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EE 110 Spring 2017 CQUPT HW 6 Solution Mlsna EE 110 Spring 2017 CQUPT Homework 6 Solution 1. Design a code converter that converts a 4-bit number from Gray code to binary code. The following table shows how these two codes relate. Let the input’s 4-bit Gray code bits be G3, G2, G1, and G0 and the output’s 4-bit binary code bits be B3, B2, B1, B0. Use K-maps to minimize your design. a. (10 points each – 40 pts total) Show your K-maps and your minimized equations. There should be 4 K-maps and 4 equations for the outputs B3, B2, B1, and B0. b. (5 points each – 20 pts total) Draw your logic diagrams for the outputs B3, B2, B1, and B0. Decimal Binary Gray 0 0000 0000 1 0001 0001 2 0010 0011 3 0011 0010 4 0100 0110 5 0101 0111 6 0110 0101 7 0111 0100 8 1000 1100 9 1001 1101 10 1010 1111 11 1011 1110 12 1100 1010 13 1101 1011 14 1110 1001 15 1111 1000 Background The Gray code has the characteristic that the code words for two consecutive numbers differ in only 1 bit. Though it is very useful in a variety of applications, we’ve already seen it in the way Karnaugh maps are structured.EE 110 Spring 2017 CQUPT HW 6 Solution Mlsna First, let’s rearrange the above table like this: Decimal Gray Binary 0 0000 0000 1 0001 0001 3 0010 0011 2 0011 0010 7 0100 0111 6 0101 0110 4 0110 0100 5 0111 0101 15 1000 1111 14 1001 1110 12 1010 1100 13 1011 1101 8 1100 1000 9 1101 1001 11 1110 1011 10 1111 1010 By inspection, it’s clear that € B3= G3. € B2= G3G2+ G3G2= G3⊕ G2 € B1= G3G2G1+ G3G2G1+ G3G2G1+ G3G2G1 = G3(G2⊕ G1)+ G3(G2⊕ G1) = G3⊕ (G2⊕ G1) = G3⊕ G2⊕ G1EE 110 Spring 2017 CQUPT HW 6 Solution Mlsna € B0= G3G2G1G0+ G3G2G1G0+ G3G2G1G0+ G3G2G1G0 + G3G2G1G0+ G3G2G1G0+ G3G2G1G0+ G3G2G1G0 = G3G2(G1⊕ G0)+ G3G2(G1⊕ G0)+ G3G2(G1⊕ G0)+ G3G2(G1⊕ G0) = (G3⊕ G2)(G1⊕ G0)+ (G3⊕ G2)(G1⊕ G0) = (G3⊕ G2) ⊕ (G1⊕ G0) = G3⊕ G2⊕ G1⊕ G0EE 110 Spring 2017 CQUPT HW 6 Solution Mlsna 2. (15 points) Complete the following table showing the 8-bit binary patterns corresponding to the provided decimal integer value. Show your work! That means show how you determine each value. Decimal 8-bit sign-magnitude format 8-bit one’s complement format 8-bit two’s complement format –1 –57 57 –125 Decimal 8-bit sign-magnitude format 8-bit one’s complement format 8-bit two’s complement format –1 10000001 11111110 11111111 –57 10111001 11000110 11000111 57 00111001 00111001 00111001 –125 11111101 10000010 10000011 57 = 00111001, therefore -57 in sign-magnitude = 10111001 -57 in 1’s complement = one’s complement of 00111001 = 11000110 -57 in 2’s complement = 11000110 + 1 = 11000111 125 = 01111101, therefore -125 in sign-magnitude = 11111101 -125 in 1’s complement = ones’s complement of 01111101 = 10000010 -125 in 2’s complement = 10000010 + 1 = 10000011EE 110 Spring 2017 CQUPT HW 6 Solution Mlsna 3. (15 points) Complete the following table. Show your work! Decimal 8-bit sign-magnitude format 8-bit one’s complement format 8-bit two’s complement format 10000000 10000000 10010110 11111100 Decimal 8-bit sign-magnitude format 8-bit one’s complement format 8-bit two’s complement format –128 Does not exist Does not exist 10000000 –127 11111111 10000000 10000001 –22 10010110 11101001 11101010 –4 10000100 11111011

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