What is the poll capacity of the cell?

– Information Rate (Ri) = 12.2 kbps

– Chip Rate (Rc) = 3.84 Mcps

– Required Eb/Nt = 4 dB

– Average interference factor due to other cells = 0.5

– Channel activity Factor = 0.65

– Interference Margin = 3 db

Mumbai University > Electronics and Telecommunication > Sem8 > Wireless Networks

Marks: 10M

Load-factor per voice user = $(1+0.5). \frac{1}{1+ \frac{3840}{12.2}. \frac{1}{2.512}. \frac{1}{0.65}}=000774$

Required interface margin=$3dB = 2 = \frac{1}{1= \rho}$

Cell loading = $\rho$ =0.5

Number of voice users= $\frac{0.5}{0.00774} \approx$ per cell

$\therefore M_{max} \approx \frac{G_p}{d}. \frac{1}{(1+\beta)}. \frac{\eta_c}{vf};$

where:

$\hspace{1cm}$ $\eta_c$ =power control efficiency (often 80% to 80%)

$\hspace{1cm}$ $\beta$=interference factor due to other cell

$\hspace{1cm}$ $d=(E_b/I_t)_{reqd}$

$\hspace{1cm}$ $G_p$=processing gain

$\hspace{1cm}$ $N_o$=noise density

$\hspace{1cm}$ R=information rate

$\hspace{1cm}$ $M_{max}$ is the poll capacity of the cell.

$\hspace{1cm}$ assuming $\eta_c$=1

$M_{max}= \frac{3.84 × 10^6}{12.2 × 10^3}. \frac{1}{0.65}. \frac{1}{1+0.5}. \frac{1}{(2.52)} = 128$