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DISCRETE MATHEMATICS Coding Theory Chapter 2 Encoding and Decoding

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DISCRETE MATHEMATICS Coding Theory Chapter 2 Encoding and Decoding DISCRETE MATHEMATICS Coding Theory Chapter 2 Encoding and Decoding : m n An Encoding function is an one-to-one function E: Z 2 → Z 2 which provides a means to detect OR correct Errors occurred during transmission of signals(0’s and 1’s) The above process is known as Encoding process. n m A Decoding function is an onto function E: Z 2 → Z 2 which provides a means to recapture the transmitted word The process is known as Decoding process. The word after encoding is –code word. The rule given by encoding function(E) is known as CODE EXAMPLES ( Coding function of ex. 1 & 2 are known as (m,m+1) parity-check code) 8 9 8 1. An Encoding function E: Z 2 → Z 2 is defined as each w1 w2 w3w4 w5 w6 w7 w8 ∈ Z 2 ,E(w) = w1 w2 w3w4 w5 w6 w7 w8 w9 where w9 = Σ wi i =1 to 8.(Sum wrt modulo 2) Discuss the significance of E(w). FURTHER if p = 0.002 is the probability of incorrect transformation of signals find the probability that a code word (say 110010010 ) is received with at most one error Solution 1:the significance of E(w). E(w) = w1 w2 w3w4 w5 w6 w7 w8 w9 where w9 = Σ wi i =1 to 8.(Sum wrt modulo 2) if w = w 1 w2 w3w4 w5 w6 w7 w8 (1) has even number of 1’s then w9 is 0 (2) has odd number of 1’s then w9 is 1 In both the cases coded word contains even number of 1’s Next we find probability. As p = 0.002 probability of receiving word with at the most one error = p0(1-p) 9 +p 1(1-p) 8 0 10 1 8 = 0.002 (1-0.002) + 0.002 (1-0.002) m m+1 8 Q2. An Encoding function E: Z 2 → Z 2 is defined as each w1 w2 w3w4 w5 w6 w7 w8 ∈ Z 2 ,E(w) = E(w1 w2 w3…w m) = w1 w2 w3…w mwm+1 wm+1 = 0, if w contains even number of 1’s =1, if w contains odd number of 1’s m+1 m and Decoding function D: Z 2 → Z 2 with D(w) = D(r1 r2 r3…r mrm+1 ) = r1 r2 r3…r m 1. encode (find codeword) of 3 000,001,011,100,110,101,010 ∈ Z 2 2. Dcode 0000,0001,0101,1111,1010,1100,1101,1001 Ans(2) By the definition of E E(000)=0000 E(100)=1000 E(110)=1101 E(001)=0011 E(101)=1011 E(011)=0111 E(010)=0100 2. Dcoding D(0000)=000 D(0001)=000 D(0101)= 010 D(1010)=101 D(1111)=111 D(1100)=110 D(1101)=110 D(1001)=100 (3m,m) Triple Repetition code m 3m Q3. An Encoding function E: Z 2 → Z 2 is defined as 8 each w1 w2 w3w4 w5 w6 w7 w8 ∈ Z 2 ,E(w) = E(w1 w2 w3…w m) = w1 w2 w3…w mw1 w2 w3…w mw1 w2 w3…w m and Decoding function 3m m D: Z 2 → Z 2 D(w) = D(r1 r2 r3…r mrm+1 r2 r3…r 2m r2m+1 r2 r3…r m3m ) = s1 s2 s3…s m Where s i = 1 if majority of r i+m ri +2m ri+3m =0 if majority of r i+m ri +2m ri+3m (a) Encode ( Or find Code words) of 3 6 E: Z 2 → Z 2 000,011,010,100,001,101,110,111 6 2 (b) Decode E: Z 2 → Z 2 111111,101010, 010101,100100,0100011,110110,010111,000111 Ans 3 a) Encoding( Or find Code words) E(000) = 000 000 000; E( 011 ) =011 011 011 E(010) = 010 010 010; E(100)= 100 100 100 E (001) = 001 001 001; E(110) = 110 110 110 E(101) = 101 101 101; E( 111) = 111 111 111 b) Decoding D( 111111)=11; D (101010) = 10; D(010101)=01; D (100100) = 00; D(010011)=01; D (110110) = 11;D(010111)= 01; D (000111) = 01; 3 9 Q4. An Encoding function is E: Z 2 → Z 2 Use triple repetition(9,3) code to Decode (a)111101100,000100011,010011111, 001 110011 (b)find THREE different received words r for which D( r ) = 000 9 3 Ans4.D : Z 2 → Z 2 triple repetition(9,3) code to Decode are si = 1 if majority of r i+m ri +2m ri+3m =0 if majority of r i+m ri +2m ri+3m a) D(111101100)=1010 D(000100011) =000 D(010011111)= 011 D(001 110011) = 011 (b) D(000) = 000000000 D(000) = 010000100 D(000) = 100010001 .
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