2-1数制

nullChapter 2 Number Systems and codes (数制和编码) P25Chapter 2 Number Systems and codes (数制和编码) P25Digital system are build from circuit that process binary digits -0s and 1s； Real-life numbers,event,and conditions binary digits Binary number system and binary arithmetic Encode Keystone 2.1,2.2,2.3,2.5,2.6,2.10,2.11,2.16Introduction To Chapter 2:Digital system are build from circuit that process binary digits -0s and 1s； Real-life numbers,event,and conditions binary digits Binary number system and binary arithmetic Encode Keystone 2.1,2.2,2.3,2.5,2.6,2.10,2.11,2.162.1 Positional Number Systems (p26)2.1 Positional Number Systems (p26)Positional Number Systems: a number is represented by a string of digits where each digit position has an associated weight. （按位计数制：按各个数码的位置规定了该数所具有的数值．） For example：a Decimal number 　１９８５ 10 is called the base or radix of the number system. (10是记数制的基数；)the radix may be any integer r≧2. Digital position（位序号） from left to right is :3、2、1、0； position i has weight 10i 。(位序号为i位的权是10i)null1. Decimal Number D Can Be Represented as the Following (任意十进制小数 D 可表示如下):D = dp-1 dp-2 ... d1 d0 . d-1 d-2 ... d-nThe value of the number is the sum of each digit multiplied by the corresponding Power of the radix. null1. Decimal Number D Can Be Represented as the Following (任意十进制小数 D 可表示如下):D = dp-1 dp-2 ... d1 d0 . d-1 d-2 ... d-n推广： B2 = ∑ b i × 2i H16= ∑ hi × 16i nullD = dp-1 dp-2 ... d1 d0 . d-1 d-2 ... d -n10进制r=10；共有10个数码，0、1、2、3、4、5、6、7、8、9；逢10进1。 二进制r=？，共有？个数码，逢？进1 ； 8进制r=？， 共有？个数码，逢？进1 ； 16进制r=？，共有？个数码，逢？进1。 。 Example:00185.6300=185.63 Most Significant digit （MSD, 最高有效数字） -the leftmost digit dp-1 Least Significant digit （LSD, 最低有效数字） -the rightmost digit d -n 2、Binary Number(二进制) P26 2、Binary Number(二进制) P26Binary digits(bits)：0、1 Radix：2 weight:2i Most Significant bit （MSB, 最高有效位） -the leftmost bit Least Significant bit（LSB, 最低有效位） -the rightmost bitExample:（1011101）2 = （1×26+0×25+1×24+1×23+1×22+0×21+1×20）10 =（64+0+16+8+4+0+1）10 =（93）10NOTES:NOTES:bit（位） 1 byte(1个字节)=8bits MEMORY:256M=256M Bytes 1KB=1024 Bytes 1MB=1024*1024 Bytes 1GB=1024*1024*1024 Bytes There are only 10 types of people in the world:Those who understand binary and those who don’t.The represent of Binary number ： (二进制计数的表示)The represent of Binary number ： (二进制计数的表示)0、1 00、01、10、11 000、001、010、011、100、101、110、111 0000、0001、0010、0011、0100、0101、0110、0111、1000、1001、1010、1011、1100、1101、1110、1111What is the range of representable number,if you have n-bit binary number ?The represent of Binary number ： (二进制计数的表示)The represent of Binary number ： (二进制计数的表示)0、1 00、01、10、11 000、001、010、011、100、101、110、111 0000、0001、0010、0011、0100、0101、0110、0111、1000、1001、1010、1011、1100、1101、1110、1111What is the range of representable number,if you have n-bit binary number ? 0~2n-1 3. Octal Number(八进制)P273. Octal Number(八进制)P27Octal digits： 0~7 (0,1,2,3,4,5,6,7) Radix：8 (powers of 2= 23) Weight:8i eg. （127）8=（1×82+2×81+7×80）10 =（64+16+7）10 =（87）10 4. Hexadecimal Number(十六进制)P274. Hexadecimal Number(十六进制)P27Hexadecimal digits： 0~F (0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F) Radix：16(powers of 2= 24) weight:powers of 16 ,16ieg. （5D）16=（5×161+13×160）10 =（80+13）10 =（93）10 nullD = dp-1 dp-2 ... d1 d0 . d-1 d-2 ... d -n10进制r=10；共有10个数码，0、1、2、3、4、5、6、7、8、9；逢10进1。 二进制r=2，共有2个数码，0 ，1, 逢2进1 ；0-1-10-11-100……. 8进制r=8， 共有8个数码，0－－7，逢8进1 ；0-1-2-3-4-5-6-7-10－－-17…. 16进制r=16，共有16个数码，逢16进1。0-1-2-3-4-5-6-7-8-9-A-B-C-D-E-F- 10---1F,…… Most Significant digit （MSD, 最高有效数字） -the leftmost digit dp-1 Least Significant digit （LSD, 最低有效数字） -the rightmost digit d -n2.3 General Positional-Number-System Conversions (常用数制的转换)P292.3 General Positional-Number-System Conversions (常用数制的转换)P29A number in any radix  a number in radix 10 method： (利用位权展开) Example 1：( 101.01 )2 = ( )10 ( F1AC )16 = ( )10 Example 1： ( 101.01 )2 = (1 22 + 1 20 +1 2-2 )10=5.2510 ( F1AC )16 = (15 163 + 1162+10161+12160)10=6186810 D = dp-1 dp-2 ... d1 d0 . d-1 d-2 ... d -nnull Example 2： ( F1AC )16 = (15 163 + 1 162 +10161 +12160)10=6186810 Shortcut(捷径) for covering whole numbers to radix 10. ( F1AC )16 = ( ( ( 15)  16 +1 )  16 + 10)  16 + 12=6186810 Shortcut(捷径) for covering a decimal number D to a radix r.61868÷16,the quotient (商)will be(( 1516 ) +1 )  16 +10 ), the remainder 余数 is 12(C);(LSB) Next:( 1516 ) +1 )  16 + 10) ÷16, the quotient will be ( 1516 ) +1 ) , the remainder is 10(A); Again: ( 1516 ) +1 ) ÷16, the quotient will be ( 1516 ), the remainder 余数 is 1; Finally: ( 1516 )÷16, the result is 15(F).(MSB)nulla number in radix 10  A number in Binary radix (十进制数转换成二进制) Example：perform the following number system conversions: （217）10 =（　　　　）2 Principle of the integer conversion:successive divisions by 2 yield successive digits of D from right to left,until all the digits of D have been derived.(P30) 整数部分的转换：除以2取余,倒着数。 nulla number in radix 10  A number in Binary radix Example1：perform the following number system conversions: （217）10 =（　　　　）2 ∵ 2∣217 ………… ...remainder 1 b0 (LSB) 2∣108 ………….remainder 0 b1 2∣54 ………….remainder 0 b2 2∣27 ………… remainder 1 b3 2∣13 ………… remainder 1 b4 2∣6 ………… remainder 0 b5 2∣3 ………… remainder 1 b6 2∣1 ………… remainder 1 b7 0∴（217）10 =（11011001）2nullDecimal to binary For integer numbers : divided by 2, get its remainders;Examples for 8-bit binary numbers:217—108 —54—27—13—6—3—1 1 0 0 1 1 0 1 1nullExample2:（0.3125）10 =（　　 　　）2 ∵0.3125 × 2 = 0.625 …………整数为0 b- 1 0.625 × 2 = 1.25 …………整数为1 b- 2 0.25 × 2 = 0. 5 …………整数为0 b- 3 0. 5 × 2 = 1.0 …………整数为1 b- 4 说明：有时可能无法得到0的结果，这时应根据转换精度的要求适当取一定位数。Additional content: (十进制小数转换成二进制) A decimal fraction  A number in Binary radix 小数部分的转换：乘2取整,正着数。∴（0.3125）10 =（0.0101）2nullFor decimal fraction numbers : multiplied by radix, and get its integers; Examples for 8-bit binary numbers:0.17—0.34 —0.68—1.36—0.72—1.44—0.88—1.76—1.52Decimal to binary Analyze 分析 Analyze 分析Because 210=1024;(1KB) >103; So 10-3 >2-10 a binary number which has p digits to the left of the radix point and n to the right. If the precision (精度)must achieve 2-10, then n≥10. If the precision must achieve 10-2,because 27>102>26,so 2-6>10-2>2-7,then n ≥7. If the precision must achieve 0.01%,because 0.01%= 10-4, 10-4约为2-13.3,then n ≥14. B = bp-1 bp-2 ... b1 b0 . b-1 b-2 ... b -nExamples for the conversion of the decimal fraction :Examples for the conversion of the decimal fraction :example 1:convert 0.3910 to a number in radix 2.The precision must achieve 10%. (例1、将0.3910转换为二进制数,要求精度达10%.) example 2:convert 0.3910 to a number in radix 8.The precision must achieve 0.1%.example 1:convert 0.3910 to a number in radix 2.The precision must achieve 10%. (例1、将0.3910转换为二进制数,要求精度达10%.) answer: 2-4=1/16< 10%<2-3 So the binary number must has 4 bits to the right. ∵0.39 × 2 = 0.78 …………整数为0 b- 1 0.78 × 2 = 1. 56 ……… 整数为1 b- 2 0.56 × 2 = 1. 12 ……… 整数为1 b- 3 0.12 × 2 = 0.24 …………整数为0 b- 4 ∴（0.39）10 =（0.0110）2 example 2:convert 0.3910 to a number in radix 8.The precision must achieve 10%. (例1、将0.3910转换为8进制数,要求精度达10%.) answer: 8-2< 10%<8-1 So the octal number must has 2 bits to the right. ∵0.39 × 8 = 3.12 …………整数为3 b- 1 0.12 × 8 = 0. 96 ……… 整数为0 b- 2 ∴（0.39）10 =（0.30）8 Examples for the conversion of the decimal fraction :Examples for the conversion of the decimal fraction :example 1:convert 0.3910 to a number in radix 2.The precision must achieve 2-10. (例1、将0.3910转换为二进制数,要求精度达2-10. ) example 2:convert 0.3910 to a number in radix 8.The precision must achieve 0.1%.2. Convert a Octal number to binary, Convert a Binary number to Octal, ( P27)2. Convert a Octal number to binary, Convert a Binary number to Octal, ( P27)（1）replace each octal digit with the 3-bit string. 　　 三位二进制数对应一位八进制数。（101011100101）2 　　　　　=（101，011，100，101）2 　　　　　=（5345）8 (P27) （6574）8 =（110，101，111，100）2 　　　　　=（110101111100）2 (P28)Radix：8(powers of 2= 23)3. Convert a hexadecimal number to binary Convert a binary number to hexadecimal3. Convert a hexadecimal number to binary Convert a binary number to hexadecimalExample, （9A7E）16 =（1001 1010 0111 1110）2 =（1001101001111110）2四位二进制数对应一位十六进制数。（10111010110）2 =（0101 1101 0110）2 =（5D6）16Radix：16(powers of 2= 24)QUESTION:QUESTION:What are the octal values of the four 8-bit bytes in the 32-bit number with octal representation 123456701238? 2nibbles=1byte=8bits Table 2-1 (P28)Binary,decimal,octal,and hexadecimalnumber Table 2-1 (P28)Binary,decimal,octal,and hexadecimalnumberThink about……Think about……What can you find out? What is AD and DA? 2.4 Binary addition and subtraction 二进制的加减法 (P32)2.4 Binary addition and subtraction 二进制的加减法 (P32)arithmetic operation:两个二进制数的算术运算 addition加法： Carry (进位) 1 + 1 = 10 subtraction减法： Borrows （借位） 10 – 1 = 1 Carry input（进位输入): Cin Carry output(进位输出): Cout Sum (本位和 ): S Borrow input (借位输入): bin Borrow output(借位输出): bout Difference bit(本位差): d EXAMPLES: P(32),P(33)EXAMPLES: P(32),P(33)nullExercisesExercisesnullHome work P(74)1、2.1 (a), (d) (e) (i) (j) 2、2.2(f) 3、2.3(f) 4 、2.4 5、2.5(g) 6、2.6(a) 7、2.7 (a ),(c) 8、think about…… Please hand your home work on next Monday!