problem(2)

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Faculty of Engineering
Cairo university
Credit Hours System
Dept. computers and communications
Major communications
CCE-E
Name: Mahmoud Hossam Madian
Code: 1072078
Mobile Communications
Report
Assignment-1
Chapter (4)
Problem(2)(4.23)
Problem (2) 4.23
Radio propagation Models (large- scale)
Path loss Models
do = 1 km= 1 *(10^3) m
Pr_do = 1 microwatt = 1 *(10^-6) w
f = 1800 MHz = 1800*(10^6)Hz
ht = 40 m
hr = 3 m
Gt = Gr = 0 dB
Gt = Gr =1 w
c= 3*(10^8)
lamda=c/f= 3*(10^8)/ 1800*(10^6) =1/6 m
find the received power at distances ??? from the same Tx
d=2km , d= 5km, d=10km,d= 20km
(a) Free space Model n=2
Pr_do =Pt*Gt*Gr *((( lamda) /(4*pi*do))^2)= 1 *(10^-6) w
Pt=Pr_do*(1./G).*(((4*pi*do)/lamda)^2) = 1 *(10^-6)*(1)*(1)*(((4*3.14*1 *(10^3)/(1/6))^2)=5684.89 w
Pt=10log5684.89=37.547dB
Using refrencess distance
Pr_d=Pr_do *((do/d)^n)
n=2 for free space model
Pr_(d=2km)=Pr_do *((do/ d=2km)^n)= (10^-6)*((( 1 *(10^3)/( 2*(10^3))^2) =0.25 mw
Pr_(d=5km)=Pr_do *((do/ d=5km)^n)= (10^-6)*((( 1 *(10^3)/( 5*(10^3))^2) =0.04 mw
Pr_(d=10km)=Pr_do *((do/ d=10km)^n)= (10^-6)*((( 1 *(10^3)/( 10*(10^3))^2) =0.01 mw
Pr_(d=20km)=Pr_do *((do/ d=20km)^n)= (10^-6)*((( 1 *(10^3)/( 20*(10^3))^2) =0.0025 mw
As distance increases between Tx and Rx , Pr decrease where Pt, Gr,Gt,n, lamda are constant
(b) Urban area cellular radio n=3
Pr_(d=2km)=Pr_do *((do/ d=2km)^n)= (10^-6)*((( 1 *(10^3)/( 2*(10^3))^3) =0.125 mw
Pr_(d=5km)=Pr_do *((do/ d=5km)^n)= (10^-6)*((( 1 *(10^3)/( 5*(10^3))^3) =0.008 mw
Pr_(d=10km)=Pr_do *((do/ d=10km)^n)= (10^-6)*((( 1 *(10^3)/( 10*(10^3))^3) =0.001 mw
Pr_(d=20km)=Pr_do *((do/ d=20km)^n)= (10^-6)*((( 1 *(10^3)/( 20*(10^3))^3) =0.00125 mw
As distance increases between Tx and Rx , Pr decrease where Pt, Gr,Gt,n, lamda are constant
(c) Obstructed in building n=4;
Pr_(d=2km)=Pr_do *((do/ d=2km)^n)= (10^-6)*((( 1 *(10^3)/( 2*(10^3))^4) =0.0625 mw
Pr_(d=5km)=Pr_do *((do/ d=5km)^n)= (10^-6)*((( 1 *(10^3)/( 5*(10^3))^4) =0.0016 mw
Pr_(d=10km)=Pr_do *((do/ d=10km)^n)= (10^-6)*((( 1 *(10^3)/( 10*(10^3))^4) =0.0001 mw
Pr_(d=20km)=Pr_do *((do/ d=20km)^n)= (10^-6)*((( 1 *(10^3)/( 20*(10^3))^4) =0.0000625 mw
As distance increases between Tx and Rx , Pr decrease where Pt, Gr,Gt,n, lamda are constant
(d) Two-ray Model;
Pr_tworay_(d)==Pt*Gt*Gr*(((ht*hr)^2)/((d)^4))
ht = 40 m
hr = 3 m
Pr_tworay_(d=2km)==Pt*G*(((ht*hr)^2)/(( (d=2km)=)^4)) =5684.89*1*(((40*3)^2/(((2*10^3)^4))
= 5 mw
Pr_tworay_(d=5km)==Pt*G*(((ht*hr)^2)/(( (d=5km)=)^4)) =5684.89*1*(((40*3)^2/(((5*10^3)^4))
=0.1 mw
Pr_tworay_(d=10km)==Pt*G*(((ht*hr)^2)/(( (d=10km)=)^4))
= 5684.89*1*(((40*3)^2/(((10*10^3)^4))= 0.008 mw
Pr_tworay_(d=20km)==Pt*G*(((ht*hr)^2)/(( (d=20km)=)^4))
=5684.89*1*(((40*3)^2/(((20*10^3)^4))=0.005 mw
As distance increases between Tx and Rx , Pr decrease where Pt, Gr,Gt,ht, hr are constant
Conculsion :
Difference between the four models as (n) decreases Pr increases , So the model that has the smallest
(n) ,have the bigger Pr, where in each model As distance increases between Tx and Rx , Pr decrease
where Pt, Gr,Gt,n, lamda are constant.
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