Question

Problem I (Hamming Distance) a) Compute the Minimum Hamming Distance (MHD) for the coding (1101011, 1010110, 0000011, 0001100) b) Modify the code by changing a single bit to increment the previous MHD by 1. Problem Il (ASCII Codes) Use the ASCII translation table to convert the following binary representation to alphanumeric. #1 01000111 01101111 01101100 01100100 01100101 01101110 00100000 01000001 01110010 01100011 01101000 01100101 01110011 00100000 01000110 01101111 01101111 01100100 #2 01000011 01101000 01101001 0110110o 01100100 01110010 01100101 01101110 00100111 01110011 00100000 01010100 01010110 00100000 01010011 01101000 0110111i01110111 Problem IV (Arithmetic with BCD) Convert the number into BCD representation and perform the following operations. a) b) 11 + 26 541 -216 c) 27+36 Problem VII (Logic Simplification) Simplify the following logic expressions (A+ B)(C + B)(D+ B)(ACD + E) a) (Ā +B + T)(Ā+T + D)E + D)

Answer 1

Problem I:

1101011 , 1010110 , 0000011 , 0001100 , MHD is the number smallest Hamming distance between all possible pairs.

HD (a,b) = 5 , HD (a,c) =3 , HD (a,d) = 5 , HD (b,c) = 4 , HD (b,d) = 4 , HD (c,d) = 4 .

Therefor , MHD = 3.

To increase the MHD by 1.

1101011 should become 1101111 or 1101010. So, now

HD (a,b) = 4 , HD (a,c) =4 , HD (a,d) = 4 , HD (b,c) = 4 , HD (b,d) = 4 , HD (c,d) = 4 .

Thus, MHD = 4.

Problem II:

1. Golden Arches Food

2. Children's TV Show

Problem IV :

a. 11 = 0001 0001 and 26 = 0010 0110 , 11 + 26 = 0011 0111

b. 541 = 0101 0100 0001 and 216 = 0010 0001 0110 , 541 - 216 =0011 0010 0101

c. 27 = 0010 0111 and 36 = 0011 0110 , 27 + 36 = 0110 0011

explanation - in BCD 10,11,12,13,14,15 in binary representation are forbidden so we add 6 whenever the sum goes beyond 9.

Problem V|| :

(A+CBDE