Hi folks, I could really do with some help with something.
> I'm studying telecoms in particular T1 circuits. Currently its on about
> superframing and Extended Superframing.
> I've been reading something:
>
formatting link
> D4 Voice and Data Signaling
> The transport of signaling states is required in Switched voice or data
> (Switched 56K service). Signaling is accomplished through a "Robbed
> 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?
> Looking at the diagram on the webpage, the least significant bit in
> all channels has the last bit robbed, for frames 6 and 12, so in every
> superframe thats sent, thats 24 bits, multiply that by 8000 and i get
> 192Kbps???? Where am i going wrong.
> Could someone explain this to me.
> Thanks in advance.
How many times per second, on each channel, does 'frame 6' occur? with one bit of that frame being used for 'A' signalling, the bit rate for that bit is the same as the frequency of that frame.
I don't know where you got that 8000 multiplier from, but it is _way_ wrong.
A superframe has _12_ frames, each containing 1 sample from each channel it is 12x (192+1) bits long.
A superframe passes once every 12/8000 second. this means that there are
666.66666 ... superframes per second.`oThus, there is 1 bit per superframe *per*channel* used for 'A' signalling, and 1 bit per superframe *per channel* used for 'B' signalling.
666.666666 ... superframes /second means 666.6666 ... 'A' bits *per*channel* per second and a similar number of 'B' bits/second multiplied by 24 channels, gives 16,000 'A' bits/second aggregate, and the same for the 'B' bits.