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# Quiz1_sol - it can be implemented using NAND gates only AB...

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EE2000 Quiz 1 Write down your NAME and STUDENT ID on the answer sheets provided. Q1 (2 marks) Convert the decimal number 174.25 to octal number and demonstrate the procedures. Q2 (4 marks) Calculate A+B and A-B for the pair of numbers below assuming an 8-bit two’s complement number system. Show your procedures and indicate the decimal equivalent value of your results. A = 1010101, B = 1010 Q3 (2 marks) (a) Convert the binary numbers “1101101” to gray code (b) Convert the gray code “1100111” to binary Q4 (4 marks) Simplify the following using Boolean algebra. Present your answer in the format so that

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Unformatted text preview: it can be implemented using NAND gates only. AB + A(B+C) + C’ Ans. 1. (174.25) 10 = (256.2) 8 (2) 2. A = 01010101, B = 00001010 -B = 11110110 (1) (A+B) = 01011111 = (95) 10 (1+0.5) (A-B) = 01001011 = (75) 10 (Discard carry) (1+0.5) 3. (a) 1011011 (1) (b) 1000101 (1) 4. AB + A(B+C) + C’ = AB + AC + C’ (idempotent, distributive) = AB + A + C’ (simplification) = A + C’ (absorption) (2) = (A + C’)’’ (involution) = (A’C)’ (DM) (2)...
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