Astronomy · Finding change in the sky

Sampling intervals, missed events and aliasing

Compare two positive periodic signals with identical finite-exposure averages at six regular centres, and test what an extra observation or a completely unobserved gap can establish.

Continue in the approved chapter order. Necessary vocabulary, quantities and assumptions are explained in the lesson.

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After reading, you can

Describe cadence using actual times, exposure durations, gaps and total span.
Reproduce a finite-exposure alias example with its declared averaging response.
Distinguish a period-search statistic from a physical interpretation and a conditional false-alarm probability.

The observing window decides which changes are sampled

Cadence describes how observations are distributed in time, including exposure durations, intervals, gaps and the total span. Two programmes with the same mean spacing can have different opportunities to detect a brief event or distinguish repeating signals.

A finite exposure averages a varying source through its temporal response. Rapid variation can be reduced or reshaped in that average, while long gaps leave intervals unmeasured. The plotted centres alone do not fully describe this sampling.

Aliasing occurs when different underlying signals produce indistinguishable or closely competing sampled patterns under a particular observing window. Adding measurements can distinguish some alternatives, but the result depends on where, how and with what uncertainty those measurements are made.

The ideal uniform-sampling theorem assumes an appropriately band-limited signal and exact infinite uniform samples. It does not supply a universal Nyquist limit from half the mean cadence of every finite irregular astronomical data set. VanderPlas discusses why such shortcuts can fail.

Declare the average before comparing sampled points

For an original example, let t0 be an arbitrary positive time unit and r0 an arbitrary positive source-rate unit. Compare frequencies f₁ = 0.2/t0 and f₂ = 0.8/t0, with periods 5t0 and 1.25t0. Every exposure has uniform duration Δ = 0.2t0 about its stated centre.

Define intrinsic rates r_i(t)/r0 = 10 + a_i cos(2πf_i t), where a_i = 2/sinc(πf_iΔ). In this lesson sinc(z) means sin z/z for a radian argument, with sinc(0) = 1. This convention differs from the normalized sinc used in the Foundation interpolation formula.

The supplied amplitudes are approximately a₁ = 2.005273503 and a₂ = 2.086771110. Both intrinsic rates remain positive at every time. Define them over all times needed by the exposures, including the first interval from −0.1t0 to + 0.1t0.

Uniformly averaging each intrinsic rate over its exposure gives μ_i(t)/r0 = 10 + a_i sinc(πf_iΔ)cos(2πf_i t) = 10 + 2cos(2πf_i t). Thus the following curves and table describe expected exposure-averaged rates as functions of centre time, not instantaneous observations.

The same sampled averages can have two explanations

At centres t = n × t0 for integers n = 0 through 5, the averaged cosine arguments differ by the regular-sampling alias symmetry. The two means agree: 12, approximately 10.618033989, 8.381966011, 8.381966011, 10.618033989 and 12, in units r0.

An added exposure centred at 0.5t0, with the same duration and response, gives approximately 11.618033989r0 for the first signal and 8.381966011r0 for the second. That off-grid centre distinguishes this particular pair in the exact expected-rate calculation.

The extra point does not prove unique recovery among every possible signal, nor does it supply a noise significance. Finite data can remain compatible with other histories. Uneven sampling changes the window without automatically eliminating all aliases.

The amplitude correction is essential to this finite-duration construction. Using equal intrinsic amplitudes while claiming the same exposure averages would omit the different averaging attenuation of the two frequencies.

Original expected uniform-exposure means μ_i/r0 for Δ = 0.2t0 and the declared positive intrinsic rates. First six centres are regular; the final row is an added off-grid exposure. These are exact model expectations rounded for display, not noisy observations or recovered periods.
Exposure centre / t0	μ₁/r0, f₁t0 = 0.2	μ₂/r0, f₂t0 = 0.8
0	12	12
1	10.618033989	10.618033989
2	8.381966011	8.381966011
3	8.381966011	8.381966011
4	10.618033989	10.618033989
5	12	12
0.5, added	11.618033989	8.381966011

Finite exposures can share samples

Scroll sideways to see the full diagram. Its explanation is below.

Original synthetic finite-exposure example, not observations or a physical variability diagnosis. t0 is an arbitrary positive time unit and r0 is an arbitrary positive source-rate unit. Every exposure has uniform temporal weighting and duration Delta = 0.2 t0. The shared horizontal linear axis is exposure-centre time t/t0 from 0 to 5. The shared vertical linear axis is expected exposure-averaged source rate mu/r0 from 0 to 13. Both axes start at zero. The blue solid curve has f1 t0 = 1/5 and P1/t0 = 5. The orange dashed curve has f2 t0 = 4/5 and P2/t0 = 5/4. The curves are mu_i(t)/r0 = 10 + 2 cos(2 pi f_i t), functions of exposure centre, not the instantaneous intrinsic rates. The instantaneous inputs on all times are r_i(u)/r0 = 10 + a_i cos(2 pi f_i u), with a_i = 2/sinc(pi f_i Delta). Here sinc(z) = sin(z)/z for radian argument z, with sinc(0) = 1; this is the unnormalized sinc convention. The intrinsic amplitudes are a1 = 2.005273502803515 and a2 = 2.086771109786030. Dividing the integral of each intrinsic input over [t - Delta/2, t + Delta/2] by Delta gives the displayed common averaged amplitude of 2. The first regular exposure is centred at zero and extends from -0.1 t0 to +0.1 t0; both inputs are explicitly defined at those negative times and remain positive. The six regular exposure centres 0, 1, 2, 3, 4 and 5 in units of t0 give identical expected averages for the pair: 12; 10.618033989; 8.381966011; 8.381966011; 10.618033989; and 12, respectively. At each shared point a filled blue circle is drawn inside an orange hollow square at exactly the same coordinates; no artificial horizontal shift is used. At the one additional centre t/t0 = 0.5, a larger hollow blue circle marks mu1/r0 = 11.618033989 and a larger hollow orange square marks mu2/r0 = 8.381966011. These hollow points distinguish this pair. The table contains all six regular rows and this one additional row. Decimal precision reports model arithmetic, not observational uncertainty. No noise, period recovery, false-alarm probability or physical cause is supplied. This particular extra exposure separates these two expected-rate models; irregular sampling is not automatically free of all aliases. Finite rectangular integration produces sinc attenuation with zeros and sidelobes rather than a sharp frequency cutoff.

Check the exposure intervals, not only their centres

For a separate missed-event test, keep the six exposure intervals [n − 0.1, n + 0.1]t0 for n = 0 through 5. Add any positive extra rate confined entirely to the interval from 1.3t0 to 1.6t0, with zero extra rate outside it.

The exposure around centre 1t0 ends at 1.1t0, and the exposure around 2t0 begins at 1.9t0. The entire extra event lies between them. It overlaps none of the six integration intervals and therefore contributes no additional signal to those exposures.

A non-detection in these particular measurements does not rule out the event in the unobserved gap. No event amplitude or physical cause has been supplied. A longer exposure or another centre would need its own overlap calculation.

Finite averaging also has a frequency response with sinc attenuation, zeros and sidelobes. It is not a sharp boundary above which every higher frequency vanishes. The explicit integral, rather than a blanket cutoff slogan, defines this example.

A peak is not a physical cause or a guaranteed period

A periodogram summarizes agreement with a family of periodic models across trial frequencies. Its interpretation depends on the model, offset treatment, observing times and measurement uncertainties, including any correlations.

A false-alarm probability for a maximum peak describes a tail probability under a specified noise null and search procedure. It is not the probability that a stellar mechanism is correct or that the selected peak has the right alias. A single-frequency test and a search over many frequencies pose different statistical questions.

The Astropy interpretation documentation and VanderPlas review explain these distinctions. Their illustrative data and numerical significance examples are not reused here. No false-alarm probability has been calculated for the noiseless invented table.

Useful follow-up can target the times or observables where competing models disagree. The next lesson turns to image comparison, where calibration and image response can create apparent changes even before a physical light curve or period is inferred.

Worked examples

Test the supplied pair at a new centre

What changes when the same-duration exposure is centred at 0.5t0 instead of an integer multiple of t0?

The first expected average is 10 + 2cos(2π × 0.2 × 0.5) ≈ 11.618033989 in units r0.
The second is 10 + 2cos(2π × 0.8 × 0.5) ≈ 8.381966011 in units r0.
The added expected means differ even though all six original integer-centre means agree.

This added centre distinguishes the two stipulated signals, without proving unique recovery among all possible histories.

Test a missed event against finite intervals

Does an extra positive rate confined to 1.3–1.6t0 alter any of the six regular exposures?

The adjacent exposure intervals end at 1.1t0 and begin at 1.9t0.
The extra event is entirely inside that gap.
Its overlap duration with every supplied exposure is zero, so its additional integrated contribution is zero.

No. Those measurements cannot exclude this wholly unobserved event.

Try the reasoning

Does half the mean irregular cadence define a universal recoverable-frequency boundary?
Show the answer and reasoning
No.
The actual observation window, finite exposures and signal assumptions matter.
Are the table entries instantaneous intrinsic rates at the plotted centre times?
Show the answer and reasoning
No.
They are expected rates averaged through the declared finite uniform exposures.
Does rectangular exposure averaging erase every frequency above one sharp cutoff?
Show the answer and reasoning
No.
Its sinc response has attenuation, zeros and sidelobes rather than a universal step cutoff.
Is a period-search false-alarm probability the probability that a proposed physical cause or alias is correct?
Show the answer and reasoning
No.
It is defined under a particular noise null and search, which is a different conditional question.

What to carry forward

Sampling includes finite intervals, gaps and the full observing window.
Different positive signals can produce identical sampled exposure averages.
An extra observation can test specific alternatives while significance and physical interpretation remain separate.

References

Teaching resources and research supporting this lesson’s explanations and examples.