The eoPortal article for TESS (Transiting Exoplanet Survey Satellite) gives a fairly in-depth summary of the space telescope design and planned operation. Some further background information on its observation mode and orbit can be found in the video linked below.

Based on the caption (also shown below), I'm guessing that "star noise" is possibly related to shot noise; the counting statistics of the incoming photons or perhaps the photoelectrons thereby produced, and "sky noise" is the cumulative effect of all the other actual light present in the spatially binned area of a given star, including Zodiacal light within the solar system as well as all other sources.

If so, and even if not, I don't really understand the units of $\sigma$ as parts per million per square-root hours, and why sky noise would vary as $I^{-1}$ while star noise expressed in these units varies as $I^{-1/2}$, where $I$ is the intensity of the light from a given star.

below: Figure 12: Top: Expected 1σ photometric precision as a function of stellar apparent magnitude in the IC band. Contributions are from photon-counting noise from the target star and background (zodiacal light and unresolved stars), detector read noise (10 e-), and an assumed 60 ppm of incorrigible noise on hourly timescales. Bottom: The number of pixels in the photometric aperture that optimizes the signal-to-noise ratio (image credit: TESS Team)

Excellent background on TESS: https://youtu.be/mpViVEO-ymc

TESS photometric precision


1 Answer 1


The key formula for signal-to-noise calculations in photometry can be written as something like $${\rm SNR} = \frac{S_{\rm star}}{\sqrt{S_{\rm star} + S_{\rm sky} + \sigma_R^{2}}},$$ where $S_{\rm star}$ is the photons counted from your target star, $S_{\rm sky}$ is the amount of unrelated photons in your photometry aperture (due to sky or other stars) and $\sigma_R$ is the readout noise in your detector.

The numerator is of course your (interesting) signal and the denominator is the noise. The reason the noise terms are added in the way they are is that each term inside the square root is a variance. i.e. The variance of the detected signal from the star is equal to the number of detected photons (a property of Poissonian counting statistics).

Now for more complexity we can express the signals as being photons per unit time (I'll use a small $s$ for these), such that $S_{\rm star} = s_{\rm star} t$, where $t$ is the integration time. $${\rm SNR} = \frac{s_{\rm star} t}{\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}}},$$

Now the thing defined as $\sigma$ in your plots is the inverse of the SNR, plotted on a logarithmic axis. $$\sigma = (s_{\rm star }t)^{-1}\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}} $$ $$\log \sigma = -\log s_{\rm star} -\log t +0.5 \log(s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2})\ \ \ (1)$$ This is being plotted against the $I$ magnitude, which is proportional to $-2.5\log s_{\rm star}$.

So if we now look at each of these contributions individually, we see that the noise due to "shot noise" on the star should indeed be a straight line on your plot. From eqn (1), ignoring other sources of noise $$ \log \sigma = -0.5\log s_{\rm star} -0.5\log t$$ Thus in units of parts per million and scaled for the square root of the exposure time, this will appear as a straight line in a log-log plot with gradient of $-0.5/-2.5=+0.2$.

Now just look at the contribution due to the sky background, from eqn (1), ignoring other sources of noise $$\log \sigma = -\log s_{\rm star} + 0.5 \log s_{\rm sky} -0.5 \log t$$ So the gradient here is -1 in log space (the sky term is just a constant offset in the log-log plot) and +0.4 in terms of (log) ppm/magnitude. Hence the gradient due to the sky noise term is twice as steep.

Finally, looking at the readout noise contribution $$\log \sigma = -\log s_{\rm star} + \log \sigma_{R} -\log t$$ Therefore this has the same gradient as the sky contribution (as you can see in the plot), but the offset in the log-log plot should depend on the exposure time. This makes sense because the brightness at which readout noise will dominate does depend on how long the exposure is. So there is some unspoken (at least in the caption you show) assumption about how long the exposure time is in order to place the contribution from readout noise on the same graph, although the gradient is understood.

  • $\begingroup$ Another "Rob Jeffries special"! I really appreciate this step-by-step walk-through of the basics all the way to the answer. I was surprised to find that indeed the variance of a Poisson distribution is equal to the average rate at any rate, not just in the Gaussian limit. I'll look more closely at the issue you mention in the last paragraph tomorrow (it's late here now). Thank you! $\endgroup$
    – uhoh
    Aug 14, 2017 at 16:16
  • $\begingroup$ OK finally got to it; thanks for the thorough answer! $\endgroup$
    – uhoh
    Sep 3, 2017 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .