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HW01_sol

# Againnoticethatthe2leftmostcarrybitsatthetoparediffere

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Unformatted text preview: 0 0 0 0 1 0 0 1 0 (‐46)10 (010010)2s = (+20)10 ≠ (‐46)10 Overflow must have occurred. Again notice that the 2 leftmost carry bits at the top are different. c) 0 0 1 1 (‐25)10 (‐0 1 1 0 0 1) 1 0 0 1 1 1 (18)10 + 0 1 0 0 1 0 = + 0 1 0 0 1 0 1 1 1 0 0 1 (‐7)10 (111001)2s = ‐(000111)2 = (‐7)10 The result is correct. Notice that the 2 leftmost carry bits at the top are the same. d) 1 1 1 1 (‐12)10 (‐0 0 1 1 0 0) 1 1 0 1 0 0 (13)10 + 0 0 1 1 0 1 = + 0 0 1 1 0 1 0 0 0 0 0 1 (1)10 (000001)2s = (1)10 The result is correct. Notice that the 2 leftmost carry bits at the top are the same. e) 1 1 1 1 1 1 (‐11)10 (‐0 1 0 1 0 1) 1 0 1 0 1 1 (‐21)10 + (‐0 0 1 0 1 1) = + 1 1 0 1 0 1 1 0 0 0 0 0 (‐32)10 (100000)2s = (‐32)10 The result is correct. Notice that the 2 leftmost carry bits at the top are the same. 0 1 1 1 1 4) a) Find the value of z for the following problems using bit‐shifting. Truncate the remainder if applicable. All values represent unsigned numbers. i) (01101001)2 / (1000)2 = (00001101)2 ii) (1010)2 x (0010)2 = (00010100)2 iii) (A43F)16 / (0004)16 = (1010 0100 0011 1111)2 / (0000 0000 0000 0100)2 = (0010 1001 0000 1111)2 32‐bit => 8‐hex digits = (0000 290F)16 iv) (FFFF)16 x (20)16 = (1111 1111 111...
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