Lecture 5
Topics
Sec 1.4 Representing Information as Bit Patterns
 Representing Text
 Representing Numeric Values
 Representing Images
Sec 1.6 Storing Integers
 Two’s Complement Notation
 Excess Notation
Representing Text

Each character (letter, punctuation, etc.) is
assigned a unique bit pattern.

ASCII: Uses patterns of 7-bits to represent most
symbols used in written English text

Unicode: Uses patterns of 16-bits to represent the
major symbols used in languages world side

ISO standard: Uses patterns of 32-bits to represent
most symbols used in languages world wide
Figure 1.12 The message “Hello.”
in ASCII
Representing Numeric Values

Binary notation: Uses bits to represent a number
in base two

Limitations of computer representations of
numeric values
Overflow – occurs when a value is too big to be
represented
 Truncation – occurs when a value cannot be
represented accurately

Representing Images



Images are stored using a variety of
formats and compression techniques
The simplest representation is a bitmap
Bitmaps partition an image into a grid of
picture elements, called pixels, and then
convert each pixel into a bit pattern
Resolution refers to the sharpness or clarity
of an image
 bitmaps that are divided into smaller
pixels will yield higher resolution images
 the left image is stored using 96 pixels per
square inch, and the right image is stored
using 48 pixels per square inch
 the left image appears sharp, but has
twice the storage requirements
Storing Integers

Integers are categorized as:
Signed Integers
 Unsigned Integers

Signed are those that include positive as well as
negative numbers
 While Unsigned are those that only represent
positive numbers
 The range for Signed Numbers is reduced to half

Signed Integers
There are two ways to represent Signed Integers:

Two’s complement notation: The most
popular means of representing integer values

Excess notation: Another means of
representing integer values
Both can suffer from overflow errors.

Representing Negative Numbers

Can use “+” and “-”, as usual.
Decimal Binary
+17 +10001
-17 -10001

But requires extra symbols. How can we use bits
instead?
Figure 1.21 Two’s complement
notation systems
1-11
Two’s complement Notation

Invert all bits, then add 1.
Binary
Binary
4-bit (1 bit
sign + 3
bits
magnitude)
Decimal
4-bit (1 bit sign +
3 bits magnitude)
Decimal
0
0000
0
1
0001
-1
1111
2
0010
-2
1110
3
0011
-3
1101
4
0100
-4
1100
5
0101
-5
1011
6
0110
-6
1010
7
0111
-7
1001
-8
1000
Two’s complement – it just works!
+
=
0011 (3 decimal)
0010 (2 decimal)
0101 (5 decimal ?)
1101 (-3 decimal)
+ 1110 (-2 decimal)
11011 (-5 decimal ?)
= 
0011 (3 decimal)
+ 1110 (-2 decimal)
= 
10001 (1 decimal ?)
Problems with binary arithmetic


Fixed number of bits  can go out of range.
0011
(3 decimal)
+ 0111
(7 decimal)
= 1010
(-6 decimal???)
Overflow: When result can’t be represented within
range of bits.
Overflow occurs when the value that we compute cannot
fit into the number of bits we have allocated for the
result.
 In addition, if operands have same sign and sum doesn’t.

Excess Notation

Excess 8 notation indicates that the value for
zero is the bit pattern for 8, that is 1000

The bit patterns are 4 bits long

Positive numbers are above it in order and
negative numbers are below it in order.
Figure 1.24 An excess eight
conversion table
Excess Notation (continue)

That is the zero point for Excess 128 notation is
128; the zero point for excess 64 notation is 64; and so
forth.

For example, let's say we want to determine the pattern
for 15 in Excess 128 notation.

The decimal number would be 128 + 15, or
143. Therefore, the bit pattern would be 10001111.
Storing Fractions

In contrast to storage of Integers the storage of
fractions requires not only the representation of
the magnitude by the radix point as well

A popular way of doing this is based on
scientific notation called Floating-Point
Notation
Binary Fractions (revisited)
We use a radix point in the same role as the
decimal point in decimal notation. That is, the
digits to the left of the radix point represent the
integer part of the value. The digits to the right
represent the fractional part of the value.
 _ _ _ ._ _ _
2 = 1/2
2 = 1/4
2
1
0
-1
-2
-3
2 2 2 2 2 2
2 = 1/8
Example:
101.101 is 5 5/8 and 101.01011 is 5 11/32

-1
-2
-3
Short Cut (Bin --- Decimal)

Binary-to-decimal for the part after the point
1.
2.

Convert binary digits as though there were no point.
Divide result by 2#bits – e.g., if 7 bits converted, then
divide by 27 = 128.
E.g., .1101
–
–
1101 = 13
So, .1101 = 13  24 = 13/16 = .8125
Figure 1.26 Floating-point
notation components
Look at Scientific Notation
Let's look at some decimal numbers first, to get a feel for
what needs to be done. First we need to know how many
digits in the mantissa. Let's choose 5.
Real
5 digits and an exponent
Floating point
12001.00
12001 * 100
+.12001 * 105
-120.01
-12001 * 10-2
-.12001 * 103
0.12
12000 * 10-5
+.12000 * 100
15,555,555
15555 * 103
+.15555 * 108
The last example shows us one of the problems with
floating-point notation – truncation error. Part of the
value being stored is lost because the mantissa field is not
large enough.
Floating Point Notation

Now let's switch to binary. Although a common
standard uses 64 bits for a floating-point
number: 1 bit for the sign, 11 bits for the
exponent, and 52 bits for the mantissa, we will
use an abbreviated version. Let's use one byte: 1
bit for the sign, 3 bits for the exponent, and 4
bits for the mantissa.

The sign is easy enough 0 or 1.

The exponent has 3 bits and must be either + or
-. We could use 2's complement, but excess
notation is used instead! 3 bits means excess 4
notation. That leaves 4 bits for the mantissa.
Why excess notation? It gives the best range of
exponents and ease of wiring the circuitry.
Excess 4 Notation

000
001
010
011
100
101
110
111
0 – 4 = -4
1 – 4 = -3
2 – 4 = -2
3 – 4 = -1
4–4=0
5–4=1
6–4=2
7–4=3
How to convert ??





Determine the sign bit
Convert the whole number to binary
Convert the fraction to binary
Move the radix to the left of the most significant
digit (left-most non-zero digit)
Determine the exponent in excess 4 notation
Example:
 To store 1 1/8 we express it in binary notation
and obtain 1.001
 Copy the bit patterns into the mantissa field
from left to right.
Example 1


Examples:
+2 3/4
This is positive so our sign bit is 0
 2 is 10
 3/4 is .11
 This gives us 10.11
Move the radix two places to the left: .1011
 So the exponent is +2: 110 in excess 4
 Giving us: 0 110 1011

Example 2

-1.1875
This is negative so our sign bit is 1
 1 is 1
 .1875 is 3/16 giving us .0011
 This gives us 1.0011: Because we have 4 bits we have
to use 1.001: Move the radix one place to the left:
.1001
 So the exponent is +1: 101 in excess 4
 Giving us: 1 101 1001

Example 3

What is 10111100 in decimal?
Split it apart into its pieces: 1 011 1100
 The sign is 1 or negative
 The exponent is 011 or -1
 The mantissa is 1100. Move the radix one place to
the left: .01100 giving us 3/8
 Our number is -3/8
