EE/Ae 157a
MIDTERM/FINAL
Due Date: November 23, 2015
Problem 1 (20 points)
A telescope is orbiting Mars at 300 km altitude. The telescope lens diameter is 40 cm,
the focal length is 120 cm, and the focal plane is 4 cm wide. The pixels in the detector
are 10 microns on a side. Calculate the swath width, size of the pixels on the ground, and
the number of pixels across the swath assuming a pushbroom design. Also, calculate the
maximum integration time per pixel, and the resulting data volume per day if we acquire
20 images that are 50 km long in the along-track direction.
The instrument is now changed into a spectrometer. The incoming light is dispersed
using a grating such that the different colors are separated spatially in a direction
orthogonal to the pushbroom line array. Assume that the wavelength region 0.4 microns
to 2.4 microns is dispersed over 640 microns in the focal plane. Now assume that we
stack 64 line arrays next to each other to cover the dispersed spectrum. Calculate the
bandpass of each channel, the dwell time per spectral channel and the resulting data
volume if we acquire the same number of images as before.
Solution
Similar to the solution of the third homework, we solve for the swath width using the
imaging geometry:
𝐿
𝑙
𝐻𝑙
= →𝐿=
𝐻 𝑓
𝑓
Here 𝐻 is the orbit altitude, 𝑙 is the size of the focal plane, and 𝑓 is the focal length of the
camera. In this case, 𝐻 = 300 km, 𝑙 = 4 cm, and 𝑓 = 120 cm, which gives
𝐿 = 10 km
A 10 micron detector will then project to
𝑑 = 2.5 m
on the ground. The number of pixels across the swath is
𝑁=
40000
= 4000
10
The maximum dwell time per pixel is the time it takes the footprint of the satellite to
move the distance equal to the size of a pixel, i.e. 5.875 m. For an orbit altitude of 300
km, the satellite velocity is
𝑔 𝑅2
𝑚
𝑣𝑠 = √𝑅 𝑠 +𝐻
= 3410.6 m/s
𝑚
Here I have used the radius of Mars as 3396 km and the gravitational acceleration on
Mars as 3.7 m/s2. Assuming that the imager points to the surface underneath the
spacecraft, the velocity of the footprint is
𝑣 = 𝑣𝑠 𝑅
𝑅𝑚
𝑚 +𝐻
= 3133.75 m/s
For a pixel of 2.5 m, the integration time is
𝜏𝑚𝑎𝑥 = 0.8 𝑚sec
If the images are 50 km long in the along-track direction, it means we acquire a total of
20,000 pixels in the along-track direction, and 4000 pixels in the cross-track direction.
Therefore, we acquire a total of
𝐷𝑎𝑡𝑎 𝑉𝑜𝑙𝑢𝑚𝑒 = 20 × 20,000 × 4000 × 8 = 1.28 × 1010 bits/day
The bandpass of each spectrometer channel is simply 1/64 of the spectral range.
Therefore,
𝛿𝑏 =
2.4 − 0.4
= 0.03125 𝜇𝑚
64
The data volume is simply 64 times that of the single channel. Therefore,
𝐷𝑎𝑡𝑎 𝑉𝑜𝑙𝑢𝑚𝑒 = 64 × 20 × 20,000 × 4000 × 8 = 81.92 × 1010 bits/day
Problem 2 (20 points)
Consider the case of a planet orbiting a star as shown in the figure below. The planet
orbits the star once every 2 earth days.
Planet appears
from behind star
Planet disappears
behind star
Sun-like star
Transit
starts
Transit
ends
The star has the following characteristics
Tstar  6000 K
Rstar  7  108 m
emissivity  0.8
The planet is tidally locked to the star, so that the side facing the star (day side) is hotter
than the side facing away from the star (night side). The planet has the following
characteristics:
Tday  1200 K
Tnight  900 K
R p  89000 km
Orbit radius  47 108 m
emissivity  0.3
We are observing this system from a long distance, so we cannot resolve individual
details. Plot the relative intensity that we would observe as a function of time (i.e. as a
function of the planet position in its orbit) at a wavelength of 16 microns. Can we detect
the presence of the planet? Ignore the star light reflected from the planet. Hint: The
total intensity is the sum of the powers emitted by the sun and the planet. For part of the
time, the planet blocks out a portion of the star, which means we receive less starlight.
For a different portion of the time the planet is not visible while it is behind the sun, so
we only receive starlight. Assume that the planet orbit is in the same plane as the equator
of the star, and that we are observing in the plane of the equator of the star. Let us define
time = 0 when the planet is directly behind the star. Further assume that the temperature
of the planet varies linearly between 1200K at t = 0 to 900K at t = 24 hours, and then
back to 1200K at t= 48 hours.
By the way, this is a real-life problem of an actual star/planet combination that was
observed with the Spitzer space telescope.
Solution:
The radiation from the star when the planet is behind the star is given by:
𝐼𝑆 = 𝜀𝑠 𝑆(𝜆, 𝑇𝑠 )𝜋𝑅𝑠2
When the planet is visible, but not obscuring any part of the star, the total radiation is
𝐼𝑐 = 𝜀𝑠 𝑆(𝜆, 𝑇𝑠 )𝜋𝑅𝑠2 + 𝜀𝑝 𝑆(𝜆, 𝑇𝑝 )𝜋𝑅𝑝2
When the plant obscures part of the planet, the total radiation from both bodies is
𝐼𝑆𝑝 = 𝜀𝑠 𝑆(𝜆, 𝑇𝑠 )(𝜋𝑅𝑠2 − 𝜋𝑅𝑝2 ) + 𝜀𝑝 𝑆(𝜆, 𝑇𝑝 )𝜋𝑅𝑝2
If we want to plot the relative brightness of the star and planet combination, we can
normalize everything to when the plant is behind the star. In that case, the relative
brightness is
𝐵𝑠 =
𝐼𝑆
=1
𝐼𝑆
When both the planet and star are visible and not obscuring each other, the relative
brightness is
𝜀𝑝 𝑆(𝜆, 𝑇𝑝 )𝜋𝑅𝑝2
𝐼𝑐
𝐵𝑐 =
=1+
𝐼𝑆
𝜀𝑠 𝑆(𝜆, 𝑇𝑠 )𝜋𝑅𝑠2
Finally, when the planet obscures part of the star, the relative brightness is
𝐵𝑆𝑝
𝐼𝑆𝑝
𝑅𝑝2
𝜀𝑝 𝑆(𝜆, 𝑇𝑝 )𝜋𝑅𝑝2
=
= (1 − 2 ) +
𝐼𝑆
𝑅𝑠
𝜀𝑠 𝑆(𝜆, 𝑇𝑠 )𝜋𝑅𝑠2
So, we only need to use the linearly varying temperature of the planet in the expressions
above to find the curve of the apparent brightness as a function of time. The result is
shown in the figure below.
1.002
1.000
0.998
Relative Intensity
0.996
0.994
0.992
0.990
0.988
0.986
0.984
0.982
0
30
60
90
120
150
180
210
Planet Rotation Angle
240
270
300
330
360
Problem 3 (10 points)
We are imaging surfaces that could be one or more of the following materials: limestone,
montmorillonite, kaolinite, olivine basalt, and quartz monzonite. Our system images in
five bands as shown in the figure on the next page. Select a set of criteria that would
allow the discrimination and identification of these five materials.
Solution:
Of these five minerals, only limestone has absorption features in both bands 1 and 3.
Hence, we can identify limestone by these two features.
Next, we notice that kaolinite has features in both bands 2 and 4. But so does
montmorillonite. However, kaolinite has a much larger transmission in band 3 than
montmorillonite. Hence, absorption features in both bands 2 and 4 select these two
materials. The final identification is then made using the relative ratio of the response in
band 3 to than in band 2. Alternatively, we could look at the ratio of the responses in
bands 5 and 1 to identify which one of these two minerals is present.
The only two remaining materials are quarts monzonite and olivine basalt. To
discriminate between these two materials, consider the ratios of the responses in bands 1
and 5, and in bands 2 and 3. In both cases, olivine basalt should have ratios near 1.
Quartz monzonite, however, has ratios significantly different from 1.
To summarize, first we look for two lines in bands 1 and 3 to identify limestone. Then
we look for lines in bands 2 and 4 to find montmorillonite and kaolinite. Next, we look
at the ratio of band 1 to band 5 for those areas identified as either montmorillonite or
kaolinite – kaolinite has a ratio near 1, while montmorillonite has a ratio much less than
1. Finally, we look at the ratio of band 1 to band 5 and the ratio of band 2 to band 3 for
those areas not already identified as one of the other three materials. If both ratios are
near 1, the material is likely olivine basalt, and if the ratios are very different from 1, it is
likely quartz monzonite.