1
1.1 Data representation
Information representation
and multimedia
1
In this chapter, you will learn about
★ binary magnitudes, binary prefixes and decimal prefixes
★ binary, denary and hexadecimal number systems
★ how to carry out binary addition and subtraction
★ the use of hexadecimal and binary coded decimal (BCD) number
systems
★ the representation of character sets (such as ASCII and Unicode)
★ how data for a bit-mapped image is encoded
★ how to estimate the file size for a bit-map image
★ image resolution and colour depth
★ encoding of vector graphics
★ the representation of sound in a computer
★ the effects of changing sampling rate and resolution on sound quality
★ the need for file compression methods (such as lossy and lossless
formats)
★ how to compress common file formats (such as text files, bit-map
images, vector graphics, sound files and video files).
WHAT YOU SHOULD ALREADY KNOW
Try these four questions before you read this
chapter.
1 What are the column weightings for the
binary number system?
2 Carry out these binary additions. Convert
your answers to denary.
a) 0 0 1 1 0 1 0 1 + 0 1 0 0 1 0 0 0
b) 0 1 0 0 1 1 0 1 + 0 1 1 0 1 1 1 0
c) 0 1 0 1 1 1 1 1 + 0 0 0 1 1 1 1 0
d) 0 1 0 0 0 1 1 1 + 0 1 1 0 1 1 1 1
e) 1 0 0 0 0 0 0 1 + 0 1 1 1 0 1 1 1
f) 1 0 1 0 1 0 1 0 + 1 0 1 0 1 0 1 0
3 What are the column weightings for the
hexadecimal (base 16) number system?
4 Carry out these hexadecimal additions.
Convert your answers to denary.
a) 1 0 7 + 2 5 7
b) 2 0 8 + A 1 7
c) A A A + 7 7 7
d) 1 F F + 7 F 7
e) 1 4 9 + F 0 F
f) 1 2 5 1 + 2 5 6 7
g) 3 4 A B + C 0 0 A
h) A 0 0 1 + D 7 7 F
i) 1 0 0 9 + 9 F F 1
j) 2 7 7 7 + A C F 1
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1 InformatIon representatIon and multImedIa
1.1 Data representation
Key terms
Binary – base two number system based on the values
0 and 1 only.
Bit – abbreviation for binary digit.
One’s complement – each binary digit in a number is
reversed to allow both negative and positive numbers to
be represented.
Two’s complement – each binary digit is reversed and
1 is added in right-most position to produce another
method of representing positive and negative numbers.
Sign and magnitude – binary number system where
left-most bit is used to represent the sign (0 = + and
1 = –); the remaining bits represent the binary value.
Hexadecimal – a number system based on the value 16
(uses the denary digits 0 to 9 and the letters A to F).
Memory dump – contents of a computer memory output
to screen or printer.
Binary-coded decimal (BCD) – number system that
uses 4 bits to represent each denary digit.
ASCII code – coding system for all the characters on a
keyboard and control codes.
Character set – a list of characters that have been
defined by computer hardware and software. It is
necessary to have a method of coding, so that the
computer can understand human characters.
Unicode – coding system which represents all the
languages of the world (first 128 characters are the
same as ASCII code).
1.1.1 Number systems
Every one of us is used to the decimal or denary (base 10) number system. This
uses the digits 0 to 9 which are placed in ‘weighted’ columns.
10 000 1000 100 10 units
3 1 4 2 1
The denary number represented above is thirty-one thousand, four hundred and
twenty-one.
(Note that dealing with decimal fractions is covered in Chapter 13 since this is
slightly more complex.)
Designers of computer systems adopted the binary (base 2) number system
since this allows only two values, 0 and 1. No matter how complex the system,
the basic building block in all computers is the binary number system. Since
computers contain millions and millions of tiny ‘switches’, which must be in the
ON or OFF position, this lends itself logically to the binary system. A switch in
the ON position can be represented by 1; a switch in the OFF position can be
represented by 0. Each of the binary digits are known as bits.
1.1.2 Binary number system
The binary system uses 1s and 0s only which gives these corresponding weightings:
128 64 32 16 8 4 2 1
(2
7
) (2
6
) (2
5
) (2
4
) (2
3
) (2
2
) (2
1
) (2
0
)
A typical binary number would be:
1 1 1 0 1 1 1 0
Converting from binary to denary and from denary to binary
It is fairly straightforward to change a binary number into a denary number.
Each time a 1 appears in a column, the column value is added to the total. For
example, the binary number above is:
128 + 64 + 32 + 8 + 4 + 2 = 238 (denary)
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1.1 Data representation
The 0 values are simply ignored when calculating the total.
The reverse operation – converting from denary to binary – is slightly more
complex. There are two basic ways of doing this.
Consider the conversion of the denary number, 107, into binary …
Method 1
This method involves placing 1s in the appropriate position so that the total
equates to 107.
128 64 32 16 8 4 2 1
0 1 1 0 1 0 1 1
Method 2
This method involves successive division by 2; the remainders are then written
from bottom to top to give the binary value.
2 107
Write the remainder from
bottom to top to get the
binary number:
0 1 1 0 1 0 1 1
2 53 remainder: 1
2 26 remainder: 1
2 13 remainder: 0
2 6 remainder: 1
2 3 remainder: 0
2 1 remainder: 1
2 0 remainder: 1
0 remainder: 0
Binary addition and subtraction
Up until now we have assumed all binary numbers have positive values. There are a
number of methods to represent both positive and negative numbers. We will consider:
» one’s complement
» two’s complement.
In one’s complement, each digit in the binary number is inverted (in other
words, 0 becomes 1 and 1 becomes 0). For example, 0 1 0 1 1 0 1 0 (denary
value 90) becomes 1 0 1 0 0 1 0 1 (denary value −90).
In two’s complement, each digit in the binary number is inverted and a ‘1’ is
added to the right-most bit. For example, 0 1 0 1 1 0 1 0 (denary value 90)
becomes:
1 0 1 0 0 1 0 1
+ 1
= 1 0 1 0 0 1 1 0 (since 1 + 1 = 0, a carry of 1) = denary value −90
Throughout the remainder of this chapter, we will use the two’s complement
method to avoid confusion. Also, two’s complement makes binary addition
and subtraction more straightforward. The reader is left to investigate one’s
complement and the sign and magnitude method in binary arithmetic.
Now that we are introducing negative numbers, we need a way to represent
these in binary. The two’s complement uses these weightings for an 8-bit
number representation:
−128 64 32 16 8 4 2 1
ACTIVITY 1A
Convert these
binary numbers into
denary.
a) 0 0 1 1 0 0 1 1
b) 0 1 1 1 1 1 1 1
c) 1 0 0 1 1 0 0 1
d) 0 1 1 1 0 1 0 0
e) 1 1 1 1 1 1 1 1
f) 0 0 0 0 1 1 1 1
g) 1 0 0 0 1 1 1 1
h) 0 0 1 1 0 0 1 1
i) 0 1 1 1 0 0 0 0
j) 1 1 1 0 1 1 1 0
ACTIVITY 1B
Convert these
denary numbers into
binary (using either
method).
a) 4 1
b) 6 7
c) 8 6
d) 1 0 0
e) 1 1 1
f) 1 2 7
g) 1 4 4
h) 1 8 9
i) 2 0 0
j) 2 5 5
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1 InformatIon representatIon and multImedIa
Add 0 0 1 0 0 1 0 1 (37 in denary) and 0 0 1 1 1 0 1 0 (58 in denary).
Example 1.1
Solution
−128 64 32 16 8 4 2 1
0 0 1 0 0 1 0 1
+
0 0 1 1 1 0 1 0
=
0 1 0 1 1 1 1 1
This gives us 0 1 0 1 1 1 1 1, which is 95 in denary; the correct answer.
This means:
−128 64 32 16 8 4 2 1
1 1 0 1 1 0 1 0
0 0 1 0 0 1 1 0
The first example is: −128 + 64 + 16 + 8 + 2 = −38
The second example is: 32 + 4 + 2 = 38
The easiest way to convert a number into its negative equivalent is to use two’s
complement. For example, 104 in binary is 0 1 1 0 1 0 0 0.
To find the binary value for −104 using two’s complement:
invert the digits: 1 0 0 1 0 1 1 1 (+104 in denary)
add 1: 1
which gives: 1 0 0 1 1 0 0 0 = −104)
ACTIVITY 1C
Convert these denary numbers into 8-bit binary numbers using two’s
complement where necessary. Use these binary column weightings:
−128 64 32 16 8 4 2 1
a) +114
b) +61
c) +96
d) −14
e) −116
Binary addition
Consider Examples 1.1 and 1.2.
EXTENSION
ACTIVITY 1A
Show the column
headings for a
system that uses
16 bits to represent
a binary number.
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1.1 Data representation
Add 0 1 0 1 0 0 1 0 (82 in denary) and 0 1 0 0 0 1 0 1 (69 in denary).
Example 1.2
Solution
−128 64 32 16 8 4 2 1
0 1 0 1 0 0 1 0
+
0 1 0 0 0 1 0 1
=
1 0 0 1 0 1 1 1
This gives us 1 0 0 1 0 1 1 1, which is –105 in denary (which is clearly nonsense).
When adding two positive numbers, the result should always be positive
(likewise, when adding two negative numbers, the result should always be
negative). Here, the addition of two positive numbers has resulted in a negative
answer. This is due to the result of the addition producing a number which is
outside the range of values which can be represented by the 8 bits being used (in
this case +127 is the largest value which can be represented, and the calculation
produces the value 151, which is larger than 127 and, therefore, out of range).
This causes overf low; it is considered in more detail in Chapter 13.
Binary subtraction
To carry out subtraction in binary, we convert the number being subtracted into
its negative equivalent using two’s complement, and then add the two numbers.
Carry out the subtraction 95 – 68 in binary.
Example 1.3
Solution
1 Convert the two numbers into binary:
95 = 0 1 0 1 1 1 1 1
68 = 0 1 0 0 0 1 0 0
2 Find the two’s complement of 68:
invert the digits: 1 0 1 1 1 0 1 1
add 1: 1
which gives: 1 0 1 1 1 1 0 0 = −68
3 Add 95 and −68:
−128 64 32 16 8 4 2 1
0 1 0 1 1 1 1 1
+
1 0 1 1 1 1 0 0
=
1 0 0 0 1 1 0 1 1
The additional ninth bit is simply ignored leaving the binary number
0 0 0 1 1 0 1 1 (denary equivalent of 27, which is the correct result of the
subtraction).
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1 InformatIon representatIon and multImedIa
Measurement of the size of computer memories
The byte is the smallest unit of memory in a computer. Some computers use
larger bytes, such as 16-bit systems and 32-bit systems, but they are always
multiples of 8. 1 byte of memory wouldn’t allow you to store very much
information; so memory size is measured in these multiples. See Table 1.1.
ACTIVITY 1D
Carry out these binary additions and subtractions using these 8-bit column
weightings:
−128 64 32 16 8 4 2 1
a) 0 0 1 1 1 0 0 1 + 0 0 1 0 1 0 0 1
b) 0 1 0 0 1 0 1 1 + 0 0 1 0 0 0 1 1
c) 0 1 0 1 1 0 0 0 + 0 0 1 0 1 0 0 0
d) 0 1 1 1 0 0 1 1 + 0 0 1 1 1 1 1 0
e) 0 0 0 0 1 1 1 1 + 0 0 0 1 1 1 0 0
f) 0 1 1 0 0 0 1 1 − 0 0 1 1 0 0 0 0
g) 0 1 1 1 1 1 1 1 − 0 1 0 1 1 0 1 0
h) 0 0 1 1 0 1 0 0 − 0 1 0 0 0 1 0 0
i) 0 0 0 0 0 0 1 1 − 0 1 1 0 0 1 0 0
j) 1 1 0 1 1 1 1 1 − 1 1 0 0 0 0 1 1
Carry out the subtraction 49 – 80 in binary.
Example 1.4
Solution
1 Convert the two numbers into binary:
49 = 0 0 1 1 0 0 0 1
80 = 0 1 0 1 0 0 0 0
2 Find the two’s complement of 80:
invert the digits: 1 0 1 0 1 1 1 1
add 1: 1
which gives: 1 0 1 1 0 0 0 0 = −80
3 Add 49 and −80:
−128 64 32 16 8 4 2 1
0 0 1 1 0 0 0 1
+
1 0 1 1 0 0 0 0
=
1 1 1 0 0 0 0 1
This gives us 1 1 1 0 0 0 0 1, which is −31 in denary; the correct answer.
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1.1 Data representation
Name of memory size Equivalent denary value (bytes)
1 kilobyte (1 KB) 1 000
1 megabyte (1 MB) 1 000 000
1 gigabyte (1 GB) 1 000 000 000
1 terabyte (1 TB) 1 000 000 000 000
1 petabyte (1 PB) 1 000 000 000 000 000
▲ Table 1.1 Memory size and denary values
The system of numbering shown in Table 1.1 only refers to some storage
devices, but is technically inaccurate. It is based on the SI (base 10) system
of units where 1 kilo is equal to 1000. A 1 TB hard disk drive would allow the
storage of 1 × 10
12
bytes according to this system. However, since memory size
is actually measured in terms of powers of 2, another system has been proposed
by the International Electrotechnical Commission (IEC); it is based on the
binary system. See Table 1.2.
Name of memory size Number of bytes Equivalent denary value (bytes)
1 kibibyte (1 KiB) 2
10
1 024
1 mebibyte (1 MiB) 2
20
1 048 576
1 gibibyte (1 GiB) 2
30
1 073 741 824
1 tebibyte (1 TiB) 2
40
1 099 511 627 776
1 pebibyte (1 PiB) 2
50
1 125 899 906 842 624
▲ Table 1.2 IEC memory size system
This system is more accurate. Internal memories (such as RAM) should
bemeasured using the IEC system. A 64 GiB RAM could, therefore, store
64×2
30
bytes of data (68 719 476 736 bytes).
See Section 1.2 for examples of how to calculate the size of a file.
1.1.3 Hexadecimal number system
The hexadecimal system is very closely related to the binary system.
Hexadecimal (sometimes referred to as simply hex) is a base 16 system with the
weightings:
1 048 576 65 536 4096 256 16 1
(16
5
) (16
4
) (16
3
) (16
2
) (16
1
) (16
0
)
Because it is a system based on 16 different digits, the numbers 0 to 9 and the
letters A to F are used to represent hexadecimal digits.
A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15.
Since 16 = 2
4
, four binary digits are equivalent to each hexadecimal digit. Table
1.3 summarises the link between binary, hexadecimal and denary.
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1 InformatIon representatIon and multImedIa
Convert 1 0 1 1 1 1 1 0 0 0 0 1 from binary to hexadecimal.
Convert 1 0 0 0 0 1 1 1 1 1 1 1 0 1 from binary to hexadecimal.
Example 1.5
Example 1.6
Solution
First split it into groups of 4 bits:
1 0 1 1 1 1 1 0 0 0 0 1
Then find the equivalent hexadecimal digits:
B E 1
Solution
First split it into groups of 4 bits:
1 0 0 0 0 1 1 1 1 1 1 1 0 1
The left group only contains 2 bits, so add in two 0s to the left:
0 0 1 0 0 0 0 1 1 1 1 1 1 1 0 1
Now find the equivalent hexadecimal digits:
2 1 F D
Binary value Hexadecimal value Denary value
0 0 0 0 0 0
0 0 0 1 1 1
0 0 1 0 2 2
0 0 1 1 3 3
0 1 0 0 4 4
0 1 0 1 5 5
0 1 1 0 6 6
0 1 1 1 7 7
1 0 0 0 8 8
1 0 0 1 9 9
1 0 1 0 A 10
1 0 1 1 B 11
1 1 0 0 C 12
1 1 0 1 D 13
1 1 1 0 E 14
1 1 1 1 F 15
▲ Table 1.3 The link between binary, hexadecimal and denary
Converting from binary to hexadecimal and from hexadecimal to binary
Converting from binary to hexadecimal is a fairly easy process. Starting from
the right and moving left, split the binary number into groups of 4 bits. If the
last group has less than 4 bits, then simply fill in with 0s from the left. Take
each group of 4 bits and convert it into the equivalent hexadecimal digit using
Table 1.3.
Examples 1.5 and 1.6 show you how this works.
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1.1 Data representation
Converting from hexadecimal to binary is also straightforward. Using the data
from Table 1.3, simply take each hexadecimal digit and write down the 4 bit
code which corresponds to the digit.
ACTIVITY 1E
Convert these binary numbers into hexadecimal.
a) 1 1 0 0 0 0 1 1
b) 1 1 1 1 0 1 1 1
c) 1 0 0 1 1 1 1 1 1 1
d) 1 0 0 1 1 1 0 1 1 1 0
e) 0 0 0 1 1 1 1 0 0 0 0 1
f) 1 0 0 0 1 0 0 1 1 1 1 0
g) 0 0 1 0 0 1 1 1 1 1 1 1 0
h) 0 1 1 1 0 1 0 0 1 1 1 0 0
i) 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1
j) 0 0 1 1 0 0 1 1 1 1 0 1 0 1 1 1 0
Convert this hexadecimal number to its binary equivalent.
4 5 A
Example 1.7
Solution
Using Table 1.3, find the 4-bit code for each digit:
0 1 0 0 0 1 0 1 1 0 1 0
Put the groups together to form the binary number:
0 1 0 0 0 1 0 1 1 0 1 0
Use of the hexadecimal system
This section reviews two uses of the hexadecimal system.
Memory dumps
It is much easier to work with:
B 5 A 4 1 A F C
than it is to work with:
1 0 1 1 0 1 0 1 1 0 1 0 0 1 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 0
So, hexadecimal is often used when developing new software or when trying to
trace errors in programs. When the memory contents are output to a printer or
monitor, this is known as a memory dump.
Convert this hexadecimal number to its binary equivalent.
B F 0 8
Example 1.8
Solution
Using Table 1.3:
1 0 1 1 1 1 1 1 0 0 0 0 1 0 0 0
Then put all the digits together:
1 0 1 1 1 1 1 1 0 0 0 0 1 0 0 0
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1 InformatIon representatIon and multImedIa
00990F60 54 68 69 73 20 69 73 20 61 6E 20 65 78 61 6D 70 6C 65 20 6F 66
00990F77 61 20 6D 65 6D 6F 72 79 20 64 75 6D 70 20 66 72 6F 6D 20 20 61
00990E8E 74 79 70 69 63 61 6C 20 20 63 6F 6D 70 75 74 65 72 20 20 6D 85
00990EA5 6D 6F 72 79 20 73 68 6F 77 69 6E 67 20 74 68 65 20 20 63 6F 6E
00990EBC 74 65 6E 74 73 20 6F 66 20 61 20 6E 75 6D 62 65 72 20 20 6F 66
00990ED3 6C 6F 63 61 74 69 6F 6E 73 20 20 69 6E 20 20 68 65 78 20 20 20
00990EEA 6E 6F 74 61 74 69 6F 6E 20 20 00 00 00 00 00 00 00 00 00 00 00
▲ Table 1.4 Memory dump
A program developer can look at each of the hexadecimal codes (as shown in
Table 1.4) and determine where the error lies. The value on the far left shows
the memory location, so it is possible to find out exactly where in memory
the fault occurs. Using hexadecimal is more manageable than binary. It is a
powerful fault-tracing tool, but requires considerable knowledge of computer
architecture to be able to interpret the results.
1.1.4 Binary-coded decimal (BCD) system
The binary-coded decimal (BCD) system uses a 4-bit code to represent each
denary digit:
0 0 0 0 = 0 0 1 0 1 = 5
0 0 0 1 = 1 0 1 1 0 = 6
0 0 1 0 = 2 0 1 1 1 = 7
0 0 1 1 = 3 1 0 0 0 = 8
0 1 0 0 = 4 1 0 0 1 = 9
Therefore, the denary number 3 1 6 5 would be 0 0 1 1 0 0 0 1 0 1 1 0 0 1 0 1
in BCD format.
The 4-bit code can be stored in the computer either as half a byte or two 4-bit
codes stored together to form one byte. For example, using 3 1 6 5 again …
Method 1: four single bytes
0 0 0 0 0 0 1 1 3
0 0 0 0 0 0 0 1 1
0 0 0 0 0 1 1 0 6
0 0 0 0 0 1 0 1 5
Method 2: two bytes
0 0 1 1 0 0 0 1 3 1
0 1 1 0 0 1 0 1 6 5
ACTIVITY 1G
1 Convert these denary numbers into BCD format.
a) 2 7 1 b) 5 0 0 6 c) 7 9 9 0
2 Convert these BCD numbers into denary numbers.
a) 1 0 0 1 0 0 1 1 0 1 1 1
b) 0 1 1 1 0 1 1 1 0 1 1 0 0 0 1 0
ACTIVITY 1F
Convert these
hexadecimal
numbers into binary.
a) 6 C
b) 5 9
c) A A
d) A 0 0
e) 4 0 E
f) B A 6
g) 9 C C
h) 4 0 A A
i) D A 4 7
j) 1 A B 0
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1.1 Data representation
Uses of BCD
The most obvious use of BCD is in the representation of digits on a calculator
or clock display.
Each denary digit will have a BCD equivalent value which makes it easy to
convert from computer output to denary display.
As you will learn in Chapter 13, it is nearly impossible to represent decimal
values exactly in computer memories which use the binary number system.
Normally this doesn’t cause a major issue since the differences can be dealt
with. However, when it comes to accounting and representing monetary
values in computers, exact values need to be stored to prevent significant
errors from accumulating. Monetary values use a fixed-point notation, for
example $1.31, so one solution is to represent each denary digit as a BCD
value.
Consider adding $0.37 and $0.94 together using fixed-point decimals.
$0.37 0
0 0 0 0 0 0 0 . 0 0 1 1 0 1 1 1
+
+
$0.94 0
0 0 0 0 0 0 0 . 1 0 0 1 0 1 0 0
Expected result = $1.31
Using binary addition, this sum will produce:
0 0 0 0 0 0 0 0 . 1 1 0 0 1 0 1 1 which produces 1 1 0 0 (denary 12) and 1
0 11 (denary 11), which is clearly incorrect. The problem was caused by 3
+ 9 = 12 and 7 + 4 = 11, as neither 12 nor 11 are single denary digits. The
solution to this problem, enabling the computer to store monetary values
accurately, is to add 0 1 1 0 (denary 6) whenever such a problem arises. The
computer can be programmed to recognise this issue and add 0 1 1 0 at each
appropriate point.
If we look at the example again, we can add .07 and .04 (the two digits in the
second decimal place) first.
0 1 1 1
+
0 1 0 0
=
1 0 1 1
1 0 1 1
+
0 1 1 0
=
1 0 0 0 1
This produces 1 0 1 1 which
isn’t a denary digit; this
will flag an error and the
computer needs to add
0 1 1 0.
This now
produces a fifth
bit which is
carried to the
next decimal
digit position.
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1 InformatIon representatIon and multImedIa
Now we will add .3 and .9 together (the two digits in the first decimal place)
remembering the carry bit from the addition above:
0 0 1 1
+
1 0 0 1
+
1
=
1 1 0 1
This produces 1 1 0 1 which isn’t a denary digit; this will flag an error and the
computer again needs to add 0 1 1 0.
1 1 0 1
+
0 1 1 0
=
1 0 0 1 1
Adding 1 to 0 0 0 0 0 0 0 0 produces:
0 0 0 0 0 0 0 1
Final answer:
0 0 0 0 0 0 0 1 . 0 0 1 1 0 0 0 1
which is 1.31 in denary – the correct answer.
1.1.5 ASCII codes and Unicodes
The ASCII code system (American Standard Code for Information Interchange)
was set up in 1963 for use in communication systems and computer systems.
The newer version of the code was published in 1986. The standard ASCII
code character set consists of 7-bit codes (0 to 127 denary or 0 to 7F in
hexadecimal); this represents the letters, numbers and characters found on a
standard keyboard together with 32 control codes (which use up codes 0 to 31
(denary) or 0 to 19 (hexadecimal)).
Table 1.5 shows part of the standard ASCII code table (only the control codes
have been removed from the table).
This again produces
a fifth bit which is
carried to the next
decimal digit position.
ACTIVITY 1H
Carry out these BCD
additions.
a) 0.45 + 0.21
b) 0.66 + 0.51
c) 0.88 + 0.75
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1.1 Data representation
Dec Hex Char Dec Hex Char Dec Hex Char
32 20 <SPACE> 64 40 @ 96 60 `
33 21 ! 65 41 A 97 61 a
34 22 “ 66 42 B 98 62 b
35 23 # 67 43 C 99 63 c
36 24 $ 68 44 D 100 64 d
37 25 % 69 45 E 101 65 e
38 26 & 70 46 F 102 66 f
39 27 ‘ 71 47 G 103 67 g
40 28 ( 72 48 H 104 68 h
41 29 ) 73 49 I 105 69 i
42 2A * 74 4A J 106 6A j
43 2B + 75 4B K 107 6B k
44 2C , 76 4C L 108 6C l
45 2D - 77 4D M 109 6D m
46 2E . 78 4E N 110 6E n
47 2F / 79 4F O 111 6F o
48 30 0 80 50 P 112 70 p
49 31 1 81 51 Q 113 71 q
50 32 2 82 52 R 114 72 r
51 33 3 83 53 S 115 73 s
52 34 4 84 54 T 116 74 t
53 35 5 85 55 U 117 75 u
54 36 6 86 56 V 118 76 v
55 37 7 87 57 W 119 77 w
56 38 8 88 58 X 120 78 x
57 39 9 89 59 Y 121 79 y
58 3A : 90 5A Z 122 7A z
59 3B ; 91 5B [ 123 7B {
60 3C < 92 5C \ 124 7C |
61 3D = 93 5D ] 125 7D }
62 3E > 94 5E ^ 126 7E ~
63 3F ? 95 5F _ 127 7F <DELETE>
▲ Table 1.5 Part of the ASCII code table
Notice the storage of characters with uppercase and lowercase. For example:
a 1 1 0 0 0 0 1 hex 61 (lower case)
A 1 0 0 0 0 0 1 hex 41 (upper case)
y 1 1 1 1 0 0 1 hex 79 (lower case)
Y 1 0 1 1 0 0 1 hex 59 (uppercase)
Notice the sixth bit changes from 1 to 0 when comparing lower and uppercase
characters. This makes the conversion between the two an easy operation. It
is also noticeable that the character sets (such as a to z, 0 to 9, and so on) are
grouped together in sequence, which speeds up usability.
Extended ASCII uses 8-bit codes (128 to 255 in denary or 80 to FF in hex). This
allows for non-English characters and for drawing characters to be included.
457591_01_CI_AS & A_Level_CS_001-026.indd 13 25/04/19 9:11 AM
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1
1 InformatIon representatIon and multImedIa
Since ASCII code has a number of disadvantages and is unsuitable for some
purposes, different methods of coding have been developed over the years.
One coding system is called Unicode. Unicode allows characters in a code
form to represent all languages of the world, thus supporting many operating
systems, search engines and internet browsers used globally. There is overlap
with standard ASCII code, since the first 128 (English) characters are the same,
but Unicode can support several thousand different characters in total. As can
be seen in Tables 1.5 and 1.6, ASCII uses one byte to represent a character,
whereas Unicode will support up to four bytes per character.
The Unicode consortium was set up in 1991. Version 1.0 was published with five
goals, these were to
» create a universal standard that covered all languages and all writing systems
» produce a more efficient coding system than ASCII
» adopt uniform encoding where each character is encoded as 16-bit or 32-bit code
» create unambiguous encoding where each 16-bit or 32-bit value always
represents the same character (it is worth pointing out here that the ASCII
code tables are not standardised and versions other than the ones shown in
tables 1.5 and 1.6 exist)
» reserve part of the code for private use to enable a user to assign codes for their
own characters and symbols (useful for Chinese and Japanese character sets).
A sample of Unicode characters are shown in Table 1.7. As can be seen from
the table, characters used in languages such as Russian, Greek, Romanian and
Croatian can now be represented in a computer).
▲ Table 1.6 Extended ASCII code table
Dec Hex Char Dec Hex Char Dec Hex Char Dec Hex Char Dec Hex Char
128 154 9A
155 9B
156 9C
157 9D
158 9E
159 9F
160 A0
161 A1
162 A2
163 A3
164 A4
165 A5
166 A6
167 A7
168 A8
169 A9
170 AA
171 AB
172 AC
173 AD
174 AE
175 AF
176 B0
177 B1
178 B2
179 B3
180 B4
181 B5
182 B6
183 B7
184 B8
185 B9
186 BA
187 BB
188 BC
189 BD
190 BE
191 BF
192 C0
193 C1
194 C2
195 C3
196 C4
197 C5
198 C6
199 C7
200 C8
201 C9
202 CA
203 CB
204 CC
205 CD
206 CE
207 CF
208 D0
209 D1
210 D2
211 D3
212 D4
213 D5
214 D6
215 D7
216 D8
217 D9
218 DA
219 DB
220 DC
221 DD
222 DE
223 DF
224 E0
225 E1
226 E2
227 E3
228 E4
229 E5
230 E6
231 E7
232 E8
233 E9
234 EA
235 EB
236 EC
237 ED
238 EE
239 EF
240 F0
241 F1
242 F2
243 F3
244 F4
245 F5
246 F6
247 F7
248 F8
249 F9
250 FA
251 FB
252 FC
253 FD
254 FE
255 FF
129
130
131
153
152
151
150
149
148
147
146
145
144
143
142
141
140
139
138
137
136
135
134
133
132
80
Ç
ü
é
â
ä
à
å
ç
ê
ë
è
ï
î
ì
Ä
Å
É
æ
Æ
ô
ö
ò
û
ù
ӱ
Ö
81
82
83
99
98
97
96
95
94
93
92
91
90
8F
8E
8D
8C
8B
8A
89
88
87
86
85
84
Ü
ḉ
£
¥
₧
ƒ
á
í
ó
ú
ñ
Ñ
ᵃ
ᵒ
¿
⌐
¬
½
¼
¡
«
»
░
▒
▓
│
╬
╧
╨
╤
╥
╙
╘
╒
╓
╫
╪
┘
┌
█
▄
▌
▐
▀
α
ß
Г
п
∑
σ
μ
τ
Φ
Θ
Ω
δ
∞
ø
ε
∩
≡
±
≥
≤
⌠
⌡
÷
≈
°
▪▪
▪
√
³
²
■
□
┤
╡
╢
╖
╕
╣
║
╗
╝
╜
╛
┐
└
┴
┬
├
─
┼
╞
╟
╚
╔
╩
╦
╠
═
457591_01_CI_AS & A_Level_CS_001-026.indd 14 25/04/19 9:11 AM
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1
1.2 Multimedia
1.2 Multimedia
Key terms
Bit-map image – system that uses pixels to make up an
image.
Pixel – smallest picture element that makes up an image.
Colour depth – number of bits used to represent the
colours in a pixel, e.g. 8 bit colour depth can represent
2
8
= 256 colours.
Bit depth – number of bits used to represent the
smallest unit in, for example, a sound or image file –
the larger the bit depth, the better the quality of the
sound or colour image.
Image resolution – number of pixels that make up an
image, for example, an image could contain 4096 × 3192
pixels (12 738 656 pixels in total).
Screen resolution – number of horizontal and vertical
pixels that make up a screen display. If the screen
resolution is smaller than the image resolution, the
whole image cannot be shown on the screen, or the
original image will become lower quality.
Resolution – number of pixels per column and per row
on a monitor or television screen.
Pixel density – number of pixels per square centimetre.
Vector graphics – images that use 2D points to describe
lines and curves and their properties that are grouped to
form geometric shapes.
Sampling resolution – number of bits used to represent
sound amplitude (also known as bit depth).
Sampling rate – number of sound samples taken per
second.
Frame rate – number of video frames that make up a
video per second.
Images can be stored in a computer in two common formats: bit-map image and
vector graphic.
1.2.1 Bit-map images
Bit-map images are made up of pixels (picture elements); the image is stored
in a two-dimensional matrix of pixels.
Pixels can take different shapes, such as
or
or
or
Ơ ơ Ƣ ƣ Ƥ ƥ Ʀ Ƨ ƨ Ʃ ƪ ƫ Ƭ ƭ Ʈ
ư Ʊ Ʋ Ƴ ƴ Ƶ ƶ Ʒ Ƹ ƹ ƺ ƻ Ƽ ƽ ƾ
ǀ ǁ ǂ ǃ
DŽ
Dž dž LJ Lj lj NJ Nj nj
Ǎ
ǎ
ǐ
Ǒ
ǒ
Ǔ
ǔ
Ǖ
ǖ
Ǘ
ǘ
Ǚ
ǚ
Ǜ
ǜ ǝ
Ǟ
Ǡ
ǡ Ǣ ǣ Ǥ ǥ
Ǧ
ǧ
Ǩ ǩ
Ǫ ǫ Ǭ ǭ
Ǯ
ǰ DZ Dz dz
Ǵ
ǵ Ƕ Ƿ
Ǹ
ǹ
Ǻ
ǻ Ǽ
ǽ
Ǿ
Ȁ
ȁ
Ȃ
ȃ
Ȅ
ȅ
Ȇ
ȇ
Ȉ
ȉ
Ȋ
ȋ
Ȍ
ȍ
Ȏ
Ȑ
ȑ
Ȓ
ȓ
Ȕ
ȕ
Ȗ
ȗ Ș ș Ț ț Ȝ ȝ
Ȟ
Ƞ ȡ Ȣ ȣ Ȥ ȥ
Ȧ
ȧ Ȩ ȩ
Ȫ
ȫ
Ȭ
ȭ
Ȯ
Ȱ
ȱ Ȳ ȳ ȴ ȵ ȶ ȷ ȸ ȹ Ⱥ Ȼ ȼ Ƚ Ⱦ
ɀ Ɂ ɂ Ƀ Ʉ Ʌ Ɇ ɇ Ɉ ɉ Ɋ ɋ Ɍ ɍ Ɏ ɏ
ɐ ɑ ɒ ɓ ɔ ɕ ɖ ɗ ɘ ǝ ɚ ɛ ɜ ɝ ɞ
ɠ ɡ ɢ ɣ ɤ ɥ ɦ ɧ ɨ ɩ ɪ ɫ ɬ ɭ ɮ
ɰ ɱ ɲ ɳ ɴ ɵ ɶ ɷ ɸ ɹ ɺ ɻ ɼ ɽ ɾ
ʀ ʁ ʂ ʃ ʄ ʅ ʆ ʇ ʈ ʉ ʊ ʋ ʌ ʍ ʎ
ʐ ʑ ʒ ʓ ʔ ʕ ʖ ʗ ʘ ʙ ʚ ʛ ʜ ʝ ʞ
ʠ ʡ ʢ ʣ ʤ ʥ ʦ ʧ ʨ ʩ ʪ ʫ ʬ ʭ ʮ
ʰ
01A0
01B0
01C0
01D0
01E0
01F0
0200
0210
0220
0230
0240
0250
0260
0270
0280
0290
02A0
02B0
ʱ ʲ ʳ ʴ ʵ ʶ ʷ ʸ ʹ ʺ ʻ ʼ ʽ ʾ
Ư
0 1 2 3 4 5 6 7 8 9 A B C D E F
ƿ
Ǐ
ǟ
ǯ
ǿ
ȏ
ȟ
ȯ
ȿ
ɟ
ɯ
ɿ
ʏ
ʟ
ʯ
ʿ
▲ Table 1.7 Sample of Unicode characters
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1
1 InformatIon representatIon and multImedIa
When storing images as pixels, we have to consider
» at least 8 bits (1 byte) per pixel are needed to code a coloured image (this
gives 256 possible colours by varying the intensity of the blue, green and
red elements)
» true colour requires 3 bytes per pixel (24 bits), which gives more than one
million colours
» the number of bits used to represent a pixel is called the colour depth.
In terms of images, we need to distinguish between bit depth and colour
depth; for example, the number of bits that are used to represent a single
pixel (bit depth) will determine the colour depth of that pixel. As the bit
depth increases, the number of possible colours which can be represented also
increases. For example, a bit depth of 8 bits per pixel allows 256 (2
8
) different
colours (the colour depth) to be represented, whereas using a bit depth of
32 bits per pixel results in 4 294 967 296 (2
32
) different colours. The impact of
bit depth and colour depth is considered later.
We will now consider the actual image itself and how it can be displayed on a
screen. There are two important definitions here:
» Image resolution refers to the number of pixels that make up an image; for
example, an image could contain 4096 × 3192 pixels (12 738 656 pixels in
total).
» Screen resolution refers to the number of horizontal pixels and the number
of vertical pixels that make up a screen display (for example, if the screen
resolution is smaller than the image resolution then the whole image cannot
be shown on the screen or the original image will now be a lower quality).
We will try to clarify the difference by using an example.
Figure 1.1 has been taken by a digital camera using an image resolution of
4096 × 3192 pixels:
▲ Figure 1.1 Image taken by a digital camera
Suppose we wish to display Figure 1.1 on a screen with screen resolution of
1920 × 1080. To display this image the web browser (or other software) would
need to re-size Figure 1.1 so that it now fits the screen. This could be done by
removing pixels so that it could now be displayed, or part of the image could
be cropped (and, in this case, rotated through 90°) as shown in Figure 1.2.
EXTENSION
ACTIVITY 1B
Find out how HTML
is used to control
the colour of each
pixel on a screen.
How is HTML used in
the design stage of
a web page screen
layout?
▲ Figure 1.2 Image cropped
and rotated through 90°
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1
1.2 Multimedia
However, a lower resolution copy of Figure 1.1 (for example, 1024 × 798) would
now fit on the screen without any modification to the image. We could simply
zoom in to enlarge it to full screen size; however, the image could now become
pixelated (in other words, the number of pixels per square inch (known as the
pixel density) is smaller, causing deterioration in the image quality).
We will now consider a calculation which shows how pixel density can be
calculated for a given screen. Imagine we are using an Apple iPhone 8 which
has 5.5-inch screen size and screen resolution of 1920 pixels × 1080 pixels:
1 add together the squares of the resolution size ((1920
2
+ 1080
2
) = (3 686 400
+ 16 640) = 4 852 800)
2 find the square root
4852800 2202.907
()
=
3
divide by screen size (2202.907 ÷ 5.5 = 401)
This gives us the pixel density of 401 pixels per square inch (ppi) (which is the
same as the published figure from the manufacturer).
A pixel-generated image can be scaled up or scaled down; it is important
to understand that this can be done when deciding on the resolution. The
resolution can be varied on many cameras before taking, for example, a digital
photograph. When magnifying an image, the number of pixels that makes up
the image remains the same but the area they cover is now increased. This
means some of the sharpness could be lost. This is known as the pixel density
and is key when scaling up photographs. For example, look at Figure 1.3.
A CB D E
▲ Figure 1.3 Five images of the same car wheel
Image A is the original. By the time it has been scaled up to make image E it
has become pixelated (‘fuzzy’). This is because images A and E have different
pixel densities.
The main drawback of using high resolution images is the increase in file size. As
the number of pixels used to represent the image is increased, the size of the file
will also increase. This impacts on how many images can be stored on, for example,
a hard drive. It also impacts on the time to download an image from the internet or
the time to transfer images from device to device. Bit-map images rely on certain
properties of the human eye and, up to a point, the amount of file compression
used (see Section 1.3 File compression). The eye can tolerate a certain amount of
resolution reduction before the loss of quality becomes significant.
Calculating bit-map image file sizes
It is possible to estimate the file size needed to store a bit-map image. The file
size will need to take into account the image resolution and bit depth.
For example, a full screen with a resolution of 1920 × 1080 pixels and a bit depth
of 24 requires 1920 × 1080 × 24 bits = 49 766 400 bits for the full screen image.
Dividing by 8 gives us 6 220 800 bytes (equivalent to 6.222 MB using the SI
units or 5.933 MiB using IEE units). An image which does not occupy the full
screen will obviously result in a smaller file size.
EXTENSION
ACTIVITY 1C
Calculate the file
size needed to store
the screen image on
a UHD television.
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1
1 InformatIon representatIon and multImedIa
Note: when saving a bit-map image, it is important to include a file header; this will
contain items such as file type (.bmp or .jpeg), file size, image resolution, bit depth
(usually 1, 8, 16, 24 or 32), any type of data compression employed and so on.
1.2.2 Vector graphics
Vector graphics are images that use 2D points to describe lines and curves and
their properties that are grouped to form geometric shapes. Vector graphics
can be designed using computer aided design (CAD) software or using an
application which uses a drawing canvas on the screen. See Figure 1.4.
A vector graphic will contain a drawing list (included in a file header) that is
made up of
» the command used for each object that makes up the graphic image
» the attributes that define the properties that make up each object (for
example consider the ellipse of the robot’s mouth – this will need the
position of the two centres, the radius from centres, the thickness and style
of each line, the line colour and any fill colour used)
» the relative position of each object will also need to be included
» the dimensions of each object are not defined, but the relative positions of
objects to each other in the final graphic need to be defined; this means
that scaling up the vector graphic image will result in no loss of quality.
When printing out vector graphics it is usually necessary to first convert it into
a bit-map image to match the format of most printers.
Comparison between vector graphics and bit-map images
Vector graphic images Bit-map images
made up of geometric shapes which
require definition/attributes
made up of tiny pixels of different colours
to alter/edit the design, it is necessary
to change each of the geometric shapes
possible to alter/edit each of the pixels to
change the design of the image
they do not require large file size since
it is made up of simple geometric shapes
because of the use of pixels (which give very
accurate designs), the file size is very large
because the number of geometric shapes
is limited, vector graphics are not
usually very realistic
since images are built up pixel by pixel, the
final image is usually very realistic
file formats are usually .svg, .cgm, .odg file formats are usually .jpeg, .bmp, .png
▲ Table 1.8 Comparison between vector graphics and bit-map images
It is now worth considering whether a vector graphic or a bit-map image would
be the best choice for a given application. When deciding which is the better
method, we should consider the following:
» Does the image need to be resized? If so, a vector graphic could be the best
option.
» Does the image need to be drawn to scale? Again, a vector graphic is
probably the best option.
» Does the image need to look real? Usually bit-map images look more realistic
than vector graphics.
» Are there file restrictions? If so, it is important to consider whether vector
graphic images can be used; if not, it would be necessary to consider the
image resolution of a bit-map image to ensure the file size is not too large.
▲ Figure 1.4 Drawing of a
robot made up of a number
of geometric shapes
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1
1.2 Multimedia
For example, when designing a logo for a company or composing an ‘exploded
diagram’ of a car engine, vector graphics are the best choice.
However, when modifying photographs using photo software, the best method
is to use bit-map images.
1.2.3 Sound files
Sound requires a medium in which to travel through (it cannot travel in a
vacuum). This is because it is transmitted by causing oscillations of particles
within the medium. The human ear picks up these oscillations (changes in air
pressure) and interprets them as sound. Each sound wave has a frequency and
wavelength; the amplitude specifies the loudness of the sound.
period
high frequency wave
Time
Pressure
period
low frequency wave
Time
Pressure
▲ Figure 1.5 High and low frequency wave signals
Sound is an analogue value; this needs to be digitised in order to store sound
in a computer. This is done using an analogue to digital converter (ADC). If
the sound is to be used as a music file, it is often filtered first to remove
higher frequencies and lower frequencies which are outside the range of human
hearing. To convert the analogue data to digital, the sound waves are sampled
at a given time rate. The amplitude of the sound cannot be measured precisely,
so approximate values are stored.
Time intervals
Sound amplitude
0
10 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1
2
3
4
5
6
7
8
9
10
▲ Figure 1.6 A sound wave
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1
1 InformatIon representatIon and multImedIa
Figure 1.6 shows a sound wave. The x-axis shows the time intervals when
the sound was sampled (0 to 20), and the y-axis shows the amplitude of the
sampled sound (the amplitudes above 10 and below 0 are filtered out in this
example).
At time interval 1, the approximate amplitude is 9; at time interval 2, the
approximate amplitude is 4, and so on for all 20 time intervals. Because the
amplitude range in Figure 1.6 is 0 to 10, then 4 binary bits can be used to
represent each amplitude value (for example, 9 would be represented by the
binary value 1001). Increasing the number of possible values used to represent
sound amplitude also increases the accuracy of the sampled sound (for
example, using a range of 0 to 127 gives a much more accurate representation
of the sound sample than using a range of, for example, 0 to 10). This is known
as the sampling resolution (also known as the bit depth).
Sampling rate is the number of sound samples taken per second. The higher
the sampling rate and/or sampling resolution, the greater the file size. For
example, a 16-bit sampling resolution is used when recording CDs to give better
sound quality.
So, how is sampling used to record a sound clip?
» The amplitude of the sound wave is first determined at set time intervals
(the sampling rate).
» This gives an approximate representation of the sound wave.
» The sound wave is then encoded as a series of binary digits.
Using a higher sampling rate or larger resolution will result in a more faithful
representation of the original sound source.
Pros Cons
larger dynamic range produces larger file size
better sound quality takes longer to transmit/download sound files
less sound distortion requires greater processing power
▲ Table 1.9 The pros and cons of using a larger sampling resolution when recording sound
Recorded sound is often edited using software. Common features of such
software include the ability to
» edit the start/stop times and duration of a sample
» extract and save (or delete) part of a sample
» alter the frequency and amplitude of a sample
» fade in and fade out
» mix and/or merge multiple sound tracks or sources
» combine various sound sources together and alter their properties
» remove ‘noise’ to enhance one sound wave in a multiple of waves (for
example, to identify and extract one person’s voice out of a group of people)
» convert between different audio formats.
1.2.4 Video
This section considers the use of video and extends beyond the syllabus. While
this is not specifically mentioned in the syllabus, it has been included here
for completeness. Many specialist video cameras exist. However, most digital
cameras, smart phones and tablets are also capable of taking moving images by
‘stitching’ a number of still photos (frames) together. They are often referred to
as DV (digital video) cameras; they store compressed photo frames at a speed
of 25 MB per second – this is known as motion JPEG.
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1
1.3 File compression
In both single frame and video versions, the camera picks up the light from
the image and turns it into an electronic signal using light-sensitive sensors.
In the case of the DV cameras, these signals are automatically converted into a
compressed digital file format.
When recording video, the frame rate refers to the number of frames recorded
per second.
1.3 File compression
Key terms
Lossless file compression – file compression method
where the original file can be restored following
decompression.
Lossy file compression – file compression method
where parts of the original file cannot be recovered
during decompression, so some of the original detail
is lost.
JPEG – Joint Photographic Expert Group – a form of
lossy file compression based on the inability of the eye
to spot certain colour changes and hues.
MP3/MP4 files – file compression method used for
music and multimedia files.
Audio compression – method used to reduce the size of
a sound file using perceptual music shaping.
Perceptual music shaping – method where sounds
outside the normal range of hearing of humans, for
example, are eliminated from the music file during
compression.
Bit rate – number of bits per second that can be
transmitted over a network. It is a measure of the data
transfer rate over a digital telecoms network.
Run length encoding (RLE) – a lossless file
compression technique used to reduce text and photo
files in particular.
It is often necessary to reduce the file size of a file to either save storage
space or to reduce the time taken to stream or transmit data from one device
to another (see Chapter 2). The two most common forms of file compression are
lossless file compression and lossy file compression.
Lossless file compression
With this technique, all the data from the original file can be reconstructed
when the file is uncompressed again. This is particularly important for files
where loss of any data would be disastrous (such as a spreadsheet file of
important results).
Lossy file compression
With this technique, the file compression algorithm eliminates unnecessary
data (as with MP3 and JPEG formats, for example).
Lossless file compression is designed to lose none of the original detail from
the file (such as Run-Length Encoding (RLE) which is covered later in this
chapter). Lossy file compression usually results in some loss of detail when
compared to the original; it is usually impossible to reconstruct the original
file. The algorithms used in the lossy technique have to decide which parts of
the file are important (and need to be kept) and which parts can be discarded.
We will now consider file compression techniques applied to multimedia files.
1.3.1 File compression applications
MPEG-3 (MP3) and MPEG-4 (MP4)
MPEG-3 (MP3) uses technology known as audio compression to convert
music and other sounds into an MP3 file format. Essentially, this compression
technology will reduce the size of a normal music file by about 90%. For example,
an 80 MB music file on a CD can be reduced to 8 MB using MP3 technology.
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1
1 InformatIon representatIon and multImedIa
MP3 files are used in MP3 players, computers or mobile phones. Music files
can be downloaded or streamed from the internet in a compressed format, or
CD files can be converted to MP3 format. While streamed or MP3 music quality
cannever match the ‘full’ version found on a CD, the quality is satisfactory for
most purposes.
But how can the original music file be reduced by 90% while still retaining
most of the music quality? This is done using file compression algorithms that
use perceptual music shaping.
Perceptual music shaping removes certain sounds. For example
» frequencies that are outside the human hearing range
» if two sounds are played at the same time, only the louder one can be heard
by the ear, so the softer sound is eliminated.
This means that certain parts of the music can be removed without affecting
the quality too much. MP3 files use what is known as a lossy format, since part
of the original file is lost following the compression algorithm. This means that
the original file cannot be put back together again. However, even the quality
of MP3 files can be different, since it depends on the bit rate – this refers
to the number of bits per second used when creating the file. Bit rates are
between 80 and 320 kilobits per second; usually 200 kilobits or higher gives a
sound quality close to a normal CD.
MPEG-4 (MP4) files are slightly different to MP3 files. This format allows the
storage of multimedia files rather than just sound. Music, videos, photos and
animation can all be stored in the MP4 format. Videos, for example, could
be streamed over the internet using the MP4 format without losing any real
discernible quality (see Chapter 2 for notes on video streaming).
Photographic (bit-map) images
When a photographic file is compressed, both the file size and quality of image
are reduced. A common file format for images is JPEG, which uses lossy file
compression. Once the image is subjected to the JPEG compression algorithm,
a new file is formed and the original file can no longer be constructed. A JPEG
will reduce the raw bit-map image by a factor of between 5 and 15, depending
on the quality of the original.
Vector graphics can also undergo some form of file compression. Scalable
vector graphics (.svg) are defined in XML text files which, therefore, allows
them to be compressed.
Run-length encoding (RLE)
Run-length encoding (RLE) can be used to compress a number of different file
formats.
It is a form of lossless/reversible file compression that reduces the size of a
string of adjacent, identical data (such as repeated colours in an image).
A repeating string is encoded into two values.
The first value represents the number of identical data items (such as
characters) in the run. The second value represents the code of the data item
(such as ASCII code if it is a keyboard character).
RLE is only effective where there is a long run of repeated units/bits.
EXTENSION
ACTIVITY 1D
Find out how file
compression can
be applied to a
photograph without
noticeably reducing
its quality. Compare
this to run-length
encoding (RLE),
described below.
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1.3 File compression
Using RLE on text data
Consider the text string ‘aaaaabbbbccddddd’.
Assuming each character requires 1 byte, then this string needs 16 bytes. If we
assume ASCII code is being used, then the string can be coded as follows:
a a a a a b b b b c c d d d d d
05 97 04 98 02 99 05 100
This means we have five characters with ASCII code 97, four characters with
ASCII code 98, two characters with ASCII code 99, and five characters with
ASCII code 100. Assuming each number in the second row requires 1 byte of
memory, the RLE code will need 8 bytes. This is half the original file size.
One issue occurs with a string such as ‘cdcdcdcdcd’, where compression is not
very effective. To cope with this we use a flag. A flag preceding data indicates
that what follows are the number of repeating units (for example, 255 05 97
where 255 is the flag and the other two numbers indicate that there are five
items with ASCII code 97). When a flag is not used, the next byte(s) are taken
with their face value and a run of 1 (for example, 01 99 means one character
with ASCII code 99 follows).
Consider this example:
String aaaaaaaa bbbbbbbbbb c d c d c d eeeeeeee
Code 08 97 10 98 01 99 01 100 01 99 01 100 01 99 01 100 08 101
The original string contains 32 characters and would occupy 32 bytes of
storage.
The coded version contains 18 values and would require 18 bytes of storage.
Introducing a flag (255 in this case) produces:
255 08 97 255 10 98 99 100 99 100 99 100 255 08 101
This has 15 values and would, therefore, require 15 bytes of storage. This is a
reduction in file size of about 53%.
Using RLE with images
Black and white images
Figure 1.7 shows the letter F in a grid where each square requires 1 byte of
storage. A white square has a value 1 and a black square a value of 0.
1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 1
1 0 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 0 0 0 0 0 1 1
1 0 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 0 1 1 1 1 1 1
In compressed RLE format this becomes:
9W 6B 2W 1B 7W 1B 7W 5B 3W 1B 7W
1B 7W 1B 6W
Using W = 1 and B = 0 we get:
91 60 21 10 71 10 71 50 31 10 71 10 71
10 61
▲ Figure 1.7 Using RLE with a black and white image
The 8 × 8 grid would need 64 bytes; the compressed RLE format has 30 values,
and therefore needs only 30 bytes to store the image.
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1
1 InformatIon representatIon and multImedIa
Coloured images
Figure 1.8 shows an object in four colours. Each colour is made up of red, green
and blue (RGB) according to the code on the right.
0 0 0
255 255 2 5 5
255 0 0
0 2 5 5 0
Square
colour
Red Green Blue
Components
▲ Figure 1.8 Using RLE with a coloured image
This produces the following data:
2 0 0 0 4 0 255 0 3 0 0 0 6 255 255 255 1 0 0 0 2 0 255 0 4 255 0 0 4 0 255 0 1
255 255 255 2 255 0 0 1 255 255 255 4 0 255 0 4 255 0 0 4 0 255 0 4 255 255 255
2 0 255 0 1 0 0 0 2 255 255 255 2 255 0 0 2 255 255 255 3 0 0 0 4 0 255 0 2 0 0 0
The original image (8 × 8 square) would need 3 bytes per square (to include
all three RGB values). Therefore, the uncompressed file for this image is
8 × 8 × 3 = 192 bytes.
The RLE code has 92 values, which means the compressed file will be 92 bytes
in size. This gives a file reduction of about 52%. It should be noted that the
file reductions in reality will not be as large as this due to other data which
needs to be stored with the compressed file (such as a file header).
1.3.2 General methods of compressing files
All the above file compression techniques are excellent for very specific
types of file. However, it is also worth considering some general methods
to reduce the size of a file without the need to use lossy or lossless file
compression:
reduce the sampling rate used
reduce the sampling resolution
reduce the frame rate
movie files
image files
crop the image
decrease the colour/bit depth
reduce the image resolution
▲ Figure 1.9 General methods of compressing files
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1
1.3 File compression
ACTIVITY 1I
1 a) What is meant by lossless and lossy file compression?
b) Give an example of a lossless file format and an example of a lossy file
format.
2 a) Describe how music picked up by a microphone is turned into a
digitised music file in a computer.
b) Explain why it is often necessary to compress stored music files.
Describe how the music quality is essentially retained.
3 a) What is meant by run length encoding?
b) Describe how RLE compresses a file. Give an example in your
description.
4 a) Describe the differences between bit-map images and vector graphics.
b) A software designer needs to incorporate images into her software to
add realism.
Explain what she needs to consider when deciding between using
bit-map images and vector graphics in her software.
1 a) The following bytes represent binary integers using the two’s complement
form. State the equivalent denary values.
i) 0 1 0 0 1 1 1 1 [1]
ii) 1 0 0 1 1 0 1 0 [1]
iii) Write the integer −53 in two’s complement form. [1]
iv) Write the maximum possible range of numbers using the two’s
complement form of an 8-bit binary number.
Give your answers in denary. [2]
b) i) Write the denary integer 798 in binary-coded decimal (BCD) format. [1]
ii) Write the denary number that is represented by the following BCD number.
1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 0
[2]
c) Give one use of binary-coded decimal system. [1]
2 A software developer is using a microphone and a sound editing app to
collect and edit sounds for his new game.
When collecting sounds, the software developer can decide on the sampling
resolution he wishes to use.
a) i) State what is meant by sampling resolution. [1]
ii) Describe how sampling resolution will affect how accurate the
stored digitised sound will be. [2]
b) The software developer will include images in his new game.
i) Explain the term image resolution. [1]
ii) The software developer is using 16-colour bit-map images.
State the number of bits required to encode data for one pixel of his
image. [1]
iii) One of the images is 16 384 pixels wide and 512 pixels high.
The developer decides to save it as a 256-colour bit-map image.
Calculate the size of the image file in gibibytes. [3]
End of chapter
questions
➔
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1
1 InformatIon representatIon and multImedIa
iv) The bit-map image will contain a header.
State two items you would expect to see in the header. [2]
v) Give three features you would expect to see in the sound editing app. [3]
3 The editor of a movie is finalising the music score. They will send the final
version of the score to the movie producer by email attachment.
a) Describe how sampling is used to record the music sound clips. [3]
b) The music sound clips need to undergo some form of data compression
before the music editor can send them via email.
Identify the type of compression, lossy or lossless, they should use.
Give a justification for your answer. [3]
c) One method of data compression is known as run length encoding (RLE).
i) Explain what is meant by RLE. [3]
ii) Show how RLE would be used to produce a compressed file for
the image below.
Write down the data you would expect to see in the RLE compressed
format (you may assume that the grey squares have a code value of 85
and the white squares have a code value of 255). [4]
4 a) Write the denary numbers 60, 27 and −27 in 8-bit binary two’s
complement form. [3]
b) Show the result of the addition 60 + 27 using 8-bit binary two’s
complement form. Show all of your working. [2]
c) Show the result of the subtraction 60 − 27 using 8-bit binary two’s
complement form. [2]
d) Give the result of the following addition.
0 1 0 1 1 0 0 1
+
0 1 1 0 0 0 0 1
Explain why the expected result is not obtained. [2]
5 a) Carry out 0.52 + 0.83 using binary-coded decimal (BCD). Show all
of your working. [4]
b) i) Define the term hexadecimal. [1]
ii) Give two uses of the hexadecimal system. [2]
iii) Convert the following binary number into hexadecimal. [2]
0 1 1 1 1 1 1 0 1 1 1 1 0 0 1 0
6 a) Convert the denary number 95 into binary coded decimal (BCD). [1]
b) Using two’s complement, carry out the binary subtraction:
0 0 1 0 0 0 1 1 – 0 1 0 0 0 1 0 0
and convert your answer into denary. [3]
c) Convert the denary number 506 into hexadecimal. [1]
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