1.2.5 Meggitt decoder of cyclic codes

Meggitt decoder is most acceptable in practice in terms of substantially less complication and providing the best interference protection by comparison to a syndrome decoder.

In the Meggitt decoder, the construction of which is shown in the fig. 1.4, is used register Rg1 for memorizing of the accepted code combination and syndrome register Rg2 with the feedback for the calculation of syndrome S0, S1, S2 and S3. An additional device is a register Rg3, which is an electronic mechanism for the position-fix of error symbol.

Figure 1.4 – Flow diagram of the Meggitt decoder for (15, 11) code

By the signal of the time synchronization device (TSD) after 15 clock cycles, using the keys К0, К1, К2, К3 the content of Rg2 rewrites by the parallel method in Rg3. Recognition and correction of error are produced by the method of linear transformation of syndrome in Rg3 (G0, G1, G2, G3) at l = 14 - j shifts of Rg3, where j is a number of error in the codeword. The value l is determined by the content of Rg2 = 1 0 0 0. The Rg3 states take values 1 0 0 0 only at the single errors, when the error bit in Rg1 is in output, that determined by the decipherer (DC).

The construction of Meggitt decoder is simple and is based on the next property of cyclic (n, k) code. If the syndrome

S(x) = F(x) mod P(x),

where F(x) – the code combination registered by a demodulator, corresponds to the error vector

E(x) = x^j,

then the content of syndrome register Rg2 S(x), moved on l positions in Rg3 at the condition l = n - j, will be equal S(x) = 1, or, in binary presentation, 1000.

Functional transformations of content of registers Rg1, Rg2 and Rg3 is given below, allow to get a clear idea of principle of the Meggitt decoder work.

Example 1

States of the memory cells of Rg1 of the accepted code combination at the single error:

Rg1  0 0 0 0 1 0 0 0 0 0 0 0 0 0 0,

n = 14, j = 4, l = 14 – j = 10.

Clock cycle l	Rg1	Rg2				Rg3
		S0	S1	S2	S3	G0	G1	G2	G3
0	0	0	0	0	0	0	1	1	0
1	0	0	0	0	0	0	0	1	1
2	0	0	0	0	0	1	1	0	1
3	0	0	0	0	0	1	0	1	0
4	0	0	0	0	0	0	1	0	1
5	0	0	0	0	0	1	1	1	0
6	0	0	0	0	0	0	1	1	1
7	0	0	0	0	0	1	1	1	1
8	0	0	0	0	0	1	0	1	1
9	0	0	0	0	0	1	0	0	1
10	1	1	0	0	0	1	0	0	0
11	0	0	1	0	0	 Correction  signal
12	0	0	0	1	0
13	0	0	0	0	1
14	0	1	1	0	0
		Syndrome

So, on the clock cycle l = 10 the output of Rg3 is 1 0 0 0, in this case the correction signal from DC enters the adder modulo 2, and the single error will be corrected.

Example 2

States of the memory cells of Rg1 of the accepted code combination at the double error:

Rg1  0 0 0 0 1 0 0 0 0 1 0 0 0 0 0,

n = 14, j = 9 (1^st error), l = 14 – j = 5.

Clock cycle l	Rg1	Rg2				Rg3
		S0	S1	S2	S3	G0	G1	G2	G3
0	0	0	0	0	0	1	0	0	0
1	0	0	0	0	0	0	1	0	0
2	0	0	0	0	0	0	0	1	0
3	0	0	0	0	0	0	0	0	1
4	0	0	0	0	0	1	1	0	0
5	1	1	0	0	0	0	1	1	0
6	0	0	1	0	0	No correction signal
7	0	0	0	1	0
8	0	0	0	0	1
9	0	1	1	0	0
10	1	1	1	1	0
11	0	0	1	1	1
12	0	1	1	1	1
13	0	1	0	1	1
14	0	1	0	0	1
		Syndrome

So, on the clock cycle l = 5 the output of Rg3 is 0 1 1 0, in this case there is no correction signal from DC, and the errors won’t be corrected. However, extra errors in the codeword won’t appear.