CMPS 375
Class Notes
Chapter 2 Data Representation in Computer Systems
1. Number Systems
System
Radix Allowable Digits
--------------------------------------------------------------------Decimal
10
0,1,2,3,4,5,6,7,8,9
Binary
2
0,1
Hexadecimal
16
0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
A base-r number is represented by the following series of coefficients:
( anan-1 … a3a2a1a0 . a-1a-2a-3 ... )r
where r = 2,3,4, ..., 8,9,10, ...16, ...
Examples:
(705)10
(01011000001)2
(2C1)16
(0.8125) 10
(0.1101)2
(0.D)16
(705.8125) 10
(01011000001.1101)2
(2C1.D)16
There are two important groups of number base conversions:
1. Conversion of decimal numbers to base-r numbers
2. Conversion of base-r numbers to decimal numbers
2. Conversion of base-10 numbers to base-r numbers
(
)10 ---> (
)r
where r = 2,3,4, ..., 8, ...16, ...
A decimal number containing an integer part and a fraction part can be converted to a
base-r number by converting these two parts separately and then combining the two
answers.
2-1. Conversion of a base-10 integer to a base-r integer
The Division Algorithm (The radix divide technique)
Divide the given base-10 integer successively by the required base-r, noting the remainders at
each step. The quotient at each step becomes the new dividend for subsequent division. Stop
division when the quotient reaches zero. The series of remainders in reverse order is the
equivalent base-r integer.
Examples
a. (705)10 ---> (
705/2 =
352/2 =
176/2 =
88/2 =
44/2 =
22/2 =
11/2 =
5/2 =
2/2 =
1/2 =
)2
quotient remainder coefficient
--------- ------------- ------------352
1
a0 = 1
176
0
a1 = 0
88
0
a2 = 0
44
0
a3 = 0
22
0
a4 = 0
11
0
a5 = 0
5
1
a6 = 1
2
1
a7 = 1
1
0
a8 = 0
0
1
a9 = 1
a10 = 0
Therefore, ( 705)10 = (01011000001)2
b. ( 705)10 ---> (
)16
quotient remainder coefficient
--------- ------------- ------------705/16 = 44
1
a0 = 1
44/16 =
2
C
a1 = C
2/16 =
0
2
a2 = 2
Therefore, (705)10 = (2C1)16
2-2. Conversion of a decimal fraction to a base-r fraction
The Multiplication Algorithm (The radix multiply technique)
Multiply the base-10 fraction by the required base-r, noting the integer portion of the product at
each step. Use the fractional part of the product as the multiplicand for subsequent steps. Stop
when the fraction either reaches 0 or the required number of digits has been obtained. The
series of integer digits at each step from first to last is the equivalent base-r fraction.
Examples
a. (0.8125)10 ---> (
)2
integer fraction
-------- --------0.8125x2 = 1
.625
0.625x2 = 1
.25
0.25x2 = 0
.5
0.5x2 = 1
.0
coefficient
-----------a-1 = 1
a-2 = 1
a-3 = 0
a-4 = 1
Therefore, (0.8125)10 = (.1101)2
b. ( 705.8125)10 ---> (01011000001.1101)2
c. (0.8125)10 ---> (
)8
integer fraction
-------- --------0.8125x8 = 6
.5
0.5x8 = 4
.0
coefficient
-----------a-1 = 6
a-2 = 4
Therefore, (0.8125)10 = (.64)8
3. Conversion of base-r numbers to decimal numbers
(
)r ---> (
)10
where r = 2,3,4, ..., 8, ...16, ...
The positional value method
The decimal equivalent of a base-r number is obtained as described below:
a. Each coefficient (aj) is multiplied by its positional value (rj).
b. Add all the ajrj as shown below:
an rn + ...+ a3r3 + a2r2 + a1r1 + a0r0 + a-1r-1 + a-2 r-2 + ...+ a-m r-m
Examples
a. (011001.101)2 ----> (
)10
1x24 + 1x23 + 1x20 + 1x2-1 + 1x2-3
= 16 + 8 + 1 + 0.5 + 0.125
= 25.625
Therefore, (011001.101)2 = (25.625
)10
b. (37B)16 ----> (
)10
3x162 + 7x161 + 12x160
= 3x256 + 7x16 + 12x1
= 768 + 112 +12
= 892
Therefore, (37B)16 ----> (892)10
4. Conversion between Binary and Hexadecimal Numbers
(
)2 <----> (
)16
Conversion of a binary number to its hexadecimal equivalent or a hexadecimal
number to its binary equivalent can easily be accomplished by merely inspecting
the number. The following conversion table should be memorized for effective
use of the inspection method.
Hex
Binary
digit
digits
-----------------0
0000
1
0001
2
0010
3
0011
4
0100
5
0101
6
0110
7
0111
Hex
Binary
digit
digits
------------------8
1000
9
1001
A
1010
B
1011
C
1100
D
1101
E
1110
F
1111
4-1. Binary-to-Hexadecimal Conversion
To convert a binary number to hexadecimal, start at the binary point and work in
both directions, collecting the bits in 4-bit groups. Then convert each group to
its hexadecimal equivalent.
Examples
a.
(10101111011)2 ---> (
)16
10101111011
101 0111 1011
= 5
7
B
Therefore, (10101111011)2 = (57B)16
b. (1011100101.1010011)2 ---> (
)16
1011100101.1010011
= 0010 1110 0101 . 1010 0110
= 2
E
5 . A
6
Therefore, (1011100101.1010011)2 = (2E5.A6)16
4-2. Hexadecimal-to-Binary Conversion
To convert a hexadecimal number to binary, simply translate each hexadecimal
digit to its 4-bit binary equivalent. Then eliminate any leading zeros in integer
and any trailing zeros in the fraction.
5. Complement Systems
5.1 Radix Complement and Diminished Radix Complement
Complements are used in digital computers for simplifying the subtraction operation and logical
manipulations. Since A – B = A + (-B) and complement systems provide a convenient way of
representing negative numbers, we can convert a subtraction to an addition. When implemented
with digital hardware, it is more efficient to use complements than to use direct subtraction.
Since multiplication and division correspond to repeated addition and subtraction, respectively, it
is possible to perform the four basic arithmetic operations using only the hardware for addition
when negative numbers are represented in complement form.
There are two types of complements for each base-r system: (1) radix complement and
(2) diminished radix complement. The first is referred to as the r’s complement and the
second as the (r-1)’s complement as shown in the following examples.
In the decimal system (r=10), the radix complement is 10’s complement and the diminished
radix complement is 9’s complement. In the octal system (r=8), the radix complement is 8’s
complement and the diminished radix complement is 7’s complement. In the binary system
(r=2), the radix complement is 2’s complement and the diminished radix complement is 1’s
complement.
Given a number N in base r, or (N)r, it is easier to find the diminished complement than to find
the radix complement. The diminished complement of a base r number can easily be found
by subtracting each digit of the number from (r-1). The radix complement can then be
obtained by adding 1 to the diminished radix complement.
A. Decimal system (r=10)
The radix complement is 10’s complement and the diminished radix complement is 9’s
complement. The 9's complement of a decimal number is obtained by subtracting each
digit from 9.
B. Octal system (r=8)
The radix complement is 8’s complement and the diminished radix complement is 7’s
complement. The 7's complement of an octal number is obtained by subtracting each
digit from 7.
C. Binary system (r=2)
The radix complement is 2’s complement and the diminished radix complement is 1’s
complement. The 1's complement of a binary number is obtained by subtracting each
digit from 1. That is, inverting each bit of the number.
Examples
a. Find the 10’s complement and the 9's complement of (546700)10
(546700)10
= (453299)10
= (453300)10
9's complement
10’s complement
b. Find the 8's complement and the 7's complement of (526071)8
(526071)8
= (251706)8
= (251707)8
7's complement
8's complement
c. Find the 2's complement the 1's complement of (00011010)2
(00011010)2
= (11100101)2
= (11100110)2
1's complement
2's complement
5.2 Subtraction with Complement
The subtraction of two n-digit unsigned numbers M – N in base r can be done as follows:
1. Add the r’s complement of the subtrahend, N to the minuend, M. This performs M+(-N).
2. If M >= N, the sum will produce an end carry and the end carry is discarded.
3. If M < N, the sum does not produce an end carry. To obtain the answer in a familiar form,
take the r’s complement of the sum and place a negative sign in front.
Example 1. Using base-10 integers and10’s complement, subtract (214)10 – (39)10
214
+ (-39)
----------
214
+ 961 (-39 =10’s complement of 039 = 961)
-----------Sum = 1175
Answer= 175 (end carry is discarded)
Example 2. Using base-10 integers and 10’s complement, subtract (28)10 – (174)10
28
+ (-174)
-----------
028
+ 826 (-174 =1 0’s complement of 174 = 826)
-----------Sum = 854
Answer= -146 (10’s complement of 854 = -146)
Example 3. Using base-2 integers and 2’s complement, subtract (214) – (39)
214
+ (-39)
-------------
011010110
+ 111011001
------------------Sum = 1010101111
Answer = 010101111
= (175) 10
(-39 = 2’ complement of 39 = 111011001)
(end carry is discarded)
Example 4. Using base-2 integers and 2’s complement, subtract (28)10 – (174)10
28
+ (-174)
-------------
000011100
+ 101010010
-----------------Sum = 101101110
If M (28) < N (174), the sum does not produce an end carry. To obtain the answer, take the 2’s
complement of the sum and place a negative sign in front.
Answer = - (010010010) 2
= - 146
6. Internal Representation of Integer Type Data
6-1. Positive Numbers
a.. Decimal 26 in an 8-bit byte
(26)10 = (00011010)2
= (1A)16
b. Decimal 26 in an 16-bit byte
(26)10 = (0000000000011010)2
= (001A)16
6-2. Negative Numbers
a. Decimal -26 in an 8-bit byte
-26 = -(26)10 = -(00011010)2
= (11100101)2
= (11100110)2
= (E6)16
1's complement
2's complement
2's complement
b. Decimal -26 in a 16-bit word
-26 = -(26)10 = -(0000000000011010)2
= (1111111111100101)2 1's complement
= (1111111111100110)2 2's complement
= (FFE6)16
6-3. Interpretation of Internal Integer Type Data
The leftmost bit of a memory byte or word of a number indicates whether the number
is positive or negative. If the bit is 0, the number is positive. If the bit is 1, the
number is negative and is in the 2’s complement. The conversion of the number to its
decimal equivalent is illustrated in the examples below.
Examples
a. (7C)16 = ( ? )10
(7C)16 = (0111 1100)2
= (124)10
b. (8B)16 = ( ? )10
(8B)16 = (10001011)2
= (10001010)2
= -(01110101)2
= -(117)10
or (8B)16 = (10001011)2
= - (01110101)2
= - (117)10
(2’s complement)
(1’s complement)
(1’s complement of 10001010 is 01110101)
(2’s complement)
(2’s complement of 10001011 is 01110101)
7. Internal Representation of Floating-Point Type Data
2.5.1. A Simple Model
(Textbook: pp.56-59)
Floating-Point Representation
1 bit
5 bits
Sign bit Exponent
8 bits
Mantissa
(a) Unbiased exponent
00000 for actual exponent 20
00101 for actual exponent 25
(b) Biased exponent
10000 for actual exponent 20
100100 for actual exponent 24
It is more efficient to use a biased exponent and the biased exponent with
a bias of 16 (excess-16) will be used in this floating-point format.
Example: Internal representation of (10.25)10
(10.25)10 = (1010.01) 2
= (1010.01) 2 x 20
= (101.001) 2 x 21
:
= (.101001) 2 x 24
= (.0101001) 2 x 25
= (.00101001) 2 x 26
Internal representation of (10.25)10 is
(Un-normalized form)
.
.
(Normalized form)
(Un-normalized form)
.
0 10100 10100100
A normalized form is used for storing a floating-point number in memory.
Note that a normalized form is a floating-point representation where the
leftmost bit of the mantissa will always be a 1.
2.5.3. Floating-Point Error
8. Internal Representation of Character Type Data
ASCII Character Code
Character binary hexadecimal
------------------------------------------
.
bell
LF
FF
CR
.
.
space
.
.
0
1
.
.
A
B
.
.
a
b
.
00000111
00001010
00001100
00001101
07
0A
0C
0D
00100000
20
00110000
00110001
30
31
01000001
01000010
41
42
01100001
01100010
61
62
9. Binary Integer Arithmetic
Since A – B = A + (-B), we can accomplish subtraction by adding the complement of B to A.
The complement method is more efficient with digital hardware.
a. Binary implementation of (13)10 + (6) 10.
13
+ 6
---------19
00001101
+ 00000110
-----------------00010011
b. Binary implementation of (13)10 - (6) 10.
13
+ (-6)
----------7
00001101
+ 11111010
-----------------00000111
(-6 in 2’s complement)
Note that the overflow bit (1) is discarded in the 2’s complement arithmetic.
10. Error-Detection Code
Errors may occur whenever data are transmitted between two devices. One of the most common
ways to facilitate error detection is by means of a parity bit. A parity bit is an extra bit included
as part of the data to make the total number of 1s transmitted either odd or even.
a. Odd Parity Method
b. Even Parity Method
Examples:
8-bit Odd Parity Generator
--------------------------------Data
P
------------ -----00100000 0
00110000 1
01100101 1
8-bit Even Parity Generator
---------------------------------Data
P
----------- -----00100000 1
00110000 0
01100101 0