# CHAPTER 4 - Principles of Cellular Communication¶

## EXAMPLE 4.1 - PG NO.101¶

In [1]:
#page no. 101
A=140.
n=7.
Na=40.
C=A/n#coverage area of each cell
Nvchpercell=30./100.*Na
N=Nvchpercell*n#Number of voice channels
print '%s %d %s' %('coverage area of each cell =',C,'kmsqr')
print '%s %d %s' %('Number of voice channels =',N,'channels')
coverage area of each cell = 20 kmsqr
Number of voice channels = 84 channels

## EXAMPLE 4.2 - PG NO.102¶

In [2]:
#page no.102
K=4.
Acell=7.
Acl=K*Acell#area of cluster
Asys=1765.
Nservarea=Asys/Acl#number of clusters
N=round(Nservarea)
print'%s %d' %('Numer of times the cluster of size 4  has to be replicated is',N)
Numer of times the cluster of size 4  has to be replicated is 63

## EXAMPLE 4.3 - PG NO.103¶

In [3]:
#page no.103
import math
N=32.
Rkm=1.6
Acell=(3.*math.sqrt(3.)/2.)*(Rkm**2.)
TA=N*Acell#total service area
Tc=336.
n=7.
Ncpc=Tc/n#number of channels per cell
TSC=Ncpc*N#total sysytem capacity
N1=128.
Ahex=TA/N1
R=math.sqrt(Ahex/(1.5*math.sqrt(2.)))
NCap=Ncpc*N1
print'%s %d %s' %('total service area is =',round(TA),'kmsqr')
print'%s %d ' %('number of channels per cell =',Ncpc)
print'%s %d %s' %('total sysytem capacity in no. of channels =',TSC,'channels')
print'%s %.2f %s' %('radius of the new smaller cell is =',R,'km')
print'%s %d %s' %('new system capacity in no. of channels',NCap,'channels')
total service area is = 213 kmsqr
number of channels per cell = 48
total sysytem capacity in no. of channels = 1536 channels
radius of the new smaller cell is = 0.89 km
new system capacity in no. of channels 6144 channels

## EXAMPLE 4.5 - PG NO.107¶

In [4]:
#page no.107
N=1000.
n=20.
n1=4.
M=n/n1
TSC=N*M#system capacity
print'%s %d %s' %('the system capacity in no. of users',TSC,'users')
n2=100.
n3=4.
M1=n2/n3
NSC=N*M1#new system capacity for increased no. of cells
print'%s %d %s' %('the new system capacity for increased no. of cells in no. of users',NSC,'users')
n4=700.
n5=7.
M2=n4/n5
NSC1=N*M2#new system capacity for increased no. of cells
print'%s %d %s' %('the system capacity for increased no. of cells & also cluster size in no. of users',NSC1,'users')
the system capacity in no. of users 5000 users
the new system capacity for increased no. of cells in no. of users 25000 users
the system capacity for increased no. of cells & also cluster size in no. of users 100000 users

## EXAMPLE 4.6 - PG NO.109¶

In [5]:
#page no.109
import math
Asys=4200.#area of system
Acell=12.#area of cell
N=1001.
K=7.
Acl=K*Acell#area of cluster
M=Asys/Acl#no. of clusters
print'%s %d %s' %('no. of clusters',M,'clusters')
J=N/K#cell capacity
print'%s %d %s' %('cell capacity is =',J,'channels/cell')
C=N*M#system capacity
print'%s %d %s' %('the system capacity in no. of channels',C,'channels')
k=4.
acl=k*Acell
m=Asys/acl
m1=math.floor(m)
print'%s %d' %('no. of clusters for reduced cluster size',m1)
c=N*m1
print'%s %d %s' %('new system capacity for reduced cluster size in no. of channels',c,'channels')
no. of clusters 50 clusters
cell capacity is = 143 channels/cell
the system capacity in no. of channels 50050 channels
no. of clusters for reduced cluster size 87
new system capacity for reduced cluster size in no. of channels 87087 channels

## EXAMPLE 4.7 - PG NO.110¶

In [6]:
#page no. 110
n=16.
N=7.
M=12.
Ncpc=n*N#no. of channels per cluster
TSC=Ncpc*M#system capacity
print'%s %d %s' %('no. of channels per cluster is =',Ncpc,'channels/cluster')
print'%s %d %s' %('the system capacity in channels/system is =',TSC,'channels/system')
no. of channels per cluster is = 112 channels/cluster
the system capacity in channels/system is = 1344 channels/system

## EXAMPLE 4.9 - PG NO.114¶

In [7]:
#page no. 114
i=2.#no. of cells(centre to centre) along any chain of hexagon
j=4.#no. of cells(centre to centre) in 60deg. counterclockwise of i
K=i*i+j*j+i*j#cluster size
print'%s %.f' %('no. of cells in a cluster for i=2 & j=4 is',K)

i1=3.
j1=3.
K1=i1*i1+j1*j1+i1*j1#cluster size
print'%s %.f' %('no. of cells in a cluster for i=3 & j=3 is',K1)
no. of cells in a cluster for i=2 & j=4 is 28
no. of cells in a cluster for i=3 & j=3 is 27

## EXAMPLE 4.10 - PG NO.115¶

In [8]:
#page no. 115
q=12.#co-channel reuse ratio
D=q*R#nearest distance
print'%s %.2f %s' %('distance from the nearest cochannel cell is D =',D,'km')
distance from the nearest cochannel cell is D = 7.68 km

## EXAMPLE 4.11 - PG NO.115¶

In [9]:
#page no.115
R=.8
D=6.4
q=D/R#frquency reuse ratio
print'%s %d' %('frquency reuse ratio q is =',q)
frquency reuse ratio q is = 8