Solution For a systematic cyclic code p p X p p c k n where p is the reminder

Solution for a systematic cyclic code p p x p p c k n

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Solution For a systematic cyclic code ) ( ) ( ) ( p p X p p c k n where ) ( p is the reminder of the division ) ( ) ( p g p X p k n . Since 5 6 2 3 4 7 ) ( ) ( p p p p p p X p k n and , and the codeword polynomial shall be 1 1 ) ( ) ( 2 5 6 2 p p p p p X p p c k n . Hence codeword is 1100101.
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151227621 DIGITAL COMMUNICATIONS 25 39. 03.12.2007 2 nd Midterm For a single bit error correcting code (n,k) what would be the numbers n and k when n is required to be greater than 100 ? Solution For an integer 3 m parity bits we can find a ( n,k ) code where code bits. For an integer 100 n , m can be found to be 7 so that 127 1 2 7 n . The information codeword size for this case is to be found as 120 7 1 2 7 k . So the code is called to be (127,120) 40.03.12.2007 2nd Midterm What is the PN sequence generated by SR circuit with feedback taps of [4,1] ? Is it an m- sequence? Why? Solution . Let us assume the initial state is 0001. 0 0 0 1 1 0 0 0 1 1 0 0 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 We reached initial state after 15 clock cycles. Since 2 4 -1=15 this is an m-sequence. All possible states except 0000 are generated. The output is the 4 th column. That is; 100011110101100. 41. 10.1.2008 Final Exam A frequency hopping digital communication system operates in 2.4-2.48 GHz ISM band with 8 hopping channels (10 MHz BW each, including the guard spacing). At each channel, 8-ary FSK is employed. System operates at 2 khops/s and transmits 8 symbols at each hopping channel. Calculate the bit rate.
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151227621 DIGITAL COMMUNICATIONS 26 Solution 8-ary FSK = 3 bits/symbol Bit rate = hop_rate x symbol_per_hop x bit_per_symbol = 2000x8x3 = 48 kbits/s 42. 10.1.2008 Final Exam Make up a quaternary PAM (4-ary pulse-amplitude mod.) system yourself and draw the waveform generated by it when transmitting the sequence 01001011100111110100. Solution 43.10.1.2008 Final Exam In a (7,4) Hamming code given below, show that 2 bit errors go undetected. 1 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 G ( 'undetected' should be changed to 'uncorrected' ) Solution Since we have only 3 parity bits, we can only have 3 syndrome bits which can indicate 2 3 =8 different conditions (7 single bit error and 1 no error condition). Other possible error types would generate the same syndrome bits which, then, are confused with the above- 00 10 11 t
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151227621 DIGITAL COMMUNICATIONS 27 From the 4 th , 5 th and 6 th columns, we deduce that the parity bits are generated using 4 2 1 5 c c c c , 4 3 1 6 c c c c and 4 3 2 7 c c c c . That means the syndrome equations are 5 4 2 1 1 r r r r s , 6 4 3 1 2 r r r r s and 7 4 3 2 3 r r r r s . If bits 1 and 2 are in error then s 2 and s 3 would be affected. s1 would stay the same since it has both r 1 and r 2 canceling each others effect. However, we would affect the same syndrome bits by just changing bit 3. So the simultaneous error in bits 1 and 2 is confused with the error in bit 3.
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  • Fall '18
  • Mr. Bhullar
  • Hamming Code, Error detection and correction, Parity bit

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