(4.5) Fundamentals of data representation

Question 1

What are natural numbers? (ℕ)

Answer 1

Positive integers
(including 0)
number set ℕ = {0, 1, 2, 3, … }

Question 2

What are integers? (ℤ)

Answer 2

Whole numbers
number set ℤ = { …, -3, -2, -1, 0, 1, 2, 3, … }

Question 3

What are rational numbers? (ℚ)

Answer 3

Numbers that can be written as fractions (ratios of integers), including integers (ex 7; 7/1)

Question 4

What are irrational numbers?

Answer 4

Numbers that cannot be written as a fraction (ex √2, √3, √5, √7, √11, √13, √17, √19)

Question 5

What are real numbers? ( ℝ)

Answer 5

possible real world quantities

includes natural, rational and irrational numbers

numbers that are not real include imaginary numbers (ex e or i)

Question 6

What are ordinal numbers?

Answer 6

ordinal
numbers are used to tell an objects numerical position in a list (ex 1st, 2nd, 3rd, etc)

Question 7

What type of numbers are used for:
-counting
-measurements

Answer 7

Counting - natural numbers
Measurements - real numbers

Question 8

Why is hexadecimal (base 16) used as shorthand for binary (base 2)?

Answer 8

• large numbers can be represented using fewer digits
• easier to understand and remember

(colour values and MAC addresses are often represented in hex)

Question 9

How do you work out how many values can be represented with n bits?

Answer 9

2^n
ex if n=3, 2^3 = 8
(000 001 011 111 010 110 101 100)x8

Question 10

How can quantities of bytes be described using binary prefixes representing powers of 2
(ex 1KiB = 2^10)

Answer 10

• kibi, Ki - 2^10
• mebi, Mi - 2^20
• gibi, Gi - 2^30
• tebi, Ti - 2^40

Question 11

How can quantities of bytes be described using decimal prefixes representing powers of 10
(1kB = 10^3)

Answer 11

• kilo, k - 10^3
• mega, M - 10^6
• giga, G - 10^9
• tera, T - 10^12

Question 12

Historically innacurate use of units

Answer 12

Historically the terms kilobyte, megabyte, etc
have often been used when kibibyte, mebibyte,
etc are meant

Question 13

In unsigned binary, what are the minimum and maximum values for a given number of bits (n) ?

Answer 13

0 and (2^n)-1

ex bits = 4
Minimum value = 0
Maximum value = (2^4)-1 = 15

Question 14

Adding two (unsigned) binary integers

Answer 14

010010
100100

110110

0 + 0 = 0
1 + 0 = 1
1 + 1 = 10 (0 carry the 1)
1 + 1 + 1 = 11 (1 carry the 1)

Question 15

Multiplying two (unsigned) binary integers

Answer 15

10100
x —10

—00000
101000
0101000

0 x 0 = 0
0 x 1 = 0
1 x 1 =1

Question 16

How to represent negative and positive integers in two’s complement (Signed binary)

Answer 16

Positive
-128 64 32 16 8 4 2 1
—0—1—1—1-0-1-0-1
= (64+32+16+4+1) = 117
The most significant bit is always 0

Negative
-128 64 32 16 8 4 2 1
–1– 0—0—0–1-0-1-1
= (-128+8+2+1) = -117
The most significant bit (negative) is always 1

Question 17

Converting binary from unsigned to signed

Answer 17

Unsigned
01110110 = 177

Signed
10001010 = -117

From the least significant bit, keep the binary digit the same up to and including the first 1, after which switch a 1 to a 0/ a 0 to a 1.

Question 18

Performing subtraction using two’s complement

Answer 18

165 - 23 = ?
1) Convert the number being subtracted (23) into negative signed binary
00010111 = 23
11101001 = -23

2) Add together the positive binary number (165) and the negative number (-23)
-10100101
-11101001
110001110

3) If there is overflow with subtraction, ignore the most significant figure
so 110001110
= 10001110 = 142

165 + -23 = 142

Question 19

Minimum and maximum ranges of values that can be represented in signed and unsigned binary

Answer 19

Unsigned
Min = 0
Max = (2^n) -1

ex 3 bit
Min = 000 = 0
Max = 111 = (2^3) -1 = 7

Signed
Min = -2^(n-1)
Max = 2^(n-1) -1

ex 3 bit
Min = 100 = -2^(3-1) = -4
Max = 011 = 2^(3-1)-1 = 3

Question 20

Representing numbers with fractional points in binary

Answer 20

Fixed point form binary:

-16 8 4 2 1 . 1/2 1/4 1/8
0 0 1 1 1 1 1 0 = 7.75 (4+2+1+1/2+1/4)

Disadvantages:
-range of numbers that can be stored is limited as some bits are being used for fractional part of the number
-Some numbers cannot be stored accurately (ex 1/3, recurring, etc)

Floating point binary:

-8 4 2 1 . 1/2 1/4 1/8 1/16
-64 32 16 8 4 2 1 . 1/2
(both use 8 bits - but can represent a wider range of magnitudes using the same number of bits, or allow for more relative precision for smaller numbers in the range)

Mantissa (number being stored, always in two’s complement)

Exponent (binary point position, always in two’s complement)
ex -4 2 1
1 1 0
=move binary point 2 to the left (because negative)

Question 21

Advantages and disadvantages of fixed point and floating point

Answer 21

Range:
Floating point can represent a wider range of magnitudes than a fixed-point number using the same number of bits

Precision:
Floating point also allows for more relative precision for smaller numbers in the range.

Speed of calculation:
Fixed point can be faster and/or use less hardware than floating point

Question 22

Rounding errors

Answer 22

Both fixed point and floating point binary representation of decimal numbers may be inaccurate.

Question 23

Why was unicode introduced?

Answer 23

To represent a much wider range of different characters than ASCII, because it uses more bits and therefore combinations to represent.

However, it takes up more space.

Question 24

Parity bits (Error checking and correction)

Answer 24

ASCII characters only use 7 bits - the left over bit can be used as a parity bit (usuallly the most significant bit)

Even parity = 0 (to create an equal number of 1s and 0s)
Odd parity = 1 (to create an inequal number of 1s and 0s)

Disadvantages:
-We cannot tell which bit has been corrupted, so the whole byte has to be resent
-If two bits came corrupted it wouldn’t be detected

Question 25

Majority voting (Error checking and correction)

Answer 25

Identifies errors in data by transmitting binary digits multiple times and looking at the pattern recieved.

010 = 0 111 = 1 110 = 1 etc

If the pattern doesn’t match, majority voting checks which bit occurs most frequently and assumes it is the correct bit.

Advantages:
-Data doesn’t need to be requested again

Disadvantages:
-Requires sending x3 the bits just to receive an 8 bit character

Question 26

Checksums (Error checking and correction)

Answer 26

A mathematical algorithm applied to a block of data

-data from the block is used to create the initial checksum, which is added up and transmitted along with the original data

ex(not accurate)
01010000 01110010 10001011
=50 + 23 + 87 = checksum

The same algorithm is applied at the end. If the checksums match, it is assumed the data has been transmitted correctly.

Question 27

Check digits (Error checking and correction)

Answer 27

Redundancy check used for error detection on identification numbers, such as bank cards, where they are entered manually (where human error occurs often)

-Takes original code, each digit is assigned a weight, weights are added up, (some function, varies) produces check digit

(ex IBSN numbers on books)

Question 28

Analogue signals and digital signals

Answer 28

analogue signal: Natural sound waves, occurring in a continuous wave form. e.g. Human voice

digital signal: Discrete digital format for representing natural sound waves. e.g. CDs and DVDs

Question 29

Analogue data and digital data

Answer 29

Analogue data: continuous values
Digital data: discrete values

Question 30

The principles of operation of an analogue to digital converter (ADC)

Answer 30

An analogue to digital converter (ADC): Any device which can convert analogue signal (continuous natural sound waves) into a digital format

They are used together with analogue sensors (e.g.a microphone)

They measure and record the amplitude of the sound wave at set intervals

Question 31

The principles of operation of a digital to analogue converter (DAC)

Answer 31

A Digital to analogue converter (DAC): Any device which can convert a digital audio signal into an analogue signal (continuous natural sound waves)

Question 32

Describe sampling rate (digital representation of sound)

Answer 32

The frequency you record the amplitude of a sound wave

number of samples per second is measured in hertz (Hz)

The more often you record a sample the smoother the playback will sound

Question 33

Describe sample resolution (digital representation of sound)

Answer 33

Represents how many different gradations of amplitude can be represented in a digital wave form

sample resolution is stored it bits

For example, if a sample only measures 3 different gradations of amplitude, only 2 bits are required (2^2 = 8), however for a sample that measures 16 gradations of amplitude, 4 bits are required (2^4 = 16)

Question 34

The nyquist theorem

Answer 34

If you want to produce an accurate recording you need to use a sampling rate which is at least double that of the highest frequency in the original signal.

Question 35

Calculating sound sample sizes in bytes

Answer 35

Size of sample = (Number of samples per second) x (Number of bits per sample) x (Length of sample in seconds)

Gives answer in bits.

To find bytes, divide by 8

Question 36

Purpose of MIDI and the use of event messages in MIDI

Answer 36

MIDI is a technical standard

It allows a wide range of electronic musical intruments, computers, etc. to communicate with each other.

It uses a MIDI controller to send and receive event messages to each device. The messages specify details, such as:
- Duration of note
- Pitch
- Volume change
- Vibrato
- Tempo synchronisation

Question 37

The advantages of using MIDI files for representing music

Answer 37

• MIDI file uses far less disk space than a traditional digital recording
• Instruments can be recorded seperately and put together digitally

Question 38

How are bitmaps represented?

Answer 38

Digital bitmapped images are made up of pixels. Each pixel is represented by a binary number

Question 39

Resolution, colour depth and size in pixels for bitmaps

Answer 39

Resolution: The number of dots per inch where a dot is a pixel
Colour depth: The number of bits stored for each pixel
Size in pixels: the width of an image in pixels x height of image in pixels

Question 40

Calculating storage requirements for bitmapped images

Answer 40

Ignoring metadata:
Storage requirements = size of image x colour depth

gives size in bits

To find it in bytes, divide by 8

However, bitmap image files may also contain metadata

Question 41

Typical metadata examples

Answer 41

Metadata is data about data

It is stored along with the actual bits which make up the image and increase the overall file size.

examples:
- width
- height
- colour depth
- file name
- etc

Question 42

How do vector graphics represent images using lists of objects?

Answer 42

The properties of each geometric object/shape in the vector graphic image are stored as a list

Typical properties of objects examples:
- centre coordinates
- radius
- fill colour
- outline colour
- outline width

Question 43

Advantages and disadvantages of vector graphics vs bitmapped graphics

Answer 43

Vector graphics
Advantages:
- File size is kept relatively small, regardless of scale
- will always scale without loss of quality
- great format for logos or images with simple shapes and colours

Disadvantages:
- Cannot easily replicate an image with continuous areas of changing colour
- Individual pixels cannot be changed

Bitmapped graphics
Advantages:
- Great format for storing full coour images taken on phone/digital camera
- Can manipulate individual pixels easily
- images photos can easily be altered, retouched etc.

Disadvantages:
- generally takes up more memory and file storage
- images dont scale very well, they become pixelated the larger they get

Question 44

Why images, sound files, and text files are compressed

Answer 44

Files are compressed to reduce their size. Smaller files can be transferred faster between storage devices/over the internet

Question 45

Lossy compression, advantages and disadvantages

Answer 45

Reduces the file size by remvoing data. Original cannot be reconstructed

Advantages:
* Greatly reduced file sizes
* The extent to which the file size can be reduced is not limited

Disadvantages:
* Loss in data, original cannot be reconstructed
* Quality of file is reduced

Question 46

Lossless compression, advantages and disadvantages

Answer 46

File size is reduced in a way which results in no data loss

Advantages:
* No reduction in quality
* No loss of data

Disadvantages:
* Larger file sizes than lossy
* Limit to how much a file can be compressed

Question 47

The principles behind run length encoding (RLE) for lossless compression

Answer 47

RLE reduces the size of a file by removing repeated info and replacing it with one occurance of the repeated info, followed by the number of times it is repeated

Question 48

The principles behind dictionary-based methods for lossless compression

Answer 48

A dictionary containing repeated data is appended to the file.