SME
Signals & Systems · Problem type
Data rate and file size of sampled audio
Use this for “CD audio is sampled at 44.1 kHz with 16 bits in 2 channels. Find the bit rate and the size of one minute.”
Diagram
Ch 1 16 + Ch 2 16 = 32 bits, the whole 32-bit word.
Sampling rate fₛ = 44.1 kHz
Bit rate R = 1.411 Mb/s
This is an example. Type your own values to start a new problem.
R (Mb/s)
=
fₛ (kHz)
×
b
×
C
D (MB)
=
R (Mb/s)
×
t (min)
÷
8
F
Frame size
worked out (not typed) · bit
Formulas
F
=
b
×
C
32 = 16 × 2
R
=
fₛ
×
b
×
C
1,411,200 = 44,100 × 16 × 2
D
=
R
×
t
8
10,584,000
=
1,411,200
×
60
8
Formulas are worked in Hz, b/s, s, B. The step-by-step shows the conversions.
Assumptions
•
Uncompressed (PCM): every sample of every channel is stored with b bits.
•
Each sampling instant gives a frame of one sample per channel, b × C bits, fₛ frames a second.
•
Sizes use decimal prefixes: 1 MB = 10⁶ bytes and 1 Mb/s = 10⁶ bits a second; a byte is 8 bits.
Step-by-step
WE KNOW
fₛ = 44.1 kHz, b = 16, C = 2, t = 1 min
FIND
frame size (F), bit rate (R), file size (D)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Hz, b/s, s, B).
fₛ = 44.1 kHz = 44,100 Hz (1 kHz = 1000 Hz)
t = 1 min = 60 s (1 min = 60 s)
Step 2 · Find frame size (F)
Use F = b × C
One
sample
from
each
channel,
b
bits
each,
makes
one
frame.
F
=
16
×
2
F
=
32
bit
Step 3 · Find bit rate (R)
Use R = fₛ × b × C
Samples a second, times bits a sample, times channels: bits a second.
R
=
44,100
×
16
×
2
R
=
705,600
×
2
R
=
1,411,200
b/s
Step 4 · Find file size (D)
Use D = R × t ÷ 8
Bits a second times seconds gives bits; 8 bits make a byte.
D
=
1,411,200
×
60
8
D
=
84,672,000
8
D
=
10,584,000
B
Step 5 · Convert the answers
Change each answer back to the unit you chose.
R = 1,411,200 b/s = 1.4112 Mb/s (1 Mb/s = 10⁶ b/s)
D = 10,584,000 B = 10.584 MB (1 MB = 10⁶ B)
Check
Put the numbers back in. Both sides should match.
32 = 16 × 2 ✓
1,411,200 = 44,100 × 16 × 2 ✓
10,584,000
=
1,411,200
×
60
8
✓
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