• I have to say that so far, this is the best 802.11n whitepaper I've read. It covers many of the 802.11n enhancements from an "overview" perspective. It's a great read, and I'd like to offer my compliments to the author (whoever at Cisco that might be). The marketing hype was minimized, and the coverage of the technology itself was VERY well done. Kudos to Cisco for releasing this type of document in a timely fashion.

    I'd LOVE to see many of the other manufacturers offer the same types of whitepapers.


  • I agree, Devin. Very well written. It's certainly easier to read than the draft that I read through about five times. If I'm not careful, you and Criss will turn me into a standards quoter.


    I also recommend the book, MIMO Wireless Communications, for a better understanding of the MIMO techniques themselves of which 802.11n will be only one standard implementation.

  • The fact I don't understand on page 7 of this paper is, how come the data rate increases (65 Mbps as opposed to 54 Mbps) when the number of subcarriers increase from 48 (802.11a) 52 (802.11n).

    3rd paragraph tells that the symbol (each containing 288 bits) is spread over 48 subcarriers in 802.11a standard. Instead of 48 subcarriers 802.11n uses 52 subcarriers to transmit each symbol.

  • the key to calculate the throughput here is to know exectly the distribution of data bits and error checking bits, In case of 288 bits ( with 54Mbps) only 216 bits are data and the rest 72 bits are error checking bits , hence the throughput is 216* 250000 (symbols)= 54 Mbps
    In case of 802.11n the number of data bits get increased to 260 while rest of the 52 bits are error checking bits so the rate would come out to be 260*250000 = 65 Mbps.

  • Hi,

    From where this value = 250000 (symbols)?

    Also why on 802.11n only 52 bit error checking?

    Please advice

    Thanks alot


  • well a symbol( the RF energy) last about 4 microseconds, which means there will be 250000 symbols per second
    about the 72 bit , it came form redundant error coding 3/4 means for evey 3 data bits there will be 1 bit for redundant error coding, there are four coding schemes I know of and they are 1/2, 2/3,3/4 and 5/6 we used 5/6 with 64-QAM modulation, that gave us the best throughput.

    sorry for the delayed answer

  • sorry, forgot to explain 52 bit, we have 52 bits because for 11n, we use 5/6 channel coding, hence 260 bits of data and 52 bits o redundant error coding

  • Hi Kashif,

    Thanks alot for your great explanation. Understand now.


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