- The base is ten – there are 10 different symbols, the digits 0, 1, 2,
etc upto 9
- To represent value less than ten involves only one digit larger values need
two or more digits
 Binary system
- The base must be two, with only the digits 0 and 1 available
- To show values of two or ever require two or more binary digits
 Octal system
- Octal system has eight as its base, it uses the symbol 0, 1, 2 up to 7 only
- Two or more digits are needed for values of eight and above
 Hexadecimal system(hex)
- Hexadecimal system has sixteen as its base, it use the symbols 0, 1, 2 ,9
& A, B, C, D, E, F, to stand for the “digits” ten, eleven, twelve, thirteen,
fourteen, fifteen.
Question 2. Converting from Bases To Bases?
1. Change the decimal
- Binary:
Eg. (2559) 10
2559 1
1279 1
639 1
319 1
159 1
79 1 (2559)
10
= (10111111111)
2

39 1
19 1
9 1

7

11

1221
15

81

11

Click to buy NOW!
P
D
F
-
X
C
h
a
n
g
e

V
i
e
w
e
r


V
i
e
w
e
r
w
w
w
.
d
o
c
u
-
t
r
a
c
k
.
c
o
m C F
(6369)
10

100101101
 1
st
step change into denary
= 1.2
8
+ 1.2
5
+ 1.2
3
+ 1.2
2
+ 1.2
0

= 256 + 32 + 8 + 4 + 1
=(301)
10

 2
nd
step: convert to octal

301 8
61 37 8 (301)
10
= (455)

= 2048+ 1024 + 256 + 158 + 64 + 16 + 8 + 2 + 1
= (3547)
10

2
nd
step
3547 16
384 221 16
27 61

(3547)
10
= (CCA)
16
(110111011011)
2
= (CCA)
16

3. Convert into binary and display the answer in normalized exponential form

247 1
123 1
61 1
30 1
5

w
w
w
.
d
o
c
u
-
t
r
a
c
k
.
c
o
m
Click to buy NOW!
P
D
F
-
X
C
h
a
n
g
e

(247)
10
= (11110111)
2

= 0. 1111011 x 2
normalized exponential form
Question 3. Integer and Floating – point arithmetic?
1. Floating – point Addition
a. (0.1011 x 2
5
) + (0.1001 x 2
5
)
= (0.1011 + 0. 1001) x 2
5

= 1.0100 x 2
5

= 0.10100 x 2
6

b. (0.1001 x 2
3
) + (0.1110 x 2
5
)
= (0.001001 x 2
5

5
)
= (0.1001 x 2
8
) – ( 0.0001 x 2
8
)
= 0.1000 x 2
8

3. Floating – point multiplication
a. (0.1010 x 2
3
) x (0.1100 x 2
3
)
= (0.1010 x 0.1100) x 2
6

= 0.01111 x 2
6

= 0.1111 x 2
5

b. (0.11110 x 2
3
) x ((0.01011) x 2
4
)

) : (0.1001 x 2
4
)
= (0.110111 : 0.1001) x 2
2

= (1101.11 : 1001) x 2
2

= 1.100001 x 2
2

= 0.1100001 x 2
3Chapter 3: TYPES OF INSTRUCTION AND ADDRESSING.
Click to buy NOW!
P
D
F
-
X
C
h
a
n
g
e


C
h
a
n
g
e

V
i
e
w
e
r
w
w
w
.
d
o
c
u
-
t
r
a
c
k
.
c
o

12345
Data item 12345
Data item
Main storage
Direct addressing
Indirect addressing
3. Indexed addressing
- The main applications of this type of addressing technique is to enable to
access of sequential locations in memory that are adjacent to each other
- Each adjcent memory address has value n+1, where n is the address of the
previous location
- When the first of the location have been accessed, the next memory
location in sequence is accessed by simply increasing the add of the
present location by 1 & using accessing it
Click to buy NOW!
P
D
F
-
X
C
h
a
n

F
-
X
C
h
a
n
g
e

V
i
e
w
e
r
w
w
w
.
d
o
c
u
-
t
r
a
c
k

1. Machine language
- ML is the set of bit(0,1) that can performed considered by CPU
- Ads
 fast
 short prog
 store in small memory
- Dis
 difficult to understand & remember its code
 takes a lot of time to programming
 difficult to use
2. Low level language
- LLL is used to dercribe exactly procedure of performance of CPU at
certain time
- Features:
 Instruction is written by natrural English or natural
language
 More powerful and so the prog is shortest
 Need less instruction
 Is a one to one relationship between the written instruction
and the machine instructions
 It’s instruction tend to be machine. It runs in OS
Data 1 Data 2 Data 3 Data 4
Click to buy NOW!
P
D
F
-
X
C
h

P
D
F
-
X
C
h
a
n
g
e

V
i
e
w
e
r
w
w
w
.
d
o
c
u
-
t
r
a