Signal Attenuation and Path Loss in Wireless Communication Systems

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Assessment 2
Task and Design Project
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Contents
Introduction......................................................................................................................................2
Task 1: Free space propagation and path loss.................................................................................5
Calculation of Free space path loss and Received power............................................................5
Results..............................................................................................................................................6
Octave code.................................................................................................................................7
Graph plots...................................................................................................................................8
Task 2: Research project................................................................................................................10
Area selection for base station...................................................................................................10
Justification of design................................................................................................................10
Conclusion.....................................................................................................................................11
References......................................................................................................................................11
List of Figures
Figure 1: Free Space Path Loss.......................................................................................................4
Figure 2: Relationship of distance and received power...................................................................5
Figure 3: Octave code path loss and Pr calc....................................................................................8
Figure 4: Plots for path loss and Pr at 150 MHz..............................................................................8
Figure 5: Plots for path loss and Pr at 400 MHz..............................................................................9
Figure 6:Plots for path loss and Pr 1000 MHz................................................................................9
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Introduction
Objective of this study is to understand the concept of signal attenuation and its resulting path
loss of signal which results in the reduction of power the signal at receiver’s end. And also, to
understand the design considerations for a base station. (Chigozie Iwuji, 2018). And to calculate
the path loss we take help of the following formula
Free space pathloss=( 4 πDF
c )
2
Where
D is the distance between transmitting end and receiving end. c is the speed of light and F is
MHZ frequency.
But path loss is usually shown in dB so converting this equation into FS path loss (dB).
FS path loss ( dB )=10 log10 X FS pathloss
FS path loss ( dB )=10 log 10( 4 πDF
c )
2
Using property of logarithm
FS path loss ( dB ) =2 X 10 log 10 4 πDf
c
Again, using logarithm property
FS path loss =20log10 (4π/c) + 20log10 (D) =20log10(F)
FS path loss ( dB ) =20 log10(4π/c) + 20log [F(MHZ) * D(km)]
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Figure 1: Free Space Path Loss
The power at the receiving end is inversely proportional to the distance, the signal covers (i.e.)
the more the signal have to travel the less power is received at point B. Well, we can calculate
this power as well if we know the transmitted powerpt , gain of transmitting antennae G”, gain
of receiving antennae Gr and the distance between transmitting and receiving ends “D”.
Power Received=( pt∗¿∗Gr∗λ2)
(4 πD)2
Graphically,
Figure 2: Relationship of distance and received power
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Task 1: Free space propagation and path loss
Calculation of Free space path loss and Received power
We have to calculate the path loss of a signal travelling from transmitter to receiver covering a
distance of 30km. where the antennae gain for both transmitter and receiver is 1. The power of
the transmitted signal is 100 watts. The calculations for FS path loss (dB) and Received power at
different frequencies are as follows
a. At 150 MHz frequency
As we already know that path loss is given by
FS path loss ( dB ) =20 log10(4π/c) + 20log [F(MHZ) * D(km)]
FS path loss ( dB )=105.53 dB
Power Received=( pt∗¿∗Gr∗λ2)
(4 πD)2
We know that v= f λ
Replace v by c and rearranging
λ= c
f =2
Power Received = 2.817e -9 watts
b. At 400 MHz frequency
As we already know that path loss is given by
FS path loss ( dB ) =20 log10(4π/c) + 20log [F(MHZ) * D(km)]
FS path loss ( dB )=114.05 dB
Power Received=( pt∗¿∗Gr∗λ2)
(4 πD)2
We know that v= f λ
Replace v by c and rearranging
λ= c
f =0.75
Power Received = 3.96e – 10 watts
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c. At 1000 MHz frequency
As we already know that path loss is given by
FS path loss ( dB ) =20 log10(4π/c) + 20log [F(MHZ) * D(km)]
FS path loss ( dB )=122.01 dB
Power Received=( pt∗¿∗Gr∗λ2)
(4 πD)2
We know that v= f λ
Replace v by c and rearranging
λ= c
f =0.3
Power Received = 6.3889e -11 watts
Results Table 1: Results of path loss and Pr calculations
F(MHz) Path loss(dB) lambda Power Received (watts)
150 105.53 2 2.817e -9
400 114.05 0.75 3.96e – 10
1000 122.01 0.3 122.01
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Octave code
clc
clear all
close all
Pt =100;
¿=1;
Gr=1 ;
l=3∗108 ;
f =150∗106 ;/¿ carrier frequency
pi=3.14 ;
lambda=1 /f ;
for D=1 :31;
Pr (D)=( Pt∗¿∗Gr)/((4∗pi∗( D−1))/lambda)2 ;
FSPL( D)=(( 4∗pi∗(D−1))/lambda)2 ;
FSPLdb ( D)=10∗log(FSPL(D)) ;
end
subplot (211)
plot (Pr)
title(Pr1);
xlabel(D);
ylabel (Pr );
subplot (212)
plot ( FSPLdb )
title(FSPL @ 150MHZ) ;
xlabel(D);
ylabel (FSPL (db));
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Figure 3: Octave code path loss and Pr calc.
Graph plots
At 150 MHz frequency
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Figure 4: Plots for path loss and Pr at 150 MHz
At 400 MHz frequency
Figure 5: Plots for path loss and Pr at 400 MHz
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At 1000 MHz frequency
Figure 6:Plots for path loss and Pr 1000 MHz
Task 2: Research project
Area selection for base station
 People residing in urban area usually are high speed users which lead in bigger number of
towers than for rural areas for that city (Sheynblat, 2008)
 If we notice, we can assume that over stacked center of the cell with towers, is
representing the urban area, justifying our previous point.
 If we work on this hierarchy of tower division, the area regarded as number 12 should be
representing a suburban area as it is not as populated with towers as the urban and doesn’t
have that a smaller number of towers to be regarded as rural so we can designate this for
suburban area.
 Number 20 will be claimed as urban area
 Number 8 is claimed as the rural area, as the tower placement is casual and is least dense.
Justification of design
 The hexagonal design for the cell is the best nomination as it manages to provide the
maximum interconnection with other cells hence optimizing the linkage with adjacent
base station cells.
 Taking in consideration the ease of access, optimum area coverage and maximum inter-
linkage, hexagonal base station cells are our best choice.
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References
Chigozie Iwuji, P. &. (2018). Determination Of A Pathloss Model For Long Term Evolution
(Lte) In Yenagoa. The International Journal of Engineering and Science, 38-44
Homood, H. K. (2018). Analyzing Study of Path Loss Propagation Models in Wireless
Communication at 8.0 GHz. Journal of Physics : Conferrence Series.
Isabona, J. &. ((2013)). Radio Field Strength Propagation Data and Pathloss calculation Methods
in UMTS Network. Advances in Physics Theories and Applications.
Libor MICHALEK, M. D. (2015). Analysis of Signal Attenuation in UHF band. Information and
Communication Technologies and Services.
Sheynblat, L. &. (2008). U.S Patent No.7,319,878. Washington DC :U.S Patent and Trade mark
office.
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