Page 1290 - 5G Basics - Core Network Aspects

2 Transport aspects

The fraction p is 0.5. Inserting into equation I-3 produces:

1 ( 3824 )( ) 4 
R 16    7616  N (I-33)
2 239 ( R ) 2
238 16
Simplifying and solving for  produces:

238
  ( 15296 )  2 N 15232 (I-34)
239

As before, 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:
238 238
  ( 15296 )  15232  2 N  ( 15296 y )
239 239 (I-35)
 2 N 15232 y

The total number of fixed stuff bytes N is zero, as given in clause 19.5.4. 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-35 gives for the range for :
  . 0 6092800 for y  40 ppm

  . 0 0000000 for y 0 ppm (I-36)
   . 0 6092800 for y  40 ppm

In addition, stuff ratios of −2 and +1 are obtained for frequency offsets of −130 ppm and 65 ppm,
respectively. As above, the range of frequency offset that can be accommodated is approximately 195 ppm.

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