1 1 0, g Exercise 1. Generator polynomials of a convolutional code, given in binary form, are g
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2 Exercise Generator polynomials of a convolutional code, given in binary form, are g 0, g 2 0 ja g 3. a) Sketch the encoding circuit. b) Sketch the state diagram. c) Find the transfer function TD. d) What is the minimum free distance of the code?
3 Solution a) Encoding circuit: + + input 2 output + 3
4 b) State diagram: 0/0 0 /00 0/0 0/ / 0/0 0 /00 /00
5 c) Modified state diagram indicating the weights of transitions (node 00 split into two): D d D 2 D 3 2 D D 2 D D a b c e
6 d) State equations: X D X DX b a c X DX DX d b d X D X D X X 2 2 c b d e 2 D X c. From the second equation we get D X d X b, D which can be inserted into the third equation:
7 3 2 D X c D X b X b D D 3 D D X a DX c D X a DX c D D D D X a DX c D D X a DX c D D X a D X c D X a D X c D X a D X c D 5 3 D X a D X c. D Solving for X and inserting into the fourth equation results in c
8 T D 7 X e D X D D a 3 D D D D D D D D D D D D... From T D we see that d free 7. Minimum free distance can also be seen directly from the modified state diagram of part c. The minimumweight route a e is a b c e. The weight of this route is d 7. free...
9 Exercise 2 Calculate union bound for error probability for a Convolutional Code generated by the polynomials g 0 ja g 2.
10 Solution 2 Union bound for the error probability is evaluated as c j, d
11 d, i j
12 From the lecture material the code spectrum for this code is! " # $ % & & $ % $ &! For this code the amount of bit error is d 4. We evaluate the union bound as!
13 simulated Union bound 0-2 BER SNR
14 Exercise 3 '() *+,&-.
15 / '() kll M 0.5 m Q P ci c j Q mk M kll M m 0.5d ij Q mk M 0 M M m the trellis and making describes the probability of selecting certain path from the state in m symbol errors because of selecting this path. The
16 amount of possible symbols is M. In Viterbi equalizer from each state there are M possible outgoing branches. By selecting one branch we can do errors. If the path is l trellis sections long we could do errors in each section. Total probability of error is multiplication of probabilities for each transition along the path. kll mk M M m M. (We multiply only l-l transitions since last L transitions corresponds to correct symbols otherwise the erroneous path would not merge with correct path). For our BPSK transmission M=2 (two possible symbols) error M m is. m
17 In one errorneous path we do in averge c j, d information bit errors. As for the case of Convolutional Code for calculating the average bit error we have to average over bit errors over all the possible error paths. For simplifying our calculations we select k to be 3.
18 . For given channel there is no error path with length. 2. There is one path with length and with distance 2 Erx 2 2 d0, T Erx T.09 Erx 4 T In this path we do one error c0,2 / 2 Where, the term /.09 normalizes the total channel power to.
19 3. There is one path with length 2 and with distance 2 Erx d0, T Erx.30.7 T Erx 0.20 T in this path we do two errors c0,3 2/
20 Total error probability is bounded as & & & & & 0
21 simulated Union bound 0-3 BER SNR
22 Exercise 4 The rate / 2 convolutional code has generators g 0 2 g Represent this code as: a) Shift register. b) State diagram. c) Tree. d) Trellis. e) Matrix..
23 Solution 4 a) + +
24 b) 0/0 0/ /00 0/0 0/ /0 0/ / 0 0/ / /0 / /00 /0 /0 0/0
25 c)
26 d)
27 e) G
28 Exercise 5 Represent the code in the previous exercise as recursive systematic convolutional code. Calculate the output (coded) sequence for both: systematic and corresponding nonsystematic encoder. The information sequence at the u input is
29 Solution 5 a) recursive systematic convolutional code u z u u z u z g z z z g z z z z z z sz u0 uz u2z u3z u4z 2 3 z z z z z sz z 2 3 z z z z z z z z z z z z z z z z z z z z z
30 s s 2 00! s 00 " 0000 b) Systematic convolutional code s u z g z 2 2 s u z g z Divide both sequences with g z and you get new sequence where one contains the systematic bit and the other is generated by a recursive digital filter. s u z s 2 u z 2 g z g z
31 + +
32 Exercise 6 The encoder starts from zero state, collects three bits and adds to them a parity bit calculated based on the previous state and the bit values. The information bits and parity bits are transmitted. In next interval the process is repeated but the initial state is equal to parity bit value in the next interval. Draw the trellis for given code. (This type of code is called zigzag code).
33 Solution 7 000/0000 0/00 0/00 0/00 00/000 00/000 00/000 /0 0 00/00 00/00 00/00 / 000/000 0/0 0/0 0/0 The state diagram
34 000/0000 0/00 0/00 0/00 000/0000 0/00 0/00 0/ /00 00/00 00/00 / 0 00/00 00/00 00/00 / 0 00/000 00/000 00/000 /0 000/000 0/0 0/0 0/0 Trellis of the code
35 Exercise ' + + +
36 # 3 )4 5 & 2 # #
37 b a 0/0 0 /00 0/ / 0 c /000 0/00 d /0 0 /0
38 6576 ' # 22 # 3 # 2 ) # 3) # * 6'72( (
39 received sequence # : 4( 2
40 :6' # 2 +,&%%%&!"$;% ;$&##%&"";"& ""#;$&";#!$&& ;!%!; &"-
41 , received sequence ,6 0,3-0,4
42 Exercise 9 (Wicker, problem 2.9) Assume that the encoder from Problem 2. (see the book) is used over a symmetric memoryless channel. The bit metrics are as follows. 0 0 y y r y Find the maximum-likelihood code word corresponding to the following received sequence. r 0,0, 0,,00, 0,,00.
43 One stage of the trellis and the branch outputs are shown below.
44 ! % " " $ % % ' $ $ $ $ % $ $ $ $ " $ $ " " " $ $ $! % ' The decoding procedure is illustrated in the following figure (check it!). Dashed lines and solid lines indicate terminating branches and surviving branches, respectively. The survivor path is indicated with bold solid line. In case of ties, upper transition was chosen.!! # $! $ & '!! #!! % % &!! % # H A? A E L I A G K A? I A G K A? A
45 Exercise 0 Consider the encoder on Figure below. Describe the decoders by the state diagram. Assume that the I is a 3$ 3 block interleaver. Derive the output sequence c k of the encoder if the input data sequence is dk Give also the input sequence to the second encoder. (The data sequence after interleaver.) Describe what kind of impact the two last bits of the input sequence have on the termination of the trellis of each encoder.
46 input + I + 2 output + + & 3 The bit sequence at the interleaver output is d % d k i, k
47 The encoder state diagram is 0/0 00 / /0 0/ / 0/0 0 0/ /0 0
48 Solution 0 The encoder outputs are x x k 2 k 00 3 xk 00 Two last bits set the encoder to the 00 state. For this example this state is reached by both of the constituent encoders.
, g. Exercise 1. Generator polynomials of a convolutional code, given in binary form, are g. Solution 1.
Exerise Genertor polynomils of onvolutionl ode, given in binry form, re g, g j g. ) Sketh the enoding iruit. b) Sketh the stte digrm. ) Find the trnsfer funtion T. d) Wht is the minimum free distne of
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