SME
Signals & Systems
Sampling
Use this for “A 5 kHz tone is sampled at 8 kHz. What frequency appears?”
Diagram
Alias: |5 − 8 × 1| = 3 kHz
This is an example. Type your own values to start a new problem.
fₐ (kHz)
=
|
f (kHz)
−
fₛ (kHz)
×
round(
f (kHz)
÷
fₛ (kHz)
)|
fN
Nyquist frequency
calculated
fR
Nyquist rate
calculated
Formulas
fN
=
fₛ
2
4
=
8
2
fR
=
2
×
f
10 = 2 × 5
fₐ
=
|
f
−
fₛ
×
round(
f
fₛ
)|
3
=
|5
−
8
×
round(
5
8
)|
Assumptions
•
A pure tone, cos(2πft), sampled fₛ times a second.
•
The alias fₐ is the frequency from 0 to fₛ ÷ 2 that the samples can’t tell from f.
•
Sampling faster than the Nyquist rate 2f keeps the tone; any slower and it shows up as fₐ.
Step-by-step
WE KNOW
f = 5 kHz, fₛ = 8 kHz
FIND
Nyquist frequency (fN), Nyquist rate (fR), alias frequency (fₐ)
Step 1 · Find Nyquist frequency (fN)
Use fN = fₛ ÷ 2
The Nyquist frequency is half the sampling rate: the highest tone the samples can show.
fN
=
8
2
fN = 4 kHz
Step 2 · Find Nyquist rate (fR)
Use fR = 2 × f
To keep the tone, sample faster than twice its frequency.
fR = 2 × 5
fR = 10 kHz
Step 3 · Find alias frequency (fₐ)
Use fₐ = |f − fₛ × round(f ÷ fₛ)|
Take away the nearest whole number of fₛ: what is left is the tone the samples show.
fₐ
=
|5
−
8
×
round(
5
8
)|
round(0.625) = 1
|5 − 8| = |−3|
fₐ = 3 kHz
Check
Put the numbers back in. Both sides should match.
4
=
8
2
✓
10 = 2 × 5 ✓
3
=
|5
−
8
×
round(
5
8
)|
✓
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