class-exercises1、1011100010112=

8=

162、(156)10=()23、convert0.3910toanumberinradix2.Theprecisionmustachieve10%。

tenminutesclass-exercises1、1011100010112=

5613

8=B8B

162、(156)10=(10011100)23、將0.3910轉換為二進制數(shù),要求精度達到10%。

0.01102ReviewofthelastlessonPositionalNumberSystemsMSD，LSD，MSB，LSBGeneralPositional-Number-SystemConversionsAdecimalfraction

AnumberinBinaryradixBinaryadditionandsubtractionBinarynumberswithcertainwidth;Thenumbersalwaysbesetas0.xxxxxxxx;Wecantakethesenumbersasintegersforoperations!MSB/LSB

UnsignednumberOperationsforunsignednumbers0+0=01+0=11+1=10AdditionC:carryS:sum

OperationsforunsignednumbersAddition:Overflow?Wrong!Right!0×A=01×A=AMultiplicationShift-add:useaddersandshift-registers;Theproductwidthwillbe2n!OperationsforunsignednumbersNewruledifferfromaddMaybegetnegativenumbersubtraction

ShouldusesignednumbertoexpresspositiveandnegativeOperationsforunsignednumbersOncethesignednumbersbeused,subtractioncanbemadebyadd!OperationsforunsignednumbersSubtractionruleisnotnecessary!2.5RepresentationofNegativeNumbers

(負數(shù)的表示)2.5.1Signed-MagnitudeRepresentationP34[符號–數(shù)值表示法（原碼）]2.5.2ComplementNumberSystemsP35(補碼數(shù)制)2.5.3Radix-ComplementRepresentation

P36(基數(shù)補碼表示法)2.5.4Two’s-ComplementRepresentation

P37(二進制補碼表示法)2.6Two’s-ComplementAdditionandSubtractionP39

(二進制補碼的加減法)Table2-6

P40(十進制數(shù)與4位二進制數(shù)對照表)2.6.3Overflow

(溢出)

P412.5.1

Signed-MagnitudeRepresentationanumberconsistsofamagnitudeandasymbolindicatingwhetherthemagnitudeispositiveornegative.

MSBastheSignbit(0=plus,1=minus)[最高有效位表示符號位（0=正，1=負）]01111111＝＋12711111111＝－12700101110＝＋4610101110＝－4600000000＝＋010000000＝－02.5.1Signed-MagnitudeRepresentation[符號–數(shù)值表示法（原碼）]TwopossiblerepresentationsofZero[零有兩種表示（+0、–0）]2.Ann-bitsigned-magnitudeintegerrangeis(n位二進制整數(shù)表示范圍)：

–(2n-1–1)~+(2n-1–1)Thesigned-magnitudesystemhasanequalnumberofpositiveandnegativeintegers.

P35representtheresultwith8-bitsigned-magnitudeinteger!110-110=?thesigned-magnitudesystemnegatesanumberbychangingitssign.

ComplementNumberSystemsitnegatesanumberbytakingitscomplementasdefinedbythesystem.

2.5.22.5.2ComplementNumberSystems(補碼數(shù)制)radix–Complement(基數(shù)補碼)2.DiminishedRadix–Complement[基數(shù)減1補碼(反碼)]2.5.2ComplementNumberSystems(補碼數(shù)制)ann-bitnumberDD=dn–1dn–2···d1d0

.Theradixpointisontherightandsothenumberisaninteger.Ifanoperationproducesaresultthatrequiresmorethanndigits,wethrowawaytheextrahighorder

digit(s).IfanumberD

iscomplementedtwice,theresultis

D.(P35)

2.5.3Radix-ComplementRepresentationThecomplementofann-digitnumberisobtainedbysubtractingitfromrn.

r’scomplement=rn-D(n位數(shù)D的基數(shù)補碼等于從rn

中減去該數(shù))Example:Table2-4P362.5.3Radix-ComplementRepresentation2.5.3Radix-ComplementRepresentation

r’scomplement=rn-D

IfDisbetween1andrn

–1,thissubtractionproduces

anothernumberbetween1andrn-1.

whatistheresultofthesubtraction,IfDis0?2.5.3Radix-ComplementRepresentation

r’scomplement=rn-D

IfDisbetween1andrn

–1,thissubtractionproduces

anothernumberbetween1andrn-1.

whatistheresultofthesubtraction,IfDis0?0000000…0(n-bit)

(positive)thereisonlyonerepresentationofzeroinaradix-complementsystem.

DiminishedRadix–ComplementRepresentation

[基數(shù)減1補碼表示法（反碼）]

TheDiminishedRadix–Complementofann-digitnumberisobtainedbysubtractingitfromrn-1

[n位數(shù)的反碼等于從rn–1中減去該數(shù)]

Example:Table2-4,2-5P.36(r-1)’sComplement=rn–1-D

*2.5.6(P38)

therearetwo

representationsofzero,positivezero(00×××00)andnegativezero(11×××11).

2.5.3Radix-ComplementRepresentation2.5.3Radix-ComplementRepresentationAdvantagern–D=[(rn-1)-D]+1ThiscanbeaccomplishedbycomplementingtheindividualdigitsofD,2.5.4Two’s–ComplementRepresentation

(二進制補碼表示法)P37Two’s-Complement(二進制補碼的求取)：MSB

(thesignbit):1=minus;0=plus

WeightoftheMSB:-2n-13.Therangeofrepresentablenumbersis

-(2n-1)through+(2n-1-1).

Two’s-Complement

(二進制補碼的求取)Example1.Writethe8-bittwo’s-complementrepresentationforthedecimalnumber:-119.(若約定字長是一個字節(jié)，試求－11910的補碼表示。)+11910＝011101112，asformula(公式):2n-D=(2n-1-D)+128-1：11111111subtract(減去)+119;-0111011110001000

plus(加)1：+1

－11910：100010012Two’s-Complement

(二進制補碼的求取)Example1.Writethe8-bittwo’s-complementrepresentationforthedecimalnumber:-119.(若約定字長是一個字節(jié)，試求－11910的補碼表示。)+11910＝011101112，

Two’s-Complement

(二進制補碼的求取)Example1.Writethe8-bittwo’s-complementrepresentationforthedecimalnumber:-119.(若約定字長是一個字節(jié)，試求－11910的補碼表示。)+11910＝011101112，

100010002+1100010012=-11910帶符號位一起按位取反再+1,得到相反數(shù)的補碼.Table2-6Decimaland4-bitnumbers(P40)DecimalSigned-MagnitudeOne’scomplementTwo’scomplement-8————1000-7111110001001-6111010011010-5110110101011-4110010111100-3101111001101-2101011011110-110011110111101000或00001111或000000001000100010001200100010001030011001100114010001000100501010101010160110011001107011101110111SumupfortheComplement

(總結)Positivenumberhasthesame：

Sign-Magnitude,Ones’–Complement,andTwo’s-Complement(正數(shù)的原碼、反碼、補碼相同)

SumupfortheComplement

(總結)

ComplementNumberSystemssigned-magnitudesystem010001101111(010001)(110001)+1710-1710SumupfortheComplement

(總結)

ComplementNumberSystemssigned-magnitudesystem(010001)(101111)(010001)(110001)+1710-1710不變+1710-1710符號位改變符號位不變其余按位取反加1.連同符號位一起按位取反加1.連同符號位一起按位取反加1.符號位改變Signextensionforatwo’scomplementnumber

(符號位擴展)Wecanconvertann-bittwo’s-complementnumberXintoanm-bitone,butsomecareisneeded.Ifm>n,wemustappendm-ncopiesofX’ssignbittotheleftofX.Thatis,wepadapositivenumberwith0sandanegativeonewith1s;thisiscalledsignextension.Ifm<n,wediscard

X’sn–mleftmostbits;however,theresultisvalidonlyifallofthediscardedbitsarethesameasthesignbitoftheresult.ExampleforSignextensionPleaseconvertthesetwo’scomplementnumbers01111and1001into8-bit.01111=000011111001=11111001NOTE:theresultofextensionisintheTwo’s-complementnumbersystem.2.6Two’s–ComplementAdditionandSubtraction

(二進制補碼的加法和減法)wedefine[x]tobethetwo’scomplementrepresentationofx

[x+y]=([x]+[y])[x-y]=([x]+[-y])

2.6Two’s–ComplementAdditionandSubtraction

(二進制補碼的加法和減法)

[x+y]=([x]+[y]);[x-y]=([x]+[-y])1101+1010=?1101+1101=?

OVERFLOE?000000010010001101000101100010011010110111111110101111000111011001234589101315141112764位無符號二進制數(shù)0000000100100011010001011000100110101101111111101011110001110110+0+1+2+3+4+587631254+7+64位二進制補碼Range:-8～+7

Wecanadded+ntothatnumberbycountingupntimes,thatis,bymovingthearrownpositionsclockwise.

Wecansubtractnfromanumberbycountingdownntimes,thatis,bymovingthearrownpositionscounterclockwise..Theresultwillalwaysbecorrectsumaslongastherangeofthenumbersystemisnotexceeded.Modularcounting(模計算)-nWhatismostinterestingisthatwecan