Cell Covered

8/13/2019 Cell Covered

1/15


2/15

A

B

E

DF

CG

A

B

E

DF

CGA

B

E

DF

CGi

j

0.2.1 Cells are shaped as hexagons

What shape do the cells have?A circle would be a natural choice, but a circle does not tesselate an area (in other words

it leaves gaps or creates overlaps between cells).Geometric shapes that tesselate: triangle, square, hexagon.The cell shape determines where to place the base stations. For example, if cells are

shaped as triangles, each cell has three closest neighbors; if cells are squares, each cell hasfour neighboring cells; if cells are in the shape of hexagons, each cell has 6 neighbors.

- Choosing a cell shape determines the location of base stations.In choosing a shape, we need to address two requirements:- The transmitted power has to be sucient to reach the most distant user in the cell.

This sets the level of transmitted power.- Given the level of transmitted power, choose a shape that maximizes the area of the

cell.Prove that given a distance R to the cell boundary (most distant point in the cell), the

cell area is maximized by the hexagon.

0.2.2 User Capacity

We will discuss later why if the same frequency channels are reused in every cell, that mightresult in too much interference.

The frequency plan consists of dividing the channels among several cells. With morecells sharing the channels, each cell gets fewer channels, and also the distance betweencells using the same channels increases (relative to the distance between users and basestation in the cell of interest).

Assume there areNcells sharing the channels. These Ncells are said to form a cluster.In the gure,N= 7:

2


3/15

A

B

E

DF

CG

Say each is allocatedk dierent frequency channels. Total number of channels per clusteris given by

S=kN

Say that a geographical area (village, town, etc.) is covered by Mclusters. Then the user

capacity over the area is:

C=M kN=M S

User capacityCover the area may be increased by increasing the number of clusters M:That requires the cells to be made smaller (note that power needs to be reduced accordingly).

0.2.3 User Capacity and Bandwidth

User capacity (the number of channels per cell) is a function of bandwidth. If each channelisB Hz,

the total bandwidth isW Hz,the number of cells per cluster isN;then the number of channels per cell is given

k= W

N B

Example

A total of 33 MHz are allocated to a system which uses 2x25 kHz for full duplex (i.e.,each channel is 50 kHz). What is the number of channels per cell for reuse N= 4?

Solution:Number of channels per system

33 MHz

(2) (25) kHz= 660 channels

For reuse N = 4:660

4 = 165channels/cell

Example

3


4/15

Total bandwidth available in a city is 25 MHz. Each user requires 30 kHz. Explore theconcept of frequency reuse.

Solution:With a single BS, we can support only

25 MHz30 kHz

= 833 users

Use 20 BSs. Divide the available spectrum into M = 5 clusters of N = 4 cells each.Then, each cluster gets25 MHz.

Each cell gets25 MHz

4 cells = 6:25 MHz

Number of users per cell:6:25MHz

30 kHz = 208users

Number of users per cluster: 4 208 = 832Number of users per system

5clusters 832 users/cluster= 4; 160 users

0.3 Cellular System Layout

Oblique set of coordinates (u; v) for cellular geometryCells are hexagonalRadius of each cell is RCell centers located at (i; j), wherei and j are integers

If the radius of a hexagonal cell is R; the distance between two adjacent cells is =2R cos30o = 2R

p3

2 =

p3R

1. The cell radius is thenR = 1=p

3

Theorem: In the (u; v) coordinate system, the distance between two points (u1; v1) and(u2; v2)is given by the formula:

d12=q

(u2 u1)2

+ (u2 u1) (v2 v1) + (v2 v1)2

Proof:Figure left:

d212

= x2 + y2

Figure middle:x= (u2 u1)cos30o

4


5/15

0

30o

(u1,v

1)

0

30o

0

30o

(u2,v

2)

x

y

u

v v

u

(u1,0)

(u2,0)

v

u(u

1,v

1)

(u2,v

2)

y

v1

v2

(u2-u

1) sin30o

d12

x=(u2-u

1) cos30o

Figure right:

y= v2+ (u2 u1)sin30o

v1Plugx and y ind12:

d212

= [(u2 u1)cos30o]2 + [v2+ (u2 u1)sin30o v1]2= (u2 u1)cos2 30o + (v2 v1)2 + (u2 u1)2 sin2 30o

+2 (u2 u1) (v2 v1)sin30o= (u2 u1)2 + (v2 v1)2 + (u2 u1) (v2 v1)

0.3.1 Number of cells in cluster

Let us assume we have a cell at (0; 0)and a cochannel cell at (i; j). Then the distancebetween the centers of cells is

D=p

i2 + ij+j2

Think about the cluster as a large cell. Distance between centers of the large clustersis not unity butD:

The radius of a cluster isRcluster=

Dp3

We want to compute the number of cells per cluster.Area of the hexagon with radius R:

Area of the triangle = R

2R cos30o =

p3

4 R2

Area of the hexagon = 6

p3

4 R2 =

3p

3

2 R2

Dene reuse ratio

q=D

R

5


6/15

A

B

E

DF

CG

A

B

E

DF

CGA

B

E

DF

CGi

j

Number of cells in cluster

N = cluster area

cell area

=

3p

3

2 R2cluster

3p

3

2 R2

= D2

3R2

It follows that the reuse ratio is related to the number of cells per cluster:

q = D

R

=

pi2 + ij+j2

1=p

3

=p

3N

The cell pattern is dened in terms of i and j parameters as follows:

1. Given the frequency reuseN ;lay out one cluster.

2. For each cell in the cluster: move i cells across hexagons.

3. Turn 60o CCW and move j cells.

Decreasing the size of cells:

Larger capacity Lower power needs to be transmitted: smaller terminals

6


7/15

More complex infrastructure network

Taxonomy of cells:

Cells can be of varying sizes depending on the location and capacity requirements. A number of cell sizes may be required to provide comprehensive coverage:

Megacells: hundreds of km (satellite BS)

Macrocells: cover metropolitan areas, tens of km

Microcells: antennas below rooftops, hundreds of meters

Picocells: W-LAN range (inside a building)

Femtocells: PAN range.

0.4 Channel Assignment StrategiesWe will discuss briey how channels are assigned to cells.

Channel assignment strategies can be classied as either xed ordynamic.Fixed channel assignment

Each cell is allocated a predetermined number of channels.When all channels are occupied, additional calls are blocked.Borrowingof channels from neighboring cells may be allowed.Dynamic channel assignment

Channels are not allocated to cells permanently.For each call, the base stations requests a cell from the MSC. The MSC can allocate cells

more eciently.Channel assignment does not have to be covered by an air interface standard. Operatorsuse proprietary strategies.

0.5 Hando

Hando: is the process of transferring a call across the cell boundaries.

Handos are prioritized over new calls. Handos should be designed to be performed as infrequently as possible.

A hando is initiated when the received signal power is at some specied value above theminimum usable signal.

Pr(hando) = Pr(usable) +;where Pr(usable)~ 90 to100 dBmIf is too large unnecessary handos.If is too small call will drop before hando is completed.Careful design:

7


8/15

BS1 BS2D

handoff

threshold

Min. usable

signal

Hando should not be initiated due to a fade. Two reasons: Channel will emerge from fade sooner than the hando can be completed. So that

would be a wasted hando.

System should be designed to overcome the fade anyway without the hando.

To ensure that handos are not triggered by fading, power level monitored over time.How is received power monitored:

In 1st generation systems, each BS monitors powers of all MS.Power = radio signal strength indication (RSSI)

In addition to monitoring RSSI of calls in progress, each BS monitors power receivedfrom users in neighboring cells. All this information is reported to the MSC, whichdecides who and when needs a hando.

In 2nd generation systems mobile assisted hando (MAHO).

Mobile measures BS in own cell and in neighboring cells. Suggests to the MSC when to hando.

This distributed operation helps simplify the MSC.

Intersystem hando mobile moves between cellular systems raises issues of roamingand compatibility.

To ensure availability of channels: Guard channels reserved only for handos.

Queuing design marginsuciently large such that some time may elapse after

asking for channel, without losing signal. 1st generation hando ~10 seconds, ~6 10 dB. 2nd generation (GSM) mobile assisted hando is faster approx. 1 2 seconds, 6

dB.

Hard hando: new channel connected, old channel disconnected. Soft hando: user is connected to two base stations simultaneously.

8


9/15

0.6 Interference and System Capacity

We motivated the concept of frequency reuse through the need to separate between cellsusing the same frequency channels. Put another way, channels reused in adjacent cells maycause too much interference.

Here, we seek to develop quantitative relations that will help us with the frequencyplanning.We can consider either downlink or uplink. To be specic, consider downlink.Performance is determined by SNR.Here, we are concerned with interference (signals in other cells using the same frequency

channel as signal of interest). Let us neglect the eect of noise and focus only on interference.Evaluate signal-to-interference ratio (SIR)Are we dealing here with eects of large or small-scale fading?The actual acceptable level of SIR depends on the technology. For example:

AMPS: 18 dB NA-TDMA: 14 dB GSM, CDMA: 7-12 dB

0.6.1 SIR Calculations, 2D

A naive way to improve SNR is by increasing power. That is because what we reallycare about is SIR. Let us calculate it:

SIR = S

P6

k=1 Ik

We know that S; Ik/ dn; where n is the pathloss exponent. Then

SIR = RnP6

k=1 Dnk

whereDk is the distance between interfering cell k and the user.

Let us see the geometry. The number of interference sources is 6 due to the hexagonalgeometry.

Assume rst, that all the distances are equal.

We already saw thatq=

D

R =

p3N

Hence

SIR =(D=R)n

6 =

(3N)n=2

6 =

qn

6

Note how SIR improves with Nand withn:

9


10/15

D+R

D+R

D-R

D-R

D

D

0.6.2 Worst-Case Design

Based on the gure above, worst case SIR:

SIR = Rn

2 (D R)n + 2Dn + 2 (D+ R)n

Example:Let a system require SIR = 18 dB for proper operation. What is the frequency reuse

required if the worst-case design is used and n = 4.Solution:TryN= 7

q=p

3N=p

(3) (7)

4:6

We compute SIR and get 17.3 dB. So N= 7 is not sucient. We need N= 9:

q=p

(3) (9) 5:2Compute SIR = 19.8 dB

Example:If SIRmin = 15 dB, what is the capacity for n = 4, n = 3? Assume equal distance to

interference.Solution:(a) Letn = 4:TryN= 7:Get SIR = 18.7 dB. So it works.

(b) Let n = 3: TryN= 7:Now, SIR = 12 dB.TryN= 9;it turns out still does not work.Finally,N= 12:

q =p

(3) (12) = 6

SIR = qn

6 =

63

6= 36 = 15:6dB

10


11/15

Note that SIR increases with n:

0.6.3 Channel Planning for Wireless Systems

The simple relation between the required SIR and N; assumed a homogeneous pathloss. In

real life, the pathloss is not homogeneous, there are shadowing eects, the base station isnot located exactly in the center of the cell, and the coverage is not hexagonal. Frequencyreuse schemes are therefore matched to the environment. The required SIR is determinedby the technology employed.

For example, CDMA employs universal reuse, GSM employs reuse N= 3:Channels are part of an air interface standard.About 5% of the channels are for control. Control channels are aorded more protection

than voice channels.

0.7 Trac Capacity and Trunking

Trunkingis the process of providing access to users on demand from available pool of chan-nels.

With trunking, a small number of channels can accommodate a large number of randomusers.

Telephone companies use trunking theory to determine number of circuits requiredin a building.

Trunking theory studies how a population can be handled by a limited numberof servers.

Trunking also applies to cellular.

0.7.1 Erlang B Model

When there are a limited number of servers, the demand for service may exceed the numberof available channels. When that happens, we have two options:

- When a channel is not available, inform the user of the situation.- When a channel is not available, queue the user for some reasonable amount of time. If

a channel becomes available, the user gets serviced.Trunking theory computes the probabilities of users getting serviced or being blocked

from service. The theory depends on assumptions such as whether blocked calls are queuedor discarded.With Erlang B model, blocked calls are discarded.In trunking theory, we are interested in trac intensity. Trac intensity is measured in

the amount of time that a channel is busy.

Trac intensity is measured in Erlangs:

11


12/15

1 Erlang: is the trac in a channel completely occupied. 0.5 Erlang, channel isoccupied 30 minutes in an hour.

Let us dene some additional terms.

Grade of Service(GOS): probability that a call is blocked. Set-Up Time: time to allocate a channel. Holding Time: (H) average duration of typical call. Load: Trac intensity across the whole system. Request Rate: () average number of call requests per unit time.

Each user generates trac.Load per user is the arrival (request) ratethe holding time for each arrival:

A= H

The system load is the load for U users: A=U AIf trac is trunked intoCchannels, then the trac intensity per channel isAc = U A=C.

GOS: If blocked calls are cleared (i.e. not queued), then under some model assumptionsThe probability of a blocked call is given by the Erlang B model.

An expression is found in the book

The chart in Figure 3.6, p.81

Example:The required is GOS = 0.5%. How many users can be accommodated in a blocked

channels cleared system for C = 5 channels? It is known that each user generates 0.1Erlangs of trac.

Solution:From the chart, with GOS = 0:005 and the number of channels = 5, A (capacity in

Erlangs) = 1.13 =>

U=A=A=1:13

0:1 11users

Example:Say a call lasts 200 sec on average. A typical user makes a call every 15 min. The desired

GOS = 1%; the number of channels available is 100. How many users can be accommodatedby this system?

Solution:If 100 channels are available, then the trac oered for GOS = 1%, is 84 Erlangs. The

trac per user is:

12


13/15

4 calls/hour x 200 sec holding time = 2/9 Erlang. It follows that

84

2=9= 378

378 users can be accommodated.Example:For GOS = 1%, compare the trac intensity for 5 and 5 channels.SolutionForC= 5 channels, trac intensity is approx. 1.5 ErlangForC= 50 channels, trac intensity is approx. 38 Erlang

0.7.2 Erlang C Model

In this model, blocked calls are queued. Letbe the time delay after which the channel ismade available.

Procedure:

1. Determine Pr[ >0] = probability of a call to be delayed from the Erlang C chart.

2. Compute the probability that the delay is longer than sayt; conditioned that there isa delay

Pr [ > tj >0] = exp [ (C A) t=H]

13


14/15

3. The unconditional probability that > t

Pr [ > t] = Pr [ >0] Pr [ > tj >0]

4. The average time delay is given by the expression

D= Pr [ >0] HC A

Erlang C model is given by the gure below:Example

A hexagonal cell within a 4-cell system has radius = 1.387 km. Total # of channels insystem = 60. Load per user A = 0.029 Erlangs, = 1call/hour. Assume Erlang C systemwith required GOS (prob of delay) = 0.05. Answer the following questions:

- How many users per square km?- Average holding time?- Probability of a delay longer than 10 sec.?

Solution:Area of cell = 2.6 x (1.387)2 = 5 sq km.# of cells/cluster = 4.# of channels = 60. =># of channels /cell = 60/4 = 15 channels.a) From Erlang C chart, GOS = 0.05, C = 15 =>A = 8.8 Erlangs.# of users = total trac/ trac per user =>U= A / A= 8.8/0.029 = 303 users.Users per sq km = 303/5 = 60 users/sq km.

14


15/15

b) H = Au/ = 0.029/1 hour = 104.4 seconds.Pr (delay> 10 secjdelay) = exp[-(C-A)t /H ] =exp(-(15-8.8)10/104.4) = 52.22%c) Pr (delay> 10) = Pr (delay > 0) Pr (delay> tjdelay)= 0.05 x 0.5522 = 0.028.

Comment: The way in which channels are grouped aects the number of users that canbe handled by the system. For example:

For GOS = 0.01 and C = 10 =>A = 4.46 Erlangs.Let 2 groups of C = 5 =>A = 1.36 x 2 = 2.72 Erlangs.Thus, how channel allocation is done is critical to performance.

15

Cell Covered

Documents