Sample rate and audio quality

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Quick Questions

Shade in the A or B circle for each question.

1. What does the sample rate of a sound recording measure?

A How many measurements are taken each second

B How loud the sound is

2. Which unit is used to measure sample rate?

A Hertz (Hz)

B Pixels (px)

3. A music CD uses a sample rate of 44,100 Hz. What does this mean?

A The sound is measured 44,100 times every second

B The song is 44,100 seconds long

4. Which sample rate will usually give better sound quality?

A 8,000 Hz

B 44,100 Hz

5. True or false: A higher sample rate makes the audio file larger.

A False

B True

Match the Pairs

Draw a line connecting each item on the left to its match on the right.

Left column		Right column
144100 Hz		ASample rate is measured in hertz (Hz), not seconds. It should be 44100 Hz, meaning 44,100 samples per second.
2sample_rate = 8000		Bfile_size = sample_rate * seconds
3Sample rate is measured in hertz (Hz), which means measurements per second		CThe sample rate used for CDs: 44,100 sound measurements are taken every second
4Doubling the sample rate doubles the number of measurements, so the file gets about twice as big		Dsample_rate = 44100 Hz
5A higher sample rate makes the file smaller		EA low sample rate: only 8,000 measurements per second, so the sound is lower quality but the file is small
6sample_rate = 44100 seconds		FA higher sample rate takes more measurements each second, so the file gets bigger, not smaller.

Definitions

sample rate

Fill in the Blanks

Complete each line of code by filling in the missing parts.

sampleRate ← _____

IF sampleRate _ 8000 THEN quality ← "higher" END IF

seconds ← 10 totalSamples ← sampleRate * _______

sampleRate ← sampleRate * 2 fileSize ← fileSize _ 2

__ sampleRate = 44100 THEN DISPLAY "CD quality" END IF

Spot the Error

Shade in the letter of the line that contains an error.

Question 1

A sample_rate <- 44100

B unit <- "KB"

C seconds <- 10

D samples <- sample_rate * seconds

Question 2

A cd_rate <- 441

B unit <- "Hz"

C PRINT cd_rate, unit

Question 3

A samples <- 44100 + 2

B PRINT samples

C PRINT "samples for 2 seconds"

Question 4

A low <- 8000

B high <- 44100

C IF high < low THEN

D PRINT "Better quality"

E PRINT "Bigger file"

F END IF

Question 5

A rate <- 22050

B rate <- rate / 2 # double it

C PRINT rate

Answer Key

Sample rate and audio quality

Section A: Quick Questions

A — How many measurements are taken each second

A — Hertz (Hz)

A — The sound is measured 44,100 times every second

B — 44,100 Hz

B — True

Section B: Match the Pairs

44100 Hz → The sample rate used for CDs: 44,100 sound measurements are taken every second

sample_rate = 8000 → A low sample rate: only 8,000 measurements per second, so the sound is lower quality but the file is small

Sample rate is measured in hertz (Hz), which means measurements per second → sample_rate = 44100 Hz

Doubling the sample rate doubles the number of measurements, so the file gets about twice as big → file_size = sample_rate * seconds

A higher sample rate makes the file smaller → A higher sample rate takes more measurements each second, so the file gets bigger, not smaller.

sample_rate = 44100 seconds → Sample rate is measured in hertz (Hz), not seconds. It should be 44100 Hz, meaning 44,100 samples per second.

Section C: Definitions

sample rate — The number of sound measurements taken each second, in hertz (Hz). CDs use 44,100 Hz. A higher rate gives better quality but makes a bigger file.

Section D: Fill in the Blanks

sampleRate ← 44100

IF sampleRate > 8000 THEN
    quality ← "higher"
END IF

seconds ← 10
totalSamples ← sampleRate * seconds

sampleRate ← sampleRate * 2
fileSize ← fileSize * 2

IF sampleRate = 44100 THEN
    DISPLAY "CD quality"
END IF

Section E: Spot the Error

B — Sample rate is measured in hertz (Hz), the number of measurements per second. KB is a unit of file size. Line 2 should be unit <- "Hz".

A — CDs use a sample rate of 44,100 Hz, not 441. Line 1 should be cd_rate <- 44100.

A — The total number of samples is the sample rate multiplied by the seconds, so it should be 44100 * 2 (88,200 samples). Adding gives the wrong answer.

C — 44,100 Hz is higher than 8,000 Hz, so the test should be high > low. A higher sample rate gives better quality and a bigger file.

B — To double the sample rate you multiply by 2. Dividing by 2 would halve it. Line 2 should use rate * 2, giving 44100.