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Trans power : 20dBmChannel loss = 60dBWe know thatReceived power (dBm) = Tx Power (dBm) + Tx Antenna Gain (dBi) + Rx Antenna Gain (dBi) - loss (db) as Tx and Rx antenna gains are not mentioned we may assume that those are 0 so we have
received power (dbm) = 20 dBm - 60dB = -40 dBm this shows that condition (a) is satisfied
We have noise power at receiver = 10^-9W = 10^-6 mW and as dBm = log (mW) * 10
the noise power at receiver in dbm = 10xlog 10-6 = -60 dbm we also know
SNR(dB) = Received power (dbm) - noise power (dbm)= -40 - (-60) = 20 dB
Thus condition (b) is also satisfied. As both the conditions are satisfied the receiver will not face any problem in receiving frames.
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why does the periodic table useful to scientist
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