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Solution Manual For Digital Design, 5th Edition

Solution Manual For Digital Design, 5th Edition offers textbook solutions that are easy to follow, helping you ace your assignments.

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2CHAPTER 11.1Base-10:1617181920212223242526272829303132Octal:2021222324252627303132333435363740Hex:101112131415161718191A1B1C1D1E1F20Base-1214 151617 18191A1B2021222324252627281.2(a)32,768(b)67,108,864(c)6,871,947,6741.3(4310)5=4*53+ 3*52+ 1*51= 58010(198)12=1*122+ 9*121+ 8*120=26010(435)8= 4*82+ 3*81+ 5*80= 28510(345)6=3*62+4*61+ 5*60= 137101.416-bit binary:1111_1111_1111_1111Decimalequivalent:216-1 =65,53510Hexadecimalequivalent:FFFF161.5Let b = base(a)14/2 = (b + 4)/2 = 5, so b = 6(b)54/4 = (5*b + 4)/4 = b + 3, so 5 * b= 52–4, and b = 8(c)(2 *b + 4) + (b + 7) = 4b, so b = 111.6(x–3)(x–6) = x2–(6 + 3)x + 6*3 = x2-11x + 22Therefore:6 + 3 = b + 1m,so b = 8Also, 6*3 = (18)10= (22)81.764CD16= 0110_0100_1100_11012=110_010_011_001_101=(62315)81.8(a)Results of repeated division by 2 (quotients are followed by remainders):43110=215(1);107(1);53(1);26(1);13(0);6(1)3(0)1(1)Answer:1111_10102= FA16(b)Results of repeated division by 16:43110= 26(15);1(10)(Faster)Answer: FA = 1111_10101.9(a)10110.01012= 16 + 4 + 2 + .25 + .0625 = 22.3125(b)16.516= 16 + 6 + 5*(.0615) = 22.3125(c)26.248= 2 * 8 + 6 + 2/8 + 4/64 = 22.3125(d)DADA.B16=14*163+ 10*162+ 14*16 + 10 + 11/16 =60,138.6875

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Document Details

Subject
Electrical Engineering & Electronics
Textbook
Digital Design by M. Morris R. Mano...

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