(Switched 56K service). Signaling is accomplished through a "Robbed8000 samples per second = 8000 bps
> Bit" method where bit 8 of each channel's timeslot is "robbed" to
> indicate a signaling state in the 6th and 12th frames. Effective
> throughput for the A signaling bit (Frame 6) is 666.66 BPS. Effective
> throughput for the B signaling bit (Frame 12) is the same (666.66 BPS).
> But i cant figure out how they got to 666.66Bps?
The "A" or "B" bit is robbed in two frames out of 12, so -
8000 bps * (10/12) = 6666.66 bps
6666.66 bps = 666.66 BPS (Bytes per second), assuming asynchronous ASCII with 8 bit bytes, one start bit, and one stop bit.
HTH.
William Warren
(Filter noise from my address for direct replies)