数字逻辑设计及应用

1

数字逻辑设计及应用Chapter 2 Number Systems and codes

(数系与编码 )

介绍在数字逻辑体系中信号的表达方式、类型，不同表达方式之间的转换，运算的规则等。

Digital Logic Design and Application ( 数字逻辑设计及应用 )

2

Review of Chapter 1 (第一章内容回顾 )

Analog versus Digital (模拟与数字 )

Digital Devices (数字器件 )： Gates （门电路）、 Flip-flops（触发器）

Electronic and Software Aspects of

Digital Design

(数字设计的电子技术和软件技术 )


3

Review of Chapter 1 (第一章内容回顾 )

Integrated Circuit（ IC，集成电路）Programmable Logic

Devices（ PLA、 PLD、 CPLD、 FPGA, 可编程逻辑器件）

Application-Specific ICs（ ASIC, 专用集成电路）

Printed-Circuit Boards (PCB, 印制电路板 )


4

Chapter 2 Number Systems and codes

(数系与编码 )

Two kinds of Information

(信息主要有两类 )：

Numeric Data (数值信息 )

Nonnumeric Data (非数值信息 )


— Number Systems and their Conversions

( 数制及其转换 )

-- Nonnumeric Data Representation – Codes ( 非数值信息的表征 -- 编码 )

5

Chapter 2 Number Systems and codes

(数系与编码 )

数字系统只处理数字信号 0 , 1；需要将任意信息用 ( 0 ,1 )表达；用（ 0， 1 ）表达数量：数制二进制用（ 0， 1 ）表达不同对象：符号编码


66

How to Encode Text: ASCII, Unicode

ASCII: 7- (or 8-) bit encoding of each letter, number, or symbol

Sample ASCII encodingsSymbolEncoding

010 0000 <space>␠010 0001 !010 0010 "010 0011 #010 0100 $010 0101 %010 0110 &010 0111 '010 1000 (010 1001 )010 1010 *010 1011 +010 1100 ,010 1101 -010 1110 .010 1111 /

SymbolEncoding

100 1110 N100 1111 O101 0000 P101 0001 Q101 0010 R101 0011 S101 0100 T101 0101 U101 0110 V101 0111 W101 1000 X101 1001 Y101 1010 Z

100 0001 A100 0010 B100 0011 C100 0100 D100 0101 E100 0110 F100 0111 G100 1000 H100 1001 I100 1010 J100 1011 K100 1100 L100 1101 M

SymbolEncoding

SymbolEncoding

011 0000 0011 0001 1011 0010 2011 0011 3011 0100 4011 0101 5011 0110 6011 0111 7011 1000 8011 1001 9

110 0001 a110 0010 b ...111 1001 y111 1010 z

… …

77

How to Encode Text: ASCII, Unicode

Unicode: Increasingly popular 16-bit encodingEncodes characters from various world

languages

… …

88

How to Encode Numbers: Binary Numbers

Base ten (decimal) Ten symbols: 0, 1, 2, ..., 8,

and 9 More than 9 -- next

position

So each position power （幂） of 10

Nothing special about base 10 -- used because we have 10 fingers

104 103 102

5 2 3

101 100

• Each position represents a base quantity; symbol in position means how many of that quantity

99

How to Encode Numbers: Binary Numbers

Base two (binary)Two symbols: 0

and 1More than 1 -- next

positionSo each

position power （幂） of 2

2 4 2 3 2 2

1 0 1

2 1 2 0

Q: How much?

+

=

4

1 5

+

=

a

• Each position represents a base quantity; symbol in position means how many of that quantity

There are only 10 types of people in the world: those who understand binary, and those who don't.

1010

Useful to know powers of 2:

24 23 22 21 2029 28 27 26 25

16 8 4 2 1512256128 64 32

16 8 4 2 1512 256 12864 32

Practice counting up by powers of 2:

11

2.1 Positional Number System (按位计数制 )

Any Decimal Number D Can Be Represented as

the Following ( 任意十进制数 D 可表示如下 ):

D = dp-1 dp-2 ... d1 d0 . d-1 d-2 ... d-n

推广： D2 = ∑ d i × 2i

D16= ∑ d i × 16i

1p

ni

ii rd

Weight of i bit; Base or Radix of r Number System(第 i 位的权； r 计数制的基数 )


12


按位计数制的特点 1) 采用基数（Base or Radix）， R进制的基数是 R

2) 基数确定数符的个数如十进制的数符为：

0、 1、 2、 3、 4、 5、 6、 7、 8、 9，个数为 10

二进制的数符为： 0、 1，个数为 2

3）逢基数进一


13



Decimal and Binary

Decimal system: base is 10, the digit may be 0 to 9

101210 1011011001011.101

Binary system: base is 2, the digit may be 0 or 1

10122 212120211.101

bit: one digit in binary system;

1414

Using Digital Data in a Digital System

temperature sensor

0 0 1 10 0 0 0 "33"

A temperature sensor outputs temperature in binary

The system reads the temperature, outputs ASCII code:“F” for freezing (0-32)“B” for boiling (212 or more)“N” for normal

1515

Using Digital Data in a Digital System

temperature sensor

0 0 1 10 0 0 0 "33"

Digital System

display N

"N"1 0 00 1 1 1

if (input <= "00100000") // "32" output = "1000110" // "F"else if (input >= "11010100") // "212" output = "1000010" // "B"else output = "1001110" // "N"

A display converts its ASCII input to the corresponding letter

16

2.2 Octal and Hexadecimal Numbers

(八进制和十六进制 )

基数数码特性 Octal Number ( 八进制 ) 8 0~7 逢八进一 Binary Number ( 二进制 ) 2 0,1 逢二进一

Hexadecimal Number( 十六进制 )

16 0~9,A~F 逢十六进一


17

说明Digital Logic Design and Application ( 数字逻辑设计及应用 )

选择什么数制来表示信息，对数字系统的成本和性能影响很大，在数字电路中多使用二进制 .

Most Significant Bit（MSB, 最高有效位）Least Significant Bit（ LSB, 最低有效位）

1011100010112

MSB LSB

18


表 2.1 十进制、二进制、八进制与十六进制数

十进制二进制八进制十六进制 0 0000 0 0

1 0001 1 1

2 0010 2 2

3 0011 3 3

4 0100 4 4

5 0101 5 5

6 0110 6 6

7 0111 7 7

8 1000 10 8

9 1001 11 9

10 1010 12 A

11 1011 13 B

12 1100 14 C

13 1101 15 D

14 1110 16 E

15 1111 17 F

1919

Base Sixteen: Another Base Used by Designers

Nice because each position represents four base-two positions Compact way to write binary numbers

Known as hexadecimal, or just hex

Q: Write 11110000 in hex

F 0Q: Convert hex A01 to binary

1010 0000 0001

A:

A:

20

二进制与八进制和十六进制之间的转换位数替换法：保持小数点不变，每位八进制数对应 3位二进制数 ;

每位十六进制数对应 4位二进制数 ;

二进制转换时，从小数点开始向左右分组，在MSB前面和 LSB后面可以加 0；

转换为二进制时，MSB前面和 LSB后面的 0不写；例： 1011100010112=56138=B8B16

10111000.10112=207.548=B8.B161000110010012 = ( )8

= ( )16


2121

Hex Example: RFID TagBatteryless （无电池） tag powered（功

率） by radio fieldTransmits （发送） unique

identification（鉴定） numberExample: 32 bit id（身份证明）

8-bit province number, 8-bit city number, 16-bit animal number Tag contents are in binary But programmers use hex when writing/reading

RFIDtag

Province # City # Animal #

Province: 7 City: 160 Animal : 513

1010000000000111 00000010 00000001

A007 02 01

Tag ID in hex: 07A00201

(a)

(b)

(c)

(d)

(f)

(e)

22

2.3 General Positional-Number-System Conversion (常用按位计数制的

转换 )


A Number in any Radix to Radix 10 ( 任意进制数十进制数 )

Method: Expanding the formula using radix-10 arithmetic (方法：利用位权展开 )

Example 1 ： ( 101.01 )2 = ( )10

( 7F.8 )16 = ( )10

More easy way( 更简便的方法 ) ？ ( F1AC )16 = ( ( ( F16 ) +1 ) 16 + A ) 16 + C

23

2.3 General Positional-Number-System Conversion (常用按位计数制的转

换 )


A Number in Radix 10 to any Radix ( 十进制其它进制 )

Method：Radix Multiplication or Division (基数乘除法 )

Integer Parts (整数部分 )：

除 r 取余，逆序排列 Example 2： ( 156 )10 = ( )2 Decimal Fraction Parts (小数部分 )：

乘 r 取整，顺序排列 Example 3： ( 0.37 )10 = ( )2

24

2.3 General Positional-Number-System Conversion (常用按位计数制的

转换 )


Example 4： Require < 10-2 ， Complete the

following conversion

( 617.28 )10 = ( )2

2-n <= < 10-2

思考：任意两种进位计数制之间的转换以十进制（二进制）作为桥梁

n = 7

2525

Converting To/From Binary by Hand: Summary

BinaryDecimal

16 8 4

0 1 0

2 1

1 1

16 + 8 + 2

To decimal

To hex

1 1010

= 2610

= 1A16

To octal

11 010

= 328

2610

To binary

4

0 1

2

0

116

1

8

1

1616+8= 24 24 --24+2

= 26

2626

Divide-By-2 Method Common in Automatic Conversion

Repeatedly divide decimal number by 2, place remainder in current binary digit (starting from 1s column)

1. Divide decimal number by 2 Insert remainder into the binary number Continue since quotient (6) is greater than 0

2. Divide quotient by 2 Insert remainder into the binary number Continue since quotient (3) is greater than 0

01

18

14

02

01

14

02

01

01

02

Decimal Binary

122–12

0

6

62–6

0

3

3. Divide quotient by 2 Insert remainder into the binary number Continue since quotient (1) is greater than 0

32–2

1

1

4. Divide quotient by 2 Insert remainder into the binary number Quotient is 0, done

12–0

1

0

(current value: 12)

(current value: 4)

(current value: 0)

(current value: 0)

Note: Works for any base N—just divide by N instead

2727

Bytes, Kilobytes, Megabytes, and More

Byte: 8 bits

Common metric prefixes: kilo (thousand, or 103), mega (million, or 106), giga (billion, 千兆 or 109), and tera (trillion, 万亿 or 1012), e.g., kilobyte, or KByte

2828

Bytes, Kilobytes, Megabytes, and More

BUT, metric prefixes also commonly used inaccurately216 = 65536 commonly written as “64

Kbyte”Typical when describing memory sizes

Also watch out for “KB” for kilobyte vs. “Kb” for kilobit

2929

Example: DIP-Switch Controlled Channel

Ceiling fan receiver should be set in factory to respond to channel “73”

Convert 73 to binary, set DIP （指拨）switch accordingly

Desired value: 73

4

0 0

2

1

116

0

8

1

32

0

64

1

128

0

64 72 73sum:

(b)(a)

Q:

30

Example: DIP-Switch Controlled Channel

channel receiver

Ceiling fanmodule

0

if (InA = InB) Out = 1else

Out = 0

0 1 0 10 1 0 0"73"

DIP switch

10

0 0 1 00 0 0 1

InA InB

Out

"34"

(c)

31

2.4 Addition and Subtraction of Nondecimal Numbers (非十进制数的加法和

减法 )


Two Binary Number Arithmetic

(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

32

•Carry in ( 进位输入 ): C in (P.32)• Carry out ( 进位输出 ) C out Sum ( 本位和 ): S

2.4 Addition and Subtraction of Non-decimal Numbers ( 非十进制数的加法和减法 )


1011 1110

+ 1000 1101

33


表 2.3.1 二进制加法真值表

输入输出

被加数 X 加数 Y 输入进位 Cin 和 S 进位输出 Cout

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

3434

Adder Example: DIP-Switch-Based Adding Calculator

Goal: Create calculator that adds two 8-bit binary numbers, specified using DIP switches

DIP switch: Dual-In-line Package switch, move each switch up or down

Solution: Use 8-bit adder

35


Solution: Use 8-bit adderDIP switches

10

a7……..a0

b7……..b0

s7…………s0

8-bit carry-ripple adderco

ci 0

CALC

LEDs

3636


To prevent spurious( 假的） values from appearing at output, can place register at outputActually, the light flickers( 闪烁） from spurious values would be too fast for humans to detect—but the principle of registering outputs to avoid spurious values being read by external devices (which normally

aren’t humans) applies here.

37


DIP switches

10

a7…….a0

b7……b0

s7………s0

8-bit adder

8-bit register

coci 0

CALC

LEDs

e

clkld

38

•Borrow in ( 借位输入 ): Bin •Borrow out ( 借位输出 ): Bout •Difference bit ( 本位差 ): D

2.4 Addition and Subtraction of Non-decimal Numbers ( 非十进制数的加法和减法 )


1010 1010

– 0101 0101

39


表 2.3.2 二进制减法真值表输入输出

被减数 X 减数 Y 输入借位 Bin 差 D 输出借位 Bout

0 0 0 0 0

0 0 1 1 1

0 1 0 1 1

0 1 1 0 1

1 0 0 1 0

1 0 1 0 0

1 1 0 0 0

1 1 1 1 1

40

Subtractor Example: DIP-Switch Based

Adding/Subtracting Calculator

Extend earlier calculator exampleSwitch f indicates whether want to add

(f=0) or subtract (f=1)Use subtractor and 2x1 mux

41

Subtractor Example: DIP-Switch Based

Adding/Subtracting CalculatorDIP switches

10

8-bit register CALC

LEDs

e

f

clkld

8

8

80 08

8

8

88 2x10 110

wiciA AB B

S Sco wo8-bit adder 8-bit subtractor

42

2.5 Representation of Negative Numbers

(负数的表示 )


2.5.1 Signed-Magnitude Representation

[ – 符号数值表示法（原码） ]MSB as the Sign bit (0 = plus, 1 = minus)

[ 最高有效位表示符号位（ 0 = 正， 1 = 负） ]

01111111＝＋ 127 11111111＝－ 12700101110＝＋ 46 10101110＝－ 4600000000＝＋ 0 10000000＝－ 0

43

2.5 Representation of Negative Numbers

(负数的表示 )


2.5.1 Signed-Magnitude Representation

[ – 符号数值表示法（原码） ]Two possible representations of Zero

[零有两种表示（ + 0 – 、 0） ]An n-bit signed-magnitude integer range is

(n位二进制整数表示范围 ) ： – ( 2n-1 – 1) ~ + ( 2n-1 – 1)

44

2.5 Representation of Negative Numbers (负数的表示 )

2.5.2 Complement Number Systems

(补码数制 )

Radix – Complement

(基数补码 )Diminished Radix – Complement

[ 基数减 1 补码 (反码 ) ]


45


2.5.3 Radix – Complement Representation ( 基数补码表示法 )

The complement of an n-digit number is obtained by subtracting it from r n

(n 位数的补码等于从 r n 中减去该数 )

Example : Table 2-4 P.36


46


Diminished Radix – Complement Representation

[ 基数减 1补码表示法（反码） ]： The Diminished Radix – Complement

of an n-digit number is obtained by subtracting it from r n -1

[ n 位数的反码等于从 r n – 1 中减去该数 ]

Example : Table 2-4 2-5 P.36


4747

Ten’s Complement Before introducing two’s

complement, let’s consider ten’s complement But, be aware that computers

DO NOT USE TEN’S COMPLEMENT. Introduced for intuition （直觉） only.

Complements for each base ten number shown to right. Complement is the number that when added results in 10

987654321

123456789

4848

Ten’s Complement Nice feature of ten’s complement

Instead of subtracting a number, adding its complement results in answer exactly 10 too much

So just drop （丢下） the 1 – results in subtracting using addition only

4 6

10

7

–4 +6

0 10 20

3 13

13

3

0 10

1

2

3

4

5

6

7

8

9

9

8

7

6

5

4

3

2

1

complements

7–4=3 7+6=13 3Adding the complement results in an answer that isexactly 10 too much – dropping the tens column givesthe right answer.

49

2.5.4 Two’s – Complement Representation (二进制补码表示法 )

The MSB of a number in this system serves as the sign bit(最高有效位用做符号位 );The weight of the MSB is -2n-1 instead of + 2n-1

(MSB的权是 -2n-1 而不是 + 2n-1 )

Obtain a Two’s- Complement

( 二进制补码的求取 )： Ones’ – Complement (反码 ) + 1 （Why？？）


50



例 2.5.1 若约定字长是一个字节，试求－ 119的补码。

解：因－ 119的绝对值 119＝ 01110111，则补码可以通过下式算法得到：

全 1 码： 1 1 1 1 1 1 1 1

减去－ 119 绝对值： 0 1 1 1 0 1 1 1 － 119 的反码： 1 0 0 0 1 0 0 0 加 1 ： 1

－ 119 补码： 1 0 0 0 1 0 0 1

51

2.5.4 Two’s – Complement Representation

(二进制补码表示法 )

Only one representations of Zero ( 零只有一种表示 )

0＝ 0 0 0 0 0 0 0 0 逐位取反 1 1 1 1 1 1 1 1 ＋ 1

约定 8 位 0 0 0 0 0 0 0 0＝ 0


52


An n-bit Two’s- Complement range is (n位二进制补码表示范围 )： – 2 n-1 ~ + ( 2 n-1 – 1)

约定字长（ 8比特）后，补码表示数的范围－ 128～ 127


53


Positive number has the same ：Sign-Magnitude, Ones’ – Complement, and Two’s- Complement

( 正数的原码、反码、补码相同 )


54

表 2-6 1 位十进制数与 4 位二进制数十进制二进制原码二进制反码二进制补码

-8—— ——

1000-7 1111 1000 1001-6 1110 1001 1010-5 1101 1010 1011-4 1100 1011 1100-3 1011 1100 1101-2 1010 1101 1110-1 1001 1110 11110 1000 或 0000 1111 或 0000 00001 0001 0001 00012 0010 0010 00103 0011 0011 00114 0100 0100 01005 0101 0101 01016 0110 0110 01107 0111 0111 0111

55


2.5.6 Ones’ – Complement Representation

(二进制反码表示法 )

The Sign Bit doesn’t change, Other Bits converse based on the Signed-Magnitude. (符号位不变，其余在原码基础上按位取反 )


5656

Representing Negative Numbers: Two’s Complement

Big advantage: Allows us to perform subtraction using addition

Thus, only need adder component, no need for separate subtractor component

57

Two’s Complement Subtractor Built with an

AdderUsing two’s complement

A – B = A + (-B)

= A + (two’s complement of B)

= A + invert_bits(B) + 1So build subtractor using adder by

inverting B’s bits, and setting carry in to 1

58

Two’s Complement Subtractor Built with an

Adder

1Cin

BA

Adder

S

BA

N-bit

59

Summary of this class (本堂课小结 )

Binary, Octal, and Hexadecimal Numbers

(二进制、八进制、十六进制 )

Positional Number System

( 按位计数制 )

1p

ni

ii rdD


60

General Positional-Number-System

Conversion

(常用按位计数制的转换 )

A Number in any Radix to Radix 10 : Expanding

the formula using radix-10 arithmetic

( 任意进制数十进制数：利用位权展开 )

Digital Logic Design and Application ( 数字逻辑设计及应用 )Summary of this class (本堂课小结 )

61


Conversion


A Number in Radix 10 to any Radix : Radix

Multiplication or Division

( 十进制其它进制：基数乘除法 )

Note: Decimal Fraction Parts Conversion

[ 注意：小数部分的转换（误差） ]


62

Addition and Subtraction of Nondecimal Numbers (非十进制的加法和减法 ) （ Table 2-3)

进位输入 Cin 、进位输出 Cout 、本位和S

借位输入 Bin 、借位输出 Bout 、本位差D


63

Representation of Negative Numbers (负数的表示 )

Signed-Magnitude [ 符号－数值（原码） ]

Complement Number Systems (补码数制 )

Radix – Complement (基数补码 )

Diminished Radix – Complement

[ 基数减 1补码（基数反码） ]


64

Binary Signed-Magnitude, Ones’ – Complement, and Two’s – Complement Representation

(二进制的原码、反码、补码 )

正数的原码、反码、补码表示相同

负数的原码表示：符号位为 1

负数的反码表示：

符号位不变，其余在原码基础上按位取反

在 |D| 的原码基础上按位取反（包括符号位）

负数的补码表示：反码 + 1


65

第 2章作业（ P50～ 52）


2.1 (e) (i) 2.2 (e) 2.3 (e) 2.4 2.5 (e) (j) 2.6 (f) 补充：

(125.17)10=?2

2.7 (a)

2.8 (a)

2.9 (b)

2.10 (c)

2.11 +25 – 42

2.18 (b) (d)

2.19

66

A Class Problem

Write the 8-bit Signed-magnitude, Two’s-complement, and One’s- complement for each of these decimal numbers:

+115 -100


67

Chapter 2 Number Systems and codes (数系与

编码 )

Numeric Data － Number Systems and their Conversions

（ —— 数值信息数制及其转换） Nonnumeric Data － Codes

（ —— 非数值信息编码）


68

Review of Chapter 2 (第二章内容回顾 )

Binary, Octal, and Hexadecimal Numbers

(二进制、八进制、十六进制 )

Positional Number System

( 按位计数制 )

1p

ni

ii rdD


69



Conversion


A Number in any Radix to Radix 10 : Expanding

the formula using radix-10 arithmetic

( 任意进制数十进制数：利用位权展开 )


70



Conversion


A Number in Radix 10 to any Radix : Radix

Multiplication or Division

( 十进制其它进制：基数乘除法 )

Note: Decimal Fraction Parts Conversion

[ 注意：小数部分的转换（误差） ]


71


Addition and Subtraction of Nondecimal Numbers (非十进制的加法和减法 ) （ Table 2-3)

进位输入 Cin 、进位输出 Cout 、本位和S

借位输入 Bin 、借位输出 Bout 、本位差D


72


Representation of Negative Numbers (负数的表示 )

Signed-Magnitude [ 符号－数值（原码） ]

Complement Number Systems (补码数制 )

Radix – Complement (基数补码 )

Diminished Radix – Complement

[ 基数减 1补码（基数反码） ]


73


Binary Signed-Magnitude, Ones’ – Complement, and Two’s – Complement Representation

(二进制的原码、反码、补码 )

正数的原码、反码、补码表示相同

负数的原码表示：符号位为 1

负数的反码表示：

符号位不变，其余在原码基础上按位取反

在 |D| 的原码基础上按位取反（包括符号位）

负数的补码表示：反码 + 1


74


An n-bit Two’s- Complement range is (n位二进制补码表示范围 )： – 2 n-1 ~ + ( 2 n-1 – 1) Only one representations of Zero

( 零只有一种表示 ) Obtain a Two’s- Complement

( 二进制补码的求取 )： Ones’ – Complement (反码 ) + 1 （为什

么？？）Expanding the Sign Bit ( 符号位扩展 )


75


Example 2.5.2：Write the 8-bit signed-magnitude, two’s-complement for each of these binary numbers.

(分别写出下面二进制数的 8 –位符号数值码、补码 )

( – 1101 )2

( – 0 . 1101 )2


76



1 、 ( – 1101 )2

2 、 ( – 0 . 1101 )2

1 、 5 位二进制表示：原码反码补码1 1101 1 0010 1 0011

2 、 8 位二进制表示：原码反码补码1000 1101 1111 0010 1111 0011

[ [ D ] 反 ] 反 = D[ [ D ] 补 ] 补 = D

77

2.6 Two’s – Complement Addition and Subtraction (二进制补码的加法和减

法 )Addition Rules: Added by ordinary binary

addition (加法：按普通二进制加法相加 )P.39Subtraction Rules: Taking its two’s

complement, then add

(减法：将减数求补，再相加 )


78


法 )


＋ 2 0010 － 3 1101 ＋＋ 5 ＋ 0101 ＋－ 5 ＋ 1011 ＋ 7 0111 － 8 11000 ＋ 7 0111 ＋ 1 0001 ＋－ 4 ＋ 1100 ＋－ 6 ＋ 1010 ＋ 3 10011 － 5 1011

7979

Adder/Subtractor Example: Calculator

Previous calculator used separate adder and subtractor

DIP switches

10

8-bitregisterCALC

LEDs

e

f

clkld

8

8

80 0

8

8

8

882x10 1

10

wiciA AB B

S Sco wo8-bit adder 8-bit subtractor

8080

Adder/Subtractor Example: Calculator

Improve by using adder/subtractor, and two’s complement numbers

DIP switches10

8-bit register

8-bit adder/subtractorsub

CALC

LEDs

e

S

A Bf

clkld

10

8 8

8

8

81


法 ) Overflow（溢出）如果加法运算产生的和超出了数制表示的范围，则结果发生了溢出（ Overflow）。对于二进制补码，加数的符号相同，和的符号与加数的符号不同。（或者， C in 与 C out 不同） P.41

对于无符号二进制数，若最高有效位上发生进位或借位，就指示结果超出范围。

－ 5 1011 ＋ 7 0111 ＋－ 6 ＋ 1010 ＋＋ 3 ＋ 0011 － 11 10101 ＝＋ 5 ＋ 10 1010 ＝－ 6


8282

OverflowSometimes result can’t be represented with given

number of bitsEither too large magnitude of positive or negativeEx. 4-bit two’s complement addition of 0111+0001 (7+1=8).

But 4-bit two’s complement can’t represent number >70111+0001 = 1000 WRONG answer, 1000 in two’s

complement is -8, not +8Adder/subtractor should indicate when overflow has

occurred, so result can be discarded

8383

Detecting Overflow: Method 1

For two’s complement numbers, overflow occurs when the two numbers’ sign bits are the same but differ from the result’s sign bit If the two numbers’ sign bits are initially different, overflow is impossible

Adding positive and negative can’t exceed largest magnitude positive or negative

0 1 1 1

1 0 0 0

+ 00 0 1

sign bits

overflow(a)

1 1 1 1

0 1 1 1

+ 01 0 0

overflow(b)

1 0 0 0

1 1 1 1

+ 10 1 1

no overflow(c)

If the numbers’ sign bits have the same value, whichdiffers from the result’s sign bit, overflow has occurred.

8484

Detecting Overflow: Method 2

Even simpler method: Detect difference between carry-in to sign bit and carry-out from sign bit

0 1 11 1 1

1

10 010 0 0

+ 00 0 1

overflow(a)

1 1 10 0 0

1

0 1 1 1

+ 01 0 0

overflow(b)

1 0 00 0 0

0

1 1 1 1

+ 10 1 1

no overflow(c)

If the carry into the sign bit column differs from thecarry out of that column, overflow has occurred.

85

2.10 Binary Codes for Decimal Numbers

(十进制数的二进制编码 )


A set of n-bit strings in which different bit strings

Represent different numbers or other things.

( 用于表示不同数或其它事件的一组 n 位二进制码的集合 )

86



How to represent a 1-bit Decimal number with a 4-bit Binary code

( 如何用 4 位二进制码表示 1位十进制码 )？

—— Binary Coded Decimal (BCD码）


87



How to represent a Negative BCD number

(负的 BCD数如何表示 )？Signed-Magnitude Representation: Encoding

of the sign bit is arbitrary

(符号－数值表示：符号位的编码任意 )10’s-complement: 0000 indicates plus, 1001

indicates minus.

(十进制补码表示： 0000正， 1001负 )

Addition of BCD Digits (BCD数的加法 ) P.50


88


01011001

11100110

0100

10001000

00000110

0110

59

14

10

88

16

10＋

＋＋

＋＋

＋＋

4 1

＋6

1

1

修正

修正

01000101

1001

10011001

00100110

1000

45

9

99

18

10＋

＋＋

＋＋

＋8

1

1

修正

89


(十进制数的二进制编码 ) (Table 2-9)


BCD Code2421 CodeExcess-3 (余 3码 )Biquinary Code (二五混合码 )1-out-of-10 (10中取 1码 )

Weighted Code ( 加权码 )

Self-Complement Code 自反码

90


0000

十进制数字

000100100011010001010110011110001001

101010111100110111101111

0000000100100011010010111100110111101111

010101100111100010011010

0011010001010110011110001001101010111100

000000010010110111101111

0100001010001001001000101000011000010000011000010100010010010001010000

00000000000001000001000000110000101

…

1000000000010000000000100000000001000000000010000000000100000000001000000000010000000000100000000001

00000000000000000011000000010100000001100000000111

…

0123456789

BCD(8421)码 2421码余3码二五混合码 10中取1码

未用的码字

91

8421 code

Natural code , just like 4-bit

binary numbers;

Each digit is weighted;

It has 10 valid code words

and 6 invalid code words.

BCD codes

92

Each digit is weighted;

Self-complementing;

Use MSB to express

higher/lower part;

It has 10 valid codes and

6 invalid codes.

2421 codes

BCD codes

93

BCD codes

Excess-3 code

Its digit is not weighted;

8421 code + “0011”;

Self-complementing .

94

Examples:

use BCD code for decimal numbers A = 1946

8421 code : A = 0001 1001 0100 0110

2421 code : A = 0001 1111 0100 1100

Excess-3 code: A = 0100 1100 0111 1001

BCD codes

95

1-out-of-10 code

One hot code:

It is very useful in

control systems.

One hot codes

96

Two hot codes

Biquinary code

7-bits; two hot code;

First 2 bits is one hot code for higher/lower range;

Last 5 bits is one hot code in the range.

Error-detecting property !

97

From one code to

its neighbor, only

one bit changed, no

transition state.

Temperature code

98

2.11 Gray code（格雷码）


110

101

100 011

000111

001

010

0 0 1

101

111

110 010

000100

001

011

0 0 1

99



特点：任意相邻码字间只有一位数位变化最高位的 0和 1只改变一次最大数回到 0也只有一位码元不同

110

101

100 011

000111

001

010

0 0 1

101

111

110 010

000100

001

011

0 0 1

100



构造方法直接构造 The bits of an n-bit binary cord word are

numbered from right to left, from 0 to n-1. [ 对 n 位二进制的码字从右到左编号（ 0 ~ 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.

( 若二进制码字的第 i 位和第 i + 1 位相同，则对应的格雷码码字的第 i 位为 0，否则为 1。 )

Reflected Code（反射码）

101

Gray codes

Target: code for continues changed numbers (in

binary system) to prevent wrong code happened

in transition time;

Property : In each pair of successive code words,

only one bit changes.

102

Gray codes

From binary number to Gray code

The width is same, the MSB is same;

From left to right, if a bit in binary number is same

as its left bit, the gray code is 0, if it is different, the

gray code is 1.

Examples:

binary number: 1001 0010 0110 0011

Gray codes: 1101 1011 0101 0010

103

Error-detecting code

Information word + checking bit

104


2.12 Character Codes (字符编码 ) ASCII码（ P36 表 2-11）

ASCII code： 128 Keyboard signs , 7-

bit

Used for keyboard or display device

105


2.13 Codes for Actions, Conditions, and States

(动作、条件和状态的编码 ) 使用 b 位二进制编码来表示 n 个不同状态

Word: a digital string to represent an object

Use n bits, we can make 2n different words;

To make n words, you must use bits. n2log

106


2.16 Codes for Serial Data Transmission and Storage

(用于串行数据传输与存储的编码 )

Parallel way

use n-line to transmit an n-bits code words ;

transmit an n-bits code words in one time period;

Serial way

use one line to transmit an n-bits code words ;

transmit an n-bits code words in n time period;

107


第 2 章掌握重点正数的十进制、二进制、八进制、十六进制表示，以及它们之间的相互转换

符号数的 S-M码、补码、反码表示，以及它们之间的相互转换；

带符号数的补码的加减运算 :溢出的判断

BCD码、 GRAY码：表示方法、特点

108

第 2章作业（ P75～ 78）


2.12 (b) (c) 2.34 2.35 2.38 2.39 2.22(选做 ) 2.30 (选做 ) 2.51 (选做 )

109

Chapter 3 Digital Circuits (数字电路 )

Give a knowledge of the Electrical aspects of Digital Circuits

(介绍数字电路中的电气知识 )


110

Consider some Questions(思考几个问题 )

在模拟的世界中如何表征数字系统？如何将物理上的实际值映射为逻辑上的 0 和 1 ？什么时候考虑器件的逻辑功能；什么时候考虑器件的模拟特性？


111

A Class Problem ( 每课一题 )

Indicate whether or not overflow occurs when adding the following 8-bit two’s-complement numbers:


1101 0100

+ 1010 1011

0010 0110

+ 0101 1010

数字逻辑设计及应用

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