Q3 Consider the E time division multiplexing hierarchy i How

Q3 Consider the E time division multiplexing hierarchy. i. How many E-3 lines are multiplexed in an E-4 line? ii. How many bytes is an E-3 frame? iii. What is the bit rate for the overhead in E-3? iv. How many slots does this correspond to in part (iii)?

Solution

Ans:

i) Four, E-3 lines are multiplexed in an E-4 line.

ii) The structure of the E-3 frame has four groups and each of them has length 384 bits, so total number of bits = 4*384 = 1536 bits = 1536 / 8 bytes = 192 bytes.

iii) Group I contains the FAS, with sequence \"1111010000\"; the A-bit (remote alarm); the S-bit (spare); and 372 T-bits (tributary) transporting data.So overhead bits in Group1 = 384-372 = 12 bits

Groups II and III contain a block of four J-bits (justification control) and 380 T-bits transporting data.So overhead bits in Group II & III = 8 bits

Group IV contains a block of four J-bits, a block of R-bits (justification opportunity) one per tributary, and 376 T-bits.So overheads bits in Group IV = 8 bits.

So, overall overhead = 8+8+12 = 28 bits.

In E3 - bit rate = 34 Mbits/sec.

So, overhead bit rate = (28/1536)*34 Mbits/sec = 0.6198 Mbits/sec.

iv) Each slot having 8-bits, so 28-bits correspond to 4 slots.