Mapping; Loading A 140 Mbit/S Plesiochronous Signal In The Vc-4 - Siemens SRA 4 User Manual

1.4.1.10
1.4.1.11
SRA 4 - 6/7/8/10/11/13 GHz (WB ODU) - UMN
911-381/02C0000 - Issue 3, July 2004

Mapping

A virtual container can be seen as a frame structure capable of transporting a signal
through a synchronous network which belongs to the normal existing plesiochronous
hierarchy, accompanied by the dedicated service information POH for the transit
through this network.
Hence, the transport capacity of the VC, without POH and therefore in coincidence with
the capacity of its own container "C", must always be greater than the bit rate of the
plesiochronous signal to be transported.
In fact, time spaces must be available which must be reserved for the stuffing procedure
which is necessary to recuperate the inevitable frequency differences between the clock
of the plesiochronous signal and that of the virtual container.
In practice, other than the bits necessary for the stuffing procedure, further time space
is available in each container, which is used by overhead bits and fixed stuffing bits.

Loading a 140 Mbit/s plesiochronous signal in the VC-4

From the mapping function recommended by ITU-T for loading of a 140 Mbit/s
plesiochronous signal into the VC-4, shown in Fig. 1.10, we can deduct that the VC-4
with reference to its own clock, is capable of an information bit load which is variable
between a minimum of:
[9 x (20 x 12 + 1)x 8 + 54] x 8 x 10
a b c
obtained by imposing in each frame a fixed stuffing (S=R) for the "S" bits of Fig. 1.10,
and a maximum of:
[9 x (20 x 12 + 1)x 8 + 54 + 9] x 8 x 10
a b c
obtained by imposing in each frame the loading of information bits (S=1) for the "S" bits
of Fig. 1.10.
a) bytes per column
b) columns containing only information bits
c) bits of each byte
d) information bits in the first column of the 20th block in Fig. 1.10
e) frame frequency (STM-1)
f) total of "S" bits in each frame
Fig. 1.10 also shows that the following is loaded into the C-4 container:
(9 x 10) x 8 x 103 = 0.72 Mbit/s
3
(9 x 5) x 8 x 10 = 0.36 Mbit/s
9 x (13 x 8 + 5 x 5 + 1) x 8 x 10
3
= 139.248 Mbit/s
d e
3
= 139.320Mbit/s
d f e
3
= 9.36 Mbit/s
1 - PRELIMINARY INFORMATION
of overhead (0 in Fig. 1.10)
for transmission of the stuffing
message (C in Fig. 1.10)
for fixed stuffing (R in Fig. 1.10).
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