Question 1 · Short Answer & Fill-in-the-blank
4.5 marksComplete the following sentences regarding digital sound representation by filling in each blank with the correct technical term or value.
(a) The number of audio snapshots taken per unit time is known as the ....................................................., which is typically measured in ..................................................... .
(b) The number of bits allocated to represent the amplitude of each snapshot is the ..................................................... .
(c) An uncompressed stereo sound recording is sampled for \(10\) seconds at a frequency of \(44.1\text{ kHz}\) with a bit depth of \(16\text{ bits}\). Calculate the total file size in mebibytes (MiB), showing your working. (You may leave your final answer as an unsimplified mathematical expression or calculate it to 2 decimal places).
(a) The number of audio snapshots taken per unit time is known as the ....................................................., which is typically measured in ..................................................... .
(b) The number of bits allocated to represent the amplitude of each snapshot is the ..................................................... .
(c) An uncompressed stereo sound recording is sampled for \(10\) seconds at a frequency of \(44.1\text{ kHz}\) with a bit depth of \(16\text{ bits}\). Calculate the total file size in mebibytes (MiB), showing your working. (You may leave your final answer as an unsimplified mathematical expression or calculate it to 2 decimal places).
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Sampling rate (or sampling frequency) measures how frequently samples are taken; its standard unit is Hertz (Hz) or kilohertz (kHz).
(b) Sampling resolution (also referred to as bit depth) defines the precision/amplitude range encoded for each sample.
(c) Calculation:
• Total samples per channel = 44,100 samples/sec × 10 sec = 441,000 samples
• Total bits for 2 channels (stereo) = 441,000 × 16 bits × 2 = 14,112,000 bits
• Conversion to bytes = 14,112,000 / 8 = 1,764,000 bytes
• Conversion to mebibytes (MiB) = 1,764,000 / (1024 × 1024) = 1,764,000 / 1,048,576 ≈ 1.682 MiB (or 1.68 MiB).
(b) Sampling resolution (also referred to as bit depth) defines the precision/amplitude range encoded for each sample.
(c) Calculation:
• Total samples per channel = 44,100 samples/sec × 10 sec = 441,000 samples
• Total bits for 2 channels (stereo) = 441,000 × 16 bits × 2 = 14,112,000 bits
• Conversion to bytes = 14,112,000 / 8 = 1,764,000 bytes
• Conversion to mebibytes (MiB) = 1,764,000 / (1024 × 1024) = 1,764,000 / 1,048,576 ≈ 1.682 MiB (or 1.68 MiB).
Marking scheme
1 mark each for the following points (up to 4.5 marks max):
• MP1 (0.5 mark): sampling rate / sampling frequency
• MP2 (0.5 mark): Hertz / Hz / kHz
• MP3 (1 mark): sampling resolution / bit depth / sample size
• MP4 (1 mark): Multiplying sample rate × duration × bit depth × 2 (channels) [44100 × 10 × 16 × 2]
• MP5 (1.5 marks): Correct division by (8 × 1024 × 1024) or 8,388,608 to convert bits to MiB, yielding 1.68 MiB (or exact unsimplified fraction 14,112,000 / 8,388,608). Award 1 mark if divided by 10^6 (decimal megabyte slip) with clear method shown.
• MP1 (0.5 mark): sampling rate / sampling frequency
• MP2 (0.5 mark): Hertz / Hz / kHz
• MP3 (1 mark): sampling resolution / bit depth / sample size
• MP4 (1 mark): Multiplying sample rate × duration × bit depth × 2 (channels) [44100 × 10 × 16 × 2]
• MP5 (1.5 marks): Correct division by (8 × 1024 × 1024) or 8,388,608 to convert bits to MiB, yielding 1.68 MiB (or exact unsimplified fraction 14,112,000 / 8,388,608). Award 1 mark if divided by 10^6 (decimal megabyte slip) with clear method shown.