B similar as in part a 15 12 10001 10100 00101

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Unformatted text preview: overflow occurs)  ­18 + 11 = 0_1110 + 0_1011 = 1_1001 =  ­7  ­7 – 7 = 1_1001 + 1_1001 = 1_0010 =  ­14  ­14 + 14 = 1_0010 + 0_1110 = 0_0000 = 0 0 + 10 = 0_0000 + 0_1010 = 0_1010 = 10 10 + 4 = 0_1010 + 0_0100 = 0_1110 = 14 The final result is correct. b. Similar as in part a:  ­15  ­12 = 1_0001 + 1_0100 = 0_0101 = 5 (should be  ­27, overflow occurs)  ­27  ­7 = 0_0101 + 1_1001 = 1_1110 =  ­2 (should be  ­34, overflow occurs)  ­34 + 14 = 1_1110 + 0_1110 = 0_1100 = 12 (should be  ­20, overflow occurs)  ­20 + 11 = 0_1100 + 0_1011 = 1_0111 =  ­9  ­9 + 10 = 1_0111 + 0_1010 = 0_0001 = 1 1 + 8 = 0_0001 + 0_1000 = 0_1001 = 9 9 + 4 = 0_1001 + 0_0100 = 0_1101 = 13 13 + 1 = 0_1101 + 0_0001 = 0_1110 = 14 The final result is correct. c. For a: 1 – 15 = 0_0001 + 1_0001 = 1_0010 =  ­14  ­14  ­12 = 1_0010 + 1_0100 = 1_0000 =  ­16 (should be  ­26, saturate occurs)  ­16 + 8 = 1_0000 + 0_1000 = 1_1000 =  ­8  ­8 + 11 = 1_1000 + 0_1011 = 0_0011 = 3 3 – 7 = 0_0011 + 1_1001 = 1_1100 =  ­4  ­4 + 14 = 1_1100 + 0_1110 = 0_1010 = 10 10 + 10 = 0_1010 + 0_1010 = 0_1111 = 15 (should be 20, saturate occurs) 15 + 4 = 0_111...
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This note was uploaded on 01/13/2014 for the course EECS 452 taught by Professor Staff during the Fall '08 term at University of Michigan.

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