數(shù)字設(shè)計(jì) 第二章

Chapter2NumberSystemsandCode重點(diǎn)：1）二進(jìn)制、八進(jìn)制和十六進(jìn)制數(shù)的表達(dá)及各種數(shù)制之間的相互轉(zhuǎn)換；2）符號(hào)數(shù)的表達(dá)及其加、減運(yùn)算；3）BCD編碼和GRAY碼。4/21/20231.2.1PositionalNumberSystems1.Decimalnumber

Adecimalnumber1536.79d

digits：0、1、2、3、4、5、6、7、8、9weight：10iradix：10power：i，startfromtheleftofthedecimalpoint,increasebyoneforeachsuccessivepositionfrom0;fromtherightofthedecimalpoint,decreasebyonefrom(-1).powerweightdigitradix2.

AnarbitrarydecimalnumberDwithpdigitsinintegerandndigitsinfractiondp-1dp-2……d1d0.d-1d-2……d-nGeneralexpression：2.Binarynumbersradix：2digits：0、1，alsobecalledbit，weight：2i3.Abinarynumber1011010.1012，：※anarbitrarybinarynumberBbp-1bp-2…b1b0

.b-1…b-n，thegeneralexpressions：Thesumofeachdigitmultipliedbythecorrespondingpoweroftheradixisthebinarynumber’sdecimalvalue.4.3.OthersLSB:LeastSignificantBitMSB:MostSignificantBit

bp-1bp-2…b1b0.b-1…b-nMSBLSB5.2.2OctalandHexadecimalNumbers1.conceptOctalHexadecimalradix816weight8i16idigits0、1、2、3、4、5、6、70、1、2、3、4、5、6、7、8、9、A、B、C、D、E、FGeneralexpression6.Exp3：givethedecimalvalueofthefollowingnumbers.215.78=3AD.8h=2.2-8and2-16number’sConvensionsoctal01234567binarysection000001010011100101110111Table1relationbetweenbinaryandoctal7.hexadecimalbinarysection0000010001200103001140100501016011070111hexadecimalbinarysection8100091001A1010B1011C1100D1101E1110F1111Table2binary-hexadecimalrelations8.（a）binary——octal、hexadecimalconversionsStartingatthebinarypoint，separatethebitsintogroupsof3or4,andreplaceeachgroupwiththecorrespondingoctalorhexadecimaldigit.Iftheleftmostgroupfallshortof3or4bits,then0shouldbeaddedtotheleftoftheMSB;sodotheright。Exp：dothefollowingpositional-number-systemconversion。

1111011010.00112=(?)8=(?)hSolution:=1732.148=3DA.3h9.(b)octal,hexadecimal-binarynumberconversionReplaceeachoctalorhexadecimalwiththecorrespondinggroupof3or4bits.Exp：267.248=（？）296CA.28h=（？）210.2.3generalpositional-number-systemconversionsRadix-r-to-decimalconversion:Eachradix-rnumber’sdigitmultiplybyitsownweight,andaddthem.Exp：512.46=（？）d1、decimal-to-radix-rconversion—integerpartmethods：除基取余Theintegerofapdigitsnumberinradixris：Dividetheformulabyrsuccessively,yieldthesuccessivedigitsoftheintegerofD(radixr)fromtherighttolefttillthequotientis0.11.Exp：371d=（?）2=（?）8=（?）h2、decimal-to-radix-rconversion—fractionpartmethods：乘基取整Thefractionofapdigitsnumberinradixris：Multiplytheformulabyrsuccessively,yieldthesuccessivedigitsofthefractionofD(radixr)fromthelefttorighttilltherequireddigitsd-1

，d-2，……，d-n

areacquired.12.Exp8：0.71875d=(?)2=(?)8=(?)h3、summaryofthepositional-number-systemconversions（1）conversionbetweendecimalandradixr（2）radix-r-to-radix-jconversion(non-decimal)binary-to-octal,binary-to-hexadecimalorreverse,beconverteddirectlyconformingtothesection-replacerules.others,radix-rnumber——decimal——radix-jnumber13.2.4additionandsubtractionofnondecimalnumber1、Rulesofbinaryadditionandsubtraction（1）rulesofadditionS=x+y+CinExp：01001011+10001111=?CinxyCoutS0000000101010010111010001101101101011111initialcarrycarry000011110c14.（2）RulesofBinarySubtractiond=x-y-binExp：11001100-01011100=?binxyboutd0000000111010010110010011101101100011111initialborrowborrow011100000b15.2.additionandsubtractionofoctalandhexadecimaladdition:Iftwodigit’ssumisgreaterthantheradixoneachcolumn,thencarry1tothenextmoresignificantbit.subtraction:Iftheminuendislessthansubtrahend,thenborrow1fromthenextmoresignificantbit.16.2.5RepresentationofNegativeNumbers1、signed-magnituderepresentation

unsignednumbers：justthemagnitudeofanumberisrepresented,no‘+’or‘-’signsymbolbeforethenumber。

signednumbers：’+’or‘-’isaddedtotheleftofthenumber。representationofsignednumber：①signsymbol+numbermagnitude，like+34d、-1102、+1Dh、…

17.②inbinarynumber,signbit+magnitudeTheMSBofabitstringisusedasthesignbit：0—“+”，1—“-”thistypeofsignedbinarynumberiscalledSignedMagnitude（S-M碼，或原碼）。比如，+11101=011101-1011=1101118.③TherangeofS-M’srepresentablenumbers.ann-bitS-MnumberB:

bn-1bn-2…b1b0Itsrangeis：-（2n-1-1）~+(2n-1-1)

Includetwozero:-0and+0Exp：findtheS-Mofthefollowingsignednumbers。Howtorepresentbyusing8-bitS-M.+11101，-1011，+18，-18④thefaultofS-Mwhichisusedinarithmeticaloperation：itcan’tbecalculateddirectly.符號(hào)位數(shù)值19.2、ComplementNumberSystemradixcomplement（基數(shù)補(bǔ)碼數(shù)制）diminishedradix-complement(基數(shù)減1補(bǔ)碼數(shù)制)Ann-digitnumberDinradixr:D=dn-1……d1d0，①radix-complement②diminishedradix-complement（也稱基數(shù)反碼）Relationbetweenthesetwo：Table2-4,2-5(P.36)showssomer’sand(r-1)s’complement.20.3.Two’s-complementRepresentation

Two'scomplementisamethodforrepresentingsignedintegersasbinarynumbers.n-bitbinarynumberB=bn-1……b1b0，thetwo’scomplementis:B

2’s=2n-B

forExp.,8-bitbinarynumber two’scomplement 00000000 28-0=00000000 00000001 28-1=11111111 00000010 28-10=11111110 …… …… 11111111 0000000121.Definethesetwo’scomplementassignednumber,theMSBisservedassignbit。

MSB=0，positivenumberMSB=1，negativenumberweightofthesignbit：MSB=0,weight+2n-1

MSB=1,weignt-2n-1Rangeofthevalue：positive—0~+（2n-1-1）；

negative—（-2n-1）~-1。Onlyonezerointwo’scomplement.22.two’scomplementnumberandtheirdecimalequivalent2’scomplementdecimal00000000000000001+1…………01111111+12710000000-12810000001-127…………11111110-211111111-123.2’s-complementdecimalequivalentS-M000000000000000001000000000000001+100000001……………..01111111+1270111111110000000-128-10000001-12711111111………………11111110-21000001011111111-11000000124.propertysignbit=0，S-Mandtwo’s-complementhavethesamedecimalvalue；signbit=1，S-Mandtwo’s-complementhavedifferentdecimalvalue。Howtofindthetwo’s-complementrepresentationofanegativedecimalnumber?25.Decimalequivalent4-bits2’scomplement-81000-71001-61010-51011-41100-31101-21110-11111ItsoriginalbinarynumberDecimalequivalent1000801117011060101501004001130010200011互為補(bǔ)數(shù)Samevalue,buthasoppositesignsymbol26.So,tocalculatethe2'scomplementofannegativeinteger,invertthen-bitbinaryequivalentofthegivennumbermagnitudebitbybit,andthenadd1totheLSB.Exp:calculatethe2’scomplementof(+65)and(-65d)in8-bitform.Solution:+6501000001-65binaryequivalentof65is01000001,invertbitbybit10111110add11011111127.Ann-bitbinarynumberB=bn-1……b1b0，theones’-complementis:B

1s’=2n-1-BItisalsothemethodofrepresentingsignedbinarynumbers.MSB=0，positive,weight+(2n-1-1)MSB=1，negative，weight-(2n-1-1)。Rangeofrepresentablenumbers：negative—-（2n-1-1）~-0positive—+0~（2n-1-1）4.Ones’-complementRepresentation28.1s’-complement’spropertyN-bitpositiveintegersarerepresentedinthesamewayasn-bitsign-magnitudenotation.Theones’-complementofann-bitnegativeintegernumberisobtainedbycomplementingeachoneofthebits(then-bitbinarynumber),i.e.,a1isreplacedbya0,anda0isreplacedbya1.Exp.8-bit1s’-complementofthenumber.+18d=00010010-18d=1'scomplementof18=1110110129.decimalS-M2’scomp.1s’comp.-1100111111110-2101011101101-3101111011100-4110011001011-5110110111010-6111010101001-7111110011000-8-1000-decimalS-M2’scomp.1s’comp.7011101110111601100110011050101010101014010001000100300110011001120010001000101000100010001000001000000011110000返回Fromtheleastnumbertobiggestnumber,2’scomp.and1s’comp.aresuccessiveincreasedby1.SummaryofS-M,2’scomp.,1s’comp.30.Allofthesethreeareusedtorepresentsignedintegernumberinbinarysystem.Comparingoftherepresentingrangeofthevalue

Representationsofpositiveintegeraresame;buttherepresentationsofnegativeintegeraredifferentatall.S-M2’scomp.1s’comp.positive+0~（2n-1-1）+0~（2n-1-1）+0~（2n-1-1）negative-（2n-1-1）~-0（-2n-1）~-1-（2n-1-1）~-031.Calculatesignedinteger’sS-M,2’s-comp.,1s’-comp.①positivesignednumber：convertthegivennumberintothewantedn-bitbinaryequivalent.②negativesignednumber：firstconvertthenumberinton-bitbinaryequivalent,

S-M—letMSB=1；

1s’-complement—invertthen-bitbinaryequivalentbitbybit，getthen-bit1s’-comp.；

2’s-complement—add1totheLSBofthen-bit1s’-complementofthegivennumberExp13：FindingtheS-M,2’s-comp,1s’-compofthefollowingsignednumberin8-bit。+60，-60，+10010，-110132.5.SignextensionWhenweconvertann-bit2’scomplementnumberXintoanm-bitone:（a）ifm>n，append(m-n)copiesofX’ssignbittotheleftofX;（b）ifm<n，discardX’s（n-m）leftmostbit。Sodoto1s’-complement.33.2.6Two’s-complementAdditionandSubtraction1、RulesTwooperandsaddorsubtractdirectly

。

Exp：

3+3，4+(-5)，7-3，1-6，符號(hào)數(shù)表格34.2.AgraphicalviewModular:thebiggestnumberofquantitiesthatan-bitsystemcanrepresentis2n.Modularoperation：mMOD2n=m-i·2n（i=int（））

Exp：18MOD16=276543210-1-2-3-4-5-6-7-80111011001010100001100100001000011111110110111001011101010011000+235.3.overflowIfanadditionoperationproducesaresultthatexceedstherangeofthenumbersystem,overflowissaidtooccur.thatis,ifresult>+(2n-1-1)，or<-(2n-1)，overflowisoccurred.Detectingofoverflow：

Anadditionoverflowsiftheaddends’signsarethesamebutthesum’ssignisdifferentfromtheaddends’.Exp:judgewhethertheresultoftheadditionoverflowornot。11111101+1000000136.4.Subtractionrulesm-n=m+(2n-n)Negatethesubtrahendbytakingits2’scomplement,andthenaddittotheminuendusingthenormalrulesforaddition.

2’s-comp.ofn76543210-1-2-3-4-5-6-7-80111011001010100001100100001000011111110110111001011101010011000-2+1437.Exp：8-bit2’s-comp.38.5.Two’s-complementandUnsignedBinaryNumbersWhenn-bitbinarynumbersaretakenforunsignednumber，therulesofadditionandsubtractionareassameasthe2’s-complement.Iftheresultofadditionoperationexceedtherangeofn-bitsystem,acarryisproducedtotheleftmoresignificantbit.39.2.10BinaryCodesforDecimalNumbersEmphasis:BCD,excess-3code:asetofn-bitinwhichdifferentbitstringsrepresentdifferentnumbersofotherthingsiscalledacode.codeword:aparticularcombinationofnbit-values.1、BCDcodethedigits0~9areencodedby4-bitunsignedbinaryrepresentations,0000through1001.andthewords1010~1111arenotused.(invalidcodeword)40.BCD’spropertyBCDisweightedcode，theweightsforthebitsfromlefttorightis：23、22、21、20。alsocalled8421BCD。(1)representingnumbersbyBCDlike，11d，BCD—00010001 256d, BCD–001001010110(2)PackedBCDtwo-digits-per-bytedecimaldigits8421BCD00000100012001030011401005010160110701118100091001Donotdiscardthese041.(3)AdditionwithBCDbesimilarto4-bitunsignedbinarynumberaddition.Butacorrectionmustbemadeifthesumexceeds1001if，sum>1001，thenadd0110totheresult。Exp.：>1001makeacorrection42.2.OtherdecimalcodesP.49table2-9，Excess-3code：th