)10=104:4) = 56:29%exp(¡(C ¡A)t=H ) = exp(¡(15¡ 9)10=104:4) = 56:29%

H = Au=¸ = 0:029 hours = 104:4 sH = Au=¸ = 0:029 hours = 104:4 s

P r[delay > t] = 0:05£ 0:5629 = 2:81%P r[delay > t] = 0:05£ 0:5629 = 2:81%



Some points of interest

• Increase in N decrease the Frequency Reuse (Not

desired) Decrease the Capacity C=M*K*N

• But Small N means Decrease in Q (Not Desired)

• Service Provider: Minimum possible value of N

• Service Buyer: Max possible value of Q





The number of the cells increases four times. To keep the same received

power at mobile, the BS power should be reduced by a factor 2n



In practice, not all cells are split at the same time. Special organization is

then needed



• Sectoring – Instead of using one omnidirectional

antenna, a base station can usemultiple directional antennas, each

of which caters for a coverage areaat a particular direction.

– Reduced co-channel interference.For 7-cell network, if we partition acell into three 120° sectors, then the

number of co-channel cells arereduced from 6 to 2 in thefirst tier. Using six sectors of 60°,

we have only one co-channel cell inthe first tier.









• Micro Zone Concept – Sectoring increases the

number of handoffs andhence increases the load

of MSC. – Microcell zone concept

decreases the burden of MSCby reducing the handoffrequest.

– Same channel is used when themobile moves from one zoneto another.



Interference Reduction Techniques

• Sectorization: Directional antennas

– Require more handoff

• Smart Antennas/Beamforming

– Require multiple Antennas• Multiuser Detection

– Complexity issue

•Interference Pre-subtraction – Require Channel knowledge

• Frequency Hopping:

– change the carrier frequency according to a unique pattern



Summary

• Frequency Reuse• Channel Assignment Strategies• Handoff Strategy• Interference

– Co-channel Interference – Adjacent Channel Interference – Power Control

• Trunking Theory – Erlang B

– Erlang C• System Expanding Techniques – Cell Splitting – Sectoring – Microcell Zone Concept



Example

Consider a cellular system in which there are a total of301 radio channels available for handling traffic.Suppose the area of a cell is 6 km2 and the area of theentire system is 2100 km2. The cluster size is 7.

– (a) Calculate the system capacity – (b) How many times would the cluster of size 4 have

to be replicated in order to approximately cover theentire cellular area?

Given that in a 7 cell reuse system the traffic load peruser is 0.03 Erlangs and the average number of calls perhour per user is 1.5, for 5% probability of call blocking, – ( c)Find the traffic load per cell?– (d) The mean duration of a call

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Lecture Cellular Concept