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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Document preview page 1

Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 1

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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition

Get accurate solutions for your textbook problems with Solution Manual For Logic And Computer Design Fundamentals, 4th Edition, your go-to guide for textbook answers.

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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 1 preview image7CHAPTER 11-1.(a)(1) Calm:(2) 10 mph(3) 100 mph(b) The microcomputer requires a table or equation for converting from rotations/second to miles/hour. The pulses pro-duced by the rotating disk must be counted over a known period of time, and the table or equation used to convert thebinary count to miles per hour.1-2.–34° quantizes to –30° => 1 V => 0001+31° quantizes to +30° => 7 V => 0111+77° quantizes to +80° => 12 V => 1100+108° quantizes to +110° => 15 V => 11111-3.*1-4.1-5.220= (1,000,00010+d) whered= 48,5761Tb = 240= (220)2= (1,000,000 +d)2= (1,000,000)2+ 2(1,000,000)d+d2= 1,000,000,000,000+97,152,000,000+2,359,627,776= 1,099,511,627,7761-6.Decimal, Binary, Octal and Hexadecimal Numbers from (16)10to (31)10Dec16171819202122232425262728293031Bin1 00001 00011 00101 00111 01001 01011 01101 01111 10001 10011 10101 10111 11001 11011 11101 1111Oct20212223242526273031323334353637Hex101112131415161718191A1B1C1D1E1For96K96210×98 304Bits,==640M640220×671 088 640Bits,,==4G4230×4 294 967 296Bits,,,==111 Bits2111–⇒2047=251 Bits2251–⇒33 554 431,,=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 2 preview image
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 3 preview image8Problem Solutions – Chapter 11-7.*1-8.2|1931110000012|751110111011112|9602|37512|4802|18712|2402|9312|1202|4602|612|2312|112|11102|512|202|1102|20071111110101112|1945001001011111110102|100312|972512|50112|486202|25002|243112|12512|121512|6202|60712|3112|30312|1512|15112|712|7512|312|3712|112|18002|912|402|202|1 101-9.*1-10.*a)8|75622166120.45×8= 3.6=>38|94510.60×8= 4.8=>48|11860.80×8= 6.4=>68|1460.20×8=3.2=>334638|110(7562.45)10= (16612.3463)8b)(1938.257)10= (792.41CB)16c)(175.175)10= (10101111.001011)21-11.*a)(673.6)8=(110 111 011.110)2=(1BB.C)16b)(E7C.B)16=(1110 0111 1100.1011)2=(7174.54)8c)(310.2)4=(11 01 00.10)2=(64.4)8DecimalBinaryOctalHexadecimal369.3125101110001.0101561.24171.5189.62510111101.101275.5BD.A214.62511010110.101326.5D6.A62407.6251111001111000111.101171707.5F3C7.A1001101()226232220+++77==1010011.101()22624212021–23–+++++83.625==10101110.1001()2272523222121–24–++++++174.5625==
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 4 preview image9Problem Solutions – Chapter 11-12.a)1101b)0101c)100111×1011×1010×01101111010000100111110101011001110000000000000011010101100111100011110110010100111000000100000111011-13.+10001Quotient = 10001101)1010110−101Remainder = 1000−000001−000011−000110−10111-14.(a) 6×123+8×122+7×121+4= 11608(b) 12|7569944691212|630612|52412|4401-15.a)0123456789ABCDEFGHIJb)20|200775072020|100020|550c)1-16.*a)(BEE)r= (2699)10By the quadratic equation:r= 15 or≈–16.27ANSWER:r= 15b)(365)r= (194)10BCI.G()2011202×=122011820016201–×+×+×+4658.8()10=11r2×14+r114r0×+×2699=11r2×14+r×2685–0=3r2×6+r1×5+r0×194=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 5 preview image10Problem Solutions – Chapter 1By the quadratic equation:r= – 9 or 7ANSWER: r = 71-17.Noting the order of operations, first add (34)rand (24)rNow, multiply the result by (21)rNext, set the result equal to (1480)rand reorganize.Finally, find the roots of this cubic polynomial.Solutions are: r = 8, – 1, – 1ANSWER: The chicken has 4 toes on each foot (half of 8).1-18.*a) (0100 1000 0110 0111)BCD=(4867)10=(1001100000011)2b) (0011 0111 1000.0111 0101)BCD=(378.75)10=(101111010.11)21-19.*(694)10=(0110 1001 0100)BCD(835)10=(1000 0011 0101)BCD1011010010100+1000+0011+0101111111001001+0110+0110+0000000101011 001010011-20.*(a)1011000111 1000Move R011 11000100column > 0111Subtract 3-0011011 10010Subtract 3-001101 1001Move R0 1100110100column > 0111Subtract 3-00110 1001110Move R01001110Move R01001110Move R01001110Move R01001110 Leftmost 1 in BCD number shifted out: Finished(b)1021011000011 1001 0111Move R001 1100 10111101and 100columns > 0111Subtract 3-0011-0011001 1001 10001Move R00 1100 110001101and 100columns > 01113r2×6+r×189–0=34()r3r1×4r0×+=24()r2r1×4r0×+=34()r24()r+5r18r0×+×=2r1×1r0×+()5r1×8r0×+()×10r223r18+×+×=10r223r18+×+×1r34r28r1×+×+×=1r36r215r1×–8r0×–×–×0=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 6 preview image11Problem Solutions – Chapter 1Subtract 3-0011 -001100 1001 100101Move R0 0100 1100101100column > 0111Subtract 3-00110 0100 1001Move R0010 01001101Move R001 001001101Move R00 1001001101100 column > 0111Subtract 3-001100 0110001101Move R0 00110001101Move R000110001101Move R000110001101Leftmost 1 in BCD number shifted out: Finished1-21.(a)10210110011110001st Move L11110002nd Move L11110003rd Move L1111000100column > 100Add 30011101010004th Move L1 0101000100column > 100Add 300111 10000005th Move L11 0000006th Move L110 00000101column > 100Add 300111001 000007th Move L1 0010 0000Least significant bit in binary number moved in:Finished(b)103102101100011100101111st Move L011100101112nd Move L01 1100101113rd Move L011100101114th Move L01110010111100column > 100Add 30011101000101115th Move L101000101116th Move L10100010111100column > 100Add 30011101011101117th Move L10101110111101& 100columns > 100Add 3001100111000101001118th Move L1 000101001119th Move L10 0010100111100column > 100Add 3001110 001011001110th Move L100 010110011101&100columns > 100Add 300110011100 10001100111th Move L1001 00011001Least significant bit in binary number moved in:Finished1-22.From Table 1-5, complementing the bit B6will switch an uppercase letter to a lower case letterand vice versa.1-23.a) The name used is Brent M. Ledvina. An alternative answer: use both upper and lower caseletters.01000010B01010010R01000101E
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 7 preview image12Problem Solutions – Chapter 101001110N01010100T00100000(SP)01001101M00101110.00100000(SP)01001100L01000101E01000100D01010110V01001001I01001110N01000001Ab)010000101101001011000101010011101101010010100000010011010010111010100000110011001100010101000100010101101100100101001110010000011-24.1000111G1101111o01000001000010B1100001a1100100d1100111g1100101e1110010r1110011s0100001!1-25.*a)(11111111)2b)(0010 0101 0101)BCDc)011 0010011 0101011 0101ASCIId)0011 00101011 01011011 0101ASCII with Odd Parity1-26.1-27.Binary Numbers from (32)10to (47)10with Odd and Even ParityDecimal3233343536373839(a) Odd100000 0100001 1100010 1100011 0100100 1100101 0100110 0100111 1(b) Even100000 1100001 0100010 0100011 1100100 0100101 1100110 1100111 0Decimal4041424344454647(a) Odd101000 1101001 0101010 0101011 1101100 0101101 1101110 1101111 0(b) Even101000 0101001 1101010 1101011 0101100 1101101 0101110 0101111 1Gray Code for Hexadecimal DigitsHex0123456789ABCDEFGray0000000100110010011001110101010011001101111111101010101110011000
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 8 preview image13Problem Solutions – Chapter 11-28.As the wind direction changes, the codes change in the order of the rows of this table, assum-ing that the bottom row is “next to” the top row. From the table, the codes that result due to awind direction change always change in a single bit.1-29.+The percentage of power consumed by the Gray code counter compared to a binary codecounter equals:Number of bit changes using Gray codeNumber of bit changes using binary codeAs shown in Table 1-6, and by definition, the number of bit changes per cycle of an n-bit Graycode counter is 1 per count = 2n.Number of bit changes using Gray code = 2nFor a binary counter, notice that the least significant bit changes on every increment. The sec-ond least significant bit changes on every other increment.The third digit changes on everyfourth increment of the counter, and so on. As shown in Table 1-6, the most significant digitchanges twice per cycle of the binary counter.Number of bit changes using binary code% Power =(a) Wind Direction Gray CodeDirectionCode WordN000S110E011W101NW100NE001SW111SE010(b) Wind Direction Gray Code (directions in adjacent order)DirectionCode WordN000NE001E011SE010S110SW111W101NW1002n2n1–…21+++2ii1=n∑=2ii0=n∑1–2n1+()1–()1–2n1+2–===2n2n1+()2–------------------------100×
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 9 preview image14CHAPTER 22-1.*a)b)c)2-2.*a)=b)=1Verification of DeMorgan’s TheoremXYZXYZXYZX+Y+Z000011001011010011011011100011101011110011111100The Second Distributive LawXYZYZX+YZX+YX+Z(X+Y)(X+Z)0000000000100010010001000111111110001111101011111100111111111111XYZXYYZXZXY+YZ+XZXYYZXZXY+YZ+XZ0000000000000101010011010100101010111001001110000111001101010110011100011010111100000000XYZXYZ++=XYZ+XY+()XZ+()⋅=XYYZXZ++XYYZXZ++=XYXYXY++XY+XYXY+()XYXY+()+=X YY+()Y XX+()+=XY+=ABBCABBC+++ABAB+()=BCBC+()+B=AA+()B CC+()+
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 10 preview image15Problem Solutions – Chapter 2c)=d)=2-3.+a)=b)=c)==2-4.+Given:Prove:=BB+1=YXZXY++XYZ++Y=XYXZ++YX+()=YY+()XZ+Y=XXZ++Y=XX+()XZ+()+X=YZ++XYYZXZXYYZ++++XYXZYZ++X=YYZ XX+()XZXYYZ++++XYXYZXYZXZXYYZ+++++=XY1Z+()XYZXZXYYZ++++=XYXZ1Y+()XYYZ+++=XYXZXY ZZ+()YZ+++=XYXZXYZYZ1X+()+++=XYXZ1Y+()YZ++=XYXZYZ++=ABCBCDBCCD+++BCD+ABCABC+BCBCDBCDCD++++=AB CC+()BC DD+()BCCD+++=ABBCBCCD+++=BABCD++=BCD+=WYWYZWXZWXY+++WYWXZXYZXYZ+++WYWXYZ+()WXYZWXYZ+()WXYZWXYZ+()WXYZWXYZ+()+++=WYWXYZ+()WXYZWXYZ+()WXYZWXYZ+()WXYZWXYZ+()+++=WYWXZ YY+()XYZ WW+()XYZ WW+()+++=WYWXZXYZXYZ+++=ADABCDBC+++ABCD+++()ABCD+++()ADABCDBC+++=AD+()AB+()CD+()BC+()=ABADBD++()BCBDCD++()=ABCDABCD+=ABCD+++()=ABCD+++()ABCD+++()ABCD+++()AB⋅0AB+,1==AC+()AB+()BC+()BCABACBC++()BC+()=ABACBC++=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 11 preview image16Problem Solutions – Chapter 22-5.+Step 1:Define all elements of the algebra as four bit vectors such asA, BandC:A=(A3,A2,A1,A0)B=(B3,B2,B1,B0)C=(C3,C2,C1,C0)Step 2:Define OR1, AND1and NOT1so that they conform to the definitions of AND,OR and NOT presented in Table 2-1.a)A+B=Cis defined such that for alli,i= 0, ... ,3,Ciequals the OR1ofAiandBi.b)AB=Cis defined such that for alli,i= 0, ... ,3,Ciequals the AND1ofAiandBi.c)The element 0 is defined such that forA= “0”, for alli, i =0, ... ,3,Aiequals logical 0.d)The element 1 is defined such that forA= “1”, for alli, i =0, ... ,3,Aiequals logical 1.e)For any elementA,Ais defined such that for alli,i= 0, ... ,3,Aiequals the NOT1ofAi.2-6.a)b)c)d)e)2-7.*a)b)c)0C AB+()+=C AB+()0()=C AB+()AB+()=C ABABB++()=BC=ACABCBC++ACABCA(BCBC)+++=ACABCABC+()BC++=A(CAC)BC++ABC+==ABC++()ABC()AABCABBCABCC++=AA()BCA BB()CAB CC()++=ABCABCABC++=ABC=ABCAC+A BCC+()A BC+()==ABDACDBD++ABBAC++()D=AACB++()D=AB+()D=AB+()AC+()ABC()AB()=AC()ABC++()ABC ABC++()=0=XYXYZXY++XXYZ+XXY+()XZ+()XX+()XY+()XZ+()===XY+()XZ+()=XYZ+=XY ZXZ++()+XY ZXZ+()+XY ZX+()ZZ+()+XYZXY++===XX+()XY+()YZ+=XYYZ++XY+==WX ZYZ+()X WWYZ+()+WXZWXYZWXWXYZ+++=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 12 preview image17Problem Solutions – Chapter 2d)2-8.a)b)2-9.*a)b)c)d)2-10.*a)Sum of Minterms:Product of Maxterms:b)Sum of Minterms:Product of Maxterms:c)Sum of Minterms:Product of Maxterms:2-11.a),b),Truth Tables a, b, cXYZaABCbWXYZc000000010000000100011000100100010000101011101110011010001000010001011101001010110111000110111111111011101000010010101011011011001110111110111111WXZWXZWX++=WXWX+X==ABAB+()CDCD+()AC+ABCDABCDABCDABCDAC+++++=ABCDAC++=ACA BCD()++ACC BD()++ACBD++===FABCACAB++=FABCACAB++=ABC++()AC+()AB+()++=ABC()AC()AB()=FAB+()AB+()=FVW+()XY+()Z=FWX+YZ+()YZ+()+[]WX+YZYZ++[]=FABCAB+()CA BC+()++=XYZXYZXYZXYZ+++XYZ++()XYZ++()XYZ++()XYZ++()ABCABCABCABC+++ABC++()ABC++()ABC++()ABC++()WXYZWXYZWXYZWXYZWXYZWXYZ+++++W+XYZWXYZ+++()WXYZ+++()WXYZ+++()WXYZ+++()WXYZ+++()WXYZ+++()WXYZ+++()WXYZ+++()WXYZ+++()EΣm1 2 4 6,,,()ΠM0 3 5 7,,,()==FΣm0 2 4 7,,,()ΠM1 3 5 6,,,()==EΣm0 3 5 7,,,()=FΣm1 3 5 6,,,()=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 13 preview image18Problem Solutions – Chapter 2c),d),e),2-12.*a)s.o.p.p.o.s.b)p.o.s.s.o.p.c)p.o.s.s.o.p.2-13.2-14.2-15.*EF+Σm0 1 2 4 6 7,,,,,()=EF⋅Σm2 4,()=EXYZXYZXYZXYZ+++=FXYZXYZXYZXYZ+++=EZ XY+()XYZ+=FZ XY+()XYZ+=ABC+()BCD+()ABABCDBC++ABBC+==B AC+()=XX XY+()YZ+()+XX+()XXY+()YZ+()+()=XXY++()=XYZ++()1Y+()XYZ++()=XYZ++=ABCCD++()BEF+()ABC++()ABD++()ACD++()BEF+()=ABC++()=ABD++()ACD++()BE+()BF+()ABCCD++()BEF+()A BEF+()BC BEF+()CD BEF+()++=A=BAEFBCEFBCDCDEF++++a)b)c)CBACXZXYZDBACYWYYZWXZXYAXYZABCXYZABCa)b)c)d)111111111111111111XZXY+XYXZYZ++CAB+BCABAC++1orBCABAC++XYZABCa)b)c)ABC111111111111111XZXY+ACB+BC+
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 14 preview image19Problem Solutions – Chapter 22-16.2-17.2-18.*2-19.*a)b)c)2-20.a)b)c)ABDWXZABD1111111111111111111111111ACADABC++YZXZXYZWXZorWXY()+++BDABCACD++WXYZABCDa)b)11111111111111111FXZWYWXXYZ+++=FBDABCABCADorCD()+++=1WXYZABCDa)b)c)XYZΣm3 5 6 7,,,()Σm3 4 5 7 9 13 14 15,,,,,,,()Σm0 2 6 7 8 10 13 15,,,,,,,()11111111111111111111PrimeXZ WX XZ WZ,,,=PrimeCD AC BD ABD BC,,,,=PrimeAB AC AD BC BD CD,,,,,=EssentialXZ XZ,=EssentialAC BD ABD,,=EssentialAC BC BD,,=PrimeXY XZ WYZ WXY XYZ WXZ,,,,,WYZ,=PrimeABC ACD ABC ACD BD,,,,=EssentialXY XZ,=EssentialABC ACD ABC ACD,,,=FXYXZWXYWXZ+++=RedundantBD=FABCACDABCACD+++=PrimeYZ WY WZ WXZ XYZ WXY,,,,,=EssentialWY WZ,=RedundantYZ=FWYWZXYZ++=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 15 preview image20Problem Solutions – Chapter 22-21.2-22.*a)s.o.p.b)s.o.p.c)s.o.p.orp.o.s.p.o.s.p.o.s.2-23.a)s.o.p.b)s.o.p.orp.o.s.p.o.s.or2-24.2-25.*WXYZABCDa)b)FF111111111111111111FΣm1 5 6 7 9 12 13 14,,,,,,,()=FΣm0 2 4 5 8 10 11 12 13 14,,,,,,,,,()=FYZWXZWXY++=FBCBDADABC+++=FYZ+()WXZ++()WXY++()=FBC+()BD+()AD+()ABC++()=CDACBD++ACBDAD++BDABDA(BC++ACD)CD+()AD+()ABC++()CD+()AD+()ABC++()AB+()BD+()BCD++()ABCABDABCABD+++ZWXXY++ACDBCDACDBCD+++WXZ++()XYZ++()ABC++()ABD++()ABC++()ABD++()ACD++()BCD++()ACD++()BCD++()WXYZa)b)c)ABC11111111XXXXXFAC+=FXYWYZWXZ++ (WXYorWYZ)+=ABCD11111XXXXFBCBCDABD++=XXXXWXYZABCDa)b)c)ABC1111111111111111XXXXXXXXXXXPrimesAB AC BC ABC,,,=FABACBC++=EssentialAB AC BC,,=PrimesXZ XZ WXY WXY WYZ WYZ,,,,,=EssentialXZ=FXZWXYWXY++=PrimesAB C AD BD,,,=EssentialC AD,=FCADBDorAB()++=
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Solution Manual For Logic And Computer Design Fundamentals, 4th Edition - Page 16 preview image21Problem Solutions – Chapter 22-26.2-27.2-28.+XABCDa)(1)b)(1)XWXYZXXX0FBD=FBD+=X1XFBD=0X00A0XXBDC1XXXXWXZXYXX1F = W X +W Y + X Y ZF =(WZorX Z)+ WXY + WXYF =((W + Z) or (X + Z))XX00000a)(2)b)(2)00XXX1X1XXXXX111111+(WXYor XYZ)+ (W + X + Y) or (X + Y + Z)(W + X + Y)(W + X + Y)+ W X YABCDa)F = A B C D + ABC + A C D + ABD + ABCD + BCDb)F = A C + AB + BD + AC + AB1111111111ABCD1111111111ABCD1111111111ABCD1111111111111111There are other solutions depending on howties are resolved.There are other solutions depending on howties are resolved.ABCDF = A B D + B C D + BC + AB + ACD1111111111ABCDABCDABCD1111F andF; Cost = 181111111111111111111111EXPANDESSENTIAL PRIMESXXXXXXX1IRREDUNDANT COVER;Cost = 17ABCD111XXXXXXX1REDUCEABCD111XXXXXXX1EXPANDABD111XXXXXXX1IRREDUNDANT COVER;Cost = 13REDUCE, EXPAND,IRREDUNDANT COVER,and LAST GASP produceno lasting changes.ABD1111FINAL SOLUTION;Cost = 131111111
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