Delay in MS to MS call is one question.
Time required by a handover to get completed is another question.
you are saying that a handover usually takes 20 to 100 ms. OK AGREED.
But don’t you think somewhere we are getting the things mixing up.
I have two doubts..let me put them together.
1)How many TCH bursts needs to be stolen for completing a handover.
2) What is the delay for a MS to MS call.
For the first you have said indirectly that handover needs 20 ms to 100 ms.
For the second case you said nothing.
Coming back to first one….
Can we go like this..
One TS =.577 micro sec
Min time for handover=20ms
Max time for handover=100ms
TDMA frame=4.615 msec
four TDMA frames will be roughly equals to 20 msec. In this case iam taking the 4.615 ms-.577micro sec time as included for 29 msec.
Roughly we can say Handover needs min 4 and max 20 TCH bursts to be stolen to complete the handover?
Also for losing a complete 20 msec speech block one needs to loose all the 456 bits corresponding to the 20 ms speech. But These 456 bits are never sent on one TCH burst rather they needs 8 bursts(8 x 57=456). The right side of TCH contains 57 bits from N+1 speech block.
Hence i think we cannot say any thing with surety like a handover will result in a loss of 20 ms speech( which is common myth).
simply(as per your points) we can say that a handover takes a time of 20ms to 100 ms to gets completed and thus “how much speech” is lost during this period that is a calculation and is variable.
For the delay of 80 msec for a MS to MS call a friend of mine has put up a nice theory(seems correct to me)
20 msec of delay comes at the RPE-LTP vocoder.
20 msec of delay comes at Interleaver.
So total delay at MS=40 ms
This call when switched by MSC comes to a transcoder. There the transcoder again performs RPE-LTP vocoding and we gets a delay of 20 msec.
Then at BTS the interleaver again induces a delay of 20 msec.
Thus ignoring transmission delays it comes out to be
20+20+20+20=80 ms
Regards