written 8.4 years ago by | • modified 8.4 years ago |
Mumbai University > Electronics and Telecommunication > Sem 7 > Mobile Communication
Marks: 10 M
Year: May 2012
written 8.4 years ago by | • modified 8.4 years ago |
Mumbai University > Electronics and Telecommunication > Sem 7 > Mobile Communication
Marks: 10 M
Year: May 2012
written 8.4 years ago by | • modified 8.4 years ago |
we know,
$\frac{s}{I} = \frac{R^{-n}}{\sum_{i=1}^{io}D^{-n}}$
$\frac{s}{I} = \frac{R^{-n}}{2D^{-n} + 2(D-R)^{-n} + 2(D+R)^{-n}}$
$\frac{s}{I} = \frac{R^{-n}}{2R^{-n} \left[\left(\frac{D}{R}\right)^{-n} + (1-\frac{D}{R})^{-n} + (1+\frac{D}{R})^{-n}\right]}$
$\frac{s}{1} = \frac{1}{2 \left[(\sqrt{3N})^{-n} + (1 - \sqrt{3n})^{-n} + (1+ \sqrt{3N})^{-n}\right]} ........(\frac{D}{R}=\sqrt{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.
$\frac {s}{I} = \frac{R^{-n}}{D^{-n}+(D + R)^{-n}}$
$\frac{s}{I} = \frac{R^{-n}}{R^{-n}(\frac{D}{R})^{-n}+R^{-n}(1+ \frac{D}{R})^{-n}}$
$\frac{s}{I} = \frac{1}{(\sqrt{3N})^{-n} + (1+\sqrt{3N})^{-n}}$
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,
$\frac{Old \ \ cluster \ \ size}{New \ \ cluster \ \ size} = \frac{12}{7} = 1.714$