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).
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.