Chapter 1.1 Digital Computers : Hardware Organization.
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Transcript of Chapter 1.1 Digital Computers : Hardware Organization.
![Page 1: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/1.jpg)
Chapter 1.1
Digital Computers :
Hardware Organization
![Page 2: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/2.jpg)
The Sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
![Page 3: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/3.jpg)
Input-Output Devices
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Printer
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Process Control I/O
![Page 6: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/6.jpg)
The Sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
![Page 7: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/7.jpg)
Memories00000 00001 00010 00011
00100 00101 00110 00111
01000 01001 01010 01011
01100 01101 01110 01111
10000 10001 10010 00011
10100 10101 10110 10111
11000 11001 11010 11011
11100 11101 11110 11111
![Page 8: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/8.jpg)
The Sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
![Page 9: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/9.jpg)
Instructions Format
OPC OP1 OP2 RES NEXT
OPC OP1 OP2 NEXT1 NEXT2
Information handling instructions
Control instructions
![Page 10: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/10.jpg)
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Electronic Lock
4 5 6
1 2 3
* 0 #
7 8 9
![Page 11: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/11.jpg)
KFL =
KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
Data Memory
Arithmetic Unit
Control Unit
Program Memory
ND =
SC =
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 12: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/12.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = SC =
COPY #0 ND P2
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 13: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/13.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC =
COPY #0 ND P2
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 14: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/14.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC =
COPY #0 SC P3
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 15: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/15.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 0
COPY #0 SC P3
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 16: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/16.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 0
EQ? KFL #0 P3 P4
(KFL = 0) = TRUE
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 17: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/17.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 0
EQ? KFL #0 P3 P4
(KFL = 0) = TRUE
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 18: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/18.jpg)
KFL = 1KDA = 3
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 0
EQ? KFL #0 P3 P4
(KFL = 0) = FALSE
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 19: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/19.jpg)
KFL = 1KDA = 3
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 0
(0 * 10) = 0
MUL SC #10 SC P5
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 20: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/20.jpg)
KFL = 1KDA = 3
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 0
(0 + 3) = 3
ADD SC KDA SC P 6
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 21: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/21.jpg)
KFL = 1KDA = 3
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 3
(0 + 3) = 3
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
ADD SC KDA SC P 6
![Page 22: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/22.jpg)
KFL = 1KDA = 3
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 3
COPY #0 KFL P7
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 23: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/23.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 3
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
COPY #0 KFL P7
![Page 24: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/24.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 3
(0 + 1) = 1
ADD ND #1 ND P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 25: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/25.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 3
(0 + 1) = 1
ADD ND #1 ND P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 26: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/26.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 3
(1 # 3) = TRUE
NE? ND #3 P3 P9
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 27: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/27.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 3
(0 = 0) = TRUE
EQ? KFL #0 P3 P4
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 28: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/28.jpg)
KFL = 1KDA = 2
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 3
(1 = 0) = FALSE
EQ? KFL #0 P3 P4
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 29: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/29.jpg)
KFL = 1KDA = 2
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 3
(3 * 10) = 30
MUL SC #10 SC P5
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 30: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/30.jpg)
KFL = 1KDA = 2
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 30
(3 * 10) = 30
MUL SC #10 SC P5
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 31: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/31.jpg)
KFL = 1KDA = 2
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 30
(30 + 2) = 32
ADD SC KDA SC P6
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 32: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/32.jpg)
KFL = 1KDA = 2
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 32
(30 + 2) = 32
ADD SC KDA SC P6
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 33: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/33.jpg)
KFL = 1KDA = 2
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 32
COPY #0 KFL P7
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 34: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/34.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 32
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
COPY #0 KFL P7
![Page 35: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/35.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 1SC = 32
(1 + 1) = 2
ADD ND #1 ND P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 36: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/36.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 32
(1 + 1) = 2
ADD ND #1 ND P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 37: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/37.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 32
(2 # 3) = TRUE
NE? ND #3 P3 P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 38: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/38.jpg)
KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 32
(0 = 0) = TRUE
EQ? KFL #0 P3 P4
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 39: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/39.jpg)
KFL = 1KDA = 1
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 32
(1 = 0) = FALSE
EQ? KFL #0 P3 P4
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 40: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/40.jpg)
KFL = 1KDA = 1
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 32
(32 * 10) = 320
MUL SC #10 SC P5
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
![Page 41: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/41.jpg)
KFL = 1KDA = 1
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 320
(32 * 10) = 320
MUL SC #10 SC P5
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 1KDA = 1
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 320
(320 + 1) = 321
ADD SCKDA SC P6
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 1KDA = 1
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 321
(320 + 1) = 321
ADD SCKDA SC P6
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 1KDA = 1
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 321
COPY #0 KFL P7
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 321
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 2SC = 321
(2 + 1) = 3
ADD ND #1 ND P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 3SC = 321
(2 + 1) = 3
ADD ND #1 ND P8
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 3SC = 321
(3 # 3) = FALSE
NE? ND #3 P3 P9
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 3SC = 321
(321 # 321) = FALSE
NE? SC #321 P1 P10
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA = 1
4 5 6
1 2 3
* 0 #
7 8 9
ND = 3SC = 321
COPY #1 DDA P1
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 3SC = 321
COPY #0 ND P2
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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KFL = 0KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
ND = 0SC = 321
COPY #0 ND P2
p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6
p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1
p6 COPY #0 KFL P7
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Instructions Format
with P-Register
OPC OP1 OP2 RES
OPC OP1 OP2 NEXT
Information handling instructions
Control instructions
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KFL =
KDA =
DDA =
4 5 6
1 2 3
* 0 #
7 8 9
Data Memory
Arithmetic Unit
Control Unit
Program Memory
ND =
SC =
p1 COPY #0 NDp2 COPY #0 SCp3 EQ? KFL #0 P3p4 MUL SC #10 SCp5 ADD SC KDA SC
p7 ADD ND #1 NDp8 NE? ND #3 P3p9 NE? SC #321 P1p10 COPY #1 DDA
p6 COPY #0 KFL
p11 JUMP P1
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The
sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
![Page 56: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/56.jpg)
The
“Von Neumann” Compute
r
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
![Page 57: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/57.jpg)
Memories
00000 00001 00010 00011
00100 00101 00110 00111
01000 01001 01010 01011
01100 01101 01110 01111
10000 10001 10010 00011
10100 10101 10110 10111
11000 11001 11010 11011
11100 11101 11110 11111
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Cost of Memory
Accesstime
10-710-8 10-6 10-5 10-4 10-3 10-2 10-1 100 S
Relative cost per bit
Semiconductor memories
Magnetic memoriesOptical memories
1
1000
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Semiconductor Memories
(RAM, ROM, PROM)
083
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Semiconductor Memories
• Read access time < 100 nS.
• Cost strongly influenced by access time
• RAM (“Random Access Memory “/ “Read And Modify”):
– volatile !
– Read and write access times equal
• ROM (“Read Only Memory”):
– non volatile
– Can only be written in factory
• PROM (“Programmable Read Only Memory”):
– non volatile
– Can be written by the user
– Write access time >> read access time
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Peripheral Memories
14
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Writing on magnetic memories
i
0 0 0 0 011 11
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Reading from a magnetic memory
e
0 0 0 0 011 11 0 0 0 0 011 11
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Manchester Code
0 00 0 1 11 11i
t
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Data blocks
Header Data Block Check
0101010101...010101XXXXXXXXX
Check = f(data block)
Synchronization sequence
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Disk Organization
Sector
Track
Cylinder
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Tracks/cylinder
Cylinders
Sectors/track
Bytes/sector
Total Capacity(in bytes)
Double Density
2
80
9
512
737 280
High density
2
80
18
512
1 474 560
Format of 3.5” diskettes for PC’s.
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Tracks/cylinder
Cylinders
Sectors/track
Bytes/sector
Total Capacity(in bytes)
4
16 383
Variable
512
12 072 517 632
Format of 12 GBytes Hard Disk.
Total # sectors 23 579 136
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Hard-
disk drive (2)
17
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Compact Disk Technology
Laser
Photodetector
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Rewritable
CD Technology
Laser
Photodetector
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DVD Technology
Laser
Photodetector
Laser
Photodetector
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The
“Von Neumann” Compute
r
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
![Page 74: Chapter 1.1 Digital Computers : Hardware Organization.](https://reader035.fdocuments.us/reader035/viewer/2022062322/56649ea25503460f94ba5728/html5/thumbnails/74.jpg)
Minimal Memory Hierarchy
Registers
Central Memory
Disks
CD-ROM
Size (log scale)
SpeedMostlyVolatile
Non-Volatile
RAMIn CPU
RAM(+small ROM)
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Traditional “Von Neumann” Computer
Input-Output
Equipment
Central
Memory
Central Processing
Unit
PeripheralMemories
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Personal Computer
047
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Central Processor
76
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Processor Board
79