a Why b What would be the minimum Hamming distance for a 2 bit error correcting

# A why b what would be the minimum hamming distance

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a. Why? b. What would be the minimum Hamming distance for a 2-bit error correcting block code? Solution a) Hamming distance between two codewords is the number of different correspondent bits. When a single error occurs the resulting erroneous codeword gets closer to a neighboring code by one bit. It should still be closer than the original codeword than all other codewords so that it can be corrected. That is the distance between the erroneous codeword and other codewords should be 2 at minimum. Therefore the distance between codewords must be at least 3. b) Using the same argument, when an erroneous codeword is two bits away from the original correct codeword it must still be at least 3 bits away from all other codewords. For that, the distance between codewords must be at least 5. 93. 12.01.2012 Final Exam A 3B4B minimum disparity code will be designed. Complete the following conversion table by selecting 4-bit correspondences to 3-bit inputs. Note that positive running disparity means there were more 1’s than 0’s in the previously transmitted data. Solution input output running disparity is + running disparity is - 000 0100 1011 001 1001 010 0101 011 0011 1100 100 0010 1101 101 1010 110 0110 111 0001 or 1000 1110 or 0111
151227621 DIGITAL COMMUNICATIONS 55 94. 12.01.2012 Final Exam Draw the rate 1/2 convolutional encoder given with 0 1111 O and 1 1011 O . Solution 95. 12.01.2012 Final Exam In the following DBPSK communication schema the purpose of ‘differential’ mode is to avoid coherent carrier at the receiver. BPSK demodulator just detects phase jumps and alternates output bit (DPCM). Assuming that the binary signals Y and Y are the same what would be the transfer function (block) between Y and X ? Use stream of 1011001110001 for X if you need to test your answer. Solution Y Y X Z => Y Y X Z => 1 1 X Y Z => 1 z Y X z (transfer function in transmitter) Since , Y Y X X , we should have 1 z Y X z and finally 1 (1 ) X Y z So the transfer function we should have in the receiver should be D in D out channel 1 z X X BPSK mod. Y DBPSK transmitter BPSK demod. DPCM DBPSK DPCM PCM DBPSK receiver Y
151227621 DIGITAL COMMUNICATIONS 56 Example calculations: 1 i i i y x y X=01011001110001 Y=01101110100001=Y’ 1 i i i y x y (transmitter) X’=1011001110001 1 i i x y y   (receiver) 96. 18.10.2012 1 st Midterm Input to a convolutional encoder whose state transition graph is shown is “0110100”. What is the output? Solution Starting from 00 state, we just follow input bits on branches state-to-state and write down the bit pairs as output: 00.11.10.10.00.01.11 First 3 transitions are marked with thicker lines on the graph. 97. 18.10.2012 1 st Midterm Find the generator matrix for the systematic encoder whose generator polynomial is given with 3 ( ) 1 g p p p . Solution X BPSK demod. Y 1 z 00 01 10 11 00 01 10 11 0/00 0/11 1/11 1/00 0/01 0/10 1/10 1/01
151227621 DIGITAL COMMUNICATIONS 57 We need to find the parity bits for the input words “1000”, “0100”, “0010” and “0001”. These can be found by Rem ( )/ ( ) n k p x p g p where ( ) x p is one of the words addressed.

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• Fall '18
• Mr. Bhullar
• Hamming Code, Error detection and correction, Parity bit

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