The operational standards for the reference of the international chain of 27 500 km long

Channel (Truct) Transmission speed Mbit/s The value of B operational standards
ESR SESR
MDC 0,064 0,04 0,001
    CT От 1.5 до 5.0 0,02 0,001
От 5.0 до 15 0,025 0,001
От 15 до 55 0,0375 0,001
От 55 до 160 0,08 0,001
От 160 ….   -

Table.P-15.3.

Some of the operational standards,
taking into account the actual length and type of area the primary network

Backbone PN Intrazone PN
Length LФ Less or equal (km)   D1 Length LФ Less or equal (km)   D2
0,015 0,026
0,02 0,040
0,022 0,052
0,024 0,065
0,026 0,075
0,027    
0,029    


2. We define the boundary values of errors S1 and S2 during the selected time Тsec.

 

(Table.P-15.2)
S1 = R0 – 2х

S2 = R0 + 2х

3. Compare the measured value of the parameter S - S(ESR) or S(SESR) to the threshold values of S1 and S2.

If

S < S1, then the input of the CT in operation is allowed,

If

S > S2,the measurement of the CT for the operation is not good,

If

S1 < S < S2,for the assessment of the possibility to enter the CT in operation it is necessary to continue measuring, increasing the time of measurement.

 

 

Bit error ratio BER:

BER = (Nbit-err/ Nbit-sum) for the time of the measurements Тmeas

In accordance with the definition of NES and NSES

Nbit-err= NES + 300*NSES

There are 300 = 30% of 1000 blocks with one or more errors per second.

Nbit-sum= Тmeas (sec)* 2,048*106 bit

In its turn

NES = ESR*Тmeas(sec)

NSES = SESR*Тmeas(sec)

So

BER = (ESR*Тmeas +300*SESR*Тmeas )/ Тmeas * 2,048*106

BER = (ESR +300*SESR)* 5*10-7

BER = ESR * 5*10-7+SESR* 1,5*10-4

 

       
 
BER ≈ SESR* 1,5*10-4
   
(Table.P-15.3.)

 

  If we believe that SESR=0, taking into account only the background bugs

       
 
BER ≈ ESR * 5*10-7
 
(Table.P-15.4)  
 

 

 


Link of BBER with coefficients ESR, SESR and BER:

 

BBER = NBBE/NB

BBER = ESR*NS/(1000*NS - 300*SESR*NS)

BBER = ESR*10-3/(1- 0.3*SESR)

 
 
BBER ≈ ESR*10-3

 

 


So-as the length of the block E1 is 2048 bits, the

BER ≈ BBER*5*10-4
 
 
(Table.P-15.5)

 

 


(Table.P-15.4)
Which coincides with с

 

 

Let us consider an example-1:

Determine the long-term norms of quality indicators in the digital path E1 zone area lengthLФ = 240 km on the basis of РРСП.

For intra-zone area С = С2 = 0,075;

 

Quality factor Calculations Number
ESR 0.04 х 0,075 = 0,003 3х10-3
SESR 0.002х 0,075 + 0.0005 = 0,00065 6,5х10-4
BBER 3 ´ 10–4х 0,075 = 0,225х 10–4 2,2х10-5
BER
  taking into account only the background bugs

ES

0,225х 10–4х 5*10-4 = 1,125х 10–8 или 3х10-3х 5*10-7= 15х10-7 1,125х10-8 1.5х10-9 BER on the basis of the background error ES, and grouping error SES     6,5х10-4х1.5х10-4 = 9,75х10-8 10-7

 


 

Consider the example of an-2:

Determine the short-term (operational) norms of quality indicators in the digital path E1 example 1.

In our example

R0 (ESR)=0.02x0.075x0.5= 7,5x10-4

R0 (SESR)=0,001x0.075x0.5= 3,75x10-5

 

Boundary values of an indicator of the quality of the ESR.
Let the time of measurement ESR is 24 hours (i.e T=.24х60х60 sec = 86 400 sec).
Then

S1(ESR) = 7,5x10-4 – 2х = (7.5 - 1,75)x10-4 = 5.75х10-4

S2(ESR) = 7,5x10-4 + 2х = (7.5 + 1,75)x10-4 = 9,25х10-4

  Boundary values of an indicator of the quality of SESR To S1(SESR) > 0, it is necessary to increase the time of the measurements up to 5 days or more. At 7 daily measurements.

S1(SESR) = 3,75x10-5 – 2х = (3.75 – 1.58)x10-5 = 2.17x10-5

S2(SESR) = 3,75x10-5 + 2х = (3.75 + 1.58)x10-5 = 5.33x10-5