Page 1287 - 5G Basics - Core Network Aspects

Transport aspects 2

Inserting this into equation I-3, and using the fact that N = 64 (number of fixed stuff bytes) for this mapping
produces:
α=15168β+64–15232=15168(β–1) (I-8)

As before, the ODUk and client frequency tolerances are 20 ppm, and  ranges from 0.99996 to 1.00004.
Using this in equation I-8 gives as the range of :
 . 0 60672    . 0 60672 (I-9)

Asynchronous mapping of CBR40G (39 813 120 kbit/s) signal into ODU3
The nominal client rate is S = 16R16. The nominal ODU3 rate is (239/236)S (see clause 7.3). But the nominal
ODU3 rate is also equal to (4)(3824)/T. Then:
236
ST  4 ( )( 3824 )  15104 (I-10)
239
Inserting this into equation I-3, and using the fact that N = 128 (number of fixed stuff bytes) for this
mapping produces:
α=15104β+128–15232=15148(β–1) (I-11)

As before, the ODUk and client frequency tolerances are 20 ppm, and  ranges from 0.99996 to 1.00004.
Using this in equation I-11 gives as the range of :
 . 0 60416    . 0 60416 (I-12)

ODU1 into ODU2 multiplexing
The ODU1 nominal client rate is (see clause 7.3):
239
S  R 16 (I-13)
238

The ODU2 nominal frame time is:
( 3824 )( ) 4
T  (I-14)
239
4 ( R 16 )
237
The fraction p is 0.25. Inserting into equation I-3 produces:

239 ( 3824 )( ) 4 
R 16    3808  N (I-15)
238 239 4 ( R ) 4
237 16

Simplifying and solving for  produces:
237
  ( 15296 )  4 N 15232 (I-16)
238

Now let  = 1 + y, where y is the net frequency offset (and is very nearly equal to yc − ys for client and server
frequency offset small compared to 1). Then:

237 237
  ( 15296 )  15232  4 N  ( 15296 y )
238 238 (I-17)
 4 N  . 0 2689076  15231 . 731092 y

The number of fixed stuff bytes N is zero, as given in clause 19.5.1. The client and mapper frequency offsets
are in the range 20 ppm, as given in clause 7.3. Then, the net frequency offset y is in the range 40 ppm.
Inserting these values into equation I-17 gives for the range for :

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