This is a question about LANs and WANs Problem 5 (10 points) An 802.11 wireless

Question

This is a question about LANs and WANs
Problem 5 (10 points) An 802.11 wireless LAN modem uses QPSK and 1024-QAM for two different data rates with the same symbol transmission rate of 12Msymbol/sec (a) Give the number of bits per symbol for each of the two constellations and determine the difference in the signal technique using the 3-dB per bit rule (b) With the difference calculated in part (a), what would be the ration of coverage of the QPSK and 1024-QAM modems in an open environment with distance power gradient of 2?

Explanation / Answer

a) For 1024- QAM, the no. of bits/symbol is given by M = 2^N

where M = 1024 , N = bits/symbol

So N = Log (1024)

N = 10

For QPSK, fs = R/N

where fs is 1/ts =symbols/sec

R = bits/sec for QPSk, there are around 4 bits and 4 phases exists i.e., 00,01,10,11.

and so N= R/fs = 4* 12 = 48bits/symbol

The signal to noise ratio according to 3db per bit rule is that, for QAM, bits per sec is 10 * 12 = 120

SNR => 120/3 = 40db

The signal to noise ratio according to 3db per bit rule is that, for QPSK, bits per sec is 48 * 12 = 576

SNR => 576/3 = 192dB

the difference between two modellations is 192-40 = 152dB

b)

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