数字设计课件第二章.ppt

N-bitpositiveintegersarerepresentedinthesamewayasn-bitsign-magnitudenotation.Theones’-complementofann-bitnegativeintegernumberisobtainedbycomplementingeachoneofthebits(then-bitbinarynumber),i.e.,a1isreplacedbya0,anda0isreplacedbya1.Exp.8-bit1s’-complementofthenumber.+18d=-18d=1scomplementof18=decimalS-M2’scomp.1s’comp.-1100111111110-2101011101101-3101111011100-4110011001011-5110110111010-6111010101001-7111110011000-8-1000-decimalS-M2’scomp.1s’comp.7011101110111601100110011050101010101014010001000100300110011001120010001000101000100010001000001000000011110000Fromtheleastnumbertobiggestnumber,2’scomp.and1s’comp.aresuccessiveincreasedby1.Allofthesethreeareusedtorepresentsignedintegernumberinbinarysystem.ComparingoftherepresentingrangeofthevalueRepresentationsofpositiveintegeraresame;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）~-0①positivesignednumber：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，-1101Whenweconvertann-bit2’scomplementnumberXintoanm-bitone:ifmn，append(m-n)copiesofX’ssignbittotheleftofX;ifmn，discardX’s（n-m）leftmostbit。Sodoto1s’-complement.RulesTwooperandsaddorsubtractdirectly。Exp：3+3，4+(-5)，7-3，1-6，12数字逻辑设计及应用数字逻辑设计及应用数字逻辑设计及应用数字逻辑设计及应用*数字逻辑设计及应用*Chapter2