©j.tiberghien - ulb-vub version 2007 première partie, chap. 1, page 1 chapitre 1.1 ordinateurs :...
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©J.Tiberghien - ULB-VUBVersion 2007 1Première partie, chap. 1, page
Chapitre 1.1
Ordinateurs :
Organisation Matérielle
©J.Tiberghien - ULB-VUBVersion 2007 2Première partie, chap. 1, page
The Sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
©J.Tiberghien - ULB-VUBVersion 2007 3Première partie, chap. 1, page
Input-Output Devices
©J.Tiberghien - ULB-VUBVersion 2007 4Première partie, chap. 1, page
Printer
©J.Tiberghien - ULB-VUBVersion 2007 5Première partie, chap. 1, page
Process Control I/O
©J.Tiberghien - ULB-VUBVersion 2007 6Première partie, chap. 1, page
The Sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
©J.Tiberghien - ULB-VUBVersion 2007 7Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 8Première partie, chap. 1, page
The Sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
©J.Tiberghien - ULB-VUBVersion 2007 9Première partie, chap. 1, page
Instructions Format
OPC OP1 OP2 RES NEXT
OPC OP1 OP2 NEXT1 NEXT2
Information handling instructions
Control instructions
©J.Tiberghien - ULB-VUBVersion 2007 10Première partie, chap. 1, page
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Electronic Lock
4 5 6
1 2 3
* 0 #
7 8 9
©J.Tiberghien - ULB-VUBVersion 2007 11Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 12Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 13Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 14Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 15Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 16Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 17Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 18Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 19Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 20Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 21Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 22Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 23Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 24Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 25Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 26Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 27Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 28Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 29Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 30Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 31Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 32Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 33Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 34Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 35Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 36Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 37Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 38Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 39Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 40Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 41Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 42Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 43Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 44Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 45Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 46Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 47Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 48Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 49Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 50Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 51Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 52Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 53Première partie, chap. 1, page
Instructions Format
with P-Register
OPC OP1 OP2 RES
OPC OP1 OP2 NEXT
Information handling instructions
Control instructions
©J.Tiberghien - ULB-VUBVersion 2007 54Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 55Première partie, chap. 1, page
The
sequential Computer
Data Memory
Arithmetic Unit
ControlUnit
Program Memory
Program Interface
Inp
ut
Inte
rfac
e
Ou
tpu
t In
terf
ace
©J.Tiberghien - ULB-VUBVersion 2007 56Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 57Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 58Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 59Première partie, chap. 1, page
Semiconductor Memories
(RAM, ROM, PROM)
083
©J.Tiberghien - ULB-VUBVersion 2007 60Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 61Première partie, chap. 1, page
Peripheral Memories
14
©J.Tiberghien - ULB-VUBVersion 2007 62Première partie, chap. 1, page
Writing on magnetic memories
i
0 0 0 0 011 11
©J.Tiberghien - ULB-VUBVersion 2007 63Première partie, chap. 1, page
Reading from a magnetic memory
e
0 0 0 0 011 11 0 0 0 0 011 11
©J.Tiberghien - ULB-VUBVersion 2007 64Première partie, chap. 1, page
Manchester Code
0 00 0 1 11 11i
t
©J.Tiberghien - ULB-VUBVersion 2007 65Première partie, chap. 1, page
Data blocks
Header Data Block Check
0101010101...010101XXXXXXXXX
Check = f(data block)
Synchronization sequence
©J.Tiberghien - ULB-VUBVersion 2007 66Première partie, chap. 1, page
Disk Organization
Sector
Track
Cylinder
©J.Tiberghien - ULB-VUBVersion 2007 67Première partie, chap. 1, page
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.
©J.Tiberghien - ULB-VUBVersion 2007 68Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 69Première partie, chap. 1, page
Hard-disk drive (2)
17
©J.Tiberghien - ULB-VUBVersion 2007 70Première partie, chap. 1, page
Compact Disk Technology
Laser
Photodetector
©J.Tiberghien - ULB-VUBVersion 2007 71Première partie, chap. 1, page
Rewritable
CD Technology
Laser
Photodetector
©J.Tiberghien - ULB-VUBVersion 2007 72Première partie, chap. 1, page
DVD Technology
Laser
Photodetector
Laser
Photodetector
©J.Tiberghien - ULB-VUBVersion 2007 73Première partie, chap. 1, page
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
©J.Tiberghien - ULB-VUBVersion 2007 74Première partie, chap. 1, page
Minimal Memory Hierarchy
Registers
Central Memory
Disks
CD-ROM
Size (log scale)
SpeedMostlyVolatile
Non-Volatile
RAMIn CPU
RAM(+small ROM)
©J.Tiberghien - ULB-VUBVersion 2007 75Première partie, chap. 1, page
Traditional “Von Neumann” Computer
Input-Output
Equipment
Central
Memory
Central Processing
Unit
PeripheralMemories
©J.Tiberghien - ULB-VUBVersion 2007 76Première partie, chap. 1, page
Central Processor
76
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