- This topic has 26 replies, 1 voice, and was last updated 11 years, 7 months ago by pix.
13th March 2008 at 08:15 #51236ManiGuest
For the connection yes it might be along with other variables mentioned by Amkam
For the difference
HR standard FR channel is broken down to two i.e the channel bit size remains the sampling rate is reduced. that is user gets every alternate slot on that specific TCH of that radio. for SAACH information of the second user the idle in the TCH multiframe is used to send the SAACH information of user B.
EFR vs FR there is difference in the number of speech and CRC bits, while using the same Full TCH channel . EFR gives better customer perceived voice quality due to greater CRC capability.
AMR works for both HR and FR and again gives a varity of codecs i.e CRC vs Speech bits combination mapped WRT BER in bad BER more CRC bits are used and in good BER less CRC bits are used.30th September 2009 at 07:18 #51237giorgiGuest
about handover reversion comand who is decide? MS or BSC?30th September 2009 at 08:20 #51238pixGuest
MS2nd October 2009 at 09:22 #51239papavchikaGuest
thank you4th October 2009 at 15:53 #51240BijoyGuest
a bit correction pix
In case there is congestion,HO request takes place,but no HO cmmand from BSC takes place4th October 2009 at 20:30 #51241PixGuest
There is a bit of misundestanding… the question is related to “reversion”, not “congestion”.
When there is reversion to old channel, it means the BSC has already allocated the radio channel in the target cell. The reversion only happens when the MS tries to connect to the target TCH (after the HO CMD from BSC).
Therefore, if there is congestion, the MS will simply stay put in the serving cell : it will not even know there could have been a HO. I wouldn’t call that a “reversion”.6th October 2009 at 05:59 #51242papavchikaGuest
before BSC will send command about handover It must allocate channel for MS in target cell! and MS need some period for make handover, and if during this time second MS will try to make call on that target cell BSC will not give them that channel which is allocated for first MS. am I right?6th October 2009 at 17:57 #51243PixGuest
Yes, 100% correct.7th October 2009 at 15:18 #51244papavchikaGuest
This means its not possible to make handover reversion because of congestion!7th October 2009 at 21:20 #51245PixGuest
Well, yes, simply because there is not yet a “physical” handover of the MS : MS is still in serving cell and is not even aware there should have been a HO 🙂 Ignorance is bliss.
Did I say anything that could suggest otherwise ? I’m sorry if my answers are unclear, or confusing.9th October 2009 at 05:34 #51246papavchikaGuest
no just someone sad me that when first MS receives command about handover it has approximately 2 second to make handover, and if during this time second MS will request call from target cell BSC will give that TCH, but it’s not logical. but I don’t understand about MS according to which data or measurement deciding about handover reversion,10th October 2009 at 17:54 #51247pixGuest
the “2 seconds” are a timer in the target BSC : it waits for the incoming MS during 2s.
But the Ms itself will perform the HO immediately, at receival of HO CMD.
Now, either the MS reaches the target TCH or, if it doesn’t, it will just revert back to the old TCH.