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jitendracellplan
2011-08-19, 02:35 AM
hi
pls explain TA=63 in GSM
or why Value of Timing advance is 63.

Thanks in Advance:lol

pathloss
2011-08-19, 02:58 AM
The TA value is normally between 0 and 63, with each step representing an advance of one bit period (approximately 3.69 microseconds). With radio waves traveling at about 300,000,000 metres per second (that is 300 metres per microsecond), one TA step then represents a change in round-trip distance (twice the propagation range) of about 1,100 metres. This means that the TA value changes for each 550-metre change in the range between a mobile and the base station. This limit of 63 × 550 metres is the maximum 35 kilometres that a device can be from a base station and is the upper bound on cell placement distance.
The limit of the original range of a GSM cell site is 35km as mandated by the duration of the standard timeslots defined in the GSM specification.

Best regards,



hi
pls explain TA=63 in GSM
or why Value of Timing advance is 63.

Thanks in Advance:lol

ritgail
2011-08-19, 03:21 AM
hi
pls explain TA=63 in GSM
or why Value of Timing advance is 63.

Thanks in Advance:lol
Lets get simple....
Speed of EM wave = 3x 10-8 m/sec
Data rate of GSM channel =270.8 kbps
therefore, 270.8kbp transmitted in = 1 sec
or 1 bit transmitted in = 1/270.8x10-e
or 1 bit transmitted in = 3.69 microsec

What is guard period of AGCH/or a RACH so that it does not overlap and you give a reply, then i will reply;)