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