Digital Design:Principles and Practices

Chapter 2Number Systems and Codes

2.1 Positional Number Systems

3

Counting in Binary

With n bits, you can count from 0 to 2n – 1

For example, with four bits (n=4), you can count from 0 to 15:

24 – 1 = 16 – 1 = 15

4

Figure 2–1 Illustration of a simple binary counting application.

Counting in Binary – An Application

5

Positional Number System

• The traditional number system that we learned in school and use everyday in business.

• A number is represented by A string of digits Each digit position has an associated weight

• Example #1: 1734 = 1*1000 + 7*100 + 3*10 + 4*1 Each weight is a power of 10 corresponding to the digit’s position.

• Example #2: 5185.68 = 5*1000 + 1*100 + 8*10 + 5*1 + 6*0.1 + 8*0.01

6

Positional Number System

• In general, a number D of the form d1d0.d-1d-2 has the value:

D = d1*101 + d0*100 + d-1*10-1 + d-2*10-2

10 is called the base or radix of the number system.

radix point ( 小數點 )

• in decimal number system: decimal point 十進位的小數點

• in binary number system: binary point 二進位的小數點

7

Positional Number System

• Radix = r• General form:

• There are p digits to the left of the radix point and n to the right.

• The value:

1 2 1 0 1 2. p p nd d d d d d d

1pi

ii n

D d r

8

Positional Number System - Binary

• Radix = r = 2• General form:

• The value:

1 2 1 0 1 2. p p nb b b b b b b

1

2p

ii

i n

B b

9

Positional Number System

• Digit Binary digit (bit)

• Radix point Decimal point, Binary point

• Most Significant Bit (MSB)• Least Significant Bit (LSB)

• When dealing with binary and other non-decimal numbers, we use a subscript to indicate the radix of each number.

10

Binary to Decimal Conversion

• 100112 = 1910

100112 = 1*16 + 0*8 + 0*4 + 1*2 + 1*1 = 1910

• 1000102 = 3410

• 101.0012 = 5.12510

2.2 Octal and HexadecimalNumbers

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Octal & Hexadecimal

• The octal number system uses radix 8. 8 digits (0-7)

• The hexadecimal number system uses radix 16. 16 digits (0-9 and then A-F)

13

Binary, Decimal, Octal, & Hexadecimal

Table 2-1

14

Binary-to-Octal Conversion

• Binary-to-Octal Conversion Method (no binary point): Starting at the binary point and working left, we simply separate

the bits into groups of three and replace each group with the corresponding octal digit.

• Example #1: 1000110011102 = 100 011 001 1102 = 43168

• Example #2: 111011011101010012 = 011 101 101 110 101 0012

= 3556518

15

Binary-to-Hexadecimal Conversion

• Binary-to-Hex Conversion Method (no binary point): Starting at the binary point and working left, we simply separate

the bits into groups of four bits and replace each group with the corresponding hexadecimal digit.

• Example #1: 1000110011102 = 1000 1100 11102 = 8CE8

• Example #2: 111011011101010012 = 0001 1101 1011 1010 10012

= 1DBA98

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Binary-to-Octal Conversion &Binary-to-Hexadecimal Conversion

• Example: 10.10110010112 = 010 . 101 100 101 1002 = 2.54548

= 0010 . 1011 0010 11002 = 2.B2C16

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Octal-to-Binary Conversion &Hexadecimal-to-Binary Conversion

• Converting from octal or hexadecimal to binary is very easy. We simply replace each octal or hexadecimal digit with the corresponding 3- or 4-bit string.

• Examples: 13578 = 001 011 101 1112

2046.178 = 010 000 100 110 . 001 1112

BEAD16 = 1011 1110 1010 11012

9F.46C16 = 1001 1111 . 0100 0110 11002

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Hexadecimal Numbers

• 1 byte = 8 bits• 1 hexadecimal digital = 1 nibble = 4 bits• 8 nibbles = 32 bits = 4 bytes

• Hexadecimal numbers are often used to describe a computer’s memory address space.

• Many computer programming languages (e.g. C/C++) use the prefix “0x” to denote a hexadecimal number, for example, 0xBFC0000.

2.3 General Positional-Number-System Conversions

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Number System Conversions• In general, conversion between two radices cannot be done by

simple substitutions; arithmetic operations are required.

• In this section, we show how to convert a number in any radix to radix 10 (decimal) and vice versa.

• Radix r to radix 10 (decimal) conversion:

r: the radix of the number which is to be converted p: there are p digits to the left of the radix point n: there are n digits to the right of the radix point

1pi

ii n

D d r

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Radix r to Radix 10 ConversionExamples

• 1CE816 = 1*163 + 12*162 + 14*161 + 8*160 = 740010

• F1A316 = 15*163 + 1*162 + 10*161 + 3*160 = 6185910

• 436.58 = 4*82 + 3*81 + 6*80 + 5*8-1 = 286.62510

• 132.34 = 1*42 + 3*41 + 2*40 + 3*4-1 = 30.7510

• An alternative method:F1A316 = ( ( (15)*16 + 1)*16 + 10 )*16 + 3 = 6185910

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Binary, Decimal, Octal, & Hexadecimal

Table 2-1

23

Radix r to Radix 10 Conversion- An Alternative Method

• General Form:

• Example: F1A316 = ( ( (15)*16 + 1)*16 + 10 )*16 + 3 = 6185910

• The method can be used to convert a decimal number D to a radix r.

1 2 1 0(( (( ) ) ) )p pD d r d r r d r d

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Radix 10 to Radix r Conversion

• General Form:

• If we divide D by r, then:Quotient:

Remainder: d0

• Successive divisions by r yield successive digits of D from right to left, until all the digits of D have been derived.

1 2 1 0(( (( ) ) ) )p pD d r d r r d r d

1 2 1( (( ) ) )p pQ d r d r r d

25

Radix 10 to Radix r ConversionAn Example (r=16)

61859 ÷ 16 = 3866 …… 3 (LSB)

÷ 16 = 241 …… 10

÷ 16 = 15 …… 1

÷ 16 = 0 …… 15

(MSB)

6185910 = F1A316

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Radix 10 to Radix r ConversionAn Example (r=2)

179 ÷ 2 = 89 …… 1 (LSB)

÷ 2 = 44 …… 1

÷ 2 = 22 …… 0

÷ 2 = 11 …… 0

÷ 2 = 5 …… 1

÷ 2 = 2 …… 1

÷ 2 = 1 …… 0

÷ 2 = 0 …… 1

(MSB)

17910 = 101100112

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Conversion Methods for Common Radices (I)Table 2-1

28

Conversion Methods for Common Radices (II)

Table 2-1 (cont’d)

EXAMPLE 2-26Convert the following hexadecimal numbers to decimal.(a)1C16

(b)A8516

Hexadecimal-to-Decimal Conversion

Ans: (a) 2810 (b) 269310

Decimal-to-Hexadecimal Conversion

EXAMPLE 2-28Convert the decimal number 650 to hexadecimal.

Ans: 28A16

Octal-to-Decimal Conversion

Example: (2374)8 = ( )10

Ans: 127610

Decimal-to-Octal Conversion

Example: (359)10 = ( )8

Ans: 5478

Octal-to-Binary Conversion

EXAMPLE 2-31Convert each of the following octal numbers to binary:(a)138

(b)258

(c)1408

(d)75268

Ans: (a) 001011 (b) 010101 (c) 001100000 (d) 111101010110

Binary-to-Octal Conversion

EXAMPLE 2-32Convert each of the following binary numbers to octal:(a)110101(b)101111001(c)100110011010(d)11010000100

Ans: (a) 658 (b) 5718 (c) 46328 (d) 32048

2.4 Addition and Subtraction ofNon-decimal Numbers

36

Addition & Subtraction of Non-decimal Numbers

• Addition and subtraction of non-decimal numbers by hand uses the same technique that we learned in elementary school for decimal numbers.

• Binary Addition s: sum cin: input carry cout: output carry

• Binary Subtraction (minuend minus subtrahend yields difference) d: difference bin: input borrow bout: output borrow

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Binary Addition & Subtraction Table

Table 2-3 Binary addition and subtraction table

38

Binary Addition

39

Binary Addition• Binary Addition

s: sum cin: input carry cout: output carry

C 1 0 1 1 1 1 0 0 0

X 190 1 0 1 1 1 1 1 0

Y + 141 + 1 0 0 0 1 1 0 1

X + Y 331 1 0 1 0 0 1 0 1 1

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Binary Subtraction

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Binary Subtraction (cont’d)

• Binary Subtraction (minuend minus subtrahend yields difference) d: difference bin: input borrow bout: output borrow

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Binary Subtraction• Binary Subtraction

d: difference bin: input borrow bout: output borrow

B 0 1 1 0 1 1 0 1 0

X 210 1 1 0 1 0 0 1 0

Y - 109 - 0 1 1 0 1 1 0 1

X - Y 101 0 1 1 0 0 1 0 1

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Binary Addition & Subtraction Table

Table 2-3 Binary addition and subtraction table

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Hexadecimal AdditionC 1 1 0 0

X 1 9 B 916

Y + C 7 E 616

X+Y E 1 9 F16

C 1 1 0 0

X 1 9 11 9

Y + 12 7 14 6

X+Y 14 17 25 F

14 16+1 16+9 15

E 1 9 F

2.5 Representation of Negative Numbers

2.5.1 Signed-Magnitude Representation

47

Signed-Magnitude Representation

• Definition of the Sign Bit: The left-most bit (a.k.a. MSB) in a signed binary number is the sign bit, which tells you whether the number is positive or negativeA 0 sign bit indicates a positive numberA 1 sign bit indicates a negative number

• Two possible representations of zero: “+0” and “–0”

• An n-bit signed-magnitude integer lies within the range –(2n–1–1) through +(2n–1–1)

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Signed-Magnitude RepresentationExamples

• 010101012 = +8510

• 110101012 = –8510

• 011111112 = +12710

• 111111112 = –12710

• 000000002 = +010

• 100000002 = –010

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Signed-Magnitude RepresentationAddition & Subtraction

• Now suppose that we wanted to build a digital logic circuit that adds signed-magnitude numbers.

• If the signs are the same, it must add the magnitudes and give the result the same sign.

• If the signs are different, it must compare the magnitudes, subtract the smaller from the larger, and give the result the sign of the larger.

• Adders for complement number systems are much simpler.

2.5.2 Complement Number Systems2.5.3 Radix-Complement Representation2.5.4 Two’s-Complement Representation

2.5.5 Diminished Radix-Complement Representation

2.5.6 One’s-Complement Representation2.5.7 Excess Representations

Vocabulary

• Complement (n.) 補充、補充物、補數• Compliment (n.) (v.) 讚美

• 兩個字發音相同

1’s Complement• How do you calculate the 1’s complement of a binary

number?Ans: Change all 1s to 0s and all 0s to 1s

• For example:The 1’s complement of the binary number 10110010 is 01001101.

1 0 1 1 0 0 1 0 Binary Number↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓0 1 0 0 1 1 0 1 1’s Complement

Figure 2–2 Example of inverters used to obtain the 1’s complement of a binary number.

Thomas L. FloydDigital Fundamentals, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

1’s Complement (cont’d)

2’s Complement• How do you calculate the 2’s complement of a binary number?

Ans: (2’s complement of a binary number) = (1’s complement of the binary number) + 1

• For example:The 2’s complement of the binary number 10110010 is 01001110.

10110010 Binary Number01001101 1’s complement

+ 1 Add 101001110 2’s complement

Figure 2–3 Example of obtaining the 2’s complement of a negative binary number.

Thomas L. FloydDigital Fundamentals, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

2’s Complement (cont’d)

Converting 1’s or 2’s Complements Back to the Original Binary Numbers

• By performing 1’s or 2’s complement conversion again, you can get the original binary number!

Signed-Magnitude Representation

• For example:

(+25)10 00011001

(-25)10 10011001

0 0 0 1 1 0 0 1↑ ↑

Sign bit Magnitude bits

Signed-Magnitude Representation

• The Sign-Magnitude Representation has two binary numbers for (0)10

(+0)10 00000000

(-0)10 10000000

1’s Complement Representation

• For example:

(+25)10 00011001 (the same as the Signed-Magnitude Rep.)

(-25)10 11100110 (1’s complement of (+25)10)

• The 1’s Complement Rep. has two binary numbers for (0)10

(+0)10 00000000

(-0)10 11111111

2’s Complement Representation

• For example:

(+25)10 00011001 (the same as the Signed-Magnitude Rep.)

(-25)10 11100111 (2’s complement of (+25)10)

• The 2’s Complement Rep. has only one binary numbers for (0)10

(+0)10 00000000

(-1)10 11111111

(-128)10 10000000

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Decimal and 4-bit Numbers

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Ranges of Different Number System Representations

• m-bit number

• Two’s Complement Representation –(2m–1) to +(2m–1 – 1)

• One’s Complement Representation –(2m–1 – 1) to +(2m–1 – 1)

• Signed-Magnitude Representation –(2m–1 – 1) to +(2m–1 – 1)

• Excess-2m–1 Representation –(2m–1) to +(2m–1 – 1)

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Excess-2m-1 Representations

• In excess-B representation, an m-bit string whose unsigned integer value is M represents the signed integer M – B, where B is called the bias of the number system.

• In excess-2m-1 representations, B equals 2m-1.

• The range of the representation is exactly the same as that of m-bit two’s-complement numbers. In fact, the representations of any number in two systems are identical except for the sign bits, which are always opposite (assume that B = 2m-1).

2.6 Two’s-Complement Additionand Subtraction

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2’s-Complement Addition and Subtraction

Figure 2-3 A modular counting representation of 4-bit two’s-complement numbers.

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2’s-Complement Addition

+3 0011+ +4 + 0100

+7 0111

–2 1110

+ –6 + 1010

–8 11000

+6 0110+ –3 + 1101

+3 10011

+4 0100+ –7 + 1001

–3 1101

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2’s-Complement Addition - Overflow

–3 1101

+ –6 + 1010

–9 10111

–8 1000

+ –8 + 1000

–16 10000

+5 0101+ +6 + 0110

+11 1011

+7 0111+ +7 + 0111

+14 1110

= +7

= +0

= -5

= -2

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• If an addition operation produces a result that exceeds the range of the number, overflow is said to occur.

• A simple rule for detecting overflow in addition: An addition overflows if the addends’ signs are the

same but the sum’s sign is different from the addends’.

2’s-Complement Addition - Overflow

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2’s-Complement Subtraction

• minuend – subtrahend = difference ( 被減數 ) – ( 減數 ) = ( 差 ) For example: 9 – 6 = 3

• 2’s-Complement Subtraction Procedures: Negate the subtrahend by taking its 2’s complement Add the negated subtrahend to the minuend using

normal rules for addition

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2’s-Complement Subtraction

• Negating the subtrahend and adding the minuend can be accomplished with only one addition operation as follows: Perform a bit-by-bit complement of the subtrahend and

add the complemented subtrahend to the minuend with an initial carry (Cin) of 1 instead of 0.

See examples.

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2’s-Complement SubtractionAn Example

1 Cin

+4 0100 0100– +3 – 0011 + 1100

+1 1 0001

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2’s-Complement SubtractionAn Example

1 Cin

+3 0011 0011– –4 – 1100 + 0011

+7 0111

73

1 Cin

–5 1011 1011– 4 – 0100 + 1011

–9 1 0111

2’s-Complement Subtraction - Overflow

• Overflow in subtraction can be detected by examining the signs of the minuend and the complemented subtrahend, using the same rule as in addition.

• Example:

EXAMPLE 2-14

Express the decimal number -39 as an 8-bit binary number in the sign-magnitude, 1’s complement, and 2’s complement forms.

Determine the decimal value of this signed binary number expressed in sign-magnitude representation: 10010101.

EXAMPLE 2-15( Signed-Magnitude Rep. Decimal Value )

EXAMPLE 2-16

Determine the decimal values of the signed binary numbers expressed in 1’s complement representation:(a) 00010111(b) 11101000

( 1’s Complement Rep. Decimal Value )

EXAMPLE 2-17

Determine the decimal values of the signed binary numbers expressed in 2’s complement representation:(a) 01010110(b) 10101010

( 2’s Complement Rep. Decimal Value )

Numbers Are Added Two at a Time

EXAMPLE 2-19Add the signed numbers: 01000100, 00011011, 00001110, and 00010010.

Arithmetic Operations with Signed Numbers - Subtraction

Related Problem for EXAMPLE 2-20:

Subtract 01000111 from 01011000.

2.10 Binary Codes forDecimal Numbers

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Binary Codes for Decimal Numbers

• Code: A set of n-bit strings in which different bit strings represent different numbers or other things.

• Code word: A particular combination of n bit-values.

• A code that uses n-bit strings need not contain 2n valid code words.

• At least 4 bits are needed to represent the ten decimal digits.

• Binary-coded decimal (BCD)

82

Decimal Codes

83

Decimal Codes• BCD = 8421 code• 2421 code

Self complementing: The code word for the 9s’ complement of any digit may be obtained by complementing the individual bits of the digit’s code word.

• Biquinary quinary: 5 The first two bits in a code word indicate whether the number is in the

range 0-4 or 5-9. The last five bits indicate which of the five numbers in the selected

range is represented Error detection: can detect one bit error

• 1-out-fo-10 code Uses 10 out of 1024 possible 10-bit code words

Decimal BCD

EXAMPLE 2-33Convert each of the following decimal numbers to BCD:(a)35(b)98(c)170(d)2469

BCD Decimal

EXAMPLE 2-34Convert each of the following BCD codes to decimal:(a)10000110(b)001101010001(c)1001010001110000

BCD Addition

4 個 bits 一組，每組超過 9 (10012) 要加 6 (01002)

BCD Addition

EXAMPLE 2-35Add the following BCD numbers:(a)0011 + 0100(b)00100011 + 00010101(c)10000110 + 00010011(d)010001010000 + 010000010111

BCD Addition

EXAMPLE 2-36Add the following BCD numbers:(a)1001 + 0100(b)1001 + 1001(c)00010110 + 00010101(d)01100111 + 01010011

2.11 Gray Code

90

A Mechanical Encoding DiskUsing a 3-bit Binary Code

91

A Mechanical Encoding DiskUsing a 3-bit Gray Code

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3-bit Binary Code & Gray Code

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Constructing a Gray Code

• A 1-bit Gray code has two code words, 0 and 1.

• The first 2n code 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.

94

Constructing a Gray Code

1-bit Gray Code 2-bit Gray Code 3-bit Gray Code01

00011110

000001011010110111101100

95

Binary code word Gray-code code word

• The bits of an n-bit binary or Gray-code code word are numbered from right to left, from 0 to n – 1.

• Bit i of a Gray-code code word is 0 if bits i and i + 1 of the corresponding binary code word are the same, else bit i is 1. (When i + 1 = n, bit n of the binary code word is considered to be 0.)

96

Binary Gray Code[Floyd]

• Step 1: The most significant bit (left-most) in the Gray code is the same as the corresponding MSB in the binary number.

• Step 2:Going from left to right, add each adjacent pair of binary code bits to get the next Gray code bit. Discard carries.

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Gray Code Binary[Floyd]

• Step 1:The most significant bit (left-most) in the binary code is the same as the corresponding bit in the Gray code.

• Step 2:Add each binary code bit generated to the Gray code bit in the next adjacent position. Discard carries.

2.12 Character Codes

99

ASCII

• American Standard Code for Information Interchange

• Each character is represented by a 7-bit string

• 27 = 128 (total number of different characters)

• “Yeccch!”1011001 1100101 1100011 1100011 1100011 1101000

0100001

100

ASCII

101

ASCII (cont’d)

2.13 Codes for Actions, Conditions, and States

103

States in a Traffic-Light Controller

104

Control Structure for a Digital System