Re: Symbol period of GSM

Ed Casas (edc@ece.ubc.ca) Sat, 15 Apr 2000 11:16:46 -0700 

Date: Sat, 15 Apr 2000 11:16:46 -0700
From: Ed Casas <edc@ece.ubc.ca>
Subject: Re: Symbol period of GSM



On Wed, Apr 12, 2000 at 02:36:56PM -0700, Yue Xu wrote:

> To calculate the symbol period of GSM, the solution of 4.10 do as
> follows
>  Ts=1/(0.3*RF Data Rate)=1/(0.3*270.833*1000)              -- BT=0.3

I believe in this case the author of the solution is trying to
compare the -3dB modulation filter bandwidth (B) to the
reciprocal of the delay spread (something similar to comparing
the coherence bandwidth to the signal bandwidth).  I don't
believe this is the correct way to determine the need for
equalization.

To motivate this reasoning, consider that the baseband filter
could be removed completely (BT=infinity) without having any
effect on the ISI introduced by the channel and thus having no
bearing on the question of whether an equalizer is required or
not.

There are examples of systems where the symbol period is not
related to the bandwidth.  Examples are OFDM/DMT modulation
(where the symbol period is much less than 1/bandwidth) and
``partial response'' systems (where the symbol period is much
greater than 1/bandwidth).  In all cases it is the symbol period,
not the bandwidth that determines the need for equalization.

> While in an example given by our text book ( Page 265, Example 5.15),
>  Ts=1/ Rb = 1/(270*1000)                    -- BT=0.25
> Can anybody tell me which one is correct?  Thanks a lot.

The symbol period is, by definition, the reciprocal of the symbol
rate.

-- 
Ed Casas   edc@ece.ubc.ca   http://casas.ece.ubc.ca   +1 604 822-2592