Question

I NEED HELP IN THIS QUESTION ASAP PLEASE

A Transmitter provides 15 W to an antenna having 12 dB gain. The receiver Antenna has a gain of 3 dB and the receiver bandwidth BW is 30 kHz. If the receiver system noise figure F = 8 dB and the carrier frequency is 1800 MHz, Determine the maximum T-R separation that will ensure that a SNR of 20 dB is provided for 95% of the time. Assume n=4, 0 = 8 dB and do=1 km. Note: noise floor = K*BW*F*To, where K is Boltzmann constant and To=290°K.

Answer 1

NF='8 dB, BW= 30 KHZ Noise floor (dBm) = 10 log (KX To X1000) + NF t . - 10109,0 BW = 10 logio (1038 X10-23 x 290 x1000) +8 + 10 log,o(30,000) Noise Flool4 = -121 dBm → Min Rec. Power 3-121 dBm + SNR = -101 dBm hec powey at do Pee (do) = Pe Get Green 1 (47)² do? S do = Ikm , pt = 15W, 1= 0.167 m Git = 12 dB = 15.9 I Giga = 3 dB = 2 • Per cdo) 28.42 x 10-8 W = -70.7 dBW = -40.7 dBm Now, Pou (do) - Pe cd - 10 milog (dido) :. log () - Peu (do) - Po(d) = -1.017 - Porld) (dBm 1on But P ( Pyld) > -101 dBm) = Q (: 1.017 Tom - 0.95 oo let -101 – Peld) = a s Pald) = -101 - 80 dBm

then log (2) = -1.019 - 101-89 = 1.508 + 40 Now , 0.95 - 012) = 1 - Q(-2) = Q(-) = 0.05 o, -d = 1.65 1. Q = -1.65 és log = 1.508 +41065) = 1.178 Tio da 15.09 km