Data Representation: Images, Sound and Compression - Worksheets, Questions and Revision

20 original exam-style questions - 11 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 12)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

GCSE · Data Representation

P2.6 Data Representation: Images, Sound and Compression

OCR J277 · Calculators not allowed · about 115 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Digital images are made up of a grid of picture elements.

(a)Define the term 'pixel'.(1)

(b)A photograph has a resolution of 1200 x 800. State what the two numbers in this resolution represent.(1)

(c)Calculate the total number of pixels in the photograph described in part (b).(2)

(Total for Question 1 is 4 marks)

Colour depth affects how many different colours a bitmap image can display.

(a)Define the term 'colour depth'.(1)

(b)An image uses a colour depth of 4 bits per pixel. Calculate the maximum number of different colours that can be represented.(2)

(c)Circle the letter next to the colour depth that would allow an image to store the greatest number of different colours.(1)

A) 1 bit
B) 2 bits
C) 4 bits
D) 8 bits

(Total for Question 2 is 4 marks)

An image editing program uses a palette of 32 different shades of grey. Each shade is given an index number, which is stored using binary.

(a)Calculate the minimum number of bits needed to store a palette index for one of the 32 shades. Show your working.(2)

(b)Convert the denary palette index 19 into binary, using the number of bits found in part (a).(2)

(c)One pixel stores the binary value 11010 as its palette index. Convert this binary value to hexadecimal.(2)

(Total for Question 3 is 6 marks)

A bitmap image is 320 pixels wide and 240 pixels high. Each pixel is stored using a colour depth of 8 bits.

(a)State the number of different colours that a colour depth of 8 bits can represent.(1)

(b)Show that the uncompressed image requires 614400 bits of storage.(2)

(c)Calculate the size of the uncompressed image in kilobytes (KB). Use 1 byte = 8 bits and 1 KB = 1000 bytes.(2)

(Total for Question 4 is 5 marks)

Digital photographs usually store extra information alongside the pixel data, called metadata.

(a)State what is meant by the term 'metadata'.(1)

(b)Give two examples of metadata that might be stored with a digital photograph.(2)

(c)State one reason why storing metadata increases the overall file size of a photograph.(1)

(Total for Question 5 is 4 marks)

Sound is recorded by sampling an analogue sound wave at regular intervals.

(a)Define the term 'sample rate'.(1)

(b)Define the term 'bit depth' in the context of sound recording.(1)

(c)State the effect on the recorded sound of increasing the sample rate used.(1)

(d)State one drawback of increasing the sample rate.(1)

(Total for Question 6 is 4 marks)

A mono sound clip is recorded with a sample rate of 8000 Hz and a bit depth of 8 bits. The clip lasts 3 seconds.

(a)Calculate the total number of samples taken during the recording.(2)

(b)Calculate the file size of the recording in bits.(2)

(c)Calculate the file size of the recording in kilobytes (KB). Use 1 byte = 8 bits and 1 KB = 1000 bytes.(2)

(Total for Question 7 is 6 marks)

A sound engineer is deciding on recording settings for a podcast that will be streamed to mobile phones.

(a)Explain the effect of recording in stereo instead of mono on the file size of the podcast.(2)

(b)Explain why the sound engineer might choose a lower bit depth for the podcast than a professional music studio would use for a music recording.(2)

(Total for Question 8 is 4 marks)

Files can be compressed using lossy or lossless compression.

(a)Define 'lossy compression'.(1)

(b)Define 'lossless compression'.(1)

(c)Give one example of a file format that commonly uses lossy compression.(1)

(d)State one situation where lossless compression would be more suitable than lossy compression, and justify your answer.(2)

(Total for Question 9 is 5 marks)

A single row of a black and white bitmap image is 12 pixels wide, shown below (W = white pixel, B = black pixel):
W W W W W B B B W W W W

(a)State what run-length encoding (RLE) is used for.(1)

(b)Use run-length encoding to encode the row of pixels above. Give your answer in the form (run length)(colour).(2)

(c)State one type of image for which RLE would achieve a high compression ratio, and one type for which it would not.(2)

(Total for Question 10 is 5 marks)

A row of pixels has been run-length encoded as: 6W 2B 6W
The original (uncompressed) row used 1 byte to store the colour of each pixel. The encoded version above uses 2 bytes for each run (1 byte for the run length number and 1 byte for the colour letter).

(a)Decode this run-length encoded data to show the original row of pixels.(2)

(b)Calculate the compression ratio of the encoded row compared to the original uncompressed row, giving your answer in the form n:1 (original size : encoded size).(3)

(Total for Question 11 is 5 marks)

A text file contains the following phrase, which is to be compressed using Huffman coding: AAAABBBCCD

(a)Explain what is meant by 'dictionary coding' as a method of lossless compression.(2)

(b)Complete a frequency table showing how many times each character (A, B, C, D) appears in the phrase AAAABBBCCD.(2)

(c)Explain why characters that occur more frequently, such as 'A', are given shorter binary codes in Huffman coding.(2)

(Total for Question 12 is 6 marks)

A programmer is writing a program, in the OCR Exam Reference Language, that helps photographers estimate storage needs. The program should input the width, height and colour depth (in bits) of a photo, calculate the file size in bytes, and output the result to the user.
Write pseudocode for this program.

(Total for Question 13 is 6 marks)

Table 1 shows the file size of the same photograph saved in three different ways.

(a)Calculate the compression ratio of the lossless compressed version compared to the original, giving your answer in the form n:1.(2)

(b)Calculate the percentage reduction in file size achieved by the lossy compressed version compared to the original.(3)

(Total for Question 14 is 5 marks)

A small business owner is building a website to sell handmade jewellery. She needs to include high-quality product photographs and a short looping background music clip. Storage space and page loading speed are both important to her.
Discuss the compression choices the business owner should make for the images and the sound file on her website, and the impact of these choices on the customer's experience.

(Total for Question 15 is 6 marks)

A programmer has written the following algorithm, using the OCR Exam Reference Language, to decode run-length encoded data. Arrays are indexed from 0 and LEN(array) returns the number of items in the array.
runLengths = [4, 2, 3]
symbols = ["W", "B", "W"]
output = ""

FOR i = 0 TO LEN(runLengths) - 1
    count = 0
    WHILE count < runLengths[i]
        output = output + symbols[i]
        count = count + 1
    ENDWHILE
NEXT i

OUTPUT output

(a)Complete the trace table below to show the value of output after the inner WHILE loop finishes, for each pass of the outer FOR loop.(4)

(b)State the final value output by the program once the whole algorithm has finished running.(1)

(Total for Question 16 is 5 marks)

A file contains only the following characters, with these frequencies:
E: 8 T: 5 A: 3 O: 2
A Huffman tree is to be built to assign binary codes to each character, with the most frequent characters given the shortest codes.

(a)Describe the general method used to build a Huffman tree from a frequency table.(3)

(b)Using the frequencies given, calculate the total number of bits needed to store the whole file if a fixed-length binary code were used for every character instead of Huffman coding.(3)

(Total for Question 17 is 6 marks)

Using the same frequencies as the previous question (E: 8, T: 5, A: 3, O: 2), Huffman coding was used to generate the following variable-length codes:
E = 0 T = 10 A = 110 O = 111

(a)Calculate the total number of bits needed to store the whole file using these Huffman codes.(3)

(b)Calculate the number of bits saved by using Huffman coding compared to the fixed-length code from the previous question, and hence calculate the percentage saving in file size.(3)

(Total for Question 18 is 6 marks)

A hospital stores digital X-ray images for diagnosis, and also stores short video clips of ultrasound scans for training purposes.
Evaluate whether lossy or lossless compression would be more appropriate for each of these two types of file, justifying your answer with reference to the consequences of data loss and the practical constraints of storage and transmission.

(Total for Question 19 is 6 marks)

A video call application sends both image frames and sound at the same time.
Each video frame is 640 x 480 pixels with a colour depth of 16 bits per pixel. The application sends 2 frames every second.
The sound is sampled at 8000 Hz with a bit depth of 16 bits, recorded in stereo (2 channels).

(a)Calculate the number of bits needed to store one uncompressed video frame.(2)

(b)Calculate the total number of bits of video data sent in one second, given that 2 frames are sent per second.(1)

(c)Calculate the number of bits of sound data generated in one second.(2)

(d)Calculate the total combined number of bits per second (video and sound together) that the application must transmit, and hence explain why real-time video calls almost always use compression.(3)

(Total for Question 20 is 8 marks)

Mark scheme · P2.6 Data Representation: Images, Sound and Compression

Question 1

(a) B1 the smallest single element (a dot of one colour) that a digital image is made up of (oe)
(a) Answer: A pixel is the smallest single element (a dot of colour) that makes up a digital image.
(b) B1 the number of pixels across the width and the number of pixels down the height of the image (oe)
(b) Answer: The number of pixels wide (1200) and the number of pixels high (800).
(c) M1 correct method: 1200 x 800 (oe)
(c) A1 960000 cao
(c) Answer: 960 000 pixels

Question 2

(a) B1 the number of bits used to represent/store the colour of each pixel (oe)
(a) Answer: The number of bits used to represent the colour of each pixel.
(b) M1 recognises number of colours = 2^{colour depth}, i.e. 2⁴ (oe)
(b) A1 16 cao
(b) Answer: 16 colours
(c) B1 D cao
(c) Answer: D) 8 bits

Question 3

(a) M1 recognises 2ⁿ must be at least 32, e.g. tests 2⁴ = 16 (too small) and 2⁵ = 32 (oe)
(a) A1 5 (bits) cao
(a) Answer: 5 bits
(b) M1 correct place value working shown using place values 16, 8, 4, 2, 1 (oe)
(b) A1 10011 cao (ft from part a for number of bits)
(b) Answer: 10011
(c) M1 converts binary to denary or splits into a nibble/group for conversion, e.g. 11010 = 26 (oe)
(c) A1 1A cao (accept 0x1A oe)
(c) Answer: 1A

Question 4

(a) B1 256 cao
(a) Answer: 256 colours
(b) M1 calculates the total number of pixels: 320 x 240 = 76800 (oe)
(b) A1 76800 x 8 = 614400 shown correctly cso (answer given, working must be shown)
(b) Answer: 320 x 240 = 76800 pixels; 76800 x 8 = 614400 bits
(c) M1 converts bits to bytes: 614400 / 8 (= 76800) (oe, ft from part b)
(c) A1 76.8 (KB) cao ft
(c) Answer: 76.8 KB

Question 5

(a) B1 data that describes/gives information about other data (oe)
(a) Answer: Data that describes other data.
(b) B1 one valid example, e.g. date and time taken (oe)
(b) B1 a different valid example, e.g. camera/device model, GPS location, resolution/dimensions, colour depth, orientation, exposure settings (oe)
(b) Answer: Any two of: date and time taken; camera/device model; GPS location; image resolution; colour depth; exposure settings.
(c) B1 extra data/bits are stored in the file alongside the pixel data, so more bits/bytes are needed in total (oe)
(c) Answer: Because extra data is stored alongside the pixel data, more bits/bytes must be stored in total.

Question 6

(a) B1 the number of samples of the sound wave taken per second, measured in Hertz (Hz) (oe)
(a) Answer: The number of samples of the sound wave taken per second.
(b) B1 the number of bits used to store/represent each sample (oe)
(b) Answer: The number of bits used to store each sample.
(c) B1 the recording more closely/accurately matches the original analogue sound wave, giving higher quality sound (oe)
(c) Answer: The recording more closely matches the original sound wave, giving higher quality.
(d) B1 the file size increases, as more samples must be stored (oe)
(d) Answer: The file size increases because more samples must be stored.

Question 7

(a) M1 correct method: 8000 x 3 (oe)
(a) A1 24000 (samples) cao
(a) Answer: 24000 samples
(b) M1 multiplies number of samples by bit depth: (their samples) x 8 (oe, ft from part a)
(b) A1 192000 (bits) cao ft
(b) Answer: 192000 bits
(c) M1 converts bits to bytes: (their bits) / 8 (= 24000) (oe, ft from part b)
(c) A1 24 (KB) cao ft
(c) Answer: 24 KB

Question 8

(a) B1 stereo requires two channels of audio data to be stored instead of one (oe)
(a) B1 so the file size (approximately) doubles compared to the equivalent mono recording (oe, ft)
(a) Answer: Stereo stores two channels of audio instead of one, so the file size roughly doubles compared to mono.
(b) B1 a lower bit depth produces a smaller file size, so it downloads/streams faster and uses less of the listener's mobile data allowance (oe)
(b) B1 spoken word/podcast audio does not need the same level of quality as music, so the small loss in quality is unlikely to be noticeable on a phone speaker (oe)
(b) Answer: A lower bit depth gives a smaller file that streams faster and uses less mobile data; speech does not need studio-quality sound, so the small quality loss is not very noticeable.

Question 9

(a) B1 compression that permanently removes some data/detail from the file, so it cannot be exactly restored to the original (oe)
(a) Answer: Compression that permanently removes some data, so the original file cannot be exactly recreated.
(b) B1 compression that reduces file size without permanently losing any data, so the original file can be exactly reconstructed (oe)
(b) Answer: Compression that reduces file size without losing any data, so the original file can be exactly recreated.
(c) B1 valid lossy format, e.g. JPEG or MP3
(c) Answer: JPEG (or MP3).
(d) B1 identifies a suitable situation, e.g. compressing a text/program file, a spreadsheet, or a medical/legal document (oe)
(d) B1 justifies: because losing any data/detail could change the meaning of the file or make it unusable/inaccurate (oe, ft)
(d) Answer: Example: a legal document. Losing even small details through lossy compression could change its meaning or make it invalid, so lossless compression must be used.

Question 10

(a) B1 to compress data by storing a value and how many times it repeats (its run length), instead of storing the value once for every repeat (oe)
(a) Answer: It compresses data by storing a repeated value once along with a count of how many times it repeats.
(b) M1 identifies the three correct runs, in order: 5 white, 3 black, 4 white (oe)
(b) A1 5W 3B 4W cao (accept equivalent consistent notation, e.g. (5,W)(3,B)(4,W))
(b) Answer: 5W 3B 4W
(c) B1 images with large areas of a single/repeated colour, e.g. simple line drawings, icons or logos, compress well (oe)
(c) B1 photographs, or images with lots of colour variation/noise between adjacent pixels, compress poorly, as there are few or no long runs (oe)
(c) Answer: RLE works well on simple images with large blocks of one colour (e.g. a logo), but poorly on photographs, which have little colour repetition between adjacent pixels.

Question 11

(a) M1 correctly identifies the total number of pixels represented: 6 + 2 + 6 = 14 (oe)
(a) A1 WWWWWW BB WWWWWW cao (14 symbols in the correct order)
(a) Answer: WWWWWWBBWWWWWW (14 pixels)
(b) M1 correctly finds the original size: 14 pixels x 1 byte = 14 bytes (ft from part a)
(b) M1 correctly finds the encoded size: 3 runs x 2 bytes = 6 bytes (oe)
(b) A1 14:6, simplified to 7:3 cao (accept awrt 2.3:1, or equivalent statement that the encoded row is 42.9% of the original size)
(b) Answer: 14:6, which simplifies to 7:3 (the compressed row is about 42.9% of the original size)

Question 12

(a) B1 frequently occurring data (e.g. repeated characters, strings or words) is replaced by a shorter code/index that refers to an entry in a dictionary/table (oe)
(a) B1 the dictionary itself (matching each code to its full original value) is also stored, so the codes can be translated back to reconstruct the original data (oe)
(a) Answer: Repeated data is replaced with a shorter code that refers to an entry in a dictionary; the dictionary is stored too so the original data can be reconstructed from the codes.
(b) B1 A = 4 and B = 3 both correct
(b) B1 C = 2 and D = 1 both correct
(b) Answer: A = 4, B = 3, C = 2, D = 1
(c) B1 assigning the shortest codes to the most frequently occurring characters (oe, ft from part b)
(c) B1 reduces the total/average number of bits needed to store the whole file, because the most common symbol uses the fewest bits each time it appears (oe)
(c) Answer: Giving the most common character the shortest code means that symbol uses the fewest bits every time it appears, reducing the total number of bits needed for the whole file.

Question 13

B1 correctly inputs the width using a sensibly named variable, e.g. width = INPUT("Enter width: ")
B1 correctly inputs the height using a sensibly named variable
B1 correctly inputs the colour depth using a sensibly named variable
B1 calculates the total number of pixels, e.g. totalPixels = width * height
B1 calculates total bits (totalPixels * colourDepth) and converts this to bytes by dividing by 8, e.g. totalBytes = totalBits / 8
B1 outputs the calculated file size in bytes, e.g. OUTPUT totalBytes
Answer:width = INPUT("Enter width in pixels: ")
height = INPUT("Enter height in pixels: ")
colourDepth = INPUT("Enter colour depth in bits: ")

totalPixels = width * height
totalBits = totalPixels * colourDepth
totalBytes = totalBits / 8
```
OUTPUT totalBytes
```

Question 14

(a) M1 correct method: 4800 / 3200 (oe)
(a) A1 1.5:1 cao
(a) Answer: 1.5:1
(b) M1 finds the reduction in size: 4800 - 800 = 4000 (oe)
(b) M1 correct method: (4000 / 4800) x 100 (oe, ft)
(b) A1 awrt 83.3% cao ft
(b) Answer: 83.3% (to 3 s.f.)

Question 15

Level 1 (1-2): Basic, isolated point(s) made, e.g. that compression makes files smaller. Little or no reference to the impact on quality or loading speed. Limited use of technical terms.
Level 2 (3-4): Some relevant discussion of lossy or lossless compression with limited application to the images and/or the sound file. A link is made between file size and loading speed or quality, but this is not fully developed for both media types. Reasonable use of technical terms.
Level 3 (5-6): Sustained, well-structured discussion that applies compression concepts directly to both the images and the sound file, weighing up the trade-off between quality and loading speed/storage, and justifying suitable choices for the business owner. Accurate and consistent use of technical terminology throughout.
Indicative content:
- Product photos need to look attractive and detailed, so some quality is important; a high-quality lossy format (e.g. JPEG) is likely to give a good balance between quality and file size.
- Fully lossless images would give the best possible quality but produce much larger files, which could slow down page loading and put customers off.
- As long as the loss of quality from lossy compression is not very noticeable, using it on the product photos keeps file sizes down and pages loading quickly.
- The background music clip is short, loops repeatedly and does not need studio-quality sound, so a lossy audio format (e.g. MP3) with a lower bit rate/sample rate would be suitable.
- A smaller sound file also loads faster and uses less data, which particularly benefits customers browsing on mobile devices or slower connections.
- Overall the business owner should balance an acceptable level of quality against fast loading times, since slow-loading pages may cause customers to leave the site before making a purchase.

Question 16

(a) B1 i = 0: runLengths[0] = 4, symbols[0] = "W", output = "WWWW"
(a) B1 i = 1: runLengths[1] = 2, symbols[1] = "B", output = "WWWWBB"
(a) B1 i = 2: runLengths[2] = 3, symbols[2] = "W", output = "WWWWBBWWW"
(a) B1 all three rows fully consistent (correct index, run length and symbol values used throughout)
(a) Answer: i=0: output="WWWW"; i=1: output="WWWWBB"; i=2: output="WWWWBBWWW"
(b) B1 WWWWBBWWW cao ft (from part a)
(b) Answer: WWWWBBWWW

Question 17

(a) B1 repeatedly combine/merge the two nodes (symbols, or previously combined nodes) with the lowest frequencies into a new node (oe)
(a) B1 the new node's frequency is the sum of the two combined frequencies; this is repeated until only one node (the root) remains (oe)
(a) B1 starting from the root, a 0 or 1 is assigned to each branch to build the binary code for each character; characters that are fewer branches from the root (more frequent) end up with shorter codes (oe)
(a) Answer: Repeatedly merge the two lowest-frequency nodes into a new node whose frequency is their sum, until one root node remains; then assign 0/1 to each branch from the root to generate each character's code, giving more frequent characters shorter codes.
(b) M1 recognises that 4 different characters need 2 bits each for a fixed-length code, since 2² = 4 (oe)
(b) M1 finds the total number of characters in the file: 8 + 5 + 3 + 2 = 18 (oe)
(b) A1 36 (bits) cao
(b) Answer: 36 bits

Question 18

(a) M1 identifies the correct code length for each character: E = 1 bit, T = 2 bits, A = 3 bits, O = 3 bits (oe)
(a) M1 multiplies each frequency by its code length and finds each product: 8x1=8, 5x2=10, 3x3=9, 2x3=6 (oe)
(a) A1 33 (bits) cao
(a) Answer: 33 bits
(b) M1 finds the bits saved: 36 - 33 = 3 (bits) (oe, ft from previous question and part a)
(b) M1 correct method for percentage: (3 / 36) x 100 (oe, ft)
(b) A1 awrt 8.3% cao ft
(b) Answer: 3 bits saved; awrt 8.3% reduction in file size

Question 19

Level 1 (1-2): A vague or one-sided statement, e.g. that lossless is always best because no data is lost, with little or no reference to the two specific file types or to practical constraints.
Level 2 (3-4): Some valid points made about both the X-ray images and the ultrasound clips, with limited justification or explanation of the consequences of data loss; the discussion may be one-sided or lack a balanced conclusion.
Level 3 (5-6): A balanced, well-justified evaluation covering both file types, referencing the medical consequences of losing diagnostic detail and the practical impact of large lossless file sizes on storage/transmission, reaching a supported conclusion.
Indicative content:
- X-ray images are used to diagnose patients, so losing fine detail (e.g. a hairline fracture) through lossy compression could lead to a missed or incorrect diagnosis, risking patient harm.
- Lossless compression is therefore more appropriate for diagnostic X-rays, even though the resulting files will be larger and require more storage.
- Ultrasound training videos are used to teach general technique rather than diagnose a specific patient, so some loss of quality from lossy compression may be acceptable.
- Video files are typically much larger than still images, so using lossy compression on the ultrasound clips saves considerably more storage space and makes the files quicker to transmit between hospitals or training sites.
- Hospitals must store and transmit very large volumes of medical data, so storage costs and available network bandwidth are real practical constraints that must be balanced against the risk of losing important information.
- A supported conclusion balancing patient safety and diagnostic accuracy against storage/transmission efficiency, e.g. lossless compression for diagnostic X-rays but lossy compression acceptable for non-diagnostic training footage.

Question 20

(a) M1 correct method: 640 x 480 x 16 (oe)
(a) A1 4915200 (bits) cao
(a) Answer: 4 915 200 bits
(b) B1 9830400 (bits per second) cao ft (their part a x 2)
(b) Answer: 9 830 400 bits per second
(c) M1 correct method: 8000 x 16 x 2 (oe)
(c) A1 256000 (bits) cao
(c) Answer: 256 000 bits per second
(d) M1 adds video and sound data rates together: (their part b) + (their part c) (oe, ft)
(d) A1 10086400 (bits per second) cao ft
(d) B1 explains that this uncompressed data rate is far larger than most internet/mobile connections can transmit in real time, so compression is needed to reduce the data rate enough for smooth transmission (oe)
(d) Answer: 10 086 400 bits per second; compression is needed because most internet/mobile connections cannot transmit this much uncompressed data in real time without delays or dropped frames.

Mark my answers