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IsaacKuo
2011-Nov-09, 10:00 PM
We use radar to image asteroids using doppler-delay (http://planetary.org/blog/article/00002462/). The idea is to send a radar pulse, and display the return signal in 2 dimensions based on delay (distance) and doppler shift (east-west position). This creates a somewhat distorted 2d image of the asteroid, albeit with the northern and southern hemispheres superimposed on each other.

Has anyone studied the possibility of detecting/imaging exoplanets with this? Obviously, it's too far away for us to beam radar beams, but a pulsar could provide radar pulses down to a resolution on the order of tens of kilometers.

From our perspective, the "pulsar" would actually be a neutron star. The beam needs to hit the exoplanet's orbit, but it does not need to hit us. Therefore, the most likely scenario is that we don't see the neutron star as a pulsar. But as the planet transits across the beam cone, we see radar-like returns from it.

Typically, the planet would transit through the beam cone four times per orbit, but we may only see two of these transits. This depends on whether any of these transits occur when the planet is "near" to us (the equivalent of a new Moon). Because of the "new Moon" effect, one or two transits may be invisible to us.

So, initial detection would look like a faint radio source that winks into existence 2-4 times per cycle. This alone would provide a lot of data about the orbit of the planet and the cone angle of the "pulsar". Each transit, though, would consist of a series of pulses, and those pulses might be used to provide doppler-delay images.

In the ideal situation, the radius of the planet is less than pulsar period times c. If the radius of the planet is greater than this, then the pulse returns would actually overlap--resulting in overlapping of the doppler-delay image with itself and making automated detection of the signal more difficult.

PraedSt
2011-Nov-09, 10:20 PM
So, initial detection would look like a faint radio source that winks into existence 2-4 times per cycle.Where would we point our telescope? At pulsars?

IsaacKuo
2011-Nov-09, 10:33 PM
Where would we point our telescope? At pulsars?

Ideally, we point the telescopes at radio quiet neutron stars. That way, we don't have to deal with glare of the pulsar beam directly hitting us.

Think of the pulsar as a flashlight swinging around. Since we want to see something lit by the flashlight, it's better if we're not blinded by the beam directly shining at us.

antoniseb
2011-Nov-09, 11:10 PM
One part of astronomy is looking at enough targets that some of them are situated to tell you something. I only know of one pulsar with planets, but as you note, it is bright to us. Maybe there is a way to see the reflections of the planets between the pulses, or with a large enough interferometer to separate the signal. I wonder of the SKA will be able to do it.

IsaacKuo
2011-Nov-09, 11:21 PM
Fatal flaw with the original idea--doppler shift only works if the radio beam is narrow spectrum. So doppler-delay won't work. But ISAR could work.

PraedSt
2011-Nov-10, 11:55 AM
Fatal flaw with the original idea--doppler shift only works if the radio beam is narrow spectrum. So doppler-delay won't work. But ISAR could work.If you have time, could you explain how ISAR works in layman's terms? The wikepedia article (http://en.wikipedia.org/wiki/Inverse_synthetic_aperture_radar) is a bit too technical.

IsaacKuo
2011-Nov-10, 01:16 PM
I'll post an explanation later today. It is indeed rather nontrivial.

IsaacKuo
2011-Nov-10, 03:12 PM
I have been trying to think of a simple explanation, and so far, I think the best may be to use an analogy with a CAT scan. It uses a bunch of averaged "slices" to deduce the brightness of each individual pixel.

Imagine viewing the target object from above the north pole, with the radar beam source coming downward from above. Each data point of the return signal represents the radar returns from an entire horizontal slice--the time delay gives the vertical coordinate. As the planet rotates, these slices slice through the planet at different angles. (Imagine keeping the planet fixed, but the radar beam source revolves around the planet.)

In order to calculate the brightness of a particular pixel, you could average the data from every slice which crosses through it. The actual math is more complex than this, but that gives you the basic idea.

Notice that each pixel actually corresponds to two points on the planet--the northern and southern hemispheres are superimposed on top of each other. Unfortunately, the imagery will be overlapping...however...

This simplified explanation assumed that the axis of rotation was perpendicular to the beam source. This means that the slices are parallel to the axis of rotation. Thus, the slices which intersect with a particular point will also intersect with all points along an entire line. But in the general case, the slices won't be parallel to the axis of rotation. Thus, the slices which intersect a particular point will only intersect at that unique point. This is good news! But it does mean that you need to guess the inclination of the slices.

A couple parameters are easy to figure out. The period of the pulsar is simply the short term periodicity of the signal (maybe on the order of a few seconds or less). The rotation period of the planet is a longer term periodicity of the signal (maybe on the order of hours or days). But the slice inclination? I think that would require simply making a bunch of guesses and seeing what results in the best looking images. The resulting images would actually be 3 dimensional, and the best looking image would be the one where the bright voxels fit onto a sphere.

Another slight complication is that the slice angle is not actually perpendicular to the angle to the pulsar. Nor is it perpendicular to the angle to Earth. Instead, each slice is at the angle required for a mirror to reflect the beam from the pulsar toward Earth. This doesn't really matter too much in practical terms, I don't think, but it makes the situation more difficult to visualize.

IsaacKuo
2011-Nov-10, 03:55 PM
Thinking about it more, there are actually four parameters required.

As before, the two easy parameters are pulsar period and planetary rotation period.

The two hard parameters are effective beam angle, and center of rotation (precise phase of the slice which passes through the center of the planet). Those two parameters will need to be guessed to find the best fit with a sphere.

This technique is actually not exactly ISAR. It's something...well, it might be something that I just now made up. It's essentially similar to a CAT scan, but with planar slices instead of linear slices.

PraedSt
2011-Nov-10, 04:26 PM
Thanks, that's much more understandable than the wikipedia article. I've also found that there's plenty of accessible literature on the inverse problem (SAR), and I'm plodding my way through it. Once I get that, I should be able to get this.

IsaacKuo
2011-Nov-10, 04:41 PM
SAR and ISAR are mathematically virtually the same. The "inverse" in ISAR refers to the cause of the "motion". In traditional SAR, the radar platform moves. In inverse SAR, the radar platform is fixed, but the target rotates.

But if you use the rotating frame of reference of the target, ISAR transforms into the same thing as SAR. There's some slight difference due to "moving" along a circular arc rather than a straight line, but the principle is the same.

IsaacKuo
2011-Nov-11, 03:56 PM
Thinking about the conversion from the signal to 3d voxel data...I'm pretty sure that mathematically it doesn't entirely work. The initial data is essentially 2 dimensional. It's a raster scan of a sort of 2d image data. Conversion from a 2d array of pixels to a 3d array of voxels would generally have ambiguities.

So, instead of that I'll take a look at a direct method of calculating the 2d surface image data. We start by assuming the target is spherical, and use the simplifying assumption that the effective beam angle is on the equator.

Consider the first slice of each signal. This comes from a single point on the sphere--the closest point on the equator. We simply use the amplitude directly for this point's pixel data. As the planet rotates, this directly gives us the pixel data for that entire equator.

Now consider the next slice of each signal. This gives us a return from a circular band of the planet, including part of the equator and two points immediately north and south of the equator. Using the equator pixel data, we can subtract away the contribution from the equator, leaving just the contribution from the northern and southern points. Unfortunately, the northern and southern latitudes overlap each other. We can't tell how much of the remaining signal is contributed from the northern pixel and how much is contributed from the southern pixel.

This same principle can be used to calculate the remaining latitudes. Each successive slice gives us signal returns from a larger and larger circle. We use the previously calculated data from the lower latitudes to cancel out everything except for the two points at the new higher latitude.

The end result is an overlapped circular image in polar coordinates--the northern and southern hemispheres overlapped with each other.

PraedSt
2011-Nov-11, 05:32 PM
Fatal flaw with the original idea--doppler shift only works if the radio beam is narrow spectrum. So doppler-delay won't work. But ISAR could work.Just a step back here. Could you not use spectral lines?

IsaacKuo
2011-Nov-14, 04:48 PM
Just a step back here. Could you not use spectral lines?

I'm pretty sure that you could not. There are at least two problems. One is that the spectral lines would be swamped by the broadband radiation around it, I think. The other is that the neutron star itself has an extremely fast spin--the radiation it emits would be doppler shift smeared by a far greater degree than the tiny doppler shift differences in the relatively slow spinning planet.

PraedSt
2011-Nov-14, 08:51 PM
Oh well, worth a try. If I remember correctly, Jupiter and the Sun emit tons of radio waves. You could possibly test your methodology within the Solar System.