<p>It's a great question!</p>
<p>Delay-Doppler imaging is a technique to produce a kind of image that represents the topography of a rotating body, usually when it is so far away that there's no way to resolve the object with traditional imaging, radar or optical.</p>
<p>A powerful radar signal is sent from Earth to an asteroid or a planet, and the reflected signal returning to Earth is recorded and analyzed. The &quot;special sauce&quot; that allows the technique to work is the pseudo-random sequence encoded into the modulation of the outgoing signal. Using correlation techniques, one can extract reflected components from a range of different delays due to different distances to each part of the planet, and different Doppler shifts due to different projected (radial) velocities of each part of the planet.</p>
<p>The raw data is pretty messy, but after you record the body rotating a substantial fraction of a full turn, using modeling, you can recover a reconstructed image of the planet's topography.</p>
<p><em>The hard part</em> is that while Doppler shift gives you <em>east-west</em> discrimination since one side moves towards you while the other moves away from you (assuming the rotation axis is &quot;up-down&quot;), there's no easy way to tell the difference between the two hemispheres in the reflected signal.</p>
<p>Thus the <em>north-south</em> ambiguity means that there is no temporal difference between a signal reflected from upper hemisphere and the lower. Say at 50 degrees North and 50 degrees South. The signals would have BOTH the same delay (since they are the same distance away) and the same doppler shift (since they are both rotating at the same speed.</p>
<p>Remember that for a spherical-like object of radius R who's axis is perpendicular to us, the projected line-of-sight (radial) velocity will be:</p>
<p><span class="math-container">$$v_r = \frac{2\pi R}{T} \cos(\phi) \sin(\lambda)$$</span></p>
<p>where <span class="math-container">$T$</span> is the period, <span class="math-container">$\phi$</span> is the latitude and <span class="math-container">$\lambda$</span> is the longitude. It's symmetrical about the equator (cosine behavior) but it varies monotonically east-west (sine).</p>
<p><a href="https://i.sstatic.net/hslw7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hslw7.png" alt="delay-doppler imaging of rotating body" /></a></p>
<p><a href="https://i.sstatic.net/fshk4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fshk4.png" alt="delay-doppler imaging of rotating body" /></a></p>
<hr />
<p>There are solutions to this if the object is large enough though. If you have a huge dish and the object has an angular size on the same scale as the beam width of your dish, you can record data with the beam illuminating first one hemisphere more strongly, then the other. That provides some additional information, and with subtraction and sufficient modeling one can overcome the North-South ambiguity.</p>
<p>Of course, if the body's rotation axis is oriented obliquely towards or away from us, then the symmetry is broken and the ambiguity can also be resolved in some cases.</p>
<p>Related and possibly helpful:</p>
<ul>
<li><p>Space SE: <a href="https://space.stackexchange.com/q/22493/12102">Why do radar maps of the surface of Venus have missing slices?</a></p>
</li>
<li><p>Emily Lakdawalla's <a href="http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html" rel="nofollow noreferrer">How radio telescopes get &quot;images&quot; of asteroids</a></p>
</li>
</ul>
<hr />
<p><strong>Left:</strong> A very slow GIF from Emily Lakdawalla's <a href="http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html" rel="nofollow noreferrer">How radio telescopes get &quot;images&quot; of asteroids</a></p>
<p><strong>Right:</strong> A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. <a href="https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif" rel="nofollow noreferrer">Here is an animation</a> that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top.&quot; From <a href="http://www.skyandtelescope.com/astronomy-news/asteroid-florence-has-two-moons/" rel="nofollow noreferrer">here</a>. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question <a href="https://astronomy.stackexchange.com/q/22480/7982">What is the physical geometry of this apparent &quot;eclipse&quot; of a tiny moon of Asteroid Florence?</a></p>
<p><strong>Note:</strong> These two images do not necessarily have the same orientation.</p>
<p><a href="https://i.sstatic.net/sMeV5.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sMeV5.gif" alt="Delay-Doppler Imaging Emily Lakdawalla" /></a> <a href="https://i.sstatic.net/Pn291.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Pn291.gif" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2019-04-08T03:24:40.467Z
" there's no easy way to tell the difference between the two hemispheres in the reflected signal. Thus the north-south ambiguity" - ok, but why tho?
MinYoung Kim
CC BY-SA 4.0
2019-04-08T03:48:14.363Z
@MinYoungKim Oh, I see what you mean, thanks! I didn't really say it clearly. I'll make an edit and add a drawing, give me about 15 minutes...
uhoh
CC BY-SA 4.0
2019-04-08T04:19:58.403Z
appreciate the effort!
MinYoung Kim
CC BY-SA 4.0
2019-04-08T04:49:32.817Z
@MinYoungKim took longer than I thought (it always does!) Let me know if this is better.
uhoh
CC BY-SA 4.0
2019-04-27T07:01:26.063Z
@MinYoungKim if there's anything more I can clarify, feel free to let me know!
uhoh
CC BY-SA 4.0
2019-04-07T23:48:13.937Z
2022-10-15T00:56:37.017Z
2022-10-15T00:56:37.017Z
uhoh
<p>I'm having a hard time understanding why there would be this ambiguity. Also, why would this be not the case if the planet was viewed pole on?</p>
<p>Edit: &quot;North South ambiguity&quot; is a term used in a paper that seemed most concise but I guess not. When using radar to doppler image a planetary surface which happens to be viewed equator on, we have no way of distinguishing whether the surface feature read in return signal belongs to the northern or southern hemisphere. What causes this?</p>

CC BY-SA 4.0
2019-04-07T01:22:26.237Z
Welcome on the Space SE! Note, your question is flagged an unclear. Please elaborate it, to avoid its closure (and then, deletion).
peterh
CC BY-SA 4.0
2019-04-07T23:26:26.890Z
Folks just because you don't recognize a term doesn't mean you have to close the question.
uhoh
CC BY-SA 4.0
2019-04-07T23:55:02.630Z
I'd say if a term is obscure, the onus is on the asker to define it. It's a good reminder that there are both professionals and amateurs (like me) on this site, and just because a term has appeared in a recent scientific publication doesn't mean that we'd be familiar with it. I think the last edit has made things sufficiently clearer though, and the question should remain open.
Ingolifs
CC BY-SA 4.0
2019-04-08T00:02:21.060Z
@Ingolifs The comment was helpful, the OP responded with an edit. The three close votes were unnecessary. Insta-closing is unhelpful because it blocks answering, thus my comment "... just because you don't recognize a term doesn't mean you have to close the question."
uhoh
CC BY-SA 4.0
2019-04-08T07:27:14.720Z
First we should made clear if N and S are defined based on rotation as seen" by us". Then @uhoh answer.
Alchimista
CC BY-SA 4.0
2019-04-08T07:29:37.917Z
@Alchimista It doesn't matter. These terms only serve to indicate that rotation happens, and one can define an "instantaneous equator" that doesn't move much during the measurement
uhoh
CC BY-SA 4.0
2019-04-08T07:31:35.357Z
@uhoh I am perfectly fine (for what it might count) with your answer and up voted.
Alchimista
CC BY-SA 4.0
2019-04-08T10:04:00.243Z
"used in a paper" - there's more than one out there. Could you cite the one you read?
Mike G
CC BY-SA 4.0
2019-04-09T00:52:50.767Z
I was just searching for a term to describe this, the this paper abstract seemed to be what I was looking for. https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JB095iB07p10947
MinYoung Kim
<observational-astronomy><planet><radio-astronomy><radar>
What causes "North-South ambiguity" when doppler radar imaging a planet surface equator?
MinYoung Kim
# What causes "North-South ambiguity" when doppler radar imaging a planet surface equator?

(Asked by: username_0 on 2019-04-06)

I'm having a hard time understanding why there would be this ambiguity. Also, why would this be not the case if the planet was viewed pole on?

Edit: "North South ambiguity" is a term used in a paper that seemed most concise but I guess not. When using radar to doppler image a planetary surface which happens to be viewed equator on, we have no way of distinguishing whether the surface feature read in return signal belongs to the northern or southern hemisphere. What causes this?

**Tags:** observational-astronomy, planet, radio-astronomy, radar
> **Comments:**
> **username_1**: Welcome on the Space SE! Note, your question is flagged an unclear. Please elaborate it, to avoid its closure (and then, deletion).  
> **username_2**: Folks just because you don't recognize a term doesn't mean you have to close the question.  
> **username_3**: I'd say if a term is obscure, the onus is on the asker to define it. It's a good reminder that there are both professionals and amateurs (like me) on this site, and just because a term has appeared in a recent scientific publication doesn't mean that we'd be familiar with it. I think the last edit has made things sufficiently clearer though, and the question should remain open.  
> **username_2**: @Ingolifs The comment was helpful, the OP responded with an edit. The three close votes were unnecessary. Insta-closing is unhelpful because it blocks answering, thus my comment "... just because you don't recognize a term doesn't mean you have to close the question."  
> **username_4**: First we should made clear if N and S are defined based on rotation as seen" by us". Then @uhoh answer.  
> **username_2**: @Alchimista It doesn't matter. These terms only serve to indicate that rotation happens, and one can define an "instantaneous equator" that doesn't move much during the measurement  
> **username_4**: @uhoh I am perfectly fine (for what it might count) with your answer and up voted.  
> **username_5**: "used in a paper" - there's more than one out there. Could you cite the one you read?  
> **username_0**: I was just searching for a term to describe this, the this paper abstract seemed to be what I was looking for. https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JB095iB07p10947  

---

## 1 answers

### Answer by username_2 on 2019-04-07. Score: 8

It's a great question!

Delay-Doppler imaging is a technique to produce a kind of image that represents the topography of a rotating body, usually when it is so far away that there's no way to resolve the object with traditional imaging, radar or optical.

A powerful radar signal is sent from Earth to an asteroid or a planet, and the reflected signal returning to Earth is recorded and analyzed. The "special sauce" that allows the technique to work is the pseudo-random sequence encoded into the modulation of the outgoing signal. Using correlation techniques, one can extract reflected components from a range of different delays due to different distances to each part of the planet, and different Doppler shifts due to different projected (radial) velocities of each part of the planet.

The raw data is pretty messy, but after you record the body rotating a substantial fraction of a full turn, using modeling, you can recover a reconstructed image of the planet's topography.

*The hard part* is that while Doppler shift gives you *east-west* discrimination since one side moves towards you while the other moves away from you (assuming the rotation axis is "up-down"), there's no easy way to tell the difference between the two hemispheres in the reflected signal.

Thus the *north-south* ambiguity means that there is no temporal difference between a signal reflected from upper hemisphere and the lower. Say at 50 degrees North and 50 degrees South. The signals would have BOTH the same delay (since they are the same distance away) and the same doppler shift (since they are both rotating at the same speed.

Remember that for a spherical-like object of radius R who's axis is perpendicular to us, the projected line-of-sight (radial) velocity will be:

$$v\_r = \frac{2\pi R}{T} \cos(\phi) \sin(\lambda)$$

where $T$ is the period, $\phi$ is the latitude and $\lambda$ is the longitude. It's symmetrical about the equator (cosine behavior) but it varies monotonically east-west (sine).

[![delay-doppler imaging of rotating body](https://i.sstatic.net/hslw7.png)](https://i.sstatic.net/hslw7.png)

[![delay-doppler imaging of rotating body](https://i.sstatic.net/fshk4.png)](https://i.sstatic.net/fshk4.png)

---

There are solutions to this if the object is large enough though. If you have a huge dish and the object has an angular size on the same scale as the beam width of your dish, you can record data with the beam illuminating first one hemisphere more strongly, then the other. That provides some additional information, and with subtraction and sufficient modeling one can overcome the North-South ambiguity.

Of course, if the body's rotation axis is oriented obliquely towards or away from us, then the symmetry is broken and the ambiguity can also be resolved in some cases.

Related and possibly helpful:

* Space SE: [Why do radar maps of the surface of Venus have missing slices?](https://space.stackexchange.com/q/22493/12102)
* Emily Lakdawalla's [How radio telescopes get "images" of asteroids](http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html)

---

**Left:** A very slow GIF from Emily Lakdawalla's [How radio telescopes get "images" of asteroids](http://www.planetary.org/blogs/emily-lakdawalla/2010/2462.html)

**Right:** A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. [Here is an animation](https://cneos.jpl.nasa.gov/images/news/florence.p5us.1Hz.s382.sep01.gif) that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top." From [here](http://www.skyandtelescope.com/astronomy-news/asteroid-florence-has-two-moons/). NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question [What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?](https://astronomy.stackexchange.com/q/22480/7982)

**Note:** These two images do not necessarily have the same orientation.

[![Delay-Doppler Imaging Emily Lakdawalla](https://i.sstatic.net/sMeV5.gif)](https://i.sstatic.net/sMeV5.gif) [![enter image description here](https://i.sstatic.net/Pn291.gif)](https://i.sstatic.net/Pn291.gif)

> **Comments:**
> **username_0**: " there's no easy way to tell the difference between the two hemispheres in the reflected signal. Thus the north-south ambiguity" - ok, but why tho?  
> **username_2**: @MinYoungKim Oh, I see what you mean, thanks! I didn't really say it clearly. I'll make an edit and add a drawing, give me about 15 minutes...  
> **username_0**: appreciate the effort!  
> **username_2**: @MinYoungKim took longer than I thought (it always does!) Let me know if this is better.  
> **username_2**: @MinYoungKim if there's anything more I can clarify, feel free to let me know!  

---

