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1、2023/9/11,数字逻辑设计及应用,1,Chapter2 Number Systems and Code,重点：1）二进制、八进制和十六进制数的表达及各种数制之间的相互转换；2）符号数的表达及其加、减运算；3）BCD编码和GRAY码。,数字逻辑设计及应用,2,2.1 Positional Number Systems,1.Decimal number A decimal number 1536.79d digits：0、1、2、3、4、5、6、7、8、9weight：10iradix：10power：i，start from the left of the decimal point,in
2、crease by one for each successive position from 0;from the right of the decimal point,decrease by one from(-1).,数字逻辑设计及应用,3,An arbitrary decimal number D with p digits in integer and n digits in fractiondp-1dp-2d1d0.d-1d-2d-nGeneral expression：,2.Binary numbersradix：2digits：0、1，also be called bit，we
3、ight：2i,数字逻辑设计及应用,4,A binary number 1011010.1012，：,an arbitrary binary number Bbp-1bp-2b1b0.b-1b-n，the general expressions：The sum of each digit multiplied by the corresponding power of the radix is the binary numbers decimal value.,数字逻辑设计及应用,5,3.Others LSB:Least Significant Bit MSB:Most Significant
4、 Bit bp-1bp-2b1b0.b-1b-n,MSB,LSB,数字逻辑设计及应用,6,2.2 Octal and Hexadecimal Numbers,1.concept,数字逻辑设计及应用,7,Exp3：give the decimal value of the following numbers.215.78=3AD.8h=2.2-8 and 2-16 numbers Convensions,Table 1 relation between binary and octal,数字逻辑设计及应用,8,Table 2 binary-hexadecimal relations,数字逻辑设计
5、及应用,9,（a）binaryoctal、hexadecimal conversions,Starting at the binary point，separate the bits into groups of 3 or 4,and replace each group with the corresponding octal or hexadecimal digit.If the leftmost group fall short of 3 or 4 bits,then 0 should be added to the left of the MSB;so do the right。Exp
6、：do the following positional-number-system conversion。1111011010.00112=(?)8=(?)hSolution:=1732.148=3DA.3h,数字逻辑设计及应用,10,(b)octal,hexadecimal-binary number conversion,Replace each octal or hexadecimal with the corresponding group of 3 or 4bits.Exp：267.248=（？）2 96CA.28h=（？）2,数字逻辑设计及应用,11,2.3 general po
7、sitional-number-system conversions,Radix-r-to-decimal conversion:Each radix-r numbers digit multiply by its own weight,and add them.Exp：512.46=（？）d1、decimal-to-radix-r conversioninteger partmethods：除基取余The integer of a p digits number in radix r is：Divide the formula by r successively,yield the succ
8、essive digits of the integer of D(radix r)from the right to left till the quotient is 0.,数字逻辑设计及应用,12,Exp：371d=（?）2=（?）8=（?）h2、decimal-to-radix-r conversionfraction partmethods：乘基取整The fraction of a p digits number in radix r is：Multiply the formula by r successively,yield the successive digits of t
9、he fraction of D(radix r)from the left to right till the required digits d-1，d-2，d-n are acquired.,数字逻辑设计及应用,13,Exp8：0.71875d=(?)2=(?)8=(?)h3、summary of the positional-number-system conversions（1）conversion between decimal and radix r（2）radix-r-to-radix-j conversion(non-decimal)binary-to-octal,binar
10、y-to-hexadecimal or reverse,be converted directly conforming to the section-replace rules.others,radix-r numberdecimal radix-j number,数字逻辑设计及应用,14,2.4 addition and subtraction of nondecimal number,1、Rules of binary addition and subtraction（1）rules of additionS=x+y+CinExp：01001011+10001111=?,initial
11、carry,carry,数字逻辑设计及应用,15,（2）Rules of Binary Subtraction,d=x-y-binExp：11001100-01011100=?,initial borrow,borrow,数字逻辑设计及应用,16,2.addition and subtraction of octal and hexadecimal,addition:If two digits sum is greater than the radix on each column,then carry 1 to the next more significant bit.subtractio
12、n:If the minuend is less than subtrahend,then borrow 1 from the next more significant bit.,数字逻辑设计及应用,17,2.5 Representation of Negative Numbers,1、signed-magnitude representation unsigned numbers：just the magnitude of a number is represented,no+or-sign symbol before the number。signed numbers：+or-is ad
13、ded to the left of the number。representation of signed number：sign symbol+number magnitude，like+34d、-1102、+1Dh、,数字逻辑设计及应用,18,in binary number,sign bit+magnitudeThe MSB of a bit string is used as the sign bit：0“+”，1“-”this type of signed binary number is called Signed Magnitude（S-M码，或原码）。比如，+11101=01
14、1101-1011=11011,数字逻辑设计及应用,19,The range of S-Ms representable numbers.an n-bit S-M number B:bn-1bn-2b1b0Its range is：-（2n-1-1）+(2n-1-1)Include two zero:-0 and+0Exp：find the S-M of the following signed numbers。How to represent by using 8-bit S-M.+11101，-1011，+18，-18the fault of S-M which is used in ar
15、ithmetical operation：it cant be calculated directly.,符号位,数字逻辑设计及应用,20,2、Complement Number System,radix complement（基数补码数制）diminished radix-complement(基数减1补码数制)An n-digit number D in radix r:D=dn-1d1d0，radix-complement diminished radix-complement（也称基数反码）Relation between these two：Table 2-4,2-5(P.36)sh
16、ows some rs and(r-1)s complement.,数字逻辑设计及应用,21,3.Twos-complement Representation,Twos complement is a method for representing signed integers as binary numbers.n-bit binary number B=bn-1b1b0，the twos complement is:B 2s=2n-B for Exp.,8-bit binary number twos complement0000000028-0=000000000000000128-1
17、=111111110000001028-10=111111101111111100000001,数字逻辑设计及应用,22,Define these twos complement as signed number,the MSB is served as sign bit。MSB=0，positive number MSB=1，negative numberweight of the sign bit：MSB=0,weight+2n-1 MSB=1,weignt-2n-1Range of the value：positive 0+（2n-1-1）；negative（-2n-1）-1。Only
18、one zero in twos complement.,数字逻辑设计及应用,23,twos complement number and their decimal equivalent,数字逻辑设计及应用,24,数字逻辑设计及应用,25,property,sign bit=0，S-M and twos-complement have the same decimal value；sign bit=1，S-M and twos-complement have different decimal value。How to find the twos-complement representati
19、on of a negative decimal number?,数字逻辑设计及应用,26,Same value,but has opposite sign symbol,数字逻辑设计及应用,27,So,to calculate the 2s complement of an negative integer,invert the n-bit binary equivalent of the given number magnitude bit by bit,and then add 1 to the LSB.Exp:calculate the 2s complement of(+65)and
20、(-65d)in 8-bit form.Solution:+65 01000001-65 binary equivalent of 65 is 01000001,invert bit by bit 10111110 add 1 10111111,数字逻辑设计及应用,28,An n-bit binary number B=bn-1b1b0，the ones-complement is:B 1s=2n-1-BIt is also the method of representing signed binary numbers.MSB=0，positive,weight+(2n-1-1)MSB=1，
21、negative，weight-(2n-1-1)。Range of representable numbers：negative-（2n-1-1）-0positive+0（2n-1-1）,4.Ones-complement Representation,数字逻辑设计及应用,29,1s-complements property,N-bit positive integers are represented in the same way as n-bit sign-magnitude notation.The ones-complement of an n-bit negative intege
22、r number is obtained by complementing each one of the bits(the n-bit binary number),i.e.,a 1 is replaced by a 0,and a 0 is replaced by a 1.Exp.8-bit 1s-complement of the number.+18d=00010010-18d=1s complement of 18=11101101,数字逻辑设计及应用,30,返回,From the least number to biggest number,2s comp.and 1s comp.
23、are successive increased by 1.,Summary of S-M,2s comp.,1s comp.,数字逻辑设计及应用,31,All of these three are used to represent signed integer number in binary system.Comparing of the representing range of the value Representations of positive integer are same;but the representations of negative integer are d
24、ifferent at all.,数字逻辑设计及应用,32,Calculate signed integers S-M,2s-comp.,1s-comp.,positive signed number：convert the given number into the wanted n-bit binary equivalent.negative signed number：first convert the number into n-bit binary equivalent,S-M let MSB=1；1s-complement invert the n-bit binary equiv
25、alent bit by bit，get the n-bit 1s-comp.；2s-complement add 1 to the LSB of the n-bit 1s-complement of the given numberExp13：Finding the S-M,2s-comp,1s-comp of the following signed number in 8-bit。+60，-60，+10010，-1101,数字逻辑设计及应用,33,5.Sign extension,When we convert an n-bit 2s complement number X into a
26、n m-bit one:（a）if m n，append(m-n)copies of Xs sign bit to the left of X;（b）if m n，discard Xs（n-m）leftmost bit。So do to 1s-complement.,数字逻辑设计及应用,34,2.6 Twos-complement Addition and Subtraction,1、RulesTwo operands add or subtract directly。Exp：3+3，4+(-5)，7-3，1-6，,符号数表格,数字逻辑设计及应用,35,2.A graphical view,M
27、odular:the biggest number of quantities that a n-bit system can represent is 2n.Modular operation：m MOD 2n=m-i2n（i=int（）Exp：18 MOD 16=2,+2,数字逻辑设计及应用,36,3.overflow,If an addition operation produces a result that exceeds the range of the number system,overflow is said to occur.that is,if result+(2n-1-
28、1)，or-(2n-1)，overflow is occurred.Detecting of overflow：An addition overflows if the addends signs are the same but the sums sign is different from the addends.Exp:judge whether the result of the addition overflow or not。11111101+10000001,数字逻辑设计及应用,37,4.Subtraction rules,m-n=m+(2n-n)Negate the subtr
29、ahend by taking its 2s complement,and then add it to the minuend using the normal rules for addition.,-2,数字逻辑设计及应用,38,Exp：,数字逻辑设计及应用,39,5.Twos-complement and Unsigned Binary Numbers,When n-bit binary numbers are taken for unsigned number，the rules of addition and subtraction are as same as the 2s-co
30、mplement.If the result of addition operation exceed the range of n-bit system,a carry is produced to the left more significant bit.,数字逻辑设计及应用,40,2.10 Binary Codes for Decimal Numbers,Emphasis:BCD,excess-3code:a set of n-bit in which different bit strings represent different numbers of other things i
31、s called a code.code word:a particular combination of n bit-values.1、BCD code the digits 09 are encoded by 4-bit unsigned binary representations,0000 through 1001.and the words 1010 1111 are not used.(invalid code word),数字逻辑设计及应用,41,BCDs property,BCD is weighted code，the weights for the bits from le
32、ft to right is：23、22、21、20。also called 8421BCD。(1)representing numbers by BCDlike，11d，BCD00010001256d,BCD 0010 0101 0110(2)Packed BCDtwo-digits-per-byte,数字逻辑设计及应用,42,(3)Addition with BCD be similar to 4-bit unsigned binary number addition.But a correction must be made if the sum exceeds 1001 if，sum1
33、001，then add 0110 to the result。Exp.：,数字逻辑设计及应用,43,2.Other decimal codes,P.49 table2-9，Excess-3 code：the code word for each decimal digit is the corresponding BCD code word plus 0011.it is a self-complement code.2421 code：weighted code,each bit of the code word has assigned weight with 2,4,2 and 1 f
34、rom left to right。self-complement,数字逻辑设计及应用,44,2.11 Gray code,3-bit mechanical encoding diskstraightforward binary encoding(natural binary codes)two successive values differ in only one bit,000,001,010,011,100,101,110,111,数字逻辑设计及应用,45,Constructing an n-bit gray code,1-bit gray code:0,1The first 2n c
35、ode words of an(n+1)-bit Gray code equal the code words of an n-bit Gray code,written in order with a leading 0 appended.The last 2n code words of an(n+1)-bit Gray code equal the code words of an n-bit Gray code,but written in reverse order with a leading 1 appended.,数字逻辑设计及应用,46,Exp：reflected binary code,数字逻辑设计及应用,47,translating a binary value into the corresponding Gray code,Binary number bn-1bn-2b1b0Gray code gn-1gn-2g1g0Then gn-1=bn-1 bit i of a Gray-code word is 0 if bits i and i+1 of the corresponding binary code word are the same,else bit i is 1.101102=（）grey解：B 1 0 1 1 0 G 1 1 1 0 1,
