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Re: [STDS-802-16] Error in section 8.3.3.5.3 of 802.11-2004 STD



Hi Ambroise,

 

If there are 90 bits at the output of the randomizer, then all the steps after the randomizer are correct. But I am not clear on where does the additional 10 zeros tail come from. The original input data is 80 bits. After randomizer, the randomized data will be also 80 bits, and then we pad 8 zeros as a tail, creating 88 bit length of data. Where do the additional two zeros come from? And which section of the standard specifies such rule?

 

By the way, you mentioned about the corrigendum. May I know if that document is available publicly?

 

Thank you very much.

 

Regards,

Ernest

 


From: owner-stds-802-16@LISTSERV.IEEE.ORG [mailto:owner-stds-802-16@LISTSERV.IEEE.ORG] On Behalf Of Ambroise Popper
Sent: Thursday, March 31, 2005 4:07 PM
To: STDS-802-16@LISTSERV.IEEE.ORG
Subject: Re: [STDS-802-16] Error in section 8.3.3.5.3 of 802.11-2004 STD

 

Ernest,

 

This example is correct, provided a slight amendment that has been made in the corrigendum that makes it clear that there are 90 bits at the output of the randomizer (and not 96).

 

Ambroise

 


De : owner-stds-802-16@LISTSERV.IEEE.ORG [mailto:owner-stds-802-16@LISTSERV.IEEE.ORG] De la part de #ERNEST KURNIAWAN#
Envoyé : jeudi 31 mars 2005 08:45
À : STDS-802-16@LISTSERV.IEEE.ORG
Objet : [STDS-802-16] Error in section 8.3.3.5.3 of 802.11-2004 STD

Hi all,

 

Have any of you tried out the example in section 8.3.3.5.3? The example pads more than 8 zeros on the Randomized bit in order to fit the number of transmittable bit based on the allocated data block (5 symbol duration and 1 subchannel), which does not comply with the statement in 8.3.3.1, which says that if the amount of data to transmit does not fit the allocated data block prior to randomization stage, padding of one’s will be done at the end of the block up to the amount of allocated data block minus one byte (which is reserved for the tail zero byte).

 

And also there is a confusion in the statement on section 8.3.3.2.1, which says “when the total number of data bits in a burst is not an integer number of bytes, zero pad bits are added after the zero tail bits”. Is it true that this statement is just to enable us to display the data after randomization bit in terms of bytes? Because if we include this additional zeros onto the convolutional coder, the number of bits generated will be larger than the allocated data block. (In our case, since we use one subchannel, we expect to get 12 subcarrier x 5 symbol duration x 2 bits per symbol = 120 bits) But if we use additional zeros onto the convolutional coder, it will force the input to the convolutional coder from 90 (which is equal to 3/4 * 120) to 96, and hence the convolutional coder output will be 128 bits, which is definitely more than 120, because this will create problem if we feed the 128 bit of data to the block interleaver, since the number of input bits is not integer multiple of the block size.

 

And are there any “tested” documents which provide correct examples on the various cases (uplink, downlink, subchannelization, etc.) for both OFDM and OFDMA mode of operation?

 

Regards,

Ernest.