数字电路与逻辑设计英文教学PPT
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数字逻辑电路英文课件 (1)Introduction
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With memory and states
The arrangement of the text
Teach in 32 times: Arithmetic and device
Char.1,2,3 7 Combinational circuit
Char.4,6 10 Sequential circuit
Buffer
yx
NOT gate (inverter , INV)
y x
Equation Diagram Truth table
Basic device in digital system
AND gate
y x1 x2
equation
Truth table
diagram
Basic device in digital system
Signals in electric circuit
Real signal : voice, music, moving picture… Analog signal: voltage changed with time Digital signal : sampling values from analog
Digital system
Any inputs and outputs can only be 1 or 0 !
For each output, its state must be determined by all the inputs.
Basic device in digital system
Char.7,8 13 Memory and ADC/DAC 2
Exercise and Exams : 100
数字电路英文版第四单元PPT课件
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A(B+C )=AB+AC
第19页/共135页
9.
B
B+C
A
B C
X
A
A
C
AB X
AC
X = A( B + C ) X = A B + A C
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10.
Rules of Boolean Algebra
1. A + 0 = A 7. A * A = A
2. A + 1 = 1 8. A * A = 0
• “Don’t care” A combination of input literals that cannot occur and can be used as a 1 or a 0 on a Karnaugh map.
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• Literal A variable or the complement of a variable. • Product-of-sums (POS) A form of Boolean expression that is
第3页/共135页
• Commutative law In addition (ORing) and multiplication (ANDing) of two variables, the order in which the variables are ORed orANDed makes no diffence.
A+B=B+A
A
A+B
B
B
A
第15页/共135页
B+A
5.
Logical Multiplication
第19页/共135页
9.
B
B+C
A
B C
X
A
A
C
AB X
AC
X = A( B + C ) X = A B + A C
第20页/共135页
10.
Rules of Boolean Algebra
1. A + 0 = A 7. A * A = A
2. A + 1 = 1 8. A * A = 0
• “Don’t care” A combination of input literals that cannot occur and can be used as a 1 or a 0 on a Karnaugh map.
第5页/共135页
• Literal A variable or the complement of a variable. • Product-of-sums (POS) A form of Boolean expression that is
第3页/共135页
• Commutative law In addition (ORing) and multiplication (ANDing) of two variables, the order in which the variables are ORed orANDed makes no diffence.
A+B=B+A
A
A+B
B
B
A
第15页/共135页
B+A
5.
Logical Multiplication
数字逻辑设计及应用教学英文课件:Lec18-chap 7
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Drive output equation
MAX=Q1 · Q0· EN (3)
11
MAX=…….
Digital Logic Design and Application
Example state machine analysis
Construct state /output table
Q0*= Q0 · EN’+Q0’EN
F (Q, I)
Combinational
Derive output equation from the circuit diagram
Y=G (Q, I)
Combinational
Determine transition equation
Q*=T (F (Q, I))
Sequential
Construct transition/output table
Draw state (transition)diagram
9
Draw timing diagram (optional)
Digital Logic Design and Application
Example state machine
Mealy machine Moore machine
EN EN
F
State Memory
Output logic
G
Next state=F(current state, inputs)
Mealy machine的名字来自这个概念的提出者,在1951年写了A Method for Synthesizing Sequential Circuits的状态机的先驱 G. H. Mealy
D1=Q1 · EN’+Q1’ · Q0 · EN+ Q1 · Q0’ · EN
数字电路系统设计中英文课件教程 07 时序逻辑电路原理-Sequential Logic Design Principles (1)
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所有的时序电路对亚稳态都是敏感的
metastable 亚稳态
stable
稳态
stable
稳态
7.2 Latches and Flip-Flops (锁存器与触发器)
—— The Basic Building Blocks of most Sequential Circuits. (大多数时序电路的基本构件)
Clock Frequency: The Reciprocal of the Clock Period
(时钟频率:时钟周期的倒数。)
Clock Tick: The First Edge of Pulse in a clock period or sometimes the period itself.
DIGITAL SYSTEM DESIGN
ESHINE
eshine.li@
Chapter 7 Sequential Logic Design Principles ( 时序逻辑设计原理 )
Latches and Flip-Flops (锁存器和触发器 ) Clocked Synchronous State-Machine Analysis (同步时序分析) Clocked Synchronous State-Machine Design (同步时序设计)
Basic Concepts (基本概念)
Sequential Logic Circuit (时序逻辑电路) Clock Period: The Time between Successive transitions in the same direction.
(时钟周期:两次连续同向转换之间的时间。)
Latches(锁存器)
metastable 亚稳态
stable
稳态
stable
稳态
7.2 Latches and Flip-Flops (锁存器与触发器)
—— The Basic Building Blocks of most Sequential Circuits. (大多数时序电路的基本构件)
Clock Frequency: The Reciprocal of the Clock Period
(时钟频率:时钟周期的倒数。)
Clock Tick: The First Edge of Pulse in a clock period or sometimes the period itself.
DIGITAL SYSTEM DESIGN
ESHINE
eshine.li@
Chapter 7 Sequential Logic Design Principles ( 时序逻辑设计原理 )
Latches and Flip-Flops (锁存器和触发器 ) Clocked Synchronous State-Machine Analysis (同步时序分析) Clocked Synchronous State-Machine Design (同步时序设计)
Basic Concepts (基本概念)
Sequential Logic Circuit (时序逻辑电路) Clock Period: The Time between Successive transitions in the same direction.
(时钟周期:两次连续同向转换之间的时间。)
Latches(锁存器)
数字电路系统设计中英文课件教程 04 组合逻辑设计原理-combinational logic design principles
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4.1.4 n-Variable Theorems (n变量定理)
Generalized idempotency theorem
( 广义同一律 )
X+X+…+X=X X· · · =X X … X Shannon’s expansion theorems
( 香农展开定理 )
F ( X 1 , X 2 ,, X 1 )
1 1 0 0
1 0 1 0
1 1 1 0
Logic function and its expressions
Voting circuit
C A Switch ABC 1stands for close Truth table B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 Y 0 0 0 0 0 1 1 1
Basic Concepts (基本概念)
Two Types of Logic Circuits(逻辑电路分为两大类): Combinational Logic Circuit(组合逻辑电路)
Outputs depend only on its Current Inputs.
(任何时刻的输出仅取决与当时的输入) 电路特点:无反馈回路、无记忆元件
A 0 Ylight B 0 1 stands for light 0 Y = F (A,B,C ) = A·(B+C) 0 1 Logic A function 1 & 1 B Y ≥1 1 C
Logic diagram
Logic Expression to Truth Table (逻辑表达式 真值表)
' X 1 F (1, X 2 ,, X 1 ) X 1 F (0, X 2 ,, X 1 )
数字电路英文版PPT 第二单元
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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. 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.
BCD : Binary coded decimal; a digital
code in which each of the decimal digits, 0 through 9, is represented by group of four bits. Byte : A group of eight bits. Carry : The digit generated when the sum of two binary digits exceeds 1.
Weight [ 权 ] Carry [ 进 位 Remainder [ 余数 ] Quotient [ 商 Integer [ 整数 ] Fraction [ 小数 1’Complement [ 反码 ] 2’Complement [ 补码 ] Format [ 格式 ] Precision [ 精度 Mantissa [ 尾数 ]
Largest decimal number = 2n – 1 if n = 5 , 25 – 1 = 32 – 1 = 31 if n = 6 , 26 – 1 = 64 – 1 = 63
数字电路系统设计中英文课件教程 01 数字系统简介-introduction to Digital System Design
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Combinational Circuits(组合电路) Sequential Circuits(时序电路)
Integrated Circuits
• Classified by size
SSI MSI LSI VLSI
(small-scale integration) (medium) (large) (very large)
Analyse, Design, Test
• Analyse logic relationship between input & output Boolean algebra
Truth table, Functional diagram, Boolean expression, Waveform
The most basic digital devices (AND Gate, OR Gate, and NOT Gate or Inverter)
最基本的数字器件(与、或、非门或反相器)
Flip-flops(触发器): A device that stores either 0 or 1
一种能存储 0 或 1 的器件
• Design circuit • Test
Computer-aided design(CAD)
Computer-aided engineering (CAE)
1. Scheme & HDL
Computer-aided design(CAD)
Computer-aided engineering (CAE)
结果再现性(稳定可靠、精度更高)
– Ease of design, Flexibility, and Functionality
数字逻辑电路英文课件Minimization logic circuit Karnaugh map
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More examples of minimal sum
Some concept
Imply :a rectangular set of 1s; Prime implicant: a maxim imply; Complete sum: sum of all the prime implicant; Distinguished 1-cell: only in one prime implicant; Essential prime implicant :with distinguished 1cell in it.
Minimal product Get a minimal sum-product equation
Step 1 get minimal sum for inverse function Step 2 use DeMorgen’s theorem to get minimal product of the function
Can be minimized by karnaugh map
Байду номын сангаас
Minimal sum Get a minimal product-sum equation
Step 1 :Make rectangular sets of 1s Cover all the 1s in the Kanaugh maps; The cell-numbers in any sets must be 2i ; The number of sets must be minimal ! Each set must be maximal ! (Any 1s can be reused in the process !)
examples:
数字电路与逻辑设计英文教学PPT课件
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blows selected internal fuse links. After blowing the fuses,
the array represents the Boolean logic expression for the
desired circuit.
A
A
B
B
What function is
PALs have a one time
A
A
B
B
programmable (OTP)
array, in which fuses are
permanently blown,
X
creating the product
terms in an AND array.
Simplified AND-OR array
Summary
Programmable Logic
Programmable Logic Devices (PLDs) are ICs with a large number of gates and flip flops that can be configured with basic software to perform a specific logic function or perform the logic for a complex circuit. Major types of PLDs are:
SPLD: (Simple PLDs) are the earliest type of array logic used for fixed functions and smaller circuits with a limited number of gates. (The PAL and GAL are both SPLDs).
数字逻辑电路英文课件 (20)Sequential logic design example
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States analyze examples: combination lock
Step 3: finish the state diagram with incorrect input sequence ;
States analyze examples: combination lock
Step 4: from the state diagram to the
Sequential logic design example
Design a generator for the periodic sequence (10110) as the clk triggered.
The state diagram
The state/output table
A generator for the periodic sequence
The output equation
CO Q2'Q1'Q0'a'Q2Q1Q0 a a combination lock:
The UNLOCK should be 1 if and only if X is 0 and the sequence of inputs received on X at the preceding seven clock ticks was 0110111. The HINT output should be 1 if and only if the current value of X is the correct one to move the machine closer to being in the “unlocked” state (with UNLK = 1) .
数字电路与逻辑设计课件-Verilog语言总结
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specparam、event
数字电路与逻辑设计2016
7
Verilog HDL语言基础
Verilog HDL数据类型和变量
例: wire a,b; wire [7:0] c; reg [7:0] buffer;
数字电路与逻辑设计2016
8
Verilog HDL语言基础
基础运算符
与(&&)、或(||)、非(!) 按位与(&)、按位或(|)、按位取反(~) 异或(^)、同或(^~ 或者 ~^) 移位运算符(<<、>>) 关系运算符(>、<、>=、<=) 等式运算符(==、!=、===、!==) 拼接运算符({})
为10进制
例 10
'o7 1'b1 8'Hc5 6'hF0 6'hF 6'hZ
未定长 10进制 未定长 8进制 1 bit 2进制 8 bits 16进制 6 bits 16进制 6 bits 16进制 6 bits 16进制
说明 0...01010 (32-bits)
0...00111 (32-bits) 1 11000101 110000 (高位被截断) 001111 (高位由0填充) ZZZZZZ (高位由Z填充)
Verilog HDL数据类型和变量
连线型(Net Type)
wire、tri
wor、wand、trior、triand、tri0、tri1、supply0、supply1、 trireg
寄存器型(Reg Type)
在时序部分出现
reg
integer、time、real
其他
parameter
数字逻辑电路英文课件 (26)Synchronous design methodology
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tr tclk tsetup
How to make the tr larger ?
Choose a faster flip-flop : f 25MHz a 2MHz tclk 40
74LS74 (97) :
1.35 T0 4.8103 tsetup 20 MTBF 2.83105 s
may be decided by an exponential formula:
MTBFtr
exptr
T0 f
/
a
MTBF Example: given condition as follows (74LS74)
T0 0.4 1.5 f 107 a 105
MTBFBiblioteka 80ns exp80 /1.5
Tclk > Tsetup + Thold
The time margins in design
Hold time margin : tcq tcomb thold Setup time margin: tclk tcq tcomb tsetup
Clock skew
Hold time margin : tcq tcomb thold tskew Setup time margin: tclk tcq tcomb tsetup tskew
0.4 107 105
3.631011 s
MTBF 40ns exp40 /1.5 9.53101s
0.4 107 105
How to make the tr larger ?
tr tclk tcomb tsetup Use cascade D flip-flops to minimize tcomb
Synchronous design methodology
How to make the tr larger ?
Choose a faster flip-flop : f 25MHz a 2MHz tclk 40
74LS74 (97) :
1.35 T0 4.8103 tsetup 20 MTBF 2.83105 s
may be decided by an exponential formula:
MTBFtr
exptr
T0 f
/
a
MTBF Example: given condition as follows (74LS74)
T0 0.4 1.5 f 107 a 105
MTBFBiblioteka 80ns exp80 /1.5
Tclk > Tsetup + Thold
The time margins in design
Hold time margin : tcq tcomb thold Setup time margin: tclk tcq tcomb tsetup
Clock skew
Hold time margin : tcq tcomb thold tskew Setup time margin: tclk tcq tcomb tsetup tskew
0.4 107 105
3.631011 s
MTBF 40ns exp40 /1.5 9.53101s
0.4 107 105
How to make the tr larger ?
tr tclk tcomb tsetup Use cascade D flip-flops to minimize tcomb
Synchronous design methodology
数字逻辑电路英文课件 (19)State-machine design
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A mode 5 counter design
state diagram
state/output table
A mode 5 counter design
State assignment
binary code Gray code one-hot code
S0 0 0 0 S1 0 0 1 S2 0 1 0 S3 0 1 1 S4 1 0 0
The excitation tables for JK flip-flop
J 2 Q1Q0
J1 Q0
J 0 Q2hesis with J-K flip-flop
minimal cost
CO Q2
J 2 Q1Q0 K 2 1
J1 Q0
K1 Q0
Minimize the logic equations
one-hot code CO Q4 D4 Q3 D3 Q2
D2 Q1 D1 Q0 D0 Q4
Ring counter
The excitation tables for JK flip-flop
Characteristic equation Q* J Q'K'Q Divide each transition table into two excitation tables
Construct the transition/output table
binary code
one-hot code
Gray code
Minimize the logic equations
Excitation table for binary code
minimal cost
minimal risk
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of Sum) form for the output (5
points)
01 1 1 0 1
3. Write the minimum SOP form. Draw the circuit. (10 points)
11 0 0 x 1
4. Write the minimum POS form. Draw the circuit (10 points)
The second half-adder has inputs of 1 and 1; therefore the Sum = 0 and the Carry out = 1.
The OR gate has inputs of 1 and 0, therefore the final carry out = 1.
Cout S
00 01 01 10 01 10 10 11
S AS B Cin Cout
Symbol
Floyd, Digital Fundamentals, 10th ed
© 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved
A full-adder can be constructed from two half adders as shown:
A AS S
AS S
Sum
B
B Cout
Cin
B Cout
Cout
Inputs
A B Cin
00 0 00 1 01 0 01 1 100 101 110 111
Outputs
© 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved
Summary
Parallel Adders
Full adders are combined into parallel adders that can add binary numbers with multiple bits. A 4-bit adder is shown.
Floyd, Digital Fundamentals, 10th ed
© 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved
Summary
Full-Adder Notice that the result from the previous example can be read directly on the truth table for a full adder.
Summaryຫໍສະໝຸດ Full-Adder1
S A
S
1
AS S 0
Sum
0
B Cout 0
B Cout 1
For the given inputs, determine 1 the intermediate and final outputs
Cout
1
of the full adder.
The first half-adder has inputs of 1 and 0; therefore the Sum =1 and the Carry out = 0.
Digital Fundamentals
Tenth Edition
Floyd
Chapter 6
Floyd, Digital Fundamentals, 10th ed
© 2009 Pearson Education©, U2p0p0e8rPSeaadrdsolenREidvuecr,aNtioJn07458. All Rights Reserved
Quiz 2 (20 minutes)
Refer to Karnaugh map :
1. Write the standard SOP (Sum of
CD
Product) form for the output (5 AB
00 01 11 10
points)
2. Write the standard POS (product 00 1 0 1 1
00 01 01 10
Floyd, Digital Fundamentals, 10th ed
© 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved
Summary
Full-Adder
By contrast, a full adder has three binary inputs (A, B, and Carry in) and two binary outputs (Carry out and Sum). The truth table summarizes the operation.
10 1 0 1 1
Floyd, Digital Fundamentals, 10th ed
© 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved
Summary
Half-Adder
Basic rules of binary addition are performed by a half adder, which has two binary inputs (A and B) and two binary outputs (Carry out and Sum).
Inputs
A B Cin
00 0 00 1 01 0 01 1 100 101 110 111
Outputs
Cout S
00 01 01 10 01 10 10 11
1
S A
S
1
AS S 0
0
B Cout 0
B Cout 1
1
Sum
Cout 1
Floyd, Digital Fundamentals, 10th ed
The inputs and outputs can be summarized on a truth table.
The logic symbol and equivalent circuit are:
AS S
S
A
B Cout
B
Cout
Inputs Outputs
AB
00 01 10 11
Cout S