← Years 7–8 Data & Information

Sample rate and audio quality

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.

Understanding Sample rate and audio quality in Years 7–8 Data & Information

What it does: The Hz unit tells us how many samples are taken each second, here 44,100. File size grows in step with the sample rate, because more samples need more storage. Hertz means 'per second'. 1 Hz is one sample each second. Pixels are used for images, not sound. More samples per second capture the sound wave more closely, so the recording sounds closer to the original. Digital systems store everything as numbers. Each sound measurement becomes a binary number.

Common mistakes: 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". CDs use a sample rate of 44,100 Hz, not 441. Line 1 should be cd_rate <- 44100.

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Learning
Duo
What does the sample rate of a sound recording measure?
How loud the sound isHow many measurements are taken each second
Matches
A higher sample rate makes the file smaller
↓
A higher sample rate takes more measurements each second, so the file gets bigger, not smaller.
Applying
Meanings
Term
sample rate
Meaning
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.
Practice
sample_rate ← 44100
Testing
Gaps
sampleRate ←
Check Answer
Spots
1sample_rate <- 44100
2unit <- "KB"

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Quick Questions

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
Sample rate is the number of sound measurements (samples) taken every second, measured in hertz (Hz).
2. Which unit is used to measure sample rate?
A. Hertz (Hz) ✓
B. Pixels (px)
Hertz means 'per second'. 1 Hz is one sample each second. Pixels are used for images, not sound.
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
44,100 Hz means 44,100 samples are taken in each second of sound. It says nothing about how long the song is.
4. Which sample rate will usually give better sound quality?
A. 8,000 Hz
B. 44,100 Hz ✓
More samples per second capture the sound wave more closely, so the recording sounds closer to the original.
5. True or false: A higher sample rate makes the audio file larger.
A. False
B. True ✓
More samples each second means more numbers to store in binary, so the file gets bigger.

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.

Key Term

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.

Fill in the Blanks

1. sampleRate ← 44100
Hint: Type the number of samples per second used by CD audio, as digits only with no comma.
2. IF sampleRate > 8000 THEN quality ← "higher" END IF
Hint: Type the comparison operator that checks if the rate is greater than a number.
3. seconds ← 10 totalSamples ← sampleRate * seconds
Hint: Type the variable that holds how long the recording lasts.
4. sampleRate ← sampleRate * 2 fileSize ← fileSize * 2
Hint: Type the operator that multiplies, so a higher rate makes a bigger file.
5. IF sampleRate = 44100 THEN DISPLAY "CD quality" END IF
Hint: Type the keyword that starts a decision in pseudocode.

Spot the Error

Question 1
A sample_rate <- 44100
B unit <- "KB"
C seconds <- 10
D samples <- sample_rate * seconds
Line 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".
Question 2
A cd_rate <- 441
B unit <- "Hz"
C PRINT cd_rate, unit
Line A — CDs use a sample rate of 44,100 Hz, not 441. Line 1 should be cd_rate <- 44100.
Question 3
A samples <- 44100 + 2
B PRINT samples
C PRINT "samples for 2 seconds"
Line 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.
Question 4
A low <- 8000
B high <- 44100
C IF high < low THEN
D PRINT "Better quality"
E PRINT "Bigger file"
F END IF
Line 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.
Question 5
A rate <- 22050
B rate <- rate / 2 # double it
C PRINT rate
Line B — To double the sample rate you multiply by 2. Dividing by 2 would halve it. Line 2 should use rate * 2, giving 44100.

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