Web_Answers_1 - 5 6 3BA.25 14 = 3 × 14 2 11 × 14 1 10 ×...

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Unit 1 Problem Solutions 757.25 10 16 | 757 0.25 16 | 47 r5 16 16 | 2 r15=F 16 (4).00 0 r2 757.25 10 = 2F5.40 16 = 0010 1111 0101 .0100 0000 2 2 F 5 4 0 1.1 (a) 1.1 (b) 123.17 10 16 | 123 0.17 16 | 7 r11 16 0 r7 (2).72 16 (11).52 16 (8).32 123.17 10 = 7B.2B 16 = 0111 1011 .0010 1011 2 7 B 2 B 356.89 10 16 | 356 0.89 16 | 22 r4 16 16 | 1 r6 (14).24 0 r1 16 (3).84 16 (13).44 16 (7).04 356.89 10 = 164.E3 16 = 0001 0110 0100 .1110 0011 2 1 6 4 E 3 1.1 (c) 1.1 (d) 1063.5 10 16 | 1063 0.5 16 | 66 r7 16 16 | 4 r2 (8).00 0 r4 1063.5 10 = 427.8 16 = 0100 0010 0111 .1000 2 4 2 7 8 EB1.6 16 = E × 16 2 + B × 16 1 + 1 × 16 0 + 6 × 16 –1 = 14 × 256 + 11 × 16 + 1 + 6/16 = 3761.375 10 1110 1011 0001 .011(0) 2 E B 1 6 7261.3 8 = 7 × 8 3 + 2 × 8 2 + 6 × 8 1 + 1 + 3 × 8 –1 = 7 × 512 + 2 × 64 + 6 × 8 + 1 + 3/8 = 3761.375 10 111 010 110 001 .011 8 7 2 6 1 3 1.2 (a) 1.2(b) 59D.C 16 = 5 × 16 2 + 9 × 16 1 + D × 16 0 + C × 16 –1 = 5 × 256 + 9 × 16 + 13 + 12/16 = 1437.75 10 0101 1001 1101 .1100 16 5 9 D C 2635.6 8 = 2 × 8 3 + 6 × 8 2 + 3 × 8 1 + 5 × 8 0 + 6 × 8 –1 = 2 × 512 + 6 × 64 + 3 × 8 + 5 + 6/8 = 1437.75 10 010 110 011 101 .110 8 2 6 3
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Unformatted text preview: 5 6 3BA.25 14 = 3 × 14 2 + 11 × 14 1 + 10 × 14 + 2 × 14 –1 + 5 ×14 –2 = 588 + 154 + 10 + 0.1684 = 752.1684 10 6 | 752 0.1684 6 | 125 r2 6 6 | 20 r5 (1).0104 6 | 3 r2 6 0 r3 (0).0624 6 (0).3744 6 (2).2464 6 (1).4784 ∴ 3BA.25 14 = 752.1684 10 = 3252.1002 6 1.3 1.4 (a) 1457.11 10 16 | 1457 0.11 16 | 91 r1 16 16 | 5 r11=B 16 (1).76 0 r5 16 (12).16 ∴ 1457.11 10 = 5B1.1C 16 1.4 (c) 5B1.1C 16 = 11 23 01 .01 30 4 5 B 1 1 C 5 B 1 1 C 1.4 (b) 5B1.1C 16 = 010110110001.00011100 2 =2661.070 8 2 6 6 1 0 7 0 1.4 (d) DEC.A 16 = D × 16 2 + E × 16 1 + C × 16 + A× 16 –1 = 3328 + 224 + 12 + 0.625 =3564.625 10 1.5 (a) 1 1 1 1111 (Add) 1111 (Sub) +1010 −1010 11001 0101 1111 (Multiply) ×1010 0000 1111 11110 0000 011110 1111 10010110 See FLD p. 625 for solution. 1.6, 1.7, 1.8, 1.9 1.5 (b, c) See FLD p. 625 for solution....
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This note was uploaded on 09/03/2008 for the course EE 316 taught by Professor Brown during the Fall '08 term at University of Texas.

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Web_Answers_1 - 5 6 3BA.25 14 = 3 × 14 2 11 × 14 1 10 ×...

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