9.9 Exposure-Time Calculation Examples

In the following sections you will find a set of examples for the two different channels and for different types of sources. The examples were chosen in order to present typical objects for the two channels and also to present interesting cases as they may arise with the use of WFC3.

9.9.1 Example 1: UVIS Imaging of a Faint Point Source

What is the exposure time needed to obtain a signal-to-noise of 10 for a point source of spectral type F0 V (Castelli and Kurucz Model, T=7250 K), normalized to V = 27.5, when using the UVIS F555W filter? Assume a photometry box size of 5 × 5 pixels, detector chip 2 (generally the preferred chip; see Section 6.4.4), average standard zodiacal light and earth shine, and, to start with, 1 frame.

The WFC3 Exposure Time Calculator (ETC) gives an exposure time of 4579 sec = 76.3 min. A single long exposure ~48 min will fit the typical HST orbit, so this observation will require more than 1 orbit. One long exposure per orbit would be a poor choice because many pixels would have bad fluxes in both of the exposures due to cosmic rays (Section 5.4.10). For multi-orbit observations, taking two dithered exposures per orbit (to move bad pixels as well as to allow for cosmic ray removal) is generally sufficient. Recalculating the needed exposure time using 4 frames (more read noise, higher fluxes and flux-dependent noise components to preserve S/N=10) gives 5088 sec, or 1272 sec = 21.2 min per frame. In a typical HST orbit, there is time for two 23 min exposures, so a box pattern (for optimal sampling of the PSF: see Appendix C) will nicely fit two orbits. The background level per frame far exceeds the level where CTE losses become a concern (see Section 6.9), so post-flash is not needed and the ETC does not post an advisory.

Using the pencil-and-paper method, for filter F555W Table 9.1 gives the integral QTdλ/λ as 0.0835 and indicates that the fraction of the star's light included in the 5 × 5 pixel square aperture is 0.78. Table A.1 shows an ABν correction term of 0.03 for filter F555W for a star with an effective temperature of 7,500 K (the closest value to our star's effective temperature 7250 K). The count rate for our V =  25 mag star can then be calculated from the equation

C=2.5\times10^{11}\epsilon_f \Bigg(\int QT\ \frac{d\lambda}{\lambda}\Bigg) \times10^{-0.4(V+AB_v)}~~,


C= 2.5×1011*0.0835*0.78*10-0.4(27.5+0.03) = 0.1584 e s−1

which agrees with the ETC-returned value of 0.158 e s−1. The exposure time can then be found by using the equation

t=\frac{\Sigma^2 [C+N_{pix}(B_{sky}+B_{det})]+ \sqrt{ \Sigma^4[C+N_{pix}( B_{sky}+B_{det} ) ]^2 +4\Sigma^2 C^2 \Bigl[ \frac{N_{pix}}{N_{bin}} N_{read} R^2\Bigr] }}{2C^2}~~,

to where we use
Σ = 10
Bsky = 0.0377 e-/s/pix from Table 9.1
Bdet = 0.0015 e-/s/pix from Table 5.1
R = read noise = 3.1 e- from Table 5.1
Nread = 4
Nbin = 1
P = 0 (no post-flash, so the post-flash noise term is not included)
which gives t = 5265 sec, compared to the ETC-derived value of 5088 sec.

9.9.2 Example 2: UVIS Imaging of a Faint Source with a Faint Sky Background

Calculate the signal to noise obtained in 1200 seconds for a point source of spectral type F0 V (Kurucz & Castelli Model, T=7250 K)), normalized to Johnson V = 23.5 mag, using the UVIS F225W filter, a photometry box size of 5 × 5 pixels, 1 frame, detector chip 2, and with average standard zodiacal light, average standard earth shine and average Air Glow values.

The ETC returns SNR=5.1886, as well as a warning message indicating that the background electrons per pixel (3.22) is less than the recommended threshold of 12 electrons to avoid charge transfer efficiency effects. For a discussion on CTE effects please refer the CTE webpage.

Recalculate the SNR as before, but this time adding 9 post-flash electrons. Now the ETC returns SNR=4.17, and no warning messages. This SNR value is lower than the case without postflash because the postflash electrons contribute to the noise.

To calculate the signal to noise by hand, the procedure is to calculate the count rate using the equation given in Section 9.6.1
Σ = 10 First calculate the count rate for the object using:

C=2.5\times10^{11} \epsilon_f \Bigg(\int QT \frac{d\lambda}{\lambda} \Bigg)\times10^{-0.4(V+AB_v)}~~,

and values of:
V=23.5 mag
QTdλ/λ = 0.0169
εf = 0.71
to obtain C = 0.0789 e-/second.
Now calculate the signal-to-noise ratio, using
Npix = 25
Bsky = 0.0066 e/s/pix from Table 9.1
Bdet = 0.0015 e/s/pix from Table 5.1
R = read noise = 3.1 e per read from Table 5.1
Nread = 1
Nbin = 1
P = 9 post flash electrons

This gives a signal-to-noise ratio of 3.7. The SNR calculated this way is an estimate of the expected signal-to-noise, and results in a more pessimistic value than the ETC. Note that the count rate calculated by the ETC is ~12% greater than the result from the equation given here, which is an approximation. The ETC utilizes the model spectrum to determine the object count rate in a given filter. The equation given here calculates an estimated count rate based on approximations to the true color of a star.

9.9.3 Example 3: IR Imaging of a Faint Extended Source

What is the exposure time needed to obtain a signal-to-noise of 10 for an E0 elliptical galaxy that subtends an area of 1 arcsec2 with an integrated V-magnitude of 26.7, when using the IR F140W filter? Assume a photometry box size of 9 × 9 pixels, and average sky values. The galaxy has a diameter of 1.13 arcsec, a surface brightness of 26.7 mag/arcsec2, and fits within the 9 × 9 pixel box. For simplicity we will assume a redshift of 0.

The WFC3 Exposure Time Calculator (ETC) gives a total exposure time of 1206 sec to obtain this S/N in a single exposure. Although the non-destructive MULTIACCUM sequences on the IR channel can mitigate cosmic rays in a single read sequence, users are encouraged to dither their observations so that there are least 2 read sequences per field, to mitigate hot pixels and resample the point spread function. Re-running the calculation with 2 exposures gives an exposure time of 1320 sec. If we assume (as in Example 1) that we can fit two 1200-second exposures in an orbit, this program fits within a single orbit. Two SPARS50 sequences, with 15 samples (703 sec) per sequence should work well for this program.

Using the pencil-and-paper method, Table 9.2 gives the integral QTdλ/λ as 0.1550. We will assume that the elliptical galaxy resembles an old (10 Gyr) burst of star formation; looking in Table A.2, the ABν correction term is 1.41. We will assume that the 9 × 9 pixel box encloses all of the light for this object. The count rate can then be calculated from the equation

C=2.5\times10^{11} \epsilon_f \Bigg(\int QT \frac{d\lambda}{\lambda} \Bigg)\times10^{-0.4(V+AB_v)}~~,

or 2.5 × 1011*0.1550*1.0*10-0.4(26.7-1.41) = 2.97 e s−1, which is close to the ETC-returned value of 3.24. The exposure time can then be found by using the equation

t=\frac{\Sigma^2 [C+N_{pix}(B_{sky}+B_{det})]+ \sqrt{ \Sigma^4[C+N_{pix}( B_{sky}+B_{det} ) ]^2 +4\Sigma^2 C^2 \Bigl[ \frac{N_{pix}}{N_{bin}} N_{read} R^2\Bigr] }}{2C^2}~~,

to give t = 1364 s, which is close to the ETC-derived value of 1320 s. We have inserted the background rates of Bsky =  1.1694 (Table 9.2), Bdet = 0.05 (Chapter 5), an effective read noise of 12.5 e per read (Chapter 5, assuming we are fitting the MULTIACCUM sequence), and 2 reads.

9.9.4 Example 4: Imaging an H II region in M83 in H-alpha

What is the exposure time needed to obtain a signal-to-noise of 10 for Hα emission in an HII region which has a diameter of 2" and a flux Fλ in Hα of 5 × 10–16 ergs/cm2/s? Suppose that the H II region is in the outskirts of M83 (i.e., insignificant continuum radiation in a narrowband filter). It has a redshift of 0.0017, so Hα appears at approximately 6574 Å. From an inspection of the throughput curves in Appendix A, we find the Hα (F656N) filter cuts off at too short a wavelength, so we elect to use the Wide Hα + [N II] (F657N) filter, which has a system throughput (QT)λ of 25% at 6574 Å.

We use the equation in Section 9.4.3 to estimate the total count rate C for the emission line to be 2.28 × 1012 * 0.25 * 5 × 10–16 * 6574 = 1.87 e/s. The source covers approximately 2012 pixels, and the sky for this filter contributes 0.0029 e/s/pixel (Table 9.1) while the dark rate is 0.0015 e/s/pixel (Table 5.1). Thus the total background rate is 8.85 e/s. Assuming 2 reads, no binning and a read noise of 3.1 e per read (Table 5.1), we find using the same formula as in Section 9.9.1 that the time to reach a S/N of 10 is 1217 sec.

For comparison, using the ETC, specify the source size as 2 arcsec in diameter, use a circular 1 arcsec radius extraction region, enter the emission line flux in ergs/cm2/s/arcsec2 as 1.6 × 10–16 at 6574 Å. Assume a FWHM of 1 Å. The linewidth is not critical since the system throughput of this filter varies slowly with wavelength at 6574 Å, as can be seen in the throughput plot provided by the ETC. Assume average zodiacal light and average earthshine. The ecliptic latitude of the target is 18.4 deg; Figure 9.2 shows that this precludes very high levels of zodiacal light. To start with, assume that post-flash will not be used (0 electrons). The ETC computes that the exposure time for the two frames must sum to 1219 sec (20.3 minutes) to achieve S/N = 10. However, it gives a warning that the electrons per pixel due to background (2.6) is less than the recommended threshold of 12 e/pix to avoid poor charge transfer efficiency (see Section 6.9). A post-flash level ~10 e, to be added to each of the two frames, will be needed to bring the background up to the recommended level. Repeating the ETC calculation with Post Flash = 10 e, you find that you need a total of 1691 sec (28.2 minutes) to make exposures with optimal CTE mitigation over the entire field of view.