<p>In the case of a ground looking radar, the cross-polarized signals HV and VH generally increase with surface roughness. For example, dense vegetation is one of the more depolarizing surface types because the polarization orientation of the emitted beam is scrambled over the course of multiple scattering interactions within the canopy before returning to the detector. Smoother surfaces can be thought of as having fewer (often just one) interactions and the corresponding signals are thus reflected to the detector predominately in their initial orientation (i.e. as HH or VV).</p>

<p>The cross-polarized signal also increases as scatters suspended in the atmosphere become less spherical in shape. This is particularly applicable in the interpretation of meteorological radar observations where it is used to estimate raindrop size, distinguish between solid and liquid precipitation and differentiate biological scatterers (e.g. flocks of birds) from hydrometers. </p>

<p>For a fixed target, the return signals HV and VH will have equal intensity. This is a direct result of the <a href="https://en.wikipedia.org/wiki/Helmholtz_reciprocity" rel="nofollow noreferrer">principle of reciprocity</a>.</p>

CC BY-SA 4.0
2020-03-21T14:04:13.660Z
Thank you for this answer, it addresses 95% of my question already. Your explanation makes a lot of sense and I feel I get the idea behind cross-polarized signals now. One thing is yet remaining, im still pondering the VH & HV differences. Your answer seem to indicate that mostly HV and VH will simply be equal. However, my understanding is that a smooth tower has large HH, a rough tower has large HV. My intuition is that conversly a tower should have little VV and only slighter larger VH depending on its roughness. Hence HV and VH would not match for a tower. Would you agree?
ckrk
CC BY-SA 4.0
2020-03-21T18:51:54.923Z
@ckrk Take a vertical metal rod. Incoming V will efficiently drive a (vertical) current in this rod, causing the rod to re-radiate exclusively to V, producing no VH. Tilt the same rod several degrees. The titled rod will absorb incoming V almost as efficiently but re-radiate in a direction with a small horizontal component, resulting in non-zero VH. Now input H. The rod no longer absorbs the incoming radiation efficiently but  what is absorbed will be re-radiated close to vertical, predominately into HV. More complicated cases are hard to intuit but reciprocity always guarantees that HV=VH.
Reed Espinosa
CC BY-SA 4.0
2020-03-21T00:10:29.143Z
2020-03-21T04:06:00.697Z
2020-03-21T04:06:00.697Z
Reed Espinosa
<p>Radar is a common tool in remote sensing, it is a so-called active instrument, hence it emits and receives energy.  Modern Syntethic Aperture Radar (SAR) systems are capable of both transmitting and receiving polarized waves. The most common polarizations are horizontal <span class="math-container">$(H)$</span> and vertical <span class="math-container">$(V)$</span> polarization, although other ones exist. Using H and V leads to 4 possible combinations of emitting and receiving signals: </p>

<p><span class="math-container">$$HH, VV, HV, VH,$$</span></p>

<p>which is read as transmitted first, received later (i.e. <span class="math-container">$HV$</span> means transmitting horizontally polarized waves and receiving only vertical polarized waves).</p>

<p>I can relatively easily imagine the use of the <span class="math-container">$HH$</span> and <span class="math-container">$VV$</span> copolarized waves. They should be easier reflected by object that have a longish geometric structure. Large <span class="math-container">$VV$</span> and small <span class="math-container">$HH$</span> would probably indicate tower-like structures and small <span class="math-container">$VV$</span> and large <span class="math-container">$HH$</span> should indicate bridge-like structures.</p>

<p>However it is not obvious to me what structure would reflect a cross-polarized wave <span class="math-container">$HV$</span> or <span class="math-container">$VH$</span> stronger than <span class="math-container">$HH$</span> or <span class="math-container">$VV$</span>? Furthermore, what is the difference between an object that reflects <span class="math-container">$VH$</span> and an object that reflects <span class="math-container">$HV$</span>?</p>

CC BY-SA 4.0
2020-03-10T13:19:08.483Z
cool question!!
uhoh
CC BY-SA 4.0
2020-03-10T16:17:37.660Z
@uhoh thank you, i think its not that obvious! My collaborator seemed to use it in a very empirical fasion (ie trial and error of all possibilities) and couldnt give an intuition. I even checked a remote sensing textbook and it did not answer the question, despite mentioning the existence of the HV and VH polarizations....
ckrk
<remote-sensing><earth-observation><satellites><radar>
What is the use of HV or VH polarization in Radar?
ckrk
# What is the use of HV or VH polarization in Radar?

(Asked by: username_0 on 2020-03-09)

Radar is a common tool in remote sensing, it is a so-called active instrument, hence it emits and receives energy. Modern Syntethic Aperture Radar (SAR) systems are capable of both transmitting and receiving polarized waves. The most common polarizations are horizontal $(H)$ and vertical $(V)$ polarization, although other ones exist. Using H and V leads to 4 possible combinations of emitting and receiving signals:

$$HH, VV, HV, VH,$$

which is read as transmitted first, received later (i.e. $HV$ means transmitting horizontally polarized waves and receiving only vertical polarized waves).

I can relatively easily imagine the use of the $HH$ and $VV$ copolarized waves. They should be easier reflected by object that have a longish geometric structure. Large $VV$ and small $HH$ would probably indicate tower-like structures and small $VV$ and large $HH$ should indicate bridge-like structures.

However it is not obvious to me what structure would reflect a cross-polarized wave $HV$ or $VH$ stronger than $HH$ or $VV$? Furthermore, what is the difference between an object that reflects $VH$ and an object that reflects $HV$?

**Tags:** remote-sensing, earth-observation, satellites, radar

> **Comments:**
> **username_1**: cool question!!  
> **username_0**: @username_1 thank you, i think its not that obvious! My collaborator seemed to use it in a very empirical fasion (ie trial and error of all possibilities) and couldnt give an intuition. I even checked a remote sensing textbook and it did not answer the question, despite mentioning the existence of the HV and VH polarizations....  

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## 1 answers

### Answer by username_2 on 2020-03-21. Score: 4 | Accepted answer

In the case of a ground looking radar, the cross-polarized signals HV and VH generally increase with surface roughness. For example, dense vegetation is one of the more depolarizing surface types because the polarization orientation of the emitted beam is scrambled over the course of multiple scattering interactions within the canopy before returning to the detector. Smoother surfaces can be thought of as having fewer (often just one) interactions and the corresponding signals are thus reflected to the detector predominately in their initial orientation (i.e. as HH or VV).

The cross-polarized signal also increases as scatters suspended in the atmosphere become less spherical in shape. This is particularly applicable in the interpretation of meteorological radar observations where it is used to estimate raindrop size, distinguish between solid and liquid precipitation and differentiate biological scatterers (e.g. flocks of birds) from hydrometers.

For a fixed target, the return signals HV and VH will have equal intensity. This is a direct result of the [principle of reciprocity](https://en.wikipedia.org/wiki/Helmholtz_reciprocity).

> **Comments:**
> **username_0**: Thank you for this answer, it addresses 95% of my question already. Your explanation makes a lot of sense and I feel I get the idea behind cross-polarized signals now. One thing is yet remaining, im still pondering the VH & HV differences. Your answer seem to indicate that mostly HV and VH will simply be equal. However, my understanding is that a smooth tower has large HH, a rough tower has large HV. My intuition is that conversly a tower should have little VV and only slighter larger VH depending on its roughness. Hence HV and VH would not match for a tower. Would you agree?  
> **username_2**: @username_0 Take a vertical metal rod. Incoming V will efficiently drive a (vertical) current in this rod, causing the rod to re-radiate exclusively to V, producing no VH. Tilt the same rod several degrees. The titled rod will absorb incoming V almost as efficiently but re-radiate in a direction with a small horizontal component, resulting in non-zero VH. Now input H. The rod no longer absorbs the incoming radiation efficiently but  what is absorbed will be re-radiated close to vertical, predominately into HV. More complicated cases are hard to intuit but reciprocity always guarantees that HV=VH.  

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