Recipe

Find every alias of a frequency at a given sampling rate

Use when: you know a true frequency $f$ and a sampling rate $f_s$ , and need every frequency that produces the identical samples

Goal: the complete alias family $\{kf_s\pm f\}$ and the single apparent frequency any reconstruction actually returns

Taught in Aliasing: two frequencies, one set of samples

Quick recipe

Write the alias formula for this $f$ and $f_s$
Generate the family by sweeping $k$
Identify the apparent frequency
Don't try to break the tie with more data at the same rate

1
Plug the known $f$ and $f_s$ into $f'=kf_s\pm f$ , leaving the integer $k$ free to sweep.

Example

For $f=3$ Hz at $f_s=10$ samples/s: $f'=10k\pm3$ .

Why: this formula falls directly out of cosine's periodicity, the same identity that made two frequencies produce identical samples in the first place.
2
Try $k=0,1,2,\dots$ in turn, keeping only the nonnegative results.

Example

$k=1$ : $10-3=7$ and $10+3=13$ . $k=2$ : $20-3=17$ and $20+3=23$ . The family so far is $\{3,7,13,17,23,\dots\}$ .
- $k=0$ — gives $f$ itself; the negative result $-f$ is discarded, since frequencies here are nonnegative.
- $k\ge1$ — gives two new aliases each time, $kf_s-f$ and $kf_s+f$ .
Why: every integer $k$ names a genuinely different alias, so sweeping $k$ is what builds the whole infinite family.
3
Among every family member, pick the one lying in $[0,f_s/2]$ ; that's what any reconstruction built from these samples actually returns.

Example

Of $\{3,7,13,17,23,\dots\}$ , only $3$ lies in $[0,5]$ , so the apparent frequency is $3$ Hz.

Why: samples alone can't distinguish family members, so reconstruction always settles on this one canonical representative.
4
Recognize that recording longer at an unchanged $f_s$ leaves the entire family exactly as it was.

Example

Sampling a $7$ Hz signal at $f_s=10$ for twice as long still leaves $7=10-3$ an alias of $3$ Hz; the family $\{10k\pm3\}$ never referenced how many samples were taken.
- the sampling rate itself increases — the alias family genuinely recomputes and can change.
- only the recording duration increases at the same rate — the alias family stays identical, with no new information gained.
Why: the formula from Step 1 has no sample count in it at all, so changing only that count changes nothing the formula depends on.

Common mistakes

Forgetting the $\pm$ and computing only $kf_s-f$ , which misses half of every family.
Reporting $k=0$ 's negative result, $-f$ , as if it were a separate valid alias.
Believing a longer recording at the same rate can break the tie between a frequency and its aliases.
Calling the apparent frequency "the true frequency," when the actual source frequency can't be identified from the samples alone.