Path Loss and Received Power Calculation: Fall 2023

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19-AUG-19
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Table of Contents
TASK1-Determination of Path loss and Received power in Free Space........................................2
Executable GNU Code....................................................................................................................4
Code Simulation..........................................................................................................................5
PR (Received power) and Lf (Path loss) Graph Results..............................................................5
TASK2.............................................................................................................................................8
Design Procedure for a Cell Architecture, Deciding its location area and the Antenna-1..........8
Cellular Network design Reasoning-2.........................................................................................8
References........................................................................................................................................8
Table of Figures
Figure 1 Code in Simulation............................................................................................................6
Figure 2 PR (Received power) and Lf (Path loss) AT 150MHz.....................................................7
Figure 3 PR (Received power) and Lf (Path loss) AT 400MHz....................................................7
Figure 4 PR (Received power) and Lf (Path loss) AT 1000MHz...................................................8

TASK1-Determination of Path loss and Received power in Free Space
Given Data
 Transmit∧receiver antennaGain=¿=Gr=0 dB
 Carrier Frequency=fc=150 , 400∧1000 Mhz
 Distance=d=0−30 km
 Transmit power=Pt =100 Watts
Required Data
 GNU Octive Code=?
 P r (Received power) calculation
 L f (Path loss calculation)
Pr (Received power) calculation
 For Frequency 150 Mhz
lambda= c
f =2
Received power=( pt∗¿∗gr) /(( 4∗pi∗d )/lambda)2
¿ 2.817 e−9 watts
Lf (Path loss calculation)
For the path loss calculation we need to know the formula;
Path loss∈Free space=20 log 10[4 πDF /c ]
¿ 20 log 10[4 π /c ]+20 log10 [D]+20 log10 [F ]
¿ 32.46+20 log [F (MHz)∗D( Km)] _______eq1
 For Frequency 150 Mhz
Put the given values we have in eq1;
Path loss∈Free space=32.46+20 log [ f ( MHz )∗d ( Km ) ]
Path loss∈Free space(dB)=105.53
Pr (Received power) calculation
 For Frequency 400 Mhz
lambda=c /f =0.75
Received power= pt∗¿∗gr
( 4∗pi∗d
lambda )2 =3.96 e−10 watt

Lf (Path loss calculation)
 For Frequency 400 Mhz
Put the given values we have in eq1;
Path loss∈Free space(dB)=32.46+20 log [f (MHz)∗d ( Km)]
Path loss∈Free space(dB)=114.05
Pr (Received power) calculation
 For Frequency 1000 Mhz
Path loss∈Free space(dB)=32.46+20 log [f (MHz)∗d ( Km)]
Path loss∈Free space(dB)=122.01
Lf (Path loss calculation)
 For Frequency 1000 Mhz
Put the given values we have in eq1;
lambda=c /f =0.3
Received power=( pt∗¿∗gr) /(( 4∗pi∗d )/lambda)2
¿ 6.3389 e−11 watts

Executable GNU Code
clc
clear all
¿=2;
GR=2 ;
PT =100;
Pii=3.141 ;
Fr=150∗106
lmbd=1/ Fr ;
L=3∗108
for d=3 :33
ReceivPower (d )=( PT∗¿∗GR )/((4∗Pii∗d−1)/lmbd )2 ;
PathLoss (d )=((4∗Pii∗(d−1))/lmbd)2 ;
PathLossIndB (d )=10∗log( PathLoss (d )) ;
end
subplot (2,2,1)
plot ( ReceivPower)
title(Receive power)
xlabel(Distance)
ylabel (Receive power )
subplot (2,2,2)
plot ( PathLossIndB)
title(Path Loss In dB at given frequency)
xlabel(Distance)
ylabel (path loss in dB)

Code Simulation
Figure 1 Code in Simulation

PR (Received power) and Lf (Path loss) Graph Results
At Frequency 150Mhz
Figure 2 PR (Received power) and Lf (Path loss) AT 150MHz
At Frequency 400Mhz
Figure 3 PR (Received power) and Lf (Path loss) AT 400MHz

At Frequency 1000Mhz
Figure 4 PR (Received power) and Lf (Path loss) AT 1000MHz

TASK2
Design Procedure for a Cell Architecture, Deciding its location area and the Antenna-1
As we know about the Central part of the Urban area is surrounded by the Towers. Keeping in
mind this whole structure we came to know about the connections of the Urban areas, which are
really of higher speeds (Waddell, 2003). Keeping in mind the value of higher speed we believe
that it really needs a lot of towers in a specific city which are good in connection systems.
We know that the cell is divided and separated by number so we came to know that these
numbers represent the areas with different polarities (Albaharna, 1994, October). Likewise the
number 20 represents the Urban area. The number 12 is marked as Suburban area, which is really
denser than the rural ones and less dense than the urban ones. Sparsely populated areas
surrounded by towers are known to represent the rural areas which are marked as 8 number.
Cellular Network Design Reasoning-2
When we consider all available designs we came to know and understand that the best defined
shape for the base station is Hexagon shape.
It utilizes the area in an optimized form in which network is available on all points, like
considering the other shapes in comparison we found that the hexagonal shape is the best design
for this architecture cell designing (Tutschku, 1998, March).

References
Albaharna, O. T. (1994, October). Area and time limitations of FPGA-based virtual hardware. In
Proceedings 1994 IEEE International Conference on Computer Design: VLSI in Computers and
Processors (pp. 184-189). IEEE.
Tutschku, K. (1998, March). Demand-based radio network planning of cellular mobile
communication systems. In Proceedings. IEEE INFOCOM'98, the Conference on Computer
Communications. Seventeenth Annual Joint Conference of the IEEE Computer and
Communicatio.
Waddell, P. B. (2003). Microsimulation of urban development and location choices: Design and
implementation of UrbanSim. Networks and spatial economics, 3(1), 43-67.