Page 183 - ISCI’2017


A =   4 aB 5 i ∑  mod M .
⋅
 5PNS i= 1 i  5
Since M = M = 420, that the residual a of the control module m = m will be always a range
k
5
5
5
of possible a distorted residual of number in the SRC.
i

The general conclusion. In the process of data diagnostics introduced in NCS A = (0 ||0 ||0 ||0 ||5)
0
0
, decided not exactly distorted residual: a = and a = . Erroneous may be the residual of the bases
2 3
0
0
m , m and m , i. e. residual a = , a = and a = 5. In this case it is necessary to carry out the
5
4
1
4
5
1
correction of residual a , a and a .
5
4
1

III. It is correction of data error A 3360 = (0 ||0 ||0 ||0 ||5). As known [1] formula
  m ⋅ (1 rm ) A   
+ ⋅
a = i   a + i  i n+ 1 −    mod m , (12)
i
  m n+ 1 ⋅ m i B i  
spend correcting a , a and a of possible distorted residuals a , a and a , where r = 1, 2, 3,.
5
1
5
4
4
1
So we have that
  m ⋅ (1 rm ) A      3 (1 r ⋅ 11) 3360  
+
+ ⋅
⋅
a = 1   a + 1  1 n+ 1 −   mod m = 1  0 +   −   mod 3 (0 + =
  m n+ 1 ⋅ m 1 B 1     11 1 ⋅ 1540  
+ [3,27 2,18])mod 3 (0 [1,09])mod 3 (0 1)mod 3 1− = + = + = ;
  m ⋅ (1 rm ) A      7 12 3360  
⋅
+ ⋅
+
=
a = 4  a +  4  4 n+ 1 −    mod m = 4  0 +  −   mod7 (0 [1,9 −
  m n+ 1 ⋅ m 4 B 4     11 4 ⋅ 2640  
− 1,27])mod7 (0 [0,63])mod7 (0 0)mod7 0= + = + = ;
  m ⋅ (1 rm ) A     11 (1 11) 3360  
+
+ ⋅
⋅ 
a = 5   a + 5  n+ 1 n+ 1 −  mod m n+ 1 =   5 +   −   mod11 =
  m n+ 1 ⋅ m n+ 1 B 5     11 6 ⋅ 2520  
= (5 [2 1,3])mod11 (5 [0,7])mod11 (5 0)mod5 0+ − = + = + = .
1
0
According to the resulting residuals a = , a = and a = , rebuilding (correcting) the number
0
5
4
1

of distorted A 3360 = (0 ||0 ||0 ||0 ||5) , i.e. the correct number, will have the following form:

A cor . = (1||0 ||0 ||0 ||5) . To verify the data correction, by the known formula [1], we define the value

of the number A cor . = (1||0 ||0 ||0 ||5) as follows [4]

A cor .PNS =   5 a B⋅ i i ∑  mod M = 0 (a B⋅ 1 1 + aB⋅ 2 2 + a B⋅ 3 3 + aB⋅ 4 4 + a B⋅ 5 5 )mod M = 0
 i= 1 
= (1 1540 0 3465 0 3696 0 2640 5 2520)mod 4620 14140(mod 4620)⋅ +⋅ +⋅ +⋅ + ⋅ = = 280.

Since 280 M< = 420 , so the number A 280 = (1||0 ||0 ||0 ||5) is correct. In order to clarify the

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