4.5.6 Representing images, sound and other data

Question 1

What is the colour depth?

Answer 1

The number of bits assigned to a pixel in an image.

Question 2

How is the storage required to represent a bitmap image calculated?

Answer 2

• Multiply the number of pixels (width x height ) by the bit depth
• This produces a minimum value

Question 3

Why is a minimum value produced?

Answer 3

Because bitmap image files may also contain metadata, hence the calculated amount is a minimum amount.

Question 4

What are some examples of metadata?

Answer 4

• Width
• Height
• Date created
• Colour depth

Question 5

What are the two methods of representing images in a computer?

Answer 5

• Vector graphics

* Bitmap images

Question 6

How do vector graphics represent images?

Answer 6

• Using geometric objects and shapes such as rectangles, circles and lines.
• The properties of each geometric object or shape in the image are stored in a list

Question 7

How are the properties of vector graphics stored?

Answer 7

The properties of each geometric object or shape in the image are stored in a list.

Question 8

What is an advantage of vector graphics?

Answer 8

• Vector graphics can be scaled without losing quality, enlargement leads to no loss of clarity
• They are well suited to simple images using shapes
• Vector graphics frequently use less storage space than bitmapped graphics, as information is stored for each shape rather than every single pixel in an image.

Question 9

What are bitmap images suited to?

Answer 9

Storing photographs.

Question 10

What is a disadvantage of bitmap images?

Answer 10

• Information being stored for every single pixel in an image makes it use more space than vector graphics
• Enlarging a bitmap image results in a blurry or pixelated image, there is a loss of clarity.

Question 11

How is sound represented by computers?

Answer 11

As a sequence of samples, each of which takes a discrete digital value.

Question 12

What is the sampling rate?

Answer 12

Number of samples per second, expressed in hertz.

Question 13

What is the sample resolution?

Answer 13

Number of bits allocated to each sample.

Question 14

What does a higher sample resolution allow for?

Answer 14

• Higher sample resolutions result in greater audio quality, but also increased file size.

Question 15

How can the size of a sound sample be calculated?

Answer 15

Multiplying together the duration of the sample in seconds, the sampling rate in Hertz and the sample resolution.

duration * sampling rate * resolution

Question 16

What may increase the space required to store an audio file?

Answer 16

Meta data

Question 17

What is the nyquist theorem?

Answer 17

• The sampling rate of digital audio file must be at least twice the frequency of the sound.
• If the sampling rate is below this, the sound may not be accurately represented.

Question 18

What is MIDI used for?

Answer 18

Electronic musical instruments which can be connected to computers.

Question 19

How does MIDI work?

Answer 19

• Rather than storing samples of sounds, MIDI stores sound as a series of event messages, each of which represents an event in a piece of music.
• Midi is a series of instructions used to recreate a piece of music

Question 20

What is an event message?

Answer 20

An event in a piece of music.

Question 21

What might event messages contain?

Answer 21

• Duration of note
• Instrument with which a note is played
• How loud a note is
• if a note should be sustained

Question 22

What are the advantages of MIDI over sampled recordings?

Answer 22

• Easy manipulation of music without loss of quality
• Instruments on which notes sound can be changed
• Notes can be transposed
• Duration of notes can be altered
• MIDI files are often smaller in size than sampled audio files
• MIDI files are lossless, there is no information lost when music is stored using MIDI

Question 23

What is the disadvantage of using MIDI?

Answer 23

• Might result in less realistic sound than sampled recordings
• MIDI can’t be used for storing speech.