Tutorial 1 Solution - -) 1011 1010 a- b = 1010 ( You can...

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EE 2000 Logic Circuit Design, Semester A, 2007/08 Tutorial 1 Solution Level 1 Question 1: Decimal Binary Octal Hexadecimal 27 11011 33 1B 237.25 11101101.01 355.2 ED.4 250.8 1111 1010.1100 1100… 372.6314 6314… FA.CC… Question 2: (a) (1111 1111 1111 1111) 2 (b) 2 15 + 2 14 + … + 2 1 + 2 0 = 2 16 – 1 = 65535 Question 3: (a) 1101 0101 (1’s complement) 1101 0110 (2’s complement) (b) 0000 0000 (1’s complement) 0000 0001 (2’s complement) Question 4: (a) -128 = 1000 0000; -1 = 1111 1111; (-128) + (-1) = 1000 0000 + 1111 1111 = (1) 0111 1111 Discard carry, we get 0111 1111 (Result incorrect. Overflow because –ve + -ve becomes +ve) (b) -128 = 1 1000 0000; -1 = 1 1111 1111; (-128) + (-1) = 1 1000 0000 + 1 1111 1111 = (1) 1 0111 1111 Discard carry, we get 1 0111 1111 (= -129) (No overflow as –ve + -ve gives a –ve number)
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Level 2 Question 5: A = 10101, B = 1011 11110 10101 +) 1011 100000 a + b = 100000 10100 10101
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Unformatted text preview: -) 1011 1010 a- b = 1010 ( You can get the same result if using 2’s complement ) 10101 x) 1011 10101000 0 101010 10101 11100111 a · b = 11100111 01 1011 ) 10101 1011 1010 a / b : quotient = 1, remainder = 1010 Question 6: (211) x = (152) 8 2 · x 2 + 1 · x + 1 = 1 · 8 2 + 5 · 8 + 2 2 x 2 + x – 105 = 0 (2 x + 15)( x – 7) = 0 ∴ x = 7 ( x must be a positive integer so cannot be -7.5) Question 7: Partition the digits into groups of 2 each (one hex digit = 2 base-4 digits) Add necessary 0s to the left and right Replace each group of base-4 digits by the hex equivalent So, (12321) 4 = (01 23 21) 4 (Note: 23 4 = 2·4 1 + 3·4 = 11 = B 16 ) = (1 B 9) 16 Question 8: (a) 010.11 1’s complement: 2 3 - 010.11 - 2-2 = 101.00 2’s complement: 2 3- 010.11 = 101.01 (b) 11011.100 1’s complement: 2 5- 11011.100 - 2-3 = 00100.011 2’s complement: 2 5- 11011.100 = 00100.100...
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Tutorial 1 Solution - -) 1011 1010 a- b = 1010 ( You can...

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