written 8.8 years ago by | • modified 8.8 years ago |
Mumbai University > Electronics and Telecommunication > Sem 7 > Mobile Communication
Marks: 10 M
Year: May 2012
written 8.8 years ago by | • modified 8.8 years ago |
Mumbai University > Electronics and Telecommunication > Sem 7 > Mobile Communication
Marks: 10 M
Year: May 2012
written 8.8 years ago by | • modified 8.8 years ago |
we know,
sI=R−n∑ioi=1D−n
sI=R−n2D−n+2(D−R)−n+2(D+R)−n
sI=R−n2R−n[(DR)−n+(1−DR)−n+(1+DR)−n]
s1=12[(√3N)−n+(1−√3n)−n+(1+√3N)−n]........(DR=√3N)
For N=12 and n=4 we get S/I = 22.53 db which is greater than 18 db, hence cluster size of 12 can be used. For N=7 S/I is 17.27 db, which is less than 18 db, thus cluster size of 7 cannot be used.
As after sectoring there are only 2 co-channel cells, assume distance of these 2 cells be D and D+R.
sI=R−nD−n+(D+R)−n
sI=R−nR−n(DR)−n+R−n(1+DR)−n
sI=1(√3N)−n+(1+√3N)−n
For N=7 and n=4, S/I is found to be as 24.81 db which is less than 18 db, thus cluster size of 7 can be used with sectoring.
Increase in capacity is given by,
Old cluster sizeNew cluster size=127=1.714