10002 1000010011112 05x1680 12

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Unformatted text preview: 0 1100)2 = (111 1011 . 0010 1100)2 = (1 0110 0100 . 1110 0100)2 d) dec ‐> hex (1063.5)10 = (427.8)16 1063/16 = 66 r 7 66/16 = 4 r 2 hex ‐> bin 4/16 = 0 r 4 (427.8)16 = (0100 0010 0111 . 1000)2 = (100 0010 0111 . 1)2 0.5 x 16 = 8.0 1.2) a) (111010110001.011)2 = (111 010 110 001 . 011)2 = (7261.3)8 = (1110 1011 0001 . 0110)2 = (EB1.6)16 7(8)3 + 2(8)2 + 6(8)1 + 1(8)0 + 3(8)‐1 = (3761.375)10 14(16)2 + 11(16)1 + 1(16)0 + 6(16)‐1 = (3761.375)10 b) (10110011101.11)2 = (010 110 011 101 . 110)2 = (2635.6)8 = (0101 1001 1101 . 1100)2 = (59D.C)16 2(8)3 + 6(8)2 + 3(8)1 + 5(8)0 + 6(8)‐1 = (1437.75)10 5(16)2 + 9(16)1 + 13(16)0 + 12(16)‐1 = (1437.75)10 1.4) a) 1457/16 = 91 r 1 91/16 = 5 r 11 5/16 = 0 r 5 0.11 x 16 = 1.76 0.76 x 16 = 12.16 0.16 x 16 = 2.56 (1457.11)10 ≈ (5B1.1C2)16 ≈ (5B1.1C)16 b) (5B1.1C)16 = (0101 1011 0001 . 0001 1100)2 = (010 110 110 001 . 000 111)2 = (2661.07)8 c) Build a lookup table counting from 0 to (15)10 in hexadecimal and base 4. Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F Base 4 00 01 02 03 10 11 12 13 20 21 22 23 30 31 32 33 (5B1.1C)16 = (11 23...
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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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