assignment3 - Copy - , but somehow the first bit is lost...

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Error Control Coding Asssignment 3 Due 11:00 am Friday 20 May 2011 Value in brackets indicates marks for each question. Total value of marks is 20. This assignment counts towards 12.5% of the total assessment. Question 1 (5): The generator matrix for a (3,2,2) convolutional code is G ( D ) + ƪ 1 ę D D D 1 1 ę D 1 ƫ . (a) Draw the encoder diagram for the above code. (b) From the encoder diagram, draw the trellis diagram. (c) Starting at the all zero state find the codeword corresponding to the information sequence u = (11,01,10). Question 2 (2): Draw the minimal encoder diagram for the systematic (3,2,3) code with G ( D ) + ȧ ȱ Ȳ (1 ę D ę D 3 ) ń (1 ę D 3 ) (1 ę D 2 ę D 3 ) ń (1 ę D 3 ) 1 0 0 1 ȧ ȳ ȴ . Question 3 (7): Given the nonsystematic (2,1,2) code with G ( D ) + [1 ę D 2 , 1 ę D ę D 2 ], suppose the code is used on a BSC and the transmitted sequence for the first seven branches is v = (11,10,10,00,01,11,00). Suppose the sequence is received error free
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Unformatted text preview: , but somehow the first bit is lost and the received sequence is incorrectly synchronised giving the received sequence r = (11,01,00,00,11,10). Starting at state zero, use the Viterbi algorithm to decode this sequence. Note that when there is incorrect synchronisation, the state metrics increase rapidly. This fact can be used to detect and then correct for incorrect bit synchronisation. Question 4 (6): Given the same code as in Question 3, and a soft decision BPSK demodulator with Q = ∞ , decode the following received sequence using maximum likelihood metrics and the Viter-bi algorithm (starting at state zero), r = (–0.5,–1.2, 0.8,–1.1, 0.5,0.8, –1.0,2.0, 1.2,–1.2, 1.3,–1.2, –1.0,0.8)....
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This note was uploaded on 10/16/2011 for the course ELECTRICAL EE251202 taught by Professor Rejaei during the Spring '10 term at Sharif University of Technology.

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