Re: [STDS-802-16] Fwd: Re: 802.16 comment
Because irrespective of the basis of the log the formula works (provided
the same base is used top and bottom).
David
Dr David Castelow Airspan Communications Ltd
Principal Systems Engineer Cambridge House, Oxford Road
DSP & Systems Design Uxbridge UB8 1UN, UK
http://www.airspan.com +44 1895 467281
> -----Original Message-----
> From: owner-stds-802-16@ieee.org
> [mailto:owner-stds-802-16@ieee.org] On Behalf Of Forwarded by
> Sent: 09 June 2005 16:15
> To: STDS-802-16@listserv.ieee.org
> Subject: [STDS-802-16] Fwd: Re: 802.16 comment
>
> >I am reading 802.16-2004, section 8.3.1.1.1, page 427. The
> text below
> >figure 195 reads:
> >"The transmitter energy increases with the length of the guard time
> >while the receiver energy remains the same (the cyclic extension is
> >discarded), so there is a 10log(1 - Tg/(Tb + Tg))/log(10) dB loss in
> >Eb/No. ..."
> >Question:
> >Is the base of logarithms 10? If so, since by definition
> log(10) = 1,
> >why is division by log(10) included in the equation? Is there a
> >misprint in the text?
> >
> >Thanks,
> >Tim Holmes
> >Sr. Principal Engineer
> >Dynetics, Inc
> >990 Explorer Blvd
> >Huntsville, AL, 35806
> >256-964-4437
> >tim.holmes@dynetics.com
>