Re: [STDS-802-16] Bandwidths for WirelessMAN-OFDM PHY profiles

[David Castelow <DCastelow@AIRSPAN.COM>]

Two observations here:

(1) You have mistaken the text as stating the Tb as (73*99)/437 usec.  It is written as 73+(99/437) usec.

Go to section 8.3 to see the process of computing Tb.

(2) Tb is wrong as printed.  For the example you quote, I believe it should be as follows:

>> BW = 3e6;
>> n = 86/75;
>> Fs = floor(n*BW/8000)*8000

Fs =

     3440000

>> deltaf = Fs/256;
>> Tb = 1/deltaf

Tb =

  7.4419e-005
(= 74 + (18/43))

a number which has been fixed in Cor1/D4.

 

Hope this helps,

 

David

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From: Steven Thoen [mailto:sthoen@CHIPIDEA.COM]
Sent: 06 September 2005 10:10
To: STDS-802-16@listserv.ieee.org
Subject: [STDS-802-16] Bandwidths for WirelessMAN-OFDM PHY profiles

Hello,

 

When looking at the WirelessMAN-OFDM PHY profiles defined in the standard in Section 12.3.2, I am finding some inconsistencies which I cannot seem to resolve regarding the bandwidth of the signals.

 

For example, for the 1.75MHz channel bandwidth profile (ProfP3_1.75), table 405 defines Tb (total symbol period) as 128us. Since there are 256 subcarriers, the basic 'chiprate' would be 256/128us = 2MHz. Since of those 256 subcarriers, only 200 are actually used, the real double-sided bandwidth of the signal would be 2MHz*200/256 = 1.5625MHz. This value indeed fits within the 1.75MH channel bandwidth for this profile. No problem here nor for the 3.5MHz and 7MHz profiles.

 

If I however do the same calculation for the 3MHz channel bandwidth profile (ProfP3_3), table 408 defines Tb as 73*99/437us = 16.538us. This leads to a double sided bandwidth of 256/16.538us * (200/256) = 12.09MHz. This value does not even come close to the specified 3MHz channel bandwidth?! The same problem exists for the 5.5MHz and the 10MHz profiles: the real double sided bandwidth does not fit within the specified channel bandwidth!

 

Could anyone explain where the error in my reasoning is?

 

Best regards,

Steven Thoen