Chapter 1

1.

0101110?
AE73F9?
CHAPTER 1.1
Number
System

2.

Binary
System

3.

Why Binary System
Motivation
Any form of information needs to be converted into a binary format so that it
can be processed by a computer
• Computer contains millions and millions of tiny switches, which can be turned
on and off.
• Therefore, the binary system is chosen as the way for a computer to represent any
sort of data.
On = 1
Off = 0

4.

Explanation
Denary System
Eg. 365
102 =100
101 =10
100 =1
3
6
5
Explanation: Multiply the digit value (eg.3) by the place value (eg. 100)
(3x100) + (6x10) + (5x1) = 365

5.

Explanation
Binary
Denary
Denary
Binary

6.

Binary
Denary
Explanation
22 =4
21 =2
20 =1
1
1
1
(1x4) + (1x2) + (1x1) = 7 in denary

7.

Binary
Explanation
Denary
Eg. "1011"
23 =8
22 =4
21 =2
20 =1
1
0
1
1
Explanation: Multiply the digit value (eg.1) by the place value (eg. 8).
Then sum it all up!
(1x8) + (0x4) + (1x2) + (1x1) = 11 in denary

8.

Binary
Denary
DIY
What is the denary form of "11100"?
24 =16
23 =8
22 =4
21 =2
20 =1
1
1
1
0
0

9.

Binary
Denary
DIY
ANSWER
24 =16
23 =8
22 =4
21 =2
20 =1
1
1
1
0
0
(1x16) + (1x8) + (1x4) + (0x2) + (0x1)
= 28 in denary

10.

Denary
Binary
Explanation
Convert 5 to binary:
5
2
2
2
5
2
1
0
remainder
remainder
remainder
1
0
1
Answer: 101
Read the remainder
from bottom to top

11.

Denary
Binary
Explanation
Convert 39 to binary:
39
2
2
2
2
2
2
39
19
9
4
2
1
0
remainder
remainder
remainder
remainder
remainder
remainder
1
1
1
0
0
1
Answer: 100111
Read the remainder
from bottom to top

12.

Denary
Binary
DIY
What is the binary form of 42?
25 =32
24 =16
23 =8
22 =4
21 =2
20 =1

13.

Denary
Binary
DIY
ANSWER
Convert 42 to binary:
42
2
2
2
2
2
2
42
21
10
5
2
1
0
remainder
remainder
remainder
remainder
remainder
remainder
0
1
0
1
0
1
Read the remainder
from bottom to top
Answer: 101010

14.

Denary System
RECAP
102
101
100
7
6
5
Hundredth
Tenth
Ones
(7x100) + (6x10)
+ (5x1) = 765

15.

Denary System
RECAP
102
101
100
7
6
5
(7x100) + (6x10)
+ (5x1) = 765
Binary System
22
21
20
1
1
1
(1x4) + (1x2)
+ (1x1) = 7

16.

Binary
Denary
DIY
What is the denary form of "1010"?
23 =8
22 =4
21 =2
20 =1
1
0
1
0

17.

Binary
Denary
DIY
ANSWER
23 =8
22 =4
21 =2
20 =1
1
0
1
0
(1x8) + (1x2) = 10 in denary

18.

Denary
Binary
DIY
What is the binary form of 38?
25 =32
24 =16
23 =8
22 =4
21 =2
20 =1

19.

Denary
Binary (Method 2)
DIY
ANSWER
Convert 38 to binary:
38
2
2
2
2
2
2
38
19
9
4
2
1
0
remainder
remainder
remainder
remainder
remainder
remainder
0
1
1
0
0
1
Read the remainder
from bottom to top
Answer: 100110

20.

PA S T Y E A R Q U E S T I O N

21.

ANSWER

22.

Hexadecimal
System

23.

Motivation
Hexadecimal System
It is a base 16 system.
It uses 16 digits to represent each value
Number System
Digits used to represent each value
Denary
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Binary
0, 1
Hexadecimal
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 , A, B, C, D, E, F

24.

Explanation
Hexadecimal System
Binary System
Denary System
102 =100 101 =10 100 =1
3
6
5
(3x100) + (6x10) + (5x1) = 365
22 =4
21 =2
20 =1
1
0
1
(1x4) + (0x2) + (1x1) = 5
Hexadecimal System
?2 =?
?1 =?
?0 =?
3
E
5

25.

Explanation
Hexadecimal System
Binary System
Denary System
102 =100 101 =10 100 =1
3
6
5
Hexadecimal System
162 =256 161 =16 160 =1
E
21 =2
20 =1
1
0
1
(1x4) + (0x2) + (1x1) = 5
(3x100) + (6x10) + (5x1) = 365
3
22 =4
5

26.

Conversion
Binary
Hexadecimal
Hexadecimal
Binary

27.

Binary
Hexadecimal
Explanation
Since 16 = 24 this means that FOUR binary digits are equivalent to each
hexadecimal digit.

28.

Binary
Explanation
Hexadecimal
101111100001
101 1
111 0
0 0 01
B
E
1
ANSWER : BE1

29.

Binary
Explanation
Hexadecimal
10 000111111101
00 10
2
0001 1111
1
F
ANSWER : 21FD
1101
D

30.

Binary
Hexadecimal
What is the hexadecimal form of
0111010011100?
DIY

31.

Binary
DIY
Hexadecimal
ANSWER
0111010011100
0000 1110
0
E
1001 1100
9
C

32.

Explanation
Binary
Hexadecimal
F 9 3 5
1111
0011
0101
1001
Answer: 1111 1001 0011 0101

33.

Hexadecimal
Binary
What is the binary form of
BF08?
DIY

34.

Binary
Hexadecimal
DIY
B F 0 8
1011
0000
1000
1111
Answer: 1011 1111 0000 1000

35.

Conversion
Hexadecimal
Denary
Denary
Hexadecimal

36.

Binary
Denary
RECAP
Eg. "111"
22 =4
21 =2
20 =1
1
1
1
(1x4) + (1x2) + (1x1) = 7 in denary

37.

Hexadecimal
Denary
Explanation
Eg. "45A"
162 =256
161 =16
160 =1
4
5
A
Note: A=10
(4x256) + (5x16) + (10x1) = 1114 in denary

38.

Hexadecimal
Denary
Explanation
Eg. "C8F"
162 =256
161 =16
160 =1
C
8
F
Note: C=12, F=15
(12x256) + (8x16) + (15x1) = 3215 in denary

39.

Hexadecimal
DIY
Denary
What is the denary form of BF08?
163 =4096
162 =256
161 =16
160 =1
B
F
0
8

40.

Hexadecimal
DIY
Denary
ANSWER
163 =4096
162 =256
161 =16
160 =1
B
F
0
8
(11x4096) + (15x256) + (0x16) + (8x1) = 48904 in denary

41.

Denary
Binary (Method 2)
RECAP
Convert 5 to binary:
5
2
2
2
5
2
1
0
remainder
remainder
remainder
1
0
1
Answer: 101
Read the remainder
from bottom to top

42.

Denary
Hexadecimal
Explanation
Eg. "2004"
2004/16 = 125
remainder = 4
125/16 = 7
remainder = 13
2004
16 125 remainder 4
remainder
7
16
13
remainder
0
7
16
Note: 13=D
Answer: 7D4

43.

Denary
Hexadecimal
DIY
What is the hexadecimal form of 3179?
3179/16 = ?
3179
16 198 remainder ?
remainder
?
16
?
remainder
?
?
16

44.

Denary
DIY
Hexadecimal
What is the hexadecimal form of 3179?
3179/16 = 198
remainder = 11
198/16 = 12
remainder = 6
3179
16 198 remainder 11
16 12 remainder 6
remainder 12
0
16
Answer: C6B

45.

PA S T Y E A R Q U E S T I O N

46.

ANSWER

47.

PA S T Y E A R Q U E S T I O N

48.

ANSWER

49.

Chapter 1.2
Use of hexadecimal
system

50.

Use of hexadecimal system
Discussion Time
Binary
Hexadecimal
110101111110
1AFD39
100111001
Brainstorm time: Why is Hexadecimal used?

51.

Use of hexadecimal system
Explanation
One hex digit represents four binary digits
The hex number is far easier for humans to
remember, copy and work with
Four uses of the hexadecimal system

52.

Usage 1: Error Code
Explanation
Error codes are often shown as
hexadecimal values.
These numbers refer to the memory
location of the error.
They are generated by the computer.
The programmer needs to know how to
interpret the hexadecimal error codes.

53.

Usage 1: Error Code
Explanation

54.

Usage 2: MAC address
Explanation
Media Access Control (MAC) address refers to
a number which uniquely identifies a device on a network.
The MAC address refers to the network interface card (NIC)
which is part of the device
The MAC address is rarely changed so that a particular
device can always be identified no matter where it is.

55.

Usage 2: MAC address
Explanation
00_1C_B3_4F_25_FE
Message
00_1C_C3_4F_23_AE
Mac address uniquely identify a device on a Local Area Network

56.

Explanation
Usage 2: MAC address
Form 1
NN-NN-NN-DD-DD-DD
00-1C-B3-4F-25-FE
Form 2
2
NN:NN:NN:DD:DD:DD
00:1C:B3:4F:25:FE
Mac Address comes with 2 forms

57.

Explanation
Usage 2: MAC address
00-1C-B3
4F-25-FE
Identity number of
Serial number of
the manufacturer
a device
Eg. 00 – 14 – 22 which identifies devices made by Dell
00 – a0 – c9 which identifies devices made by Intel

58.

Usage 3: Internet
Protocol Addresses
Explanation
Each device connected to a network is given an address
known as the Internet Protocol address
An IPv4 address is a 32-bit number written in denary or
hexadecimal form e.g. 109.108.158.1 (or 77.76.9e.01 in hex)
IPv4 has recently been improved upon by the adoption of
IPv6. An IPv6 address is a 128-bit number broken down into
16-bit chunks, represented by a hexadecimal number.
Eg. a8f b:7a88:fff0:0fff:3d21:2085:66f b:f0fa

59.

Usage 4: HyperText Markup
Language (HTML) colour code
Explanation
HyperText Mark-up Language (HTML) is used when writing
and developing web pages.
It is not a programming language, but a markup language.
A mark-up language is used in the processing, definition
and presentation of text.

60.

Overview
Explanation
The 4 usages of Hexadecimals - EMIH
1. E - Error Codes
2. M - MAC Address
3. I - Internet Protocol Address
4. H - HTML Colour Code

61.

PA S T Y E A R Q U E S T I O N

62.

ANSWER

63.

PA S T Y E A R Q U E S T I O N

64.

ANSWER

65.

Chapter 1.3
Addition of
binary number

66.

Addition of binary number
Explanation
How do we perform add and carry in denary?
0+0=0
0+9=9
9+0=9
9 + 1 = 10
1
9
+1
10

67.

Explanation
Addition of binary number
How do we perform add and carry in denary?
1 1
56
+79
1 35
6+9 = 15 (>9)
1+5+7 = 13 (>9)

68.

Addition of binary number
How do we perform add and carry in binary?
0+0=0
0+1=1
1+0=1
1 + 1 = 10
Explanation

69.

Explanation
Addition of binary number
How do we perform add and carry in binary?
1 1 1
00100111
+01001010
0 1110 001

70.

Addition of binary number
How do we perform add and carry in binary?
Perform
01111110 + 00111110
DIY

71.

Addition of binary number
The overflow condition
1 11 111
01101110
+11011110
1 01 001 100
Explanation

72.

Addition of binary number
The overflow condition
Explanation
1 11 111
01101110
+11011110
1 01 001 100
The maximum denary of an 8-bit binary number (11111111)
is (28 - 1 ) = 255
The generation of a 9th bit is a clear indication that the
sum has exceeded this value.
This is known as an overflow error. The sum is too big to
be stored using 8 bits.

73.

Addition of binary number
The overflow condition
01101110 = 110
\
11011110 = 222
Explanation
1 11 111
01101110
+11011110
1 01 001 100
110 + 222 = 322
322 > 255 (overflow)
The sum is too big to be stored in a 8 bit binary.

74.

Lesson1.4:
Objectives
Chapter
Binary
Last lesson on the binary system
Shifting
BINARY SHIFTING
Multiplication
and division of
binary numbers
TWO COMPLEMENTS
Represent
negative
number in
binary

75.

BINARY SHIFTING
Binary shift is a process that a CPU uses to perform
multiplication and division.

76.

BINARY SHIFTING - MULTIPLICATION
For a CPU to multiply a binary number, the number
needs to be shifted to the left and will fill the remaining
gaps with zeros.

77.

BINARY SHIFTING -M ULTIPLICATIO N
Examples: 111 (Binary)
64
32
16
0
0
0
8
0
4
1
2
1
1
1
Examples: 1110 (Binary)
64
32
16
0
0
0
8
1
4
1
2
1
1
0
Examples: 11100 (Binary)
64
32
16
0
0
1
8
1
4
1
2
0
1
0

78.

BINARY SHIFTING - MULTIPLICATION
Examples: 111 (Binary)
64
32
16
0
0
0
8
0
4
1
2
1
1
1
Multiply by 2, shift 1 place to the left
1110
Multiply by 4, shift 2 place to the left
11100
Multiply by 8, shift 3 place to the left
111000
Multiply by 2^n, shift n place to the left

79.

BINARY SHIFTING - DIVISION
For a CPU to multiply a binary number, the number
needs to be shifted to the right.

80.

BINARY SHIFTING - DIVISION
Examples: 101100 (Binary)
32
16
8
1
0
1
4
1
2
0
1
0
8
0
4
1
2
1
1
0
Examples: 1011 (Binary)
32
16
8
0
0
1
4
0
2
1
1
1
Examples: 10110 (Binary)
32
0
16
1

81.

BINARY SHIFTING - DIVISION
Examples: 101100 (Binary)
32
16
8
1
0
1
4
1
2
0
1
0
Divide by 2, shift 1 place to the right
10110
Divide by 4, shift 2 place to the right
1011
Divide by 8, shift 3 place to the right
101
Divide by 2^n, shift n place to the right

82.

BINARY SHIFTING WITH 8-BIT BINARY NUMBERS
Registers contained within the CPU often have 8-bits limits
on the amount of data they can hold at any one time.
• The multiplying shifting process can cause bits to be lost at
one end of the register, and zeros added at the opposite
end.
• This process is known as losing the most significant bit.

83.

BINARY SHIFTING WITH 8-BIT BINARY NUMBERS
Examples: 10110101 (181 in denary)
128
1
64
0
32
1
16
1
8
0
4
1
2
0
1
1
10110101 -> 01101010
106 in denary
The bit lost is called the most significant bit, and when it
is shifted beyond the furthest-column the binary data that
is stored loses precision due to overflow.

84.

BINARY SHIFTING WITH 8-BIT BINARY NUMBERS
The same process can happen when dividing an 8-bit
binary number.
Example: 10111101 (189 in denary)
128
1
64
0
32
1
16
1
8
1
4
1
2
0
1
1
Divide this number by 32 (move 5 places to the right)
128
0
64
0
32
0
16
0
The division shift produces the
binary number 101 = 5, not
5.9 that arithmetic suggests.
8
0
4
1
2
0
Least
Significant
bit
1
1
11101

85.

Lesson1.5:
Objectives
Chapter
Two
Last lesson on the binary system
Complements
BINARY SHIFTING
Multiplication
and division of
binary numbers
TWO COMPLEMENTS
Represent
negative
number in
binary

86.

TWO COMPLEMENTS
A PROCESSOR CAN ALSO REPRESENT NEGATIVE
NUMBERS.
• ONE OF THE METHOD THAT A PROCESS
REPRESENT NEGATIVE NUMBERS IS CALLED
TWO'S COMPLEMENT.

87.

TWO COMPLEMENTS

88.

TWO COMPLEMENTS
TO REPRESENT NEGATIVE NUMBERS, IT IS IMPORTANT TO THINK
ABOUT THE PLACE VALUE OF THE FURTHEST-LEFT BIT IN A
DIFFERENT WAY.
• PROCESSOR CAN BE SET UP TO SEE THE BIT IN THE EIGHTH
COLUMN AS A SIGN BIT.
0 = POSITIVE
1 = NEGATIVE

89.

CONVERT POSITIVE BINARY INTEGER TO A TWO'S COMPLEMENT 8-BIT INTEGER
Examples:13
Step 1: Convert 13 into binary.
1101 in binary
Step 2: Put the number into the place value column
-128
0
64
0
32
0
16
0
8
1
4
1
2
0
Step 3: Ensure that the the leftmost bit is 0 (+).
1
1

90.

DIY
Convert 19 into a Two's complement
8-bit Integer

91.

CONVERT POSITIVE BINARY INTEGER TO A TWO'S COMPLEMENT 8-BIT INTEGER
Examples:19
Step 1: Convert 19 into binary.
10011 in binary
Step 2: Put the number into the place value column
-128
0
64
0
32
0
16
1
8
0
4
0
2
1
Step 3: Ensure that the the leftmost bit is 0 (+).
Answer: 00010011
1
1

92.

CONVERT TWO'S COMPLEMENT 8-BIT INTEGER TO A POSITIVE BINARY INTEGER
Examples: Convert 00010011 (two's complement) to denary
Step 1: Put the number into the place value column
-128
0
64
0
32
0
16
1
8
0
4
0
2
1
1
1
Step 2: This shows that it is a positive number, we can
just convert the binary into denary directly.
Step 3: Calculate the denary value.
(1x16) + (1x2) + (1x1) = 19

93.

DIY
Convert 01010011 (two's complement)
to denary

94.

CONVERT TWO'S COMPLEMENT 8-BIT INTEGER TO A POSITIVE BINARY INTEGER
Examples: Convert 01010011 (two's complement) to denary
Step 1: Put the number into the place value column
-128
0
64
1
32
0
16
1
8
0
4
0
2
1
1
1
Step 2: This shows that it is a positive number, we can
just convert the binary into denary directly.
Step 3: Calculate the denary value.
(1x64) + (1x16) + (1x2) + (1x1) = 83

95.

CONVERT NEGATIVE BINARY NUMBERS IN TWO'S COMPLEMENT FORMAT AND
CONVERT TO DENARY
Examples: 10010011
Step 1: Put the number into the place value column
-128
1
64
0
32
0
16
1
8
0
4
0
2
1
1
1
Step 2: The left-most bit is 1, this means that it is a
negative number.
Step 3: Compute the denary value as usual.
(1x -128) + (1x16) + (1x2) + (1x1)
= -128 + 16 + 2 + 1
= -109

96.

DIY
Convert 10110011 (Two's Complement)
to denary

97.

CONVERT NEGATIVE BINARY NUMBERS IN TWO'S COMPLEMENT FORMAT AND
CONVERT TO DENARY
Examples: 10110011
Step 1: Put the number into the place value column
-128
1
64
0
32
1
16
1
8
0
4
0
2
1
1
1
Step 2: The left-most bit is 1, this means that it is a
negative number.
Step 3: Compute the denary value as usual.
(1x -128) (1x32)+ (1x16) + (1x2) + (1x1)
= -128 + 32 + 16 + 2 + 1
= -77

98.

CONVERTING NEGATIVE DENARY NUMBERS INTO BINARY NUMBERS IN TWO’S
COMPLEMENT FORMAT
Examples: -67
Step 1: Convert the number to positive.
67
Step 2: Write the number in binary form (8 bits).
01000011
Step 3: Invert each binary value.
10111100

99.

Step 4: Add 1 to the binary number.
10111100
1
+
10111101
Step 5: This gives us -67.
-128
1
64
0
32
1
16
1
8
1
4
1
-128 + 32 + 16 + 8 + 4 + 1 = -67
2
0
1
1

100.

DIY
Convert -65 to 8 bit two's complement
binary number

101.

CONVERTING NEGATIVE DENARY NUMBERS INTO BINARY NUMBERS IN TWO’S
COMPLEMENT FORMAT
Examples: -65
Step 1: Convert the number to positive.
65
Step 2: Write the number in binary form (8 bits).
01000001
Step 3: Invert each binary value.
10111110

102.

Step 4: Add 1 to the binary number.
10111110
1
+
10111111
Step 5: This gives us -65.
-128
1
64
0
32
1
16
1
8
1
4
1
2
1
-128 + 32 + 16 + 8 + 4 + 2 + 1 = -65
1
1

103.

Summary: Convert negative denary to two's complement
Examples: -65
65
Convert to (+)
01000001
Convert to binary
10111110
Invert the digit
1
10111111
+1
Final result

104.

Chapter 1.6

105.

106.

107.

108.

109.

110.

111.

112.

113.

The number of bits used to represent sound amplitude in
digital sound recording, as known as bit depth

114.

115.

116.

117.

118.

119.

120.

121.

122.

123.

124.

Chapter 1.7
Measurement of Data
Storage and Calculation
of file size

125.

Measurement
of Data
Storage
A bit is the basic unit of all computing memory
storage terms and is either 1 or 0.
The byte is the smallest unit of memory in a
computer.
8 bits = 1 byte
4 bits = 1 nibble

126.

Memory
Size
System
Based on the SI (base 10) system of units where
1 kilo is equal to 1000.

127.

Memory
Size
As memory size is actually measured in terms of powers of 2...
System
Based on the IEC (base 2) system of units where
1 kilo is equal to 1024 (2^10).

128.

Memory
Size
System
Converting Bytes into KiB, MiB and GiB
68719476736 Bytes
= 68719476736 Bytes / 1024 = 67108864 KiB
= 67108864 KiB / 1024
= 65536 MiB
= 65536 MiB / 1024
= 64 GiB

129.

Memory
Size
System
Converting Gib, Mib, Kib into bytes
64 GiB
= 64 x 1024
= 65536 MiB
= 65536 x 1024
= 67108864 KiB
= 67108864 x 1024
= 68719476736 Bytes

130.

DIY
Convert the size of GTA-V to bytes
72 GiB
= 72 x 1024
= 73728 MiB
= 65536 x 1024
= 75497472 KiB
= 75497472 x 1024
= 77309411328 Bytes

131.

Calculation
of file
size
Image
Audio

132.

Calculation
of file
size - Image
Image Resolution - The number of pixels
that make up an image.
The higher the image resolution, the higher
the quality of the image.

133.

Calculation
of file
size - Image
Formula
image resolution (pixels) x colour depths (bits)

134.

Example 1
Calculation
of file
size - Image
2px
2px
Total pixels = 2 x 2 = 4
Calculation = (2x2) x 2
= 8 bits = 1 byte
00 01
10 11
Colour depth = 2

135.

Calculation
Example 2
of file
size - Image
Formula : image resolution (pixels) x colour depths (bits)
Question:
Image Resolution = 1024 x 1080
Colour depth = 32
Calculate the size of this image in Bytes.
Workings:
1024 x 1080 = 1105920 pixels
1105920 x 32 = 35389440 bits
Answer in byte: 35389440/8 = 4423680 bytes

136.

Calculation
of file
size - Image
Example 2
Question:
Image Resolution = 1024 x 1080
Colour depth = 32
Calculate the size of this image in Bytes. How many photograph of
this size would fit onto a memory stick of 64Gib.
Each image = 4423680 bytes
First convert 64 Gib into bytes:
64 x 1024 = 65536 MiB
65536 x 1024 = 67108864 KiB
67108864 x 1024 = 68719476736 bytes

137.

Calculation
of file
size - Image
Example 2
Question:
Image Resolution = 1024 x 1080
Colour depth = 32
Calculate the size of this image in Bytes. How many photograph of
this size would fit onto a memory stick of 64Gib.
Each image = 4423680 bytes
First convert 64 Gib into bytes = 68719476736 bytes
68719476736/4423680 = 15534 photos.

138.

DIY
Question:
Image Resolution = 2048 x 2048
Colour depth = 16
Calculate the size of this image in Bytes.

139.

DANSWER
Question:
Image Resolution = 2048 x 2048
Colour depth = 16
Calculate the size of this image in Bytes.
Answer:
2048 x 2048 x 16 = 67108864 bits
= 67108864/8
= 8388608 bytes

140.

DIY
Question:
Image Resolution = 2048 x 2048
Colour depth = 16
Calculate the size of this image in Bytes (Answer: 8388608 bytes).
What is the size of the image in MiB.

141.

DANSWER
Question:
Image Resolution = 2048 x 2048
Colour depth = 16
Calculate the size of this image in Bytes (Answer: 8388608 bytes).
What is the size of the image in MiB.
8388608 / 1024 = 8192 KiB
8192 / 1024 = 8 MiB

142.

Calculation
Formula
of file
size - Sound
Sample Rate (in Hz) x Sample Resolution (in bits) x length of sample (in seconds)

143.

Calculation
of file
size - Sound
Mono Sound vs Stereo Sound
Comparison

144.

Calculation
Example 1 - Mono Sound
of file
size - Sound
Question:
Sample Rate: 44100
Sample Resolution: 8 bits
Length of the music: 20 seconds
Calculate the size of the audio in KiB.
44100 x 8 x 20 = 7056000 bits
7056000/8 = 882000 bytes
882000 / 1024 = 861.328 KiB

145.

Calculation
of file
Example 1 - Stereo Sound
size - Sound
An audio CD has a sample rate of 44100 and a sample resolution of 16 bits. The
music being sampled uses two channels to allow for stereo recording. Calculate the
file size for a 60-minute recording.
44100 x 16 x 3600 = 2540160000 bits
2540160000 x 2 = 5080320000 bits
5080320000 / 8 = 635040000 bytes
635040000 / 1024 = 620156.25 KiB
620156.25 / 1024 = 605.62 MiB

146.

DIY
An audio CD has a sample rate of 44100 and a sample resolution of 8 bits. The
music being sampled uses two channels to allow for stereo recording. Calculate the
file size for a 25-minute recording.

147.

DIY
An audio CD has a sample rate of 44100 and a sample resolution of 8 bits. The
music being sampled uses two channels to allow for stereo recording. Calculate the
file size for a 25-minute recording in MiB.
44100 x 8 x 1500 = 529200000 bits
529200000 x 2 = 1058400000 bits
1058400000 / 8 = 132300000 bytes
132300000 / 1024 = 129199.218 KiB
129199.218 / 1024 = 126.17 MiB

148.

PA S T Y E A R Q U E S T I O N

149.

ANSWER

150.

151.

DATA COMPRESSION

152.

DATA COMPRESSION

153.

FILE SIZE OF IMAGES AND SOUND CAN BE VERY LARGE.
• THEREFORE, DATA COMPRESSION IS NEEDED TO REDUCE
THE SIZE OF A FILE.

154.

WHAT ARE SOME BENEFITS OF REDUCING
THE FILE SIZE?

155.

156.

LOSSY
FILE
COMPRESSI
ON
LOSSLESS
FILE
COMPRESSI
ON

157.

FILE COMPRESSION ALGORITHM ELIMINATES
UNNECESSARY DATA FROM THE FILE.
• ORIGINAL FILE CANNOT BE RECONSTRUCTED ONCE IT HAS
BEEN COMPRESSED.
• IMAGE - REDUCE THE RESOLUTION // COLOUR DEPTH
• SOUND - REDUCE THE SAMPLING RATE // SAMPLING
RESOLUTION
• SOME LOSSY FILE COMPRESSION ALGORITHMS ARE:
MPEG-3
MPEG-4
JPEG

158.

MPEG-3
A COMPRESSION TECHNOLOGY THAT
REDUCES THE SIZE OF A NORMAL MUSIC
FILE BY ABOUT 90%.
SECRET
• REMOVE
SOUNDS
OUTSIDE THE
HUMAN EAR
RANGE
• ELIMINATE THE
SOFTER SOUND
- PERCEPTUAL
MUSIC SHAPING

159.

ALLOWS STORAGE OF MULTIMEDIA FILES
RATHER THAN JUST SOUND.
MPEG-4
SECRET
• MOVIES CAN BE
STREAMED
USING THE MP4
FORMAT
WITHOUT
LOSING ANY
REAL
DISCERNIBLE
QUALITY

160.

A LOSSY COMPRESSION ALGORITHM USED
FOR BITMAP IMAGES. ORIGINAL FILE CAN NO
LONGER BE CONSTRUCTED
JPEG
SECRET
• REMOVE
COLOUR
SHADES
(HUMAN CAN'T
NOTICE THEM)

161.

162.

163.

ALL THE DATA FROM THE ORIGINAL
UNCOMPRESSED FILE CAN BE
RECONSTRUCTED
• THIS IS IMPORTANT FOR FILES WHERE LOSS
OF DATA WOULD BE DISASTROUS
(COMPLEX SPREADSHEET).
• LOSSLESS FILE COMPRESSION IS DESIGNED
SO THAT NONE OF THE ORIGiNAL DETAIL
FROM THE FILE IS LOST.
LOSSLESS
FILE
COMPRESSI
ON

164.

1. CAN BE USED FOR LOSSLESS COMPRESSION OF A NUMBER OF
DIFFERENT FILE FORMATS.
RUNLENGTH
ENCODING
2. IT REDUCES THE SIZE OF A STRING OF ADJACENT, IDENTICAL DATA.
3. A REPEATED STRING IS ENCODED INTO TWO VALUES.
NUMBER OF
IDENTICAL
DATA
DATA ITEM

165.

RUNLENGTH
ENCODING
IN
ACTION
16
bytes

166.

Does not work well when no repeated data!
RUN-LENGTH
ENCODING
IN
ACTION
Each digit - 1 byte
Total size = 8 bytes (50% reduction in size)

167.

3,2,5,2,4
2,4,3,4,3
1.6.1.6.2
0, 15,1
TO BE
CONTINUE

168.

169.

REDUCE
COLOUR
DEPTH
REDUCE
IMAGE
RESOLUTION

170.

A LOSSY COMPRESSION ALGORITHM USED
FOR BITMAP IMAGES. ORIGINAL FILE CAN NO
LONGER BE CONSTRUCTED
JPEG
SECRET
• REMOVE
COLOUR
SHADES
(HUMAN CAN'T
NOTICE THEM)

171.

PA S T Y E A R Q U E S T I O N

172.

ANSWER

173.

PA S T Y E A R Q U E S T I O N

174.

ANSWER