Assuming a 20 MHz band, it is -62 dBm for the data and -82 dBm for the preamble. Source: August 2013 doc.: IEEE 802.11-13/0994r0
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The preamble is longer. This is the threshold of detection. – Tony Stewart EE75 Jan 05 '19 at 00:20
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I can't see the relation between having a longer preamble needing less energy to detect, unless the detection process function is like an accumulator. Some other sources mentioned "If you would "sleep" at -62dBm you would consume more power during idle state compared to -82dBm." so it's because to save energy, but I dn't understand the logic in this case too. – user3656142 Jan 05 '19 at 01:42
1 Answers
Although the preamble is longer, it is due to a lower bit rate that it has a lower threshold. In 802.11g, the PHY layer combines a DSSS preamble with OFDM payload transmission.
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The 2013 standard indicates that the Synchronization and Start Frame Delimiter (SFD) fields are transmitted at 1 Mbps. The rest of the header is sent at 2 Mbps (DQPSK) and the data payload is sent using DQPSK at 2 Mbps or CCK at 5.5 or 11 Mbps or more.
The threshold depends on the data rate which demonstrates the Hartley's Law tradeoff between SNR and Bit rate for the the same bit error rate... * Hartley's Law compared to Shannon–Hartley theorem.
(and I did not mean the Arizona's Shannon Law that forbid's shooting a gun up in the air)
The 20dB rise in Rx level is needed to achieve either the max rate in the 20MHz Band using higher-order modulation such as 256QAM.
This is why Modems will automatically reduce bit rates when packet error rates exceed some threshold. Normally -70 dBm is more than enough for 54Mbps.
Anecdotal
Although from my experience when Rician Fading (Echo interference) is dominant , and the RSSI is between -75 and lower levels, that you get higher thruput using 11Mb/s than leaving it on 54Mbps which often drops near 1Mbps rates when it momentarily drops to -80dBm.
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