I 011010012s10002s000011012s ii

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Unformatted text preview: 1 1111)2 x (0010 0000)2 = (1111 1111 1110 0000)2 64‐bit => 16‐hex digits = (0000 0000 0000 FFE0)16 b) Repeat part a) assuming signed 2’s complement numbers for all values. i) (01101001)2s / (1000)2s = (00001101)2s ii) (1010)2s x (0010)2s = (11110100)2s Remember to sign extend when resizing iii) (A43F)16s / (0004)16s = (1010 0100 0011 1111)2s / (0000 0000 0000 0100)2s = (1110 1001 0000 1111)2s = (FFFF E90F)16s iv) (FFFF)16s x (20)16s = (1111 1111 1111 1111)2s x (0010 0000)2s = (1111 1111 1110 0000)2s = (FFFF FFFF FFFF FFE0)16s c) Calculate the percent error due to truncation for parts a)i) and b)iii) (typo in HW). a)i) (105)10 / (8)10 = (13.125)10 (13 ‐ 13.125)/13.125 x 100% = ‐0.95% b)iii) Caution: (A43F)16s and (FFFF E90F)16s are both negative according to 2’s Complement Representation (A43F)16s = (1010 0100 0011 1111)2s = ‐(0101 1011 1100 0001)2 = (‐23489)10 (FFFF E90F)16s = (1111 … 1111 1110 1001 0000 1111)2s = ‐(0000 … 0000 0001 0110 1111 0001)2 = (‐5873)10 (‐23489)10 / (4)10 = (‐5872.25)10 (‐5873 – (‐5872.25))/(5872.25) x 100% = 0.013% 5) Express the values from problem 1.2 in BCD. a) (111010110001.011)2 = (3761.375)10 = (0011 0111 0110 0001 . 0011 0111 0101)BCD b) (10110011101.11)2 = (1437.75)10 = (0001 0100 0011 0111 . 0111 0101)BCD...
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This note was uploaded on 10/16/2013 for the course ECE 2100 taught by Professor Khan during the Spring '13 term at Ohio State.

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