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The path-loss model for the received power at a mobile user in a cell is given by:

$P_{RX}=P_{TX}d^{-v}$

where $P_{RX}$ is received power, $P_{TX}$ is the transmitted power from all base stations, $d$ is the distance from the transmitter to the receiver and $v$ is the path-loss exponent.

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Part A:

Total signal power:

$S=P_{TX}d^{-v}$

Total received power:

$I=\\sum^{M}_{i=1}P_{TX}d_i^{-v}=MP_{TX}X^{-v}$

Therefore $S/I$:

$\\frac{S}{I}=\\frac{P_{TX}d^{-v}}{MP_{TX}X^{-v}}=\\frac{1}{M}\\left(\\frac{X}{d}\\right)^{-4}=\\frac{1}{M}\\left(\\frac{X}{d}\\right)^{-4}$

Part B:

$\\frac{S}{I}_{linear}=10^{\\frac{SNR_{dB}}{10}}=57.54$

$\\frac{S}{I}=\\frac{1}{M}\\left(\\frac{X}{d}\\right)^{-4}$

$\\frac{X}{d}\\longrightarrow$ co-channel re-use distance $\\longrightarrow\\frac{X}{d}=\\sqrt{3K}$

$\\frac{S}{I}=\\frac{1}{M}\\left(\\sqrt{3K}\\right)^v=\\frac{1}{M}\\left(3K\\right)^{\\frac{2}{v}}$

Rearranging for $K$:

$K=\\frac{1}{3}\\left(\\frac{MS}{I}\\right)^{\\frac{2}{v}}=\\frac{1}{3}\\left(6\\cdot57.54\\right)^{0.5}=6.19$ clusters, therefore minimum cluster size, $K=7$.

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If a mobile user also receives signals from $M$ interferers that are all at a distance $X$ from the mobile user, derive the signal-to-interference ratio ($\\frac{S}{I}$) of the mobile user, given a distance $d$ and that all interferers have the same transmission power, $P_{TX}$, and path-loss exponent, $v$.

SIR=[[0]]

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For satisfactory performance of an arbitrary cellular system, a signal-to-interference ratio of $17.6$ dB is required. Find the minimum cluster size, $K$, needed to achieve this, considering $v=4$ and there are $M=6$ equidistant interferers.

K=[[0]]

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