2.2.2 Coding the sequence by inner convolutional code (7, 5)

Coding is multiplication of polynomial information sequence A(D) by generator polynomials G⁽ⁱ⁾(D). Let's consider coding process of the sequence A(D) by the code (7, 5) with the generator polynomials G⁽¹⁾ = D² + D + 1 and G⁽²⁾ = D² + 1. So, by formula (1.22), get:

B⁽¹⁾(D) = (D²²+D²¹+D²⁰+D¹⁹+D¹⁸+D¹⁷+D¹⁵+D¹³+D¹²+D) (D²+D+1) = = D²⁴+D²³+D²²+D²¹+D²⁰+D¹⁹+D¹⁷+D¹⁵+D¹⁴+D³+ +D²³+D²²+D²¹+D²⁰+D¹⁹+D¹⁸++D¹⁶+D¹⁴+D¹³+D²+ +D²²+D²¹+D²⁰+D¹⁹+D¹⁸+D¹⁷+D¹⁵+D¹³+D¹²+D = = D²⁴+D²²+D²¹+D²⁰+D¹⁹+D¹⁶+D¹²+D³+D²+D  001011110010001000000001110.

B⁽²⁾(D) = (D²²+D²¹+D²⁰+D¹⁹+D¹⁸+D¹⁷+D¹⁵+D¹³+D¹²+D) (D²+1) = = D²⁴+D²³+D²²+D²¹+D²⁰+D¹⁹+D¹⁷+D¹⁵+D¹⁴+D³+ +D²²+D²¹+D²⁰+D¹⁹+D¹⁸+D¹⁷+D¹⁵+D¹³+D¹²+D = = D²⁴+D²³+D¹⁸+D¹⁴+D¹³+D¹²+D³+D  001100001000111000000001010.

Let's write down now the common sequence on a coder output, reading out in pairs B⁽¹⁾B⁽²⁾:

B = 00.00.11.01.10.10.10.01.00.10.10.00.01.01.11.00.00.00.00.00.00.00.00.11.10.11.00

2.2.3 Calculating the parameters of concatenated codes

For calculating correction ability of concatenated codes by formula (1.24) it is necessary to know the minimum code distance. To calculate it, use the formula (1.30). As d₀ of outer code is 3, and d_f of inner code is 5, get

d_min1,2 = 3 × 5 = 15.

t_c = [(15 – 1)/2] = 7.

Redundancy of the concatenated code is calculated as the relation of the amount of check bits to length of a code combination. The length of a code combination of concatenated code n_c is 54. The informational part of it k_c = 16, so, the amount of check bits is r_c = 54 – 16 = 38. The redundancy is

ρ = r_c/n_c = 38/54 = 0.703.

2.2.4 Decoding of sequence by Viterbi algorithm

It is necessary to enter two errors correspond to last figures of student’s credit book in the sequence B and decode it by the Viterbi algorithm. The decoding process is considered earlier in item 1.3.2 in fig. 1.10.

2.3 Hdlc procedure informational frame construction

Objective: To learn frames structure and to learn I-frames construction algorithm.

Task to the practical seminar:

1. To learn items 1.5.1, 1.5.2, 1.5.4 of this teaching manual.

2. To composite an information frame according to following data: the station address – 2 last figures of the credit book in the binary form; number of the sent information frame – group number; information sequence – the sequence received in item 2.2.2; a serial number of expected information frame – 0.

3. To make bit stuffing in I-frame.

2.3.1 I-frame construction

Opening and closing flags are described in item 1.5.2 and equal 01111110.

To generate an address field, it is necessary to transform two last figures of the credit book in the binary form and to write down the received sequence in 7 senior bits. The younger bit is 1 because of address field consists of 1 byte.

Let the station address will be equal 102. We will transform this value in the binary form 1100110 and we will add younger bit 1. The address field will be the following: 11001101. As the address will be less than 64, first bits can take values 0.

Consider the construction of the control field in basic format according to figure 1.18. The number of sent I-frames consists of 3 bits. According to the task, it is student's group number. For example, for seventh group N_S = 111. The number of received frame is equal 0, therefore N_R = 000. Poll/finish bit at the given stage can take any value, for example 1.

So, the control field for basic format is 000 1 111 0. For extended format N_R = 0000000, N_S = 0000111 and control field is 0000000 1 0000111 0.

The information field consists of sequence, got in item 2.2.2: 000011011010100100101000010111000000000000000011101100.

The control sequence is calculated by formula (1.4), using address, control and information fields for coding. For basic format it is the next sequence in polynomial form (multiplied by x¹⁶): x⁸⁵+x⁸⁴+x⁸¹+x⁸⁰+x⁷⁸+x⁷⁴+x⁷³+x⁷²+x⁷¹+x⁶⁵+x⁶⁴+x⁶²+x⁶¹+x⁵⁹+ +x⁵⁷+x⁵⁴+x⁵¹+x⁴⁹+x⁴⁴+x⁴²+x⁴¹+x⁴⁰+x²³+x²²+x²¹+x¹⁹+x¹⁸ divided by polynomial P(x) = x¹⁶+x¹²+x⁵+1. Process of dividing is similar to that in item 2.1.3. The remainder of dividing (16 bits) is check field. R(x) = x¹⁴ + x¹³ + x⁹ + x⁸ + 1 → 01100011 00000001.

The total frame will look as follows:

Opening flag	Address field	Control field	Information field	Check field	Closing flag
01111110	11001101	00011110	000011011010100100101000010 111000000000000000011101100	01100011 00000001	01111110