SME
Measurements & Instrumentation · Problem type
Sampling rate and aliasing
Use this for “A 700 Hz signal is sampled at 1000 Hz. What frequency shows up, and what rate would avoid aliasing?”
Diagram
Alias: |700 − 1,000 × 1| = 300 Hz
This is an example. Type your own values to start a new problem.
f
Signal frequency
example · Hz
fₛ
Sampling rate
example · Hz
fN
Nyquist frequency
worked out (not typed) · Hz
fₐ
Apparent frequency
worked out (not typed) · Hz
fR
Least sampling rate
calculated · Hz
Formulas
fN
=
fₛ
2
500
=
1,000
2
fR
=
2
×
f
1,400 = 2 × 700
fₐ
=
|
f
−
fₛ
×
round(
f
fₛ
)|
300
=
|700
−
1,000
×
round(
700
1,000
)|
Assumptions
•
One sine frequency, sampled at evenly spaced instants fₛ times a second.
•
The samples can’t tell f from f − k fₛ for any whole k, so they show the nearest to 0.
•
An anti-alias low-pass filter before the converter would remove anything above fₛ ÷ 2.
Step-by-step
WE KNOW
f = 700 Hz, fₛ = 1,000 Hz
FIND
Nyquist frequency (fN), least sampling rate (fR), apparent frequency (fₐ)
Step 1 · Find Nyquist frequency (fN)
Use fN = fₛ ÷ 2
The Nyquist frequency is half the sampling rate: the highest frequency the samples can show.
fN
=
1,000
2
fN = 500 Hz
Step 2 · Find least sampling rate (fR)
Use fR = 2 × f
To keep the signal, sample faster than twice its frequency.
fR = 2 × 700
fR = 1,400 Hz
Step 3 · Find apparent frequency (fₐ)
Use fₐ = |f − fₛ × round(f ÷ fₛ)|
Take away the nearest whole number k of sampling rates: what is left is what the samples show.
fₐ
=
|700
−
1,000
×
round(
700
1,000
)|
k = round(0.7) = 1
fₐ = |700 − 1 × 1,000|
fₐ = |−300|
fₐ = 300 Hz
Note
f
=
700
Hz
is
not
below
fₛ
2
=
500
Hz:
the
record
shows
fₐ,
not
f
.
Sample
faster
than
1,400
Hz.
Check
Put the numbers back in. Both sides should match.
500
=
1,000
2
✓
1,400 = 2 × 700 ✓
300
=
|700
−
1,000
×
round(
700
1,000
)|
✓
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