Ch3

1Data Representation
Computer Organization Computer Architectures Lab
DATA REPRESENTATION
Data Types
Complements
Fixed Point Representations
Floating Point Representations
Other Binary Codes
Error Detection Codes

DATA REPRESENTATION
Information that a Computer is dealing with
* Data
- Numeric Data
Numbers( Integer, real)
- Non-numeric Data
Letters, Symbols
* Relationship between data elements
- Data Structures
Linear Lists, Trees, Rings, etc
* Program(Instruction)
Data Types

NUMERIC DATA REPRESENTATION
R = 10 Decimal number system, R = 2 Binary
R = 8 Octal, R = 16 Hexadecimal
Radix point(.) separates the integer
portion and the fractional portion
Data
Numeric data - numbers(integer, real)
Non-numeric data - symbols, letters
Number System
Nonpositional number system
- Roman number system
Positional number system
- Each digit position has a value called a weight
associated with it
- Decimal, Octal, Hexadecimal, Binary
Base (or radix) R number
- Uses R distinct symbols for each digit
- Example AR = an-1 an-2 ... a1 a0 .a-1…a-m
- V(AR ) =
Data Types
∑
−
−=
1n
mi
i
i Ra

WHY POSITIONAL NUMBER SYSTEM IN DIGITAL COMPUTERS ?
Major Consideration is the COST and TIME
- Cost of building hardware
Arithmetic and Logic Unit, CPU, Communications
- Time to processing
Arithmetic - Addition of Numbers - Table for Addition
* Non-positional Number System
- Table for addition is infinite
--> Impossible to build, very expensive even
if it can be built
* Positional Number System
- Table for Addition is finite
--> Physically realizable, but cost wise
the smaller the table size, the less
expensive --> Binary is favorable to Decimal
0 1
0 0 1
1 1 10
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 6 7 8 9 10
2 2 3 4 5 6 7 8 9 1011
3 3 4 5 6 7 8 9 101112
4 4 5 6 7 8 9 10111213
5 5 6 7 8 9 1011121314
6 6 7 8 9 101112131415
7 7 8 9 10111213141516
8 8 9 1011121314151617
9 9 101112131415161718
Binary Addition Table
Decimal Addition Table
Data Types

REPRESENTATION OF NUMBERS - POSITIONAL NUMBERS
Decimal Binary Octal Hexadecimal
00 0000 00 0
01 0001 01 1
02 0010 02 2
03 0011 03 3
04 0100 04 4
05 0101 05 5
06 0110 06 6
07 0111 07 7
08 1000 10 8
09 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F
Binary, octal, and hexadecimal conversion
1 0 1 0 1 1 1 1 0 1 1 0 0 0 1 1
1 2 7 5 4 3
A F 6 3
Octal
Binary
Hexa
Data Types

CONVERSION OF BASES
Decimal to Base R number
Base R to Decimal Conversion
V(A) = Σ ak Rk
A = an-1 an-2 an-3 … a0 . a-1 … a-m
(736.4)8 = 7 x 82
+ 3 x 81
+ 6 x 80
+ 4 x 8-1
= 7 x 64 + 3 x 8 + 6 x 1 + 4/8 = (478.5)10
(110110)2 = ... = (54)10
(110.111)2 = ... = (6.785)10
(F3)16 = ... = (243)10
(0.325)6 = ... = (0.578703703 .................)10
- Separate the number into its integer and fraction parts and convert
each part separately.
- Convert integer part into the base R number
→ successive divisions by R and accumulation of the remainders.
- Convert fraction part into the base R number
→ successive multiplications by R and accumulation of integer
digits
Data Types

EXAMPLE
Convert 41.687510 to base 2.
Integer = 41
41
20 1
10 0
5 0
2 1
1 0
0 1
Fraction = 0.6875
0.6875
x 2
1.3750
x 2
0.7500
x 2
1.5000
x 2
1.0000
(41)10 = (101001)2 (0.6875)10 = (0.1011)2
(41.6875)10 = (101001.1011)2
Convert (63)10 to base 5: (223)5
Convert (1863)10 to base 8: (3507)8
Convert (0.63671875)10 to hexadecimal: (0.A3)16
Exercise
Data Types

COMPLEMENT OF NUMBERS
Two types of complements for base R number system:
- R's complement and (R-1)'s complement
The (R-1)'s Complement
Subtract each digit of a number from (R-1)
Example
- 9's complement of 83510 is 16410
- 1's complement of 10102 is 01012(bit by bit complement operation)
The R's Complement
Add 1 to the low-order digit of its (R-1)'s complement
Example
- 10's complement of 83510 is 16410 + 1 = 16510
- 2's complement of 10102 is 01012 + 1 = 01102
Complements

FIXED POINT NUMBERS
Binary Fixed-Point Representation
X = xnxn-1xn-2 ... x1x0. x-1x-2 ... x-m
Sign Bit(xn): 0 for positive - 1 for negative
Remaining Bits(xn-1xn-2 ... x1x0. x-1x-2 ... x-m)
Numbers: Fixed Point Numbers and Floating Point Numbers

SIGNED NUMBERS
Signed magnitude representation
Signed 1's complement representation
Signed 2's complement representation
Example: Represent +9 and -9 in 7 bit-binary number
Only one way to represent +9 ==> 0 001001
Three different ways to represent -9:
In signed-magnitude: 1 001001
In signed-1's complement: 1 110110
In signed-2's complement: 1 110111
In general, in computers, fixed point numbers are represented
either integer part only or fractional part only.
Need to be able to represent both positive and negative numbers
- Following 3 representations

CHARACTERISTICS OF 3 DIFFERENT REPRESENTATIONS
Complement
Signed magnitude: Complement only the sign bit
Signed 1's complement: Complement all the bits including sign bit
Signed 2's complement: Take the 2's complement of the number,
including its sign bit.
Maximum and Minimum Representable Numbers and Representation of Zero
X = xn xn-1 ... x0 . x-1 ... x-m
Signed Magnitude
Max: 2n
- 2-m
011 ... 11.11 ... 1
Min: -(2n
- 2-m
) 111 ... 11.11 ... 1
Zero: +0 000 ... 00.00 ... 0
-0 100 ... 00.00 ... 0
Signed 1’s Complement
Max: 2n
- 2-m
011 ... 11.11 ... 1
Min: -(2n
- 2-m
) 100 ... 00.00 ... 0
Zero: +0 000 ... 00.00 ... 0
-0 111 ... 11.11 ... 1
Signed 2’s Complement
Max: 2n
- 2-m
011 ... 11.11 ... 1
Min: -2n
100 ... 00.00 ... 0
Zero: 0 000 ... 00.00 ... 0

2’s COMPLEMENT REPRESENTATION WEIGHTS
• Signed 2’s complement representation follows a “weight” scheme
similar to that of unsigned numbers
– Sign bit has negative weight
– Other bits have regular weights
X = xn xn-1 ... x0
 V(X) = - xn × 2n
+ xi × 2i
i = 0
Σ
n-1

ARITHMETIC ADDITION: SIGNED MAGNITUDE
[1] Compare their signs
[2] If two signs are the same ,
ADD the two magnitudes - Look out for an overflow
[3] If not the same , compare the relative magnitudes of the numbers and
then SUBTRACT the smaller from the larger --> need a subtractor to add
[4] Determine the sign of the result
6 0110
+) 9 1001
15 1111 -> 01111
9 1001
- ) 6 0110
3 0011 -> 00011
9 1001
-) 6 0110
- 3 0011 -> 10011
6 0110
+) 9 1001
-15 1111 -> 11111
6 + 9 -6 + 9
6 + (- 9) -6 + (-9)
Overflow 9 + 9 or (-9) + (-9)
9 1001
+) 9 1001
(1)0010overflow

ARITHMETIC ADDITION: SIGNED 2’s
COMPLEMENT
Example
6 0 0110
9 0 1001
15 0 1111
-6 1 1010
9 0 1001
3 0 0011
6 0 0110
-9 1 0111
-3 1 1101
-9 1 0111
-9 1 0111
-18 (1)0 1110
Add the two numbers, including their sign bit, and discard any carry out of
leftmost (sign) bit - Look out for an overflow
overflow9 0 1001
9 0 1001+)
+) +)
+) +)
18 1 0010
2 operands have the same sign
and the result sign changes
xn-1yn-1s’n-1 + x’n-1y’n-1sn-1 = cn-1⊕ cn
x’n-1y’n-1sn-1
(cn-1 ⊕ cn)
xn-1yn s’n-1
(cn-1 ⊕ cn)

ARITHMETIC ADDITION: SIGNED 1’s
COMPLEMENT
Add the two numbers, including their sign bits.
- If there is a carry out of the most significant (sign) bit, the result is
incremented by 1 and the carry is discarded.
6 0 0110
-9 1 0110
-3 1 1100
-6 1 1001
9 0 1001
(1) 0(1)0010
1
3 0 0011
+) +)
+)
end-around carry
-9 1 0110
-9 1 0110
(1)0 1100
1
0 1101
+)
+)
9 0 1001
9 0 1001
1 (1)0010
+)
overflow
Example
not overflow (cn-1 ⊕ cn) = 0
(cn-1 ⊕ cn)

COMPARISON OF REPRESENTATIONS
* Easiness of negative conversion
S + M > 1’s Complement > 2’s Complement
* Hardware
- S+M: Needs an adder and a subtractor for Addition
- 1’s and 2’s Complement: Need only an adder
* Speed of Arithmetic
2’s Complement > 1’s Complement(end-around C)
* Recognition of Zero
2’s Complement is fast

ARITHMETIC SUBTRACTION
Take the complement of the subtrahend (including the sign bit)
and add it to the minuend including the sign bits.
( ± A ) - ( - B ) = ( ± A ) + B
( ± A ) - B = ( ± A ) + ( - B )
Arithmetic Subtraction in 2’s complement

FLOATING POINT NUMBER REPRESENTATION
* The location of the fractional point is not fixed to a certain location
* The range of the representable numbers is wide
F = EM
mn ekek-1 ... e0 mn-1mn-2 … m0 . m-1 … m-m
sign exponent mantissa
- Mantissa
Signed fixed point number, either an integer or a fractional number
- Exponent
Designates the position of the radix point
Decimal Value
V(F) = V(M) * RV(E) M: Mantissa
E: Exponent
R: Radix
Floating Point Representation

FLOATING POINT NUMBERS
0 .1234567 0 04
sign sign
mantissa exponent
==> +.1234567 x 10+04
Example
A binary number +1001.11 in 16-bit floating point number representation
(6-bit exponent and 10-bit fractional mantissa)
0 0 00100 100111000
0 0 00101 010011100
Example
Note:
In Floating Point Number representation, only Mantissa(M) and
Exponent(E) are explicitly represented. The Radix(R) and the position
of the Radix Point are implied.
Exponent MantissaSign
or

CHARACTERISTICS OF FLOATING POINT NUMBER REPRESENTATIONS
Normal Form
- There are many different floating point number representations of
the same number
→ Need for a unified representation in a given computer
- the most significant position of the mantissa contains a non-zero digit
Representation of Zero
- Zero
Mantissa = 0
- Real Zero
Mantissa = 0
Exponent
= smallest representable number
which is represented as
00 ... 0
← Easily identified by the hardware

INTERNAL REPRESENTATION AND EXTERNAL REPRESENTATION
CPU
Memory
Internal
Representation Human
Device
Another
Computer
External
Representation
External
Representation
External
Representation

EXTERNAL REPRESENTATION
Decimal BCD Code
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
Numbers
Most of numbers stored in the computer are eventually changed
by some kinds of calculations
→ Internal Representation for calculation efficiency
→ Final results need to be converted to as External Representation
for presentability
Alphabets, Symbols, and some Numbers
Elements of these information do not change in the course of processing
→ No needs for Internal Representation since they are not used
for calculations
→ External Representation for processing and presentability
Example
Decimal Number: 4-bit Binary Code
BCD(Binary Coded Decimal)
External Representations

OTHER DECIMAL CODES
Decimal BCD(8421) 2421 84-2-1 Excess-3
0 0000 0000 0000 0011
1 0001 0001 0111 0100
2 0010 0010 0110 0101
3 0011 0011 0101 0110
4 0100 0100 0100 0111
5 0101 1011 1011 1000
6 0110 1100 1010 1001
7 0111 1101 1001 1010
8 1000 1110 1000 1011
9 1001 1111 1111 1100 d3 d2 d1 d0: symbol in the codes
BCD: d3 x 8 + d2 x 4 + d1 x 2 + d0 x 1
⇒ 8421 code.
2421: d3 x 2 + d2 x 4 + d1 x 2 + d0 x 1
84-2-1: d3 x 8 + d2 x 4 + d1 x (-2) + d0 x (-1)
Excess-3: BCD + 3
Note: 8,4,2,-2,1,-1 in this table is the weight
associated with each bit position.
BCD: It is difficult to obtain the 9's complement.
However, it is easily obtained with the other codes listed above.
→ Self-complementing codes
External Representations

GRAY CODE
* Characterized by having their representations of the binary integers differ
in only one digit between consecutive integers
* Useful in some applications
Decimal
number
Gray Binary
g3 g2 g1 g0 b3 b2 b1 b0
0 0 0 0 0 0 0 0 0
1 0 0 0 1 0 0 0 1
2 0 0 1 1 0 0 1 0
3 0 0 1 0 0 0 1 1
4 0 1 1 0 0 1 0 0
5 0 1 1 1 0 1 0 1
6 0 1 0 1 0 1 1 0
7 0 1 0 0 0 1 1 1
8 1 1 0 0 1 0 0 0
9 1 1 0 1 1 0 0 1
10 1 1 1 1 1 0 1 0
11 1 1 1 0 1 0 1 1
12 1 0 1 0 1 1 0 0
13 1 0 1 1 1 1 0 1
14 1 0 0 1 1 1 1 0
15 1 0 0 0 1 1 1 1
4-bit Gray codes
Other Binary codes

GRAY CODE - ANALYSIS
Letting gngn-1 ... g1 g0 be the (n+1)-bit Gray code
for the binary number bnbn-1 ... b1b0
gi = bi ⊕ bi+1 , 0 ≤ i ≤ n-1
gn = bn
and
bn-i = gn ⊕ gn-1 ⊕ . . . ⊕ gn-i
bn = gn
0 0 0 0 00 0 000
1 0 1 0 01 0 001
1 1 0 11 0 011
1 0 0 10 0 010
1 10 0 110
1 11 0 111
1 01 0 101
1 00 0 100
1 100
1 101
1 111
1 010
1 011
1 001
1 101
1 000
The Gray code has a reflection property
- easy to construct a table without calculation,
- for any n: reflect case n-1 about a
mirror at its bottom and prefix 0 and 1
to top and bottom halves, respectively
Reflection of Gray codes
Note:
Other Binary codes
ε

CHARACTER REPRESENTATION ASCII
ASCII (American Standard Code for Information Interchange) Code
Other Binary codes
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
NUL
SOH
STX
ETX
EOT
ENQ
ACK
BEL
BS
HT
LF
VT
FF
CR
SO
SI
SP
!
“
#
$
%
&
‘
(
)
*
+
,
-
.
/
0
1
2
3
4
5
6
7
8
9
:
;
<
=
>
?
@
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
[

]
m
n
‘
a
b
c
d
e
f
g
h
I
j
k
l
m
n
o
P
q
r
s
t
u
v
w
x
y
z
{
|
}
~
DEL
0 1 2 3 4 5 6 7
DLE
DC1
DC2
DC3
DC4
NAK
SYN
ETB
CAN
EM
SUB
ESC
FS
GS
RS
US
LSB
(4 bits)
MSB (3 bits)

CONTROL CHARACTER REPRESENTAION (ACSII)
NUL Null
SOH Start of Heading (CC)
STX Start of Text (CC)
ETX End of Text (CC)
EOT End of Transmission (CC)
ENQ Enquiry (CC)
ACK Acknowledge (CC)
BEL Bell
BS Backspace (FE)
HT Horizontal Tab. (FE)
LF Line Feed (FE)
VT Vertical Tab. (FE)
FF Form Feed (FE)
CR Carriage Return (FE)
SO Shift Out
SI Shift In
DLE Data Link Escape (CC)
(CC) Communication Control
(FE) Format Effector
(IS) Information Separator
Other Binary codes
DC1 Device Control 1
NAK Negative Acknowledge (CC)
SYN Synchronous Idle (CC)
ETB End of Transmission Block (CC)
CAN Cancel
EM End of Medium
SUB Substitute
ESC Escape
FS File Separator (IS)
GS Group Separator (IS)
RS Record Separator (IS)
US Unit Separator (IS)
DEL Delete

ERROR DETECTING CODES
Parity System
- Simplest method for error detection
- One parity bit attached to the information
- Even Parity and Odd Parity
Even Parity
- One bit is attached to the information so that
the total number of 1 bits is an even number
1011001 0
1010010 1
Odd Parity
- One bit is attached to the information so that
the total number of 1 bits is an odd number
1011001 1
1010010 0
Error Detecting codes

Parity Bit Generation
For b6b5... b0(7-bit information); even parity bit beven
beven = b6 ⊕ b5 ⊕ ... ⊕ b0
For odd parity bit
bodd = beven ⊕ 1 = beven
PARITY BIT GENERATION

PARITY GENERATOR AND PARITY CHECKER
Parity Generator Circuit (even parity)
b6
b5
b4
b3
b2
b1
b0
beven
Parity Checker
b6
b5
b4
b3
b2
b1
b0
beven
Even Parity
error indicator
Error Detecting codes

Ch3

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