Problem 1 a) The frequency band allocated to a certain cellular standard enables 60 traffic

channels/cell. When the whole band is licensed to one operator, what is the amount of traffic the operator can serve with 2 % blocking probability? The traffic is assumed to be Erlang-B distributed.

b) Competition is introduced by sharing the frequency band between three operators, each getting 20 traffic channels/cell. What is the amount of traffic the three operators can together serve with 2% blocking? Each operator will have about the same network configuration, and subscribers can only be served by their own operator.

Solution a) The Erlang B-table gives T1 = 49.64 Erlang b) The Erlang B-table gives for one operator (20 TCHs) T1 = 13.18 Erlang → T3 = 3⋅13.18 = 39.54 Erlang By giving all frequency resources to one operator instead of one third to

each of three operators would give 25.5 % higher total capacity.

Problem 2 a) A GSM900 operator has 36 carriers which are used with the reuse factor 9.

How large traffic can the operator serve with 2 % blocking probability. The network is ideal with spatially homogeneously distributed traffic and equi-sized cells.

b) To meet the increasing capacity demand the operator opens a GSM1800 network with 108 carriers given the same coverage using same base stations as the GSM900 network and having the same reuse factor. How large traffic the operator can now serve with 2 % blocking probability?

c) Calculate the traffic capacity with 2 % blocking probability if all user terminals are dual-band able to use both frequency band. How many % larger the capacity is compared to the situation in b) where only single-band terminals are used?

Solution a) Under the given conditions there are 36/9 = 4 carriers/cell. Assuming that in

each cell two timeslots are used for signalling there are 4⋅8 − 2 = 30 traffic channels/cell. From Erlangs B-table the traffic capacity for 2% blocking probability is 21.93 Erlang/cell.

b) If it is assumed that the number of timeslots is proportional to the number of carriers, there will be 90 traffic channels/cell in the GSM1800 system. Applying again Erlang’s B-table the traffic capacity in the new system is 78.31 Erlang/cell

Now the total traffic capacity is 21.93 + 78.31 = 100.24 Erlang/cell. c) When all terminals are dual-band each terminal will have 120 channels

available. Erlang’s B-table gives then a traffic capacity of 107.40 Erlang/cell.

Compared to case b) where only single-band terminals are used 7.1% higher traffic capacity is obtained

Problem 3. a) A mobile phone subscriber generates on average 20 mErlang traffic during

the busy hour. (Occupies a channel for 1.2 minutes.) How many subscribers in a cell having 30 traffic channels will cause a blocking probability of i) 0.5%, ii) 5%, iii) 50%?

b) How many subscribers can be served at the different blocking levels? c) What is the average channel load at the different blocking levels? Solution a) The traffic in a cell is obtained from the Erlang B-table. The number of

subscribers that can be served is obtained by dividing the cell traffic with the average subscriber traffic.

B T Erlang N

B T Erlang N

B T Erlang N

= → = → = = →

= → = → = = →

= → = → = = →

05% 19 0319 030 02

9515 951

5% 24 8024 800 02

1240 0 1240

50% 58115811002

29055 2905

. ...

.

..

..

..

..

b) The portion of subscribers being blocked cannot be served. Therefore

( )( )( )

0.5% 1 (1 0.005)951.5 946.7 946

5% 1 (1 0.05)1240 1178 1178

50% 1 (1 0.5)2905.5 1452.75 1452

served

served

served

B N B N

B N B N

B N B N

= → = − = − = →

= → = − = − = →

= → = − = − = →

c) The average channel load at a given blocking level is the served traffic

divided by the number of channels

BB T

N

BB T

N

BB T

N

= → =−

=−

=

= → =−

=−

=

= → =−

=−

=

05%1 1 0 005 19 03

300631

5%1 1 0 05 24 80

300 785

50%1 1 05 5811

300 969

.. .

.

. ..

. ..

η

η

η

a f a f

a f a f

a f a f

One can see that allowing a higher blocking level more subscribers can be

served and the spectral efficiency is increased. However, a high blocking level might not be a good marketing argument!

Problem 4 1 000 000 people are living uniformly distributed in the service area of a cellular operator. The traffic can be modeled with the Erlang-B distribution and there are 50 traffic channels/cell. Without considering coverage matters, how many cells must the operator use to obtain 2% blocking in the following cases.

a) The cellular phone penetration is 10% and each subscriber generates 25 mErlang traffic during the busy hour.

b) The cellular phone penetration has increased to 80% and each subscriber still generates 25 mErlang traffic during the busy hour.

c) The cellular phone penetration has increased to 80% but the new subscribers coming after the first 10% will produce a traffic inversely proportional to the penetration during the busy hour (the first new subscriber 25 mErlang).

d) If the population density is 100 inhabitants/km2, what would be the cell size in the three previous cases based on pure capacity considerations?

Solution

a) T = Population⋅Penetration⋅User traffic = 610 0.1 0.025 2500 Erlang⋅ ⋅ = With 50 traffic channels and 2 % blocking target the Erlang B table gives the

maximum cell traffic 50,0.02 40.24 ErlangT = . Thus the needed number of

cells is

50,0.02

250062.18 63

40.24cellT

NT

= = = →

b) T = Population⋅Penetration⋅User traffic = 610 0.8 0.025 20000 Erlang⋅ ⋅ =

50,0.02

20000497.01 498

40.24cellT

NT

= = = →

c) Now the user traffic vs. penetration model can be written in the form

0.1

0.025oTp

= ⋅

and the total traffic is

( ) ( )

0.80.80.1

0.10.1

0.1

2500 1 ln 2500 1 ln 8

7698.60 Erlang

pp

TT T dp p

p

⌠⌡=

=

= + = + = + =

50,0.02

7698.60191.32 192

40.24cellT

NT

= = = →

d) With a population density of 100 inhabitants/km2, one million population is spread over 1000000/100 = 10000 km2. Then a simple estimate of cell size is the whole area divided by the number of cells:

2 2

2

10000 10000158.7 km 20.1 km

63 49810000

52.1 km192

a b

c

A A

A

→ = = = =

= =

Problem 5 A cellular network operator has been granted a 2×10 MHz bandwidth. He intends to use an FDMA/TDMA/FDD-system with 200 kHz carrier spacing and 8 timeslots/carrier in both down-link and up-link. The co-channel reuse factor is 7. a) How many traffic channels are available in each base station? It is assumed

that all signaling is multiplexed on the traffic channels. b) How much traffic can be served in a base station, if the blocking probability

target is 2 %, and the traffic obeys Erlang-B? c) The frequency regulator requires an average channel load of 90%. To which

approximate value will the blocking probability increase?

Solution a) With FDD (Frequency Division Duplex) e.g. the down-link can use half of

the available frequency resource, so the total number of carriers is

total DL bandwidth 1050

carrier spacing 0.2carrierN = = =

The number of carriers in a base station is obtained by dividing the total number of carriers with the co-channel reuse factor, and taking the nearest smaller integer

( )_50

int int 7.143 77BS carrierN = = =

As each carrier will contain 8 timeslots, the number of traffic channels in a base station is _ 8 7 56BS TCHN = ⋅ =

b) The amount of served traffic is given by

( )1 0.98 45.88 44.96 Erlangserved offeredT B T= − = ⋅ =

The numerical of the offered traffic is obtained from the Erlang B table.

c) The average fractional channel load is ( )

_

1 offeredTCH

BS TCH

B T

Nη

−=

Insertion of different offered traffic values in the Erlang-B expression

0

!

!

N

kN

k

T

NBT

k=

=∑

It appears that an offered traffic of 55.9 Erlang gives a fractional channel load of 0.9, and the blocking probability is 9.84 %.

Toffered B η 55 0.0896 0.8941 55.1 0.0906 0.8948 55.2 0.0916 0.8955 55.3 0.0925 0.8961 55.4 0.0935 0.8968 55.5 0.0945 0.8974 55.6 0.0955 0.8981 55.7 0.0965 0.8987 55.8 0.0974 0.8993 55.9 0.0984 0.9000 56 0.0994 0.9006

Toffered B η 50 0.0458 0.8520 51 0.0537 0.8618 52 0.0621 0.8710 53 0.0709 0.8793 54 0.0801 0.8870 55 0.0896 0.8941 56 0.0994 0.9006 57 0.1094 0.9065 58 0.1195 0.9120 59 0.1297 0.9169 60 0.1399 0.9215

Problem 6 In a cellular service area the offered traffic density is 4 Erlang/km2. The number of traffic channels in a base station is 36, and the blocking probability target is i) 1%, ii) 2%, iii) 5%, iv) 10 %. Omnidirectional base station antennas are used a) If the cell size can be dimensioned for maximum capacity, what is the cell

area for the given blocking probability targets? b) Calculate the maximum path length for the given blocking probability targets,

1) with circular cell shape, 2) with hexagonal cell shape, and 3) with quadratic cell shape.

Solution a) The cell size (area) is obtained by dividing the traffic values for the different

blocking probabilities with the traffic density: ,36Bcell

TA

τ= . Numerical

results are given in Table 6.1 Table 6.1

B = 1% B = 2% B = 5% B = 10% TB,36 25.51 27.34 30.66 34.50 Acell/km2 6.378 6.835 7.665 8.625

R

RR

R R

R

b) The area of a cell with circular shape is 2A R R Aπ π= → =

The area of a cell with hexagonal shape is 3

62 2 6.75

R AA R R= ⋅ → =

The area of a cell with quadratic shape is ( )22 2A R R A= → =

Table 6.2

B = 1% B = 2% B = 5% B = 10% Acell/km2 6.378 6.835 7.665 8.625 Rcicrle/km 1.425 1.475 1.562 1.657 Rhexagon/km 1.567 1.622 1.718 1.822 Rquadrate/km 1.786 1.849 1.958 2.077

1

P9. In a cellular system a gain term (e.g. BS power or antenna gain) is increased with 3 dB. This increase is fully utilised to increase coverage. How many percent will i) the cell radius, ii) the cell area increase under the assumption of a single slope average path loss model with the path loss exponent n = 2, 3, 4, or 5?

( )( ) ( ) ( )( )1 0 1 2

2 1 2 12 0 2

10 log10 log log 10 log

10 log

L L n r rL L L n r r n

L L n r r

= + → − = ∆ = − = = +

( )210 10 52 2

1 110 10 10L n L n L nr A

r A∆ ∆ ∆= → = =

n 2 3 4 5 2 1r r 1.413 → 41.3% 1.259 → 25.9% 1.188 → 18.8% 1.148 → 14.8% 2 1A A 1.995 → 99.5% 1.585 → 58.5% 1.413 → 41.3% 1.318 → 31.8%

2

down-link

up-link

The figure shows the system parameters of a cellular phone radio link that should be considered in the radio link budget. The average path loss as function of the distance d is 133,8 33,8 lg( )cL d= + .

a) Determine the allowed radio channel loss in both down-link and up-link. b) How large is the cell radius based on the average path loss (giving 50 %

coverage probability at the cell border)? Consider both down-link and up-link.

BS, Pbs= 40 dBm

comb. filt. L = 3 dB

MS, SMS=

-100 dBm

MS, Pms 29 dBm

ant+feed. G = 0 dB

radio ch. L = L p

BS ant. G=10 dB

feeder L = 4 dB

div. comb. G = 4 dB

BS, SMS= -102 dBm

ant+feed. G = 0 dB

radio ch. L = L p

BS ant. G=10 dB

ant+feed. L = 4 dB

3

SOLUTION a) In the down-link

. . ,

, . .

40 3 4 10 0 100 143

bs comb filt feed bs p dl ms ms

p dl bs comb filt feed bs ms ms

P L L G L G S

L P L L G G S

dB

− − + − + ≥

→ ≤ − − + + −

≤ − − + + + =

In the up-link

,

,

29 0 10 4 4 102 141

ms ms p ul bs feed div bs

p ul ms ms bs feed div bs

P G L G L G S

L P G G L G S

dB

+ − + − + ≥

→ ≤ + + − + −

≤ + + − + + =

b) Based on the allowed down-link and up-link path loss

( )143 133.8 33.8133.8 33.8 log 143 10 1.87dl dlR R km−+ = → = =

( )141 133.8 33.8133.8 33.8 log 141 10 1.63ul dlR R km−

+ = → = =

4

P10. In this task the indoor coverage probability at the cell border should be estimated when the corresponding outdoor coverage probability is known.

The received power level is obtained from the radio link budget, and it si given by the expression rx tx tx tx p rx rxP S SFM P L G L G L= + = − + − + − ,

where S is the receiver sensitivity level, , , and tx tx txP L G are the transmit power level, antenna feeder sysstem loss, and antenna gain respectively,

and rx rxG L are the receiver antenna gain and feeder loss. The slow fade margin is INVQ(1 )SFM p σ= − ⋅ , and the average path loss with the actual radio link parameters 127.1 35.2logp w hmsL L A r= + − + and

2.6 3.9hms msA h= − . The function INVQ( ) gives the argument of the Q-function when the value of the Q-function is known. Lw is the outer wall average penetration loss.

The outdoor MS antenna height is 1.5 m, shadow fading standard deviation σ = 6 dB, and the independent log-normal average wall penetration loss is 10 dB and standard deviation 8 dB. All other parameters than SFM, Ahms and average path loss remain unchanged in the indoor case. Calculate the indoor coverage probability on the first floor (MS antenna height 5 m) at the cell border, where the outdoor coverage probability is p = 0.90.

5

SOLUTION In the outdoor situation

INVQ(0.1) 6 1.28 6 7.68

2.6 1.5 3.9 0hms

SFM dB

A dB

= ⋅ = ⋅ =

= ⋅ − =

In the indoor situation

2 2

2.6 5.0 3.9 9.1

10 9.1 0.9

7.7 0.9 6.8 INVQ(1 ) INVQ(1 ) 6 8

10 INVQ(1 )

6.8INVQ(1 ) 0.68 1 0.248 0.752

10

hms

tot

A dB

L dB

SFM dB p p

p

p p p

σ

= ⋅ − =∆ = − =

→ = − = = − ⋅ = − ⋅ += ⋅ −

→ − = = → − = → =

The Q- and INVQ-function values can be obtained from the graph in the next slide

6

Q(x)

0 1 x 2 3QFUNCT3.dsf

1

0.1

0.01

0.001

0.2

0.5

0.002

0.005

0.02

0.05

7

P11. Based on log-normal shadow fading with 8 dB standard deviation, determine the shadow fading margin needed in radio link budget calculations in a single microcell with the cell coverage probability target a) 90%, b) 95%. Obtain the path loss exponent using the COST231 Walfisch-Ikegami average path loss model when hr = 24 m and hbs = 8 m.

Solution From the COST 231 Walfisch-Ikegami average path loss model the terms containing log(d) must be identified, and the sum of the coefficient divided by 10 gives the path loss exponent. There should not be any other terms containing d, so this approach is not valid for d < 0.5 km

0.1(20 ) 0.1 20 18 15

24 80.1 38 15 4.8 1.67

24

roof msd

roof

h hPLE n k

h

n

σ

−= = + = + + − = + = → =

Based on the normalised values SFM/σ in the figure on next page

(90%) 8 0.544 4.35 dB

(95%) 8 0.963 7.70 dB

SFM

SFM

= ⋅ =

= ⋅ =

8

1.0

0.8Fu = 0.95

Fu = 0.90

Fu = 0.85

Fu = 0.80

σSF/n1 1.5 2 2.5

0

0.6

0.4

0.2

−0.2

−0.4

1.2

SFM/σSF

Normalised shadow fade margin as function of propagation parameters for different cell coverage probabilities

SFM_outage.dsf

9

P12. In the lecture material the effective gain of a mast-top amplifier was found to be 9.08 dB when the antenna noise temperature equals the

standard temperature To = 290 K. The equipment parameters are dB 4 dB, 10 dB, 2 dB, 12 ==== fsrxmtamta LFFG

To which value is the effective gain reduced when the antenna noise temperature due to man-made noise is 10 To?

SOLUTION The general expression for the BS receiver total input noise is

( ) ( )BkTBkT

L

LBTTk

LG

P rxfsfs

fsmtaa

fs

mtarxn +

−++=

1_

and the gain reduction equals the noise increase due to the mta effective gain is

_

_10 lg n rx

mtan rxo

PG

P

∆ =

10

( ) ( )( )

( ) ( )( )

( )( ) ( ) ( )

1

10 lg1

1

10 lg1

110 1 1

10 lg10

fsmtaa mta fs rx

fs fs

fsafs rx

fs fs

fsmtaa mta fs rx

fs fs

fsafs rx

fs fs

fsmtao mta o o rx o

fs fs

o

LGk T T B kT B kT B

L L

LkT BkT B kT B

L L

LGT T T T

L L

LTT T

L L

LGT F T T F T

L L

T

− + + + = − + + − + + + = − + +

−+ − + + −= ( ) ( )

( ) ( ) ( )( ) ( )

11

9 1 110 lg

10 1 1

fso rx o

fs fs

mta mta fs rx fs

fs rx fs

LT F T

L L

G F L F L

L F L

− + + − + + − + − = + − + −

11

( )9 110 lg

9mta mta rx fs

rx fs

G F F L

F L

+ + −= +

( )1.2 0.2 0.4

0.4

10 10 9 10 10 110 lg 10.88 dB

9 10 10

+ + ⋅ − = = + ⋅

The resulting effective gain is

_ 12 10.88 1.12 dBmta eff mta mtaG G G= − ∆ = − =

Conclusion: In high background noise environments a large part of the mta advantages is lost

12

P13 Based on the single cell coverage probability target, derive an expression of the coverage probability as a function of the distance normalised to the cell radius and draw the graph when σ = 6 dB, n = 4, and the cell coverage probability is i) 0.90, ii) 0.95.

SOLUTION With log-normal shadow fading the coverage probability on distance r is

( ) ( ) ( ) ( )cov

10 log 10logQ QrxS P R n r R r R SFM

P rnσ σ σ

− + = = −

( ) ( )( )cov covQ 1 Q INVQ 1SFM SFM SFM

P R P Rσ σ σ

→ = − = − → = −

( ) ( ) ( )( )cov10log

Q INVQ 1covr R

P r P Rnσ

→ = − −

From the graph of ( ), ( )covSFM f n P Rσ σ= we get

( )( )1.5,0.9 0.494

1.5,0.95 0.923

SFM f

SFM f

σ

σ

= ≈

= ≈

13

1.0

0.8Fu = 0.95

Fu = 0.90

Fu = 0.85

Fu = 0.80

σSF/n1 1.5 2 2.5

0

0.6

0.4

0.2

−0.2

−0.4

1.2

SFM/σSF

Normalised shadow fade margin as function of propagation parameters for different cell coverage probabilities

SFM_outage.dsf

0.494

0.923

14

0 1 2 43 5r/R

10-0

10-1

10-2

10-3

10-4

10-5

Pcov(r)

SFM_outage.dsf

Fu=0.95

Fu=0.90

15

P14 Cell coverage probability planning target is i) 90 %, ii) 95 %. The planning approach is based on j) connection to a dedicated base station (single cell), jj) connection to the best base station in a 7 cell cluster. The path loss exponent is 4, and the following four log-normal shadow fading cases are considered: k) σ = 4 dB, kk) σ = 6 dB, kkk) σ = 8 dB, and kkkk) σ = 10 dB.

a) What is the obtained coverage in case jj), if the coverage planning has been based on case j)?

b) How many times can the number of base stations be reduced, if the coverage planning can be based on case jj) instead of case j)?

SOLUTION a) The coverage probability will depend on the shadow fade margin (SFM)

which depends on the single cell border coverage probability and on the parameter σ/n.

Using the graph ( ), _ ( ),u single cell covF f P R nσ= we get the value for

( )coverageP R , and the using this value the asked cell coverage probability

can be obtained from the graph ( ), _ ( ),u multi cell covF f P R nσ=

16 0.2

0.2

0.4 0.6 0.8

1.0

0.4

0.6

0.8

Single cell coverage probability

1.0

Fu

Pcov(R)Cell_coverage_prob.dsf

σ/n = 1.01.11.21.31.41.51.61.7

1.81.92.02.12.22.32.42.5= σ/n

0.614

0.694 0.737

0.7650.772

0.822 0.850

0.868

17

0.614

0.694 0.737

0.7650.772

0.822 0.850

0.8680.2

0.2

0.4 0.6 0.8

1.0

0.4

0.6

0.8

Middle cell coverage probability in a 7 cell cluster

1.0

Fu

Pcov(R)Cell_coverage_prob.dsf

σ/n = 1.01.11.21.31.41.5

1.61.71.81.92.02.12.22.32.42.5= σ/n

0.9900.979

0.9990.999

18

The results obtained from the graphs are collected into the following table

σ/n = 4/4 = 1

σ/n = 6/4 = 1.5

σ/n = 8/4 = 2

σ/n = 10/4 = 2.5

Pcov(R) 0.614 0.694 0.737 0.765 Fu1 = 0.90 Fu7 0.979 0.990 0.999 0.999

Pcov(R) 0.772 0.822 0.850 0.868 Fu1 = 0.95 Fu7 0.999 0.999 0.999 0.999

b) Using the graph ( ), _ ( ),u multi cell covF f P R nσ= we get the value for

( )covP R , and the using these values the SFM/σ-values are obtained by

( )INVQ 1 ( )covSFM P Rσ = −

The reduction of the necessary SFM is obtained as ( )1 7SFM SFM SFM σ∆ = − and as the number of cells is

( )2_ 1 7 7 1 7 1cell serv area cell cell cell cell cellN A A N N A A R R= → = =

From the average path loss we obtain

19

( ) ( )22 /(10 ) / 207 1 10 10SFM n SFMR R ∆ ∆

= =

σ/n = 4/4

= 1 σ/n = 6/4 = 1.5

σ/n = 8/4 = 2

σ/n = 10/4 = 2.5

Pcov1(R) 0.614 0.694 0.737 0.765

SFM1/σ 0.290 0.507 0.634 0722

Pcov7(R) 0.345 0.405 0.417 0.422

SFM7/σ – 0.399 – 0.240 – 0.210 – 0.197 ∆SFM 2.756 4.482 6.752 9.190

Fu1 = 0.90

N1/N7 1.373 1.675 2.176 2.881

Pcov1(R) 0.772 0.822 0.850 0.868

SFM1/σ 0.745 0.923 1.036 1.117

Pcov7(R) 0.467 0.502 0.519 0.523

SFM7/σ – 0.083 0.005 0.048 0.058 ∆SFM 3.312 5.508 7.904 10.590

Fu1 = 0.95

N1/N7 1.464 1.185 2.484 3.385

20

0.3450.4050.4170.422

0.4670.5020.5190.523

0.2

0.2

0.4 0.6

1.0

0.4

0.6

0.8

Middle cell coverage probability in a 7 cell cluster

1.0

Fu

Pcov(R)Cell_coverage_prob.dsf

σ/n = 1.01.11.21.31.41.5

1.61.71.81.92.02.12.22.32.42.5= σ/n