3
This presentation is the property of its rightful owner.
Sponsored Links
1 / 69

第3章 计算机的基本器件 PowerPoint PPT Presentation


  • 73 Views
  • Uploaded on
  • Presentation posted in: General

第3章 计算机的基本器件. 下一页. 目 录. 3.1 逻辑代数与逻辑电路 3.2 组合逻辑电路 3.3 时序逻辑电路 3.4 总线缓冲器和总线控制器 3.5 时钟发生器. 上一页. 下一页. 3.1 逻辑代数与逻辑电路. 3.1.1 逻辑代数 3.1.2 基本逻辑电路. 上一页. 下一页. 3.1.1 逻辑代数.

Download Presentation

第3章 计算机的基本器件

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


3

3


3

3.1

3.2

3.3

3.4

3.5


3

3.1

3.1.1

3.1.2


3 1 1

3.1.1

Boolean algebra


3 1 11

3.1.1

01


3

+ABC1F1ABC0F0

F ABCABC+

ABC


3

ABC1F10

F ABCABC

ABC


3

A10A01

F

A


3 1 12

3.1.1

F=f(A1,A2,,Ai,,An)


3

ABF10

F=f(A,B)=


3


3 1 13

3.1.1

F=A+B F=AVB


3 1 14

3.1.1

F=AB F=A B


3 1 15

3.1.1


3 1 16

3.1.1

P57P58


3 1 17

3.1.1


3


3


3 1 2

3.1.2


3 1 21

3.1.2


3 1 22

3.1.2


3 1 23

3.1.2


3


3

10

10


3

3.2


3

3.2

3.2.1

3.2.2

3.2.3

3.2.4


3 2 1

3.2.1

:AiBiCi


3

Si=AiBi

Ci= AiBi


3

Ci-1


3

SI=AIBICI-1CI=AIBI+BICI-1+AICI-1


3

3.n

4

n


3 2 2

3.2.2

ALUArithmetic Logic UnitALUALUALU


3 2 3

3.2.3


3 2 31

3.2.3

1.

3-838

3x2x1x08y7y6y5y4y3y2y1y0


3


3

3-8


3

74LS1383-8(a)(b)

G10G211


3 2 4

3.2.4

MUX (Multiplexor/Selector)


3 2 41

3.2.4

41MUX4ABCDZ()S1S0


3

3.3


3

3.3

3.3.1

3.3.2

3.3.3


3 3 1

3.3.1

(flip-flop)

-

R-SDJ-K


3

1.R-S

100R=0S=1QQ0

R=1 S=0Q Q1

R=1 S=1

R=0 S=0


3

2.R-S

R-SCPR-S

R0 S1CPQ(t)Q(t+1)


3

3.D

D

D

RD0SD1(RDSD)D()DD=11D=00


4 j k

4.J-K

RD0SD1K0J1

J=0K=0CP

J=0K=1CP0

J=1K=0CP1

J=1K=1CP


3 3 2

3.3.2

nn


3 3 21

3.3.2

D4CP


3

CKQD


3

CP


3

74LS3738D8D


3

nnD

R0DRD Q1=Q2=Q3==Qn=0

DINCPDDIN =1CPFF1FF1CPD


3

DDDCPD

DQ

2()


3 3 3

3.3.3


3 3 31

3.3.3

1.

2.

3.


3

4

4:

DCPDQQCP


3

R0Q3Q2Q1Q00000CPQ01Q010CPQ00Q001Q11CPQ3Q2Q1Q00000


3


3

3.4

3.4.1

3.4.2


3 4 1

3.4.1

74LS24474LS245

74LS244 83-18

74LS245 83-19


3

3-18 74LS244


3

3-19 74LS245


3 4 2

3.4.2

8288

3-20 8288


3

3.5

3.5.1 8284

3.5.2 8284CPU


3 5 1 8284

3.5.1 8284

828418


8284 5

82845

1.OSC

2.CLKCLKCPU

3.RESETCPU

4.PCLKCLK2

5.READY8284RDY1AEN1RDY2AEN2ASYNC


3 5 2 8284 cpu

3.5.2 8284CPU

PC/XT8284


3

PWR GOOD RRESET8088

DMA WAIT DMA()DMADMA()

RDY/WAITI/O CH RDYIORIOW8284READY


3 the end

3 The End


  • Login