Digital Logic Design and Application (数字逻辑设计及应用)
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。
1
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 1 (第一章内容回顾)
 Analog
 Digital
versus Digital (模拟与数字)
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)
2
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 1 (第一章内容回顾)
 Integrated
Circuit（IC，集成电路）
Logic Devices（PLA、PLD、
CPLD、FPGA, 可编程逻辑器件）
 Programmable

Application-Specific ICs（ASIC, 专用集成电路
）

3
Printed-Circuit Boards (PCB, 印制电路板)
Digital Logic Design and Application (数字逻辑设计及应用)
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)
— Number Systems and their Conversions
(数制及其转换)

Nonnumeric Data (非数值信息)
-- Nonnumeric Data Representation – Codes
(非数值信息的表征 -- 编码)
4
Digital Logic Design and Application (数字逻辑设计及应用)
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
5
How to Encode Text: ASCII,
Unicode
 ASCII:
7- (or 8-) bit encoding of each
letter, number, or symbol
Encoding Symbol
Sample ASCII encodings
010 0000 <space>␠
Encoding Symbol
010 0001
!
100 0001
A
010 0010
"
100 0010
B
010 0011
#
100 0011
C
010 0100
$
100 0100
D
010 0101
%
100 0101
E
010 0110
&
100 0110
F
010 0111
'
100 0111
G
010 1000
(
100 1000
H
010 1001
)
100 1001
I
010 1010
*
100 1010
J
010 1011
+
100 1011
K
010 1100
,
100 1100
L
010 1101
100 1101
M
010 1110
.
010 1111
/
6
Encoding Symbol
100 1110
N
100 1111
O
101 0000
P
101 0001
Q
101 0010
R
101 0011
S
101 0100
T
101 0101
U
101 0110
V
101 0111
W
101 1000
X
…
101 1001
Y
101 1010
Z
Encoding Symbol
110 0001
110 0010
...
111 1001
111 1010
a
b
011 0000
011 0001
011 0010
011 0011
011 0100
011 0101
011 0110
…
011 0111
011 1000
011 1001
0
1
2
3
4
5
6
7
8
9
y
z
6
How to Encode Text: ASCII,
Unicode
 Unicode:
Increasingly popular 16-bit
encoding
 Encodes
characters from various world
languages
……
7
7
How to Encode Numbers: Binary
Numbers
• Each position represents a base quantity; symbol in
position means how many of that quantity
Base
ten (decimal)
symbols: 0, 1, 2, ..., 8,
and 9
 More than 9 -- next
position
5
2
3
 Ten
10
4
10
3
10
2
10
1
10
0
 So
each position
power （幂） of 10
 Nothing
special about
base 10 -- used because
we have 10 fingers
8
8
How to Encode Numbers: Binary
Numbers
• Each position represents a base quantity; symbol in
position means how many of that quantity
Base
two
(binary)
 Two
symbols: 0
and 1
 More than 1 -- next
position
 So each
position power
（幂） of 2
9
1 0 1 Q: How much?
24 23 22 21 20
+
=
a
4
+ 1 = 5
There are only 10 types of people in the world: those
who understand binary, and those who don't.
9
Useful to know powers of 2:
29
28
27
26
25
24
23
22
21
20
512 256 128 64
32
16
8
4
2
1
Practice counting up by powers of 2:
512
256 128 64 32 16 8 4 2 1
10
10
Digital Logic Design and Application (数字逻辑设计及应用)
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

p 1
d
i  n
i
r
i
Weight of i bit; Base or
Radix of r Number System
(第i位的权； r 计数制的基数)
11
推广：
D2 = ∑ d i × 2i
D16= ∑ d i × 16i
Digital Logic Design and Application (数字逻辑设计及应用)
2.1 Positional Number System
(按位计数制)
 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
12
Digital Logic Design and Application (数字逻辑设计及应用)
2.1 Positional Number System
(按位计数制)
Decimal and Binary
Decimal system: base is 10, the digit may be 0 to 9
101.110  110
2
 0 10  110  110
1
0
1
Binary system: base is 2, the digit may be 0 or 1
101.12  1 22  0  21  1 20  1 21
13
bit: one digit in binary system;
Using Digital Data in a Digital
System
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
temperature sensor
0
0
1
0
0
0
0
1 "33"
14
14
Using Digital Data in a Digital
System
A
display
converts its ASCII
input to the
corresponding
letter
temperature sensor
0 0 1 0 0 0 0 1 "33"
Digital System
if (input <= "00100000") // "32"
output = "1000110" // "F"
else if (input >= "11010100") // "212"
output = "1000010" // "B"
else
output = "1001110" // "N"
1 0 0 1 1 1 0
display
"N"
N
15
15
Digital Logic Design and Application (数字逻辑设计及应用)
2.2 Octal and Hexadecimal Numbers
(八进制和十六进制)
数码
特性
8
0~7
逢八进一
2
0,1
逢二进一
基数
Octal Number
(八进制)
Binary Number
(二进制)
Hexadecimal
Number(十六进制)
16
16
0~9,A~F
逢十六进一
Digital Logic Design and Application (数字逻辑设计及应用)
说 明
 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.
Significant Bit（MSB, 最高有效位）
 Least Significant Bit（LSB, 最低有效位）
 Most
 1011100010112
MSB
17
LSB
Digital Logic Design and Application (数字逻辑设计及应用)
表2.1 十进制、二进制、八进制与十六进制数
18
十进制
二进制
八进制
十六进制
0
1
2
0000
0001
0010
0
1
2
0
1
2
3
4
5
0011
0100
0101
3
4
5
3
4
5
6
7
0110
0111
6
7
6
7
8
9
10
1000
1001
1010
10
11
12
8
9
A
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
A:
F0
Q: Convert hex A01 to binary
A:
1010 0000 0001
19
19
Digital Logic Design and Application (数字逻辑设计及应用)
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.B16
20
1000110010012 = (
= (
)8
)16
Hex Example: RFID Tag
 Batteryless（无电池）
tag powered（功率
） by radio field
 Transmits（发送）
unique identification（
鉴定） number
 Example:
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
(b)
Province #
(c) Province: 7 City: 160
(d) 00000111 10100000
(e)
07
A0
(a)
RFID
tag
21
City #
(f)
Animal #
Animal : 513
00000010 00000001
02 01
Tag ID in hex: 07A00201
21
Digital Logic Design and Application (数字逻辑设计及应用)
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 = (
( 7F.8 )16 = (
)10
)10
More easy way(更简便的方法)？
22
( F1AC )16 = ( ( ( F16 ) +1 )  16 + A )  16 + C
Digital Logic Design and Application (数字逻辑设计及应用)
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
23
Digital Logic Design and Application (数字逻辑设计及应用)
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-n <= < 10-2
)2
n=7
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
24
Converting To/From Binary by
Hand: Summary
25
25
Divide-By-2 Method Common in
Automatic Conversion
 Repeatedly divide decimal number by
2, place remainder in current binary
digit (starting from 1s column)
26
1. Divide decimal number by 2
Insert remainder into the binary number
Continue since quotient (6) is greater than 0
Decimal
6
2 12
–12
0
Binary
0
1
(current value: 0)
2. Divide quotient by 2
Insert remainder into the binary number
Continue since quotient (3) is greater than 0
3
2 6
–6
0
0 0
2 1
(current value: 0)
3. Divide quotient by 2
Insert remainder into the binary number
Continue since quotient (1) is greater than 0
1
2 3
–2
1
1 0 0
4 2 1
(current value: 4)
4. Divide quotient by 2
Insert remainder into the binary number
Quotient is 0, done
0
2 1
–0
1
1 1 0 0
8 4 2 1
(current value: 12)
Note:
Works for
any base
N—just
divide by
N instead
26
Bytes, Kilobytes, Megabytes, and
More
 Byte:
8 bits
 Common
metric prefixes:
(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
 kilo
27
27
Bytes, Kilobytes, Megabytes, and
More
 BUT,
metric prefixes also commonly
used inaccurately
= 65536 commonly written as “64
Kbyte”
 Typical when describing memory sizes
 216
watch out for “KB” for kilobyte vs.
“Kb” for kilobit
 Also
28
28
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
01001001
(a)
Q
:
128 64 32 16 8 4 2 1
sum: 64
72
(b)
73
29
29
Example: DIP-Switch Controlled
Channel
DIP switch
1
0
channel receiver
"34" 0 0 1 0 0 0 1 0 "73" 0 1 0 0 1 0 0 1
InA
if (InA = InB)
Out = 1
else
Ceiling fan Out = 0
module
Out
0
30
InB
(c)
Digital Logic Design and Application (数字逻辑设计及应用)
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)
31
10 – 1 = 1
Digital Logic Design and Application (数字逻辑设计及应用)
2.4 Addition and Subtraction of
Non-decimal Numbers (非十进制数的加法和减法)
•Carry in (进位输入): C in (P.32)
• Carry out ( 进位输出 ) C out
Sum ( 本位和 ): S
1011 1110
+ 1000 1101
32
Digital Logic Design and Application (数字逻辑设计及应用)
表2.3.1 二进制加法真值表
输 入
33
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
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
1
0
1
0
0
1
0
0
0
1
0
1
1
1
Adder Example: DIP-Switch-Based
Adding Calculator
 Goal:
Create calculator that adds two 8bit binary numbers, specified using DIP
switches
DIP
switch: Dual-In-line Package
switch, move each switch up or down
Solution:
Use 8-bit adder
34
34
Adder Example: DIP-Switch-Based
Adding Calculator
Solution: Use 8-bit adder
DIP switches
1
0
a7……..a0 b7……..b0
8-bit carry-ripple adder ci
co s7…………s0
0
CALC
LEDs
35
Adder Example: DIP-Switch-Based
Adding Calculator
 To prevent spurious(假的） values from
appearing at output, can place register
at output
 Actually,
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.
36
36
Adder Example: DIP-Switch-Based
Adding Calculator
DIP switches
1
0
a7…….a0 b7……b0
ci 0
8-bit adder
co s7………s0
e
clk
ld 8-bit register
CALC
LEDs
37
Digital Logic Design and Application (数字逻辑设计及应用)
2.4 Addition and Subtraction of
Non-decimal Numbers (非十进制数的加法和减法)
•Borrow in ( 借位输入 ): Bin
•Borrow out ( 借位输出 ): Bout
•Difference bit ( 本位差 ): D
1010 1010
– 0101 0101
38
Digital Logic Design and Application (数字逻辑设计及应用)
表2.3.2 二进制减法真值表
输 入
39
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
Subtractor Example: DIP-Switch
Based Adding/Subtracting Calculator
 Extend
earlier calculator example
 Switch
f indicates whether want to add
(f=0) or subtract (f=1)
 Use subtractor and 2x1 mux
40
Subtractor Example: DIP-Switch
Based Adding/Subtracting Calculator
DIP switches
1
0
8
1
0
clk
f
e
8
A
B ci
8-bit adder
co S
8
ld
8
8
0
A
B wi
8-bit subtractor
wo S
0
8
0 2x1 1
8
8-bit register
CALC
8
LEDs
41
Digital Logic Design and Application (数字逻辑设计及应用)
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
 00101110＝＋46
 00000000＝＋0
42
11111111＝－127
10101110＝－46
10000000＝－0
Digital Logic Design and Application (数字逻辑设计及应用)
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)
43
Digital Logic Design and Application (数字逻辑设计及应用)
2.5 Representation of Negative
Numbers (负数的表示)
 2.5.2
Complement Number Systems
(补码数制)
– Complement
(基数补码)
Diminished Radix – Complement
[ 基数减1补码 (反码) ]
Radix
44
Digital Logic Design and Application (数字逻辑设计及应用)
2.5 Representation of Negative
Numbers (负数的表示)
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
45
: Table 2-4 P.36
Digital Logic Design and Application (数字逻辑设计及应用)
2.5 Representation of Negative
Numbers (负数的表示)
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]
Diminished
 Example
46
: Table 2-4 2-5 P.36
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
1
2
3
4
5
6
7
8
9
9
8
7
6
5
4
3
2
1
47
47
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
complements
1
9
2
8
3
7
4
6
5
5
6
4
7
3
8
2
9
1
0
48
10
10
4
6
7
0
3
–4
10
+6
20
13
13
3
7–4=3
7+6=13
3
Adding the complement results in an answer that is
exactly 10 too much – dropping the tens column gives
the right answer.
48
Digital Logic Design and Application (数字逻辑设计及应用)
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 + 2n1 (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？？）
49
Digital Logic Design and Application (数字逻辑设计及应用)
2.5.4 Two’s – Complement Representation
(二进制补码表示法)
 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：
11111111
 减去－119绝对值： 0 1 1 1 0 1 1 1

－119的反码：
10001000

加1：
1

－119补码：
10001001
50
Digital Logic Design and Application (数字逻辑设计及应用)
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 )

 逐位取反

 约定8位
51
0＝0 0 0 0 0 0 0 0
11111111
＋1
0 0 0 0 0 0 0 0＝0
Digital Logic Design and Application (数字逻辑设计及应用)
2.5.4 Two’s – Complement Representation
(二进制补码表示法)
 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
52
Digital Logic Design and Application (数字逻辑设计及应用)
2.5.4 Two’s – Complement Representation
(二进制补码表示法)
Positive
number has the same：
Sign-Magnitude,
 Ones’ – Complement,
 and Two’s- Complement
( 正数的原码、反码、补码相同)
53
54
十进制
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
表2-6 1位十进制数与4位二进制数
二进制原码
二进制反码
二进制补码
——
——
1000
1111
1000
1001
1110
1001
1010
1101
1010
1011
1100
1011
1100
1011
1100
1101
1010
1101
1110
1001
1110
1111
0000
1000或0000
1111或0000
0001
0001
0001
0010
0010
0010
0011
0011
0011
0100
0100
0100
0101
0101
0101
0110
0110
0110
0111
0111
0111
Digital Logic Design and Application (数字逻辑设计及应用)
2.5 Representation of Negative
Numbers (负数的表示)
 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 SignedMagnitude.
(符号位不变，其余在原码基础上按位取反)
55
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
56
56
Two’s Complement Subtractor
Built with an Adder
 Using
two’s complement
A – B = A + (-B)
= A + (two’s complement of B)
= A + invert_bits(B) + 1
 So
build subtractor using adder by
inverting B’s bits, and setting carry in
to 1
57
Two’s Complement Subtractor
Built with an Adder
A
B
N-bit
A
B
Adder
S
58
1
Cin
Digital Logic Design and Application (数字逻辑设计及应用)
Summary of this class (本堂课小结)
 Positional
Number System
(按位计数制)
D
p 1
i
d

r
 i
i  n
 Binary,
Octal, and Hexadecimal Numbers (
二进制、八进制、十六进制)
59
Digital Logic Design and Application (数字逻辑设计及应用)
Summary of this class (本堂课小结)
 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)
60
Digital Logic Design and Application (数字逻辑设计及应用)
Summary of this class (本堂课小结)
 General
Positional-Number-System
Conversion
(常用按位计数制的转换)
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）]
61
Digital Logic Design and Application (数字逻辑设计及应用)
Summary of this class (本堂课小结)

Addition and Subtraction of Nondecimal Numbers
(非十进制的加法和减法) （Table 2-3)
62

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
Digital Logic Design and Application (数字逻辑设计及应用)
Summary of this class (本堂课小结)

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）]
63
Digital Logic Design and Application (数字逻辑设计及应用)
Summary of this class (本堂课小结)

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
64
+1
Digital Logic Design and Application (数字逻辑设计及应用)
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)
65

2.8 (a)

2.9 (b)

2.10 (c)

2.11 +25 – 42

2.18 (b) (d)

2.19
Digital Logic Design and Application (数字逻辑设计及应用)
A Class Problem
 Write
the 8-bit Signed-magnitude,
Two’s-complement, and One’scomplement for each of these decimal
numbers:
+115
66
-100
Digital Logic Design and Application (数字逻辑设计及应用)
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）
67
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 2 (第二章内容回顾)
 Positional
Number System
(按位计数制)
D
p 1
i
d

r
 i
i  n
 Binary,
Octal, and Hexadecimal Numbers (
二进制、八进制、十六进制)
68
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 2 (第二章内容回顾)
 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)
69
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 2 (第二章内容回顾)
 General
Positional-Number-System
Conversion
(常用按位计数制的转换)
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）]
70
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 2 (第二章内容回顾)

Addition and Subtraction of Nondecimal Numbers
(非十进制的加法和减法) （Table 2-3)
71

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
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 2 (第二章内容回顾)

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）]
72
Digital Logic Design and Application (数字逻辑设计及应用)
Review of Chapter 2 (第二章内容回顾)

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
73
+1
Digital Logic Design and Application (数字逻辑设计及应用)
2.5.4 Two’s – Complement Representation
(二进制补码表示法)
 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 )
74
Digital Logic Design and Application (数字逻辑设计及应用)
2.5 Representation of Negative
Numbers (负数的表示)
 Example
2.5.2：Write the 8-bit signedmagnitude, 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
75
Digital Logic Design and Application (数字逻辑设计及应用)
2.5 Representation of Negative
Numbers (负数的表示)
1、( – 1101 )2
2、( – 0 . 1101 )2
[ [ D ] 反 ]反 = D
1、5位二进制表示：
原码 反码
补码
1 1101 1 0010 1 0011
[ [ D ] 补 ]补 = D
76
2、8位二进制表示：
原码
反码
补码
1000 1101 1111 0010 1111 0011
Digital Logic Design and Application (数字逻辑设计及应用)
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.39
 Subtraction Rules: Taking its two’s
complement, then add
(减法：将减数求补，再相加)
77
Digital Logic Design and Application (数字逻辑设计及应用)
2.6 Two’s – Complement Addition and
Subtraction (二进制补码的加法和减法)



＋2
＋ ＋5
＋7
0010
＋ 0101
0111
－3
1101
＋ －5 ＋ 1011
－8 11000
＋7
＋ －4
＋3
0111
＋ 1100
10011
＋1
0001
＋ －6 ＋ 1010
－5
1011




78
Adder/Subtractor Example:
Calculator
 Previous
calculator used separate adder
and subtractor
DIP switches
1
0
8
A
1
0
clk
8
B
8-bit adder
co
S
f
8
e
ld
8
ci 0
0
8
A
B wi 0
8-bit subt ractor
wo S
2x 1
8
1
8
8-bit register
CALC
8
LEDs
79
79
Adder/Subtractor Example:
Calculator
 Improve
by using adder/subtractor, and
two’s complement numbers
DIP switches
1
0
8
f
1
0
e
clk
8
A
B
sub 8-bit adder/subtractor
S
8
ld
8-bit register
CALC
8
LEDs
80
80
Digital Logic Design and Application (数字逻辑设计及应用)
2.6 Two’s – Complement Addition and
Subtraction (二进制补码的加法和减法)

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
81
Overflow
 Sometimes
result can’t be represented
with given number of bits
 Either
82
too large magnitude of positive or
negative
 Ex. 4-bit two’s complement addition of
0111+0001 (7+1=8). But 4-bit two’s
complement can’t represent number >7
0111+0001 = 1000 WRONG answer, 1000 in
two’s complement is -8, not +8
 Adder/subtractor should indicate when
overflow has occurred, so result can be
82
discarded
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
sign bits
0 1 1 1
1 1 1 1
1 0 0 0
+0 0 0 1
+1 0 0 0
+0 1 1 1
1 0 0 0
0 1 1 1
1 1 1 1
overflow
overflow no overflow
(a)
(b)
(c)
If the numbers’ sign bits have the same value, which
differs from the result’s sign bit, overflow has occurred.83
83
Detecting Overflow: Method 2

Even simpler method: Detect difference
between carry-in to sign bit and carry-out
from sign bit
000
000
111
0111
1111
1000
+0 0 0 1 +1 0 0 0 +0 1 1 1
0 1 0 0 0 10 1 1 1 01 1 1 1
overflow overflow no overflow
(a)
(b)
(c)
If the carry into the sign bit column differs from the
carry out of that column, overflow has occurred.
84
84
Digital Logic Design and Application (数字逻辑设计及应用)
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)
85
Digital Logic Design and Application (数字逻辑设计及应用)
2.10 Binary Codes for Decimal Numbers
(十进制数的二进制编码)

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)？
 ——
86
Binary Coded Decimal (BCD码）
Digital Logic Design and Application (数字逻辑设计及应用)
2.10 Binary Codes for Decimal Numbers
(十进制数的二进制编码)
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
Digital Logic Design and Application (数字逻辑设计及应用)
0101
5
＋ 9 ＋ 1001
14
1110
＋ 0110
10 ＋ 4 1 0100
8
1000
＋ 8 ＋ 1000
16 1 0000
＋ 0110
10 ＋ 6 1 0110
88
0100
4
＋ 5 ＋ 0101
9
1001
修正
修正
9
1001
＋ 9 ＋ 1001
18 1 0010
＋ 0110
10 ＋ 8 1 1000
修正
Digital Logic Design and Application (数字逻辑设计及应用)
2.10 Binary Codes for Decimal Numbers
(十进制数的二进制编码) (Table 2-9)
 BCD
Code
Weighted Code (加权码)
 2421 Code
Self-Complement Code自反码
 Excess-3
(余3码)
 Biquinary Code (二五混合码)
 1-out-of-10 (10中取1码)
89
Digital Logic Design and Application (数字逻辑设计及应用)
十进制数字
BCD(8421)码
2421码
余3码
二五混合码
10中取1码
0
1
2
3
4
5
6
7
8
9
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
0000
0001
0010
0011
0100
1011
1100
1101
1110
1111
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
0100001
0100010
0100100
0101000
0110000
1000001
1000010
1000100
1001000
1010000
1000000000
0100000000
0010000000
0001000000
0000100000
0000010000
0000001000
0000000100
0000000010
0000000001
0000000
0000001
0000010
0000011
0000101
…
0000000000
0000000011
0000000101
0000000110
0000000111
…
未用的码字
1010
1011
1100
1101
1110
1111
90
0101
0110
0111
1000
1001
1010
0000
0001
0010
1101
1110
1111
BCD codes
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.
91
BCD codes
2421 codes
Each digit is weighted;
Self-complementing;
Use MSB to express
higher/lower part;
It has 10 valid codes and
6 invalid codes.
92
BCD codes
Excess-3 code
Its digit is not weighted;
8421 code + “0011”; Selfcomplementing .
93
BCD codes
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
94
One hot codes
1-out-of-10 code
One hot code:
It is very useful in
control systems.
95
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 !
96
Temperature code
From one code to
its neighbor, only
one bit changed, no
transition state.
97
Digital Logic Design and Application (数字逻辑设计及应用)
2.11 Gray code（格雷码）
111
000
100
001
110
000
001
101
001
101
010
100
98
011
001
111
011
110
010
Digital Logic Design and Application (数字逻辑设计及应用)
2.11 Gray code（格雷码）
 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
111
000
100
001
110
000
001
101
0 0 1
101
010
100
99
011
0 0 1
111
011
110
010
Digital Logic Design and Application (数字逻辑设计及应用)
2.11 Gray code（格雷码）
 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）
100
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.
101
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:
102
binary number: 1001 0010
0110 0011
Gray codes:
0101 0010
1101 1011
Error-detecting code
Information word + checking bit
103
Digital Logic Design and Application (数字逻辑设计及应用)
 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
104
Digital Logic Design and Application (数字逻辑设计及应用)
 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 log n bits.
2
105
Digital Logic Design and Application (数字逻辑设计及应用)
 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;
106
Digital Logic Design and Application (数字逻辑设计及应用)
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
107
Digital Logic Design and Application (数字逻辑设计及应用)
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)
108
Digital Logic Design and Application (数字逻辑设计及应用)
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)
109
Digital Logic Design and Application (数字逻辑设计及应用)
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？

110
Digital Logic Design and Application (数字逻辑设计及应用)
A Class Problem ( 每课一题 )
 Indicate
whether or not overflow occurs
when adding the following 8-bit two’scomplement numbers:
111
1101 0100
0010 0110
+ 1010 1011
+ 0101 1010