Case Study On Wireless Data Link

Added on - 30 Sep 2019

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Case Studya)* For distance 5 km and frequency 2.4 GHzd = 5km, f = 2.4 GHz = 2400 MHzFSPL(dB)= 20log10d + 20log10f + 32.45=20log105 + 20log102400 + 32.45=13.97 + 67.6 +32.45=114.02 dBd = 5km, f = 5.8 GHz = 5800 MHzFSPL(dB)= 20log10d + 20log10f + 32.45=20log105 + 20log105800 + 32.45=121.69 dBd = 10km, f = 2.4GHz= 2400MHzFSPL(dB)= 20log10d + 20log10f + 32.45=20log1010 + 20log102400 + 32.45=120.05 dBd = 10km, f = 5.8GHz= 5800MHzFSPL(dB)= 20log1010 + 20log105800 + 32.45=20log1010 + 20log102400 + 32.45=127.71 dBFSPL(dB)Distance2.4 GHz5.8 GHz5114.02121.6910120.05127.71b)Received power = Transmitted power(dBm) + Transmitter antenna gain (dB) + receiverantenna gain (dB) – losses(dB)Here losses (dB) = FSPL(dB) + other lossesOther losses is given negligibleAnd from previous calculations FSPL(dB)=121.69 dBNow received power = 23+24+24-121.69= -50.69dBmLink margin = Received power – Receiver sensitivity= -50.69 – (-72)= 21.31 dBc)From the given table we can conclude that for the calculated link margin at 5km theavailability of link will be 99%.
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