數(shù)字電路英文版PPT第二單元

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1、 2: NUMBER SYSTEMS, OPERATIONS, AND CODESLSB 最 低 位 MSB 最 高 位 Byte 字 節(jié) Octal 八 進(jìn) 制 Floating-point number 浮 點(diǎn) 數(shù) Hexdecimal 十 六 進(jìn) 制 BCD 二 - 十 進(jìn) 制 碼 Parity 奇 偶 性 Weight 權(quán) Carry 進(jìn) 位 Remainder 余 數(shù) Quotient 商 Integer 整 數(shù) Fraction 小 數(shù) 1Complement 反 碼 2Complement 補(bǔ) 碼 Format 格 式 Precision 精 度 Mantissa 尾 數(shù) KEY

2、 TERMSAlphanumeric : Consisting of numerals, letters, and other characters.ASCII :American Standard Code for Information Interchange; the most widely used alphanumeric code. BCD : Binary coded decimal; a digital code in which each of the decimal digits, 0 through 9, is represented by group of four b

3、its.Byte : A group of eight bits.Carry : The digit generated when the sum of two binary digits exceeds 1. Complement : The inverse or opposite of a number.Decimal : Describes a number system with a base of ten.Digit : A symbol used to express a quantity.Exponent : The part of a floating-point number

4、 that represents the number of places that the decimal point is to be moved. Floating-point number: A number representation based on scientific notation in which the number consists of an exponent and a mantissa.Gray code : An unweighted digital code characterized by a single bit change between adja

5、cent code number in a sequence. Hexadecimal : Describes a number system with a base of 16.Integer : A whole number.Least significant bit (LSB) : Generally, the right most bit in a binary whole number or code.Mantissa : The magnitude of a floating-point number. Most significant bit (MSB) : The left m

6、ost bit in a binary whole number or code.Octal : Describes a number system with a base of eight.Overflow: The condition that occurs when the number of bits in a sum exceeds the number of bits in each of the numbers added. Parity : In relation to binary codes, the condition of evenness or oddness of

7、the number of 1s in a code group.Sign bit : The left-most bit of a binary number that designates whether the number is positive (0) or negative (1).Weight : The value of a digit in a number based on its position in the number. You are familiar with the decimal number system because you use decimal n

8、umbers every day. Although decimal numbers are commonplace, their weighted structure is often not understood. In this section, the structure of decimal numbers is reviewed. This review will help more easily understand the structure of the binary number system, which is important in computers and dig

9、ital electronics.In the decimal number system each of the ten digits, 0 through 9, represents a certain quantity.Stated anther way: A base10 number system.3. EXAMPLE 2-1 Express the number 47 as a sum of the values of each digit. Solution The digit 4 has a weight of 101, as indicated by its position

10、. The digit 7 has a weight of 1, which is 100, as indicated by its position. 47 = ( 4 X 101) + ( 7 X 100 ) = ( 4 X 10 ) + ( 7 X 1) = 40 + 7 Related Problem Determine the value of each digit in 939. The binary number system is simply another way to represent quantities. The binary system is less comp

11、licated than the decimal system because it has only two digits. It may seem more difficult at first because it is unfamiliar to you. The decimal system with its ten digits is a baseten system; 4. the binary system with its two digits is a basetwo system. The two binary digits ( bits ) are 1 and 0. T

12、he position of a 1 or 0 in a binary number indicates its weight, or value within the number, just as the position of a decimal digit determines the value of that digit. The weights in a binary number are based on powers of two. 5. Counting in Binary Decimal Number Binary Number 0123456781011912 1314

13、15 0 0 00 00 0 10 0 10 110 0 0 1 0 1 0 1 1 0 0 1 1 1 0 1 0 0 1 1 1 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 0 0 0 6. As you have seen in above table, four bits are required to count from zero to 15. In general, with n bits you can count up to a number equal to 2n 1. Largest decimal number

14、 = 2n 1 if n = 5 , 25 1 = 32 1 = 31 if n = 6 , 26 1 = 64 1 = 63 The Weighting Structure of Binary Numbers The right-most bit is the least significant bit in a binary whole number and has a weight of 20 = 1. The left-most bit is the most significant bit; its weight depends on the size of binary numbe

15、r. Binary weight Positive Power of Two Negative Power of Two 28 27 26 25 24 23 22 21 20 2-1 2-2 2-3 2-4 2-5 2-6 256 128 64 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 0.5 0.25 0.125 0.0625 0.03125 0.015625 Binary-to-Decimal Conversion The decimal value of any binary number can be found by adding the we

16、ights of all bits that are 1 and discarding the weights of all bits that are 0. EXAMPLE 2-3 Convert the binary whole number 1101101 to decimal. Weight: 26 25 24 23 22 21 20 Binary number: 1 1 0 1 1 0 1 1101101 = 26+ 25 + 23 + 22 + 20 = 64 + 32 +8 + 4 +1 = 109 Related Problem Convert the binary numbe

17、r 10010001 to decimal. In Section 22 you learned how to convert a binary number to the equivalent decimal number. Now you will learn two ways of converting from a decimal number to a binary number. 7. 1. Sum- of- Weights MethodList of eight binary weight128, 64 , 32, 16, 8, 4, 2, 1 27 , 26, 25, 24,

18、23, 22, 21, 20 9 = 8 + 1 1 0 0 1125 = 64+32+16+8+4+0+1 1 1 1 1 1 0 1 EXAMPLE 2-5 Convert the following decimal numbers to binary: (a) 12 (b) 25 (c) 58 (d) 82 Solution (a) 12 = 8 + 4 = 23 + 22 1100 (b) 25 = 16 + 8 + 1 = 24 + 23 + 20 11001 (c) 58 = 32 + 16 + 8 + 2 = 25 + 24 + 23 + 21 111010 (d) 82 = 6

19、4 + 16 + 2 = 26 + 24 + 21 1010010Related Problem Convert the decimal number 125 to binary. 2. Repeated Division-by-2 Method 2 12 0 LSB 2 6 0 2 3 1 2 1 1 MSB 0 Binary arithmetic is essential in all digital computers and in many other types of digital systems. To understand digital system, you must kn

20、ow the basics of binary addition, subtraction, multiplication, and division. 8. This section provides an introduction that will be expanded in later sections.Remember in binary 1 + 1 = 10, not 2.Remember in binary 10 - 1 = 1 , not 9. Binary Addition The four basic rules for adding binary digits are

21、as follows: 0 + 0 = 0 Sum of 0 with a carry of 0 0 + 1 = 1 Sum of 1 with a carry of 0 1 + 0 = 1 Sum of 1 with a carry of 0 1 + 1 = 10 Sum of 0 with a carry of 1 EXAMPLE 2-7 Add the following binary numbers:(a) 11 + 11 (b) 100 + 10 (c) 111 + 11 Solution (a) 11 3 ( b) 100 4 (c) 111 7 + 11 + 3 + 10 + 2

22、 + 11 + 3 110 6 110 6 1010 10 Related Problem Add 1111 and 1100 Binary Subtraction The four basic rules for subtracting bits are as follows: 0 - 0 = 0 1 - 1 = 0 1 - 0 = 1 10 - 1 = 1 0 1 with a borrow of 1 EXAMPLE 2-8(9) Perform the following binary subtraction:(a) 11 - 01 (b) 11 - 10 (c) 101 - 011 S

23、olution (a) 11 3 ( b) 11 3 (c) 101 5 - 01 - 1 - 10 - 2 - 11 - 3 10 2 01 1 010 2 Related Problem Subtract 100 from 111 . Subtract 101 from 110 . Binary Multiplication The four basic rules for multiplying bits are as follows: 0 0 = 0 0 1 = 0 1 0 = 0 1 1 = 1 Multiplication is performed with binary numb

24、ers in the same manner as with decimal numbers. EXAMPLE 2-10 Perform the following binary multiplications: (a) 11 11 (b) 101 111 Solution (a) 11 3 (b) 111 7 11 3 101 5 11 9 111 35 + 11 000 1001 + 111 100011 Related Problem Multiply 1101 1010 ( 13 10 = 130 ) Binary Division Division in binary follows

25、 the same procedure as division in decimal. Perform the following binary division: (a) 110 11 (b) 110 10 Solution (a) 10 2 11 3 11110 3 6 10110 26 11 6 10 6 000 0 10 0 10 00 Related problem Divide 1100 by 100 ( 12 4 = 3 ) The 1s complement and the 2s complement of a binary number are important becau

26、se they permit the representation of negative numbers. The method of 2s complement arithmetic is commonly used in computers to handle negative numbers. 9. Finding the 1s complement of a Binary number 1 0 1 1 0 0 1 0 Binary number 0 1 0 0 1 1 0 1 1s complement ( to use parallel inverters , NOT gate)

27、Finding the 2s Complement of a Binary Number 2s complement = ( 1s complement ) + 1 EXAMPLE 2-12 Find the 2s complement of 10110010. Solution 10110010 Binary number 01001101 1s complement + 1 Add 1 01001110 2s complement Related Problem Determine the 2s complement of 11001011. An alternative method o

28、f finding the 2s complement of a binary number is as follows: 1. Start at the right with the LSB and write the bits as they are up to and including the first 1. 2. Take the 1s complements of the remaining bits. EXAMPLE 2-13 Find the 2s complement of 10111000 using the alternative method. Solution 10

29、111000 Binary number 1s complements 01001000 2s complement of original bits These bits stay the same. Related Problem Find the 2s complement of 11000000. Figure 2-3 Example of obtaining the 2s complement of a negative binary number. Thomas L. FloydDigital Fundamentals, 8e Copyright 2003 by Pearson E

30、ducation, Inc.Upper Saddle River, New Jersey 07458All rights reserved. Digital system, such as the computer, must be able to handle both positive and negative numbers. A signed binary number consists of both sign and magnitude information. The sign indicates whether a number is positive or negative

31、and the magnitude is the value of the number. 10. There are three ways in which signed integer numbers can be represented in binary form: sign-magnitude, 1s complement, and 2s complement. Noninteger and very large or small numbers can be expressed in floating-point format. 11. The left-most bit in a

32、 signal binary number is the sign bit, which tells you whether the number is positive or negative. A 0 is for positive, and a 1 is for negative. 00011001 25 10011001 -25 In the sign-magnitude system, a negative number has the same magnitude bits as the corresponding positive number but the sign bit

33、is a 1 rather than 0. 00011001 25 10011001 -25 1s Complement System Positive numbers in the 1s complement system are represented the same way as the positive sign-magnitude numbers. Negative numbers, however, are the 1s complements of the corresponding positive numbers. +25 00011001 00011001 1s comp

34、lement -25 10011001 11100110 1s complement 2s Complement System Positive numbers in the 2s complement system are represented the same way as in the sign-magnitude and 1s complement systems. Negative numbers are the 2s complements of the corresponding positive numbers. +25 00011001 00011001 2s comple

35、ment -25 10011001 11100111 2s complement EXAMPLE 2-14 Express the decimal number -39 as an 8-bit number in the sign-magnitude, 1s complement, and 2s complement. Solution + 39 00100111 - 39 10100111 sign-M 11011000 1s Com 11011001 2s Com Related Problem Express +19 and -19 in sign-magnitude, 1s compl

36、ement, and 2s complement. Related Problem Express +19 and -19 in sign-magnitude, 1s complement, and 2s complement. Solution + 19 00010011 - 19 10010011 sign-M 11101100 1s Com 11101101 2s Com The Decimal Value of Signed Numbers EXAMPLE 2-15 Determine the decimal value of this signed binary number exp

37、ressed in sign-magnitude: 10010101 = - 21 Related Problem Determine the decimal value of the sign-magnitude number 01110111. ( 11910 ) EXAMPLE 2-16 Determine the decimal value of this signed binary numbers expressed in 1s complement: (a) 00010111 (b) 11101000Solution (a) 23 (b) -128 + 64 +32 + 8 = -

38、24 -24 + 1 = -23 Related Problem Determine the decimal value of the 1s complement number 11101011. ( -2010 ) EXAMPLE 2-17 Determine the decimal value of this signed binary numbers expressed in 2s complement: (a) 01010110 (b) 10101010Solution (a) 86 (b) -128 + 32 +8 + 2 = -86 Related Problem Determin

39、e the decimal value of the 2s complement number 11010111. ( -4110 ) In the last section, you learned how signed numbers are represented in three different systems. In this section, you will learn how signed numbers are added, subtracted, multiplied, and divided. 12. Because the 2s complement system

40、for representing signed numbers is the most widely used in computers and microprocessorbased systems, the coverage in this section is limited to 2s complement arithmetic. The processes covered can be extended to the other systems if necessary. 13. Addition 1. Both numbers positive2. Positive number

41、with magnitude larger than negative number3. Negative number with magnitude larger than positive number4. Both numbers negative Both numbers positive: 00000111 7 + 00000100 + 4 00001011 1 1Positive number with magnitude larger than negative number: 00001111 1 5 + 11111010 + - 6 1 00001001 9Discard c

42、arry Negative number with magnitude larger than positive number: 00010000 1 6 + 11101000 + - 2 4 11111000 - 8Both numbers negative: 11111011 - 5 + 11110111 + - 9 1 11110010 - 1 4Discard carry EXAMPLE 2-19 Add the signed numbers: 01000100, 00011011, 00001110, and 00010010. Solution 68 01000100 + 27 +

43、 00011011 95 01011111 + 14 + 00001110 109 01101101 + 18 + 00010010 127 01111111 Related Problem Add 00110011, 10111111, and 01100011. These are signed numbers. 51 00110011 - 65 + 10111111 - 14 11110010 + 99 + 01100011 85 1 01010101 Discard carry Subtraction : It is a special case of addition. The si

44、gn of a positive or negative binary number is changed by taking its 2s complement. To subtract two signed numbers, take the 2s complement of the subtrahend and add. Discard any final carry bit. EXAMPLE 2-20 Perform each of the following subtraction of the signed numbers: (a) 00001000 00000011 (b) 00

45、001100 11110111 (c) 11100111 00010011 (d) 10001000 11100010 Solution (a) In this case, 8 3 = 8 + (-3) = 5 00001000 Minuend ( +8) + 11111101 2s Com of subtrahend Discard carry 100000101 Difference (+5) (b) In this case, 12 (-9) = 12 + 9 = 21 00001100 Minuend ( +12) + 00001001 2s Com of subtrahend (9)

46、 00010101 Difference (+21) (c) In this case, -25 (+19) = -25 + (-19) = -44 11100111 Minuend ( -25) + 11101101 2s Com of subtrahend (-19) Discard carry 111010100 Difference (-44) (d) In this case, -120 (-30) = -120 + 30 = -90 10001000 Minuend ( -120) + 00011110 2s Com of subtrahend ( 30 ) 10100110 Di

47、fference (-90)Related Problem Subtract 01000111 from 01011000 Related Problem Subtract 01000111 from 01011000 In this case, 88 71 = 88 + (-71) = 17 01001000 Minuend ( 88) + 10111001 2s Com of subtrahend ( -71 ) Discard carry 100010001 Difference (-90) The hexadecimal number system has sixteen digits

48、 and is used primarily as a compact way of displaying or writing binary numbers because it is very easy to convert between binary and hexadecimal. As you are probably aware, long binary numbers are difficult to read and write because it is easy to drop or transpose a bit. 14. Since computers and mic

49、roprocessors understand only 1s and 0s, it is necessary to used these digits when you program in “ machine language.” Imagine writing a sixteen bit instruction for a microprocessor system in 1s and 0s. It is much more efficient to use hexadecimal or octal; octal numbers are covered in Section 29 . H

50、exadecimal is widely used in computer and microprocessor application. 15. Decimal Binary Hexadecimal 0 0000 0 1 0001 1 2 0010 2 3 0011 3 4 0100 4 5 0101 5 6 0110 6 7 0111 7 22. Decimal Binary Hexadecimal 8 1000 8 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 1110 E 15 1111 F Counting in Hexadc

51、imal 0 to 9 A, B, C, D, E, F, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 2A, 2B, 2C, 2D, 2E, 2F, 30, 31, 32, . EXAMPLE 2-24 Convert the following binary numbers to hexadecimal: (a) 1100101001010111 (b) 111111000101101001 Solution (a) 11001

52、01001010111 (b) 00111111000101101001 C A 5 7 3 F 1 6 9 Related Problem Convert the binary number 1001111011110011100 to hexadecimal. Related Problem Convert the binary number 1001111011110011100 to hexadecimal. Solution 01001111011110011100 4 F 7 9 C Hexadecimal-to-Binary Conversion To reverse the p

53、rocess EXAMPLE 2-25 Determine the binary numbers for the following hexadecimal numbers: (a) 10A416 (b) CF8E16 ( c ) 974216 Solution (a) 1 0 A 4 (b) C F 8 E 1 000010100100 1100111110001110 ( c ) 9 7 4 2 1001011101000010 Related Problem Convert the hexadecimal number 6BD3 to binary. Solution 6 B D 3 1

54、10101111010011 Hexadecimal-to-Decimal Conversion One way to find the decimal equivalent of a hexadecimal number is to first convert the hexadecimal number to binary and then convert from binary to decimal. EXAMPLE 2-26 Convert the following hexadecimal numbers to decimal: (a) 1C16 (b) A8516 Solution

55、 ( a ) 1 C 00011100 = 24+ 23 + 22 = 16 + 8 + 4 = 2810 (b) A 8 5 101010000101 = 211+ 29 + 27+ 22 + 20 = 2048 + 512 + 128 + 4 + 1 = 2693 10 Like the hexadecimal system, the octal system provides a convenient way to express binary numbers and codes. However, it is used less frequently than hexadecimal

56、in conjunction with computers and microprocessors to express binary quantities for input and output purposes. 16. The OCTAL number system is composed of eight digits, which are 0, 1, 2, 3, 4, 5, 6, 7To count above 7, begin another column and start over: 10, 11, 12, 13, 14, 15, 16, 17, 20, 21 and so

57、on. A base8 number system. 17. Binary coded decimal ( BCD ) is a way to express each of the decimal digits with a binary code. Since there are only ten code groups in the BCD system, it is very easy to convert between decimal and BCD. Because we like to read and write in decimal, the BCD code provid

58、es an excellent interface to binary systems. Examples of such interfaces are keypad inputs and digital readouts. 18. Decimal/BCD conversionDecimal Digit 0 1 2 3 4BCD 0000 0001 0010 0011 0100Decimal Digit 5 6 7 8 9BCD 0101 0110 0111 1000 100119. Decimal BCD 2421code excess-3 code 0 0000 0000 0011 1 0

59、001 0001 0100 2 0010 0010 0101 3 0011 0011 0110 4 0100 0100 0111 5 0101 1011 1000 6 0110 1100 1001 7 0111 1101 1010 8 1000 1110 1011 9 1001 1111 1100 There are many specialized codes used in digital systems. You have just learned about the BCD code; now lets look at a few others. Some codes are stri

60、ctly numeric, like BCD, and others are alphanumeric; that is, they are used to represent numbers, letters, symbols, and instructions. The codes introduced in this section are the Gray code and ASCII code. Also, the detection of errors in codes using a parity bit is covered. 20. The Gray CodeThe Gray

61、 Code is unweighted and is not an arithmetic code; that is, there are no specific weights assigned to the bit positions. The important feature of the Gray code is that it exhibits only a single bit change from one code number to the next. 21. The Gray Code Decimal Binary Gray Code 0 0000 0000 1 0001

62、 0001 2 0010 0011 3 0011 0010 4 0100 0110 5 0101 0111 6 0110 0101 7 0111 0100 22. Decimal Binary Gray Code 8 1000 1100 9 1001 1101 10 1010 1111 11 1011 1110 12 1100 1010 13 1101 1011 14 1110 1001 15 1111 1000 23. Binary-to-Gray Code Conversion1. The most significant bit ( left-most ) in the Gray cod

63、e is the same as the corresponding MSB in the binary number.2. Going from left to right, add each adjacent pair of binary code bits to get the next Gary code bit. Discard carries.For example:1 + 0 + 1 + 1 + 0 Binary1 1 1 0 1 Gray24. Gray-to-Binary Conversion1. The most significant bit ( left-most )

64、in the binary code is the same as the corresponding MSB in the Gray code.2. Add each binary code bit generated to the Gray code bit in the next adjacent position. Discard carries.For example:1 1 0 1 1 Gray + + + +1 0 0 1 0 Binary25. ASCIIThe American Standard Code for Information Interchange pronoun

65、ced “ askey “. 26. Chapter 2: Number Systems, Operation, and CodesTrue/False1. The decimal number system consists of the digits 010.2. A binary number with four digits has a maximum value of 15.3. Repeated division-by-10 is used to convert decimal numbers to binary numbers. 27. 4. The binary number

66、system is fundamental to all digital computers.5. The 1s complement of 101112 is 010002.6. The sign ( + or - ) of a binary number is shown by setting the LSB to 0 or 1.7. 111010002 is the 2s complement representation of 24.8. The largest single digit in the octal ( meaning 8 ) number system is 8. 28. 9. 3C1D16 = 11110000011101210. The BCD equivalent of 73 is 010010012.Multi Choice11.In the decimal number 481, the power-of-ten weight of the digit 4 is . a. 400 b. 102 c. 104 d. 10012.The two digit