Assignment 1 Solution - EE 2000 Logic Circuit Design,...

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EE 2000 Logic Circuit Design, Semester A, 2007/08 Assignment 1 Solution Question 1: (2925) 10 = (0000 1011 0110 1101) 2cns (-2925) 10 = (1111 0100 1001 0011) 2cns (16850) 10 = (0100 0001 1101 0010) 2cns (-16850) 10 = (1011 1110 0010 1110) 2cns A + B = (-2925) 10 + (16850) 10 : 1000 0001 0010 0100 Carries 1111 0100 1001 0011 A +) 0100 0001 1101 0010 B (1)0011 0110 0110 0101 Result = (0011 0110 0110 0101) 2cns = (13925) 10 No overflow. In 2’s complement number system, a negative number plus a positive number will never produce overflow. A - B = (-2925) 10 + (-16850) 10 : 1111 1000 0111 1100 Carries 1111 0100 1001 0011 A +) 1011 1110 0010 1110 - B (1)1011 0010 1100 0001 Result = (0011 0110 0110 0101) 2cns = (19775) 10 No overflow. Compare the sign bits, a negative number plus a negative number produce a negative number. Question 2: The r ’s complement of n -digit number a is r n a The r ’s complement of r n a = r n – ( r n a ) = a Question 3: Note: need not to find the optimal implementation
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This note was uploaded on 01/11/2011 for the course EE 2000 taught by Professor Vancwting during the Fall '07 term at City University of Hong Kong.

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Assignment 1 Solution - EE 2000 Logic Circuit Design,...

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