Analogue/digital, ADC/DAC, bitmap and vector graphics, sound and MIDI – A-level Computer Science exam questions with mark scheme

1Core4 marks

A weather station uses an analogue temperature sensor connected to a digital computer.

(a) Explain the difference between analogue data and digital data. [2]

(b) Explain why an analogue-to-digital converter (ADC) is needed between the sensor and the computer. [1]

(c) The computer stores the bit pattern 01000001. Explain why it is not possible to say what this bit pattern represents just by looking at it. [1]

Analogue data varies continuously / can take any value within a range (accept: is continuous)(1)
Digital data can only take discrete / a fixed set of values (accept: represented using a finite number of values, e.g. binary 0s and 1s)(1)
The sensor produces an analogue signal / continuously varying voltage, but the computer can only process digital (binary) data, so it must be converted (do not accept just "to convert it" without saying the computer needs digital/binary)(1)
The meaning depends on how the computer/program interprets it / the context or data type, e.g. it could be the integer 65, the character 'A', part of a pixel colour or a sound sample (any two interpretations, or reference to context/data type, scores the mark)(1)
Answer: (a) analogue = continuous range of values; digital = discrete values (b) sensor output is analogue but the computer processes only digital/binary data (c) the same bits can represent different data (65, 'A', a pixel, a sample), so meaning depends on context
Examiner tip: "Analogue is continuous" needs the contrast "digital is discrete" for both marks: one-sided answers score 1.

2Core4 marks

A bitmap image is 1024 pixels wide and 768 pixels high and has a colour depth of 4 bits.

(a) Calculate the size of the image in pixels. [1]

(b) Calculate the minimum storage needed for the image, ignoring metadata. Give your answer in kibibytes (KiB). [2]

1024 × 768 = 786 432 pixels(1)
786 432 × 4 = 3 145 728 bits (accept ÷ 8 = 393 216 bytes)(1)
393 216 ÷ 1024 = 384 KiB (correct answer with no working scores 2; allow follow-through from their (a); do not accept 393 or 393.216 as KiB because 1 KiB = 1024 bytes)(1)
2⁴ = 16 colours(1)
Answer: (a) 786 432 pixels (b) 384 KiB (c) 16
Examiner tip: A kibibyte is 1024 bytes, not 1000: dividing by 1000 here gives 393.2, which loses the final mark.

3Standard5 marks

A bitmap file stores an image that is 800 pixels wide and 600 pixels high. Ignoring metadata, the image data occupies exactly 1 440 000 bytes.

(a) Calculate the colour depth of the image. Show your working. [2]

(b) Calculate how many different colours can be represented with this colour depth. [1]

(c) The image is printed at a resolution of 200 dots per inch, with one dot per pixel. Calculate the width of the printed image in inches. [1]

(d) The file also stores the width of the image in pixels as metadata. Explain why the width must be stored. [1]

Total bits = 1 440 000 × 8 = 11 520 000 and number of pixels = 800 × 600 = 480 000 (or bytes per pixel = 1 440 000 ÷ 480 000 = 3)(1)
Colour depth = 24 bits (accept 3 bytes; correct answer with no working scores 1 only)(1)
2²⁴ = 16 777 216 colours (accept 2²⁴; follow through from their (a))(1)
800 ÷ 200 = 4 inches(1)
The pixel data is stored as one continuous sequence of bits, so without the width the program could not tell where each row ends / how to arrange the pixels into rows to display the image correctly (accept: to reconstruct the image correctly)(1)
Answer: (a) 24 bits (b) 16 777 216 (c) 4 inches (d) so the pixels can be arranged into the correct rows when displayed
Examiner tip: Working backwards needs bits ÷ pixels: forgetting to convert bytes to bits gives 3, which is bytes per pixel, not the colour depth in bits.

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4Standard5 marks

A journalist records a 4-minute interview using a microphone. The recording is mono, with a sampling rate of 16 000 Hz and a sample resolution of 12 bits.

(a) Calculate the size of the recording in megabytes (1 MB = 1 000 000 bytes). Ignore metadata. Show your working. [3]

(b) The journalist changes the sample resolution to 16 bits, keeping the same sampling rate. Explain the effect this has on the recording. [2]

Use of 16 000 × 12 × 240 (time must be converted to 240 seconds)(1)
= 46 080 000 bits (or ÷ 8 = 5 760 000 bytes)(1)
= 5.76 MB (correct answer with no working scores 3; allow follow-through for correct conversion of an incorrect bit total)(1)
More quantisation levels are available (65 536 instead of 4096), so each sample is a closer/more accurate match to the analogue amplitude / quantisation error is reduced, so quality is improved (accept: more accurate reproduction of the original sound)(1)
File size increases (by one third / to 7.68 MB) because more bits are stored per sample (must be linked to more bits per sample or quantified)(1)
Answer: (a) 5.76 MB (b) more accurate samples / better quality, but larger file (7.68 MB)
Examiner tip: Most lost marks come from using 4 instead of 240 seconds: the formula needs time in seconds.

5Standard5 marks

In a digital telephone call, the caller's voice is captured by a microphone and the listener hears it through a loudspeaker.

(a) Describe how an analogue-to-digital converter (ADC) converts the signal from the microphone into digital data. [3]

(b) State the purpose of the digital-to-analogue converter (DAC) at the listener's end. [1]

(c) Explain why the sound heard by the listener is never an exact copy of the original sound, even if no data is lost in transmission. [1]

The amplitude/voltage of the analogue signal is measured/sampled at regular (fixed) time intervals(1)
Each measured value is rounded / quantised to the nearest of a fixed number of levels (determined by the sample resolution)(1)
Each quantised value is stored/output as a binary number (bit pattern) of the sample resolution length(1)
Converts the received digital/binary values back into an analogue signal/varying voltage that can drive the loudspeaker (do not accept "converts it into sound" without reference to analogue signal/voltage)(1)
Values are rounded to the nearest level (quantisation error) and/or changes between samples are not recorded, so the digital version is only an approximation of the original analogue signal(1)
Answer: (a) sample amplitude at regular intervals; quantise to nearest level; store as binary (b) convert digital values to an analogue voltage for the speaker (c) quantisation/sampling means only an approximation is stored
Examiner tip: Say what is measured (amplitude) and when (at regular intervals): "it samples the sound" alone is too vague for a mark.

6Hard5 marks

The graph shows part of an analogue sound wave. The wave is sampled at the times marked with dots, and each sample is stored using a sample resolution of 3 bits (quantisation levels 0 to 7).

(a) Write down the four 3-bit binary values stored for the samples taken at 0.5 ms, 1.0 ms, 1.5 ms and 2.0 ms. [2]

(b) Calculate the sampling rate in hertz. [1]

(d) Explain one effect of increasing the sample resolution to 4 bits on how accurately this wave is represented. [1]

First two values correct: 110, 111(1)
Last two values correct: 110, 011 (values must be 3-bit binary; denary 6, 7, 6, 3 scores max 1 of the 2 marks)(1)
Sample interval 0.5 ms, so 1 ÷ 0.0005 = 2000 Hz (accept 2 kHz)(1)
1000 Hz (accept 1 kHz; follow through as half their (b))(1)
16 levels instead of 8, so the gap between levels is smaller and each sample is closer to the true amplitude / smaller quantisation error (accept: more detail of the amplitude is captured)(1)
Answer: (a) 110 111 110 011 (b) 2000 Hz (c) 1000 Hz (d) 16 levels, so samples are closer to the actual amplitude
Examiner tip: Nyquist says sample at least twice the highest frequency, so the highest recordable frequency is half the sampling rate, not double it.

7Standard4 marks

The diagram shows a simple vector drawing made of three objects, A, B and C, on a grid. Coordinates are given as (x, y) with the origin at the top-left.

The drawing is stored as a list of objects. The entry for object A has been completed:

Object	Properties
A: circle	centre (3, 3), radius 2, line colour black, fill colour yellow
B: rectangle	…
C: line	…

(a) Rectangle B has a black outline and a blue fill. Line C is red. Complete the properties for objects B and C. [3]

(b) The drawing is enlarged to five times its size for a poster. Explain why its quality does not decrease. [1]

B: position given correctly, e.g. top-left corner (5, 5) (accept any correctly identified corner, or centre (7, 6.5))(1)
B: width 4 and height 3 (accept opposite corner (9, 8)) and fill colour blue, line colour black(1)
C: start point (1, 9) and end point (4, 6) (either order) and colour red (ignore line thickness)(1)
The objects are redrawn/recalculated from their properties (coordinates/radius scaled) at the new size, so there are no pixels to enlarge and no pixelation/blockiness occurs(1)
Answer: (a) B: top-left (5, 5), width 4, height 3, fill blue, line black; C: from (1, 9) to (4, 6), colour red (b) shapes are redrawn from their properties, so no pixelation
Examiner tip: Every object needs enough properties to redraw it exactly: a rectangle with a position but no width/height cannot be drawn.

8Hard5 marks

An estate agent's website shows the company logo (a house outline and the company name in two colours) on every page, and photographs of houses for sale.

(a) Explain why the logo should be stored as a vector graphic. [2]

(b) Explain why the photographs should be stored as bitmap graphics. [2]

(c) Describe a situation in which a vector graphic would require more storage than a bitmap of the same image. [1]

The logo is made of simple shapes/lines/text, so it can be described by a small number of objects and properties, giving a small file size(1)
The logo is shown at different sizes (e.g. header, mobile, print) and a vector can be scaled without loss of quality / pixelation(1)
A photograph has continuously varying colours/complex detail that cannot be described by simple geometric shapes(1)
A bitmap records the colour of every pixel so captures this detail exactly; a vector version would need a huge number of objects (accept: photographs are captured by cameras as pixels)(1)
When the image is very complex/detailed (e.g. a photo-realistic drawing) so that a very large number of objects, each with several properties, must be stored (accept: an image with many small shapes displayed at a small size)(1)
Answer: (a) few simple shapes, small file, scales without pixelation (b) continuous colour variation and detail suit per-pixel storage (c) a highly complex image needing very many objects
Examiner tip: Each "explain" mark needs a feature of the image linked to a property of the format: "vectors are smaller" on its own is not enough.

9Hard6 marks

A composer writes a 2-minute piece of music. Stored as a MIDI file it contains 1800 MIDI messages, each 3 bytes long.

(a) Explain how a MIDI file represents music differently from a sampled sound file. [1]

(b) State two items of data that can be included in a MIDI event message for a note. [2]

(c) The same piece is also recorded as a mono sampled file at 44 100 Hz with a sample resolution of 16 bits. Calculate how many times larger the sampled file is than the MIDI file. Ignore metadata. [2]

(d) Give one advantage of MIDI for the composer, other than file size. [1]

MIDI stores a list of event messages / instructions describing how to play the music (which note, when, how loud) rather than samples of the sound wave's amplitude(1)
Any one from: the note/pitch; the velocity/volume (how hard the note is played); the channel; whether the note starts or stops (note on/note off); the instrument (program change)(1)
A second, different item from the list above (accept duration/timing of the note as an item)(1)
MIDI: 1800 × 3 = 5400 bytes and sampled: 44 100 × 16 × 120 ÷ 8 = 10 584 000 bytes(1)
10 584 000 ÷ 5400 = 1960 times larger (correct answer with no working scores 2)(1)
Any one from: individual notes can easily be edited/corrected; the music can be transposed or the tempo changed without loss of quality; the instrument used can be changed; the music can be shown as a score/notation (do not accept "smaller file" or "better quality")(1)
Answer: (a) MIDI stores instructions (event messages), not samples (b) e.g. pitch, velocity (c) 1960 times (d) e.g. notes/instrument/tempo easily changed
Examiner tip: In (c) both files must be in the same unit: compare bytes with bytes, or the ratio is out by a factor of 8.

10Stretch6 marks

A digital sign displays images of 1920 × 1080 pixels. Each image uses exactly 300 different colours.

(a) State the minimum colour depth needed and explain why a colour depth of 8 bits is not enough. [2]

(b) Calculate the size of one image in bytes, using the minimum colour depth. Ignore metadata. [2]

(c) Each stored file also contains 4000 bytes of metadata. Calculate the maximum number of complete image files that fit on a 64 GB memory card (1 GB = 10⁹ bytes). [2]

8 bits gives only 2⁸ = 256 colours, which is fewer than 300(1)
Minimum colour depth 9 bits (2⁹ = 512 ≥ 300) (do not accept 10 or 16)(1)
1920 × 1080 × 9 = 18 662 400 bits(1)
÷ 8 = 2 332 800 bytes (follow through from their colour depth)(1)
Size per file 2 332 800 + 4000 = 2 336 800 bytes and 64 × 10⁹ ÷ 2 336 800 (= 27 387.9…)(1)
27 387 files (must be rounded down; follow through from their (b) if rounded down)(1)
Answer: (a) 9 bits (8 bits gives only 256 colours) (b) 2 332 800 bytes (c) 27 387 files
Examiner tip: Number of files must be rounded down: a part-file cannot be stored, so 27 388 loses the final mark.

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