Multipath and Polarization

"Randy Yates" <yates@ieee.org> wrote in message 
news:m3bqwjw4tz.fsf@ieee.org...
> Hi Clay,
> Elliptical polarization? I'm not getting it. I thought the EM wave has the
> E field orthogonal to the B field, and both of these are orthogonal to the
> direction of travel. So where does an ellipse come in? Or do you just mean
> something between horizontal and vertical (e.g., 78 degrees)?

Nope - it's somewhere between circular and linear. 78 degrees would be slant 
polarisation (although normally slant polarisation is +/-45 degrees).

Cheers, Howard 

"Howard Long" <howard@howardlongxxx.com> writes:

> "Randy Yates" <yates@ieee.org> wrote in message 
> news:m3bqwjw4tz.fsf@ieee.org...
>> Hi Clay,
>> Elliptical polarization? I'm not getting it. I thought the EM wave has the
>> E field orthogonal to the B field, and both of these are orthogonal to the
>> direction of travel. So where does an ellipse come in? Or do you just mean
>> something between horizontal and vertical (e.g., 78 degrees)?
>
> Nope - it's somewhere between circular and linear. 78 degrees would be slant 
> polarisation (although normally slant polarisation is +/-45 degrees).

Yes, a few minutes of reading in my old "Field and Wave Electromagnetics" (Chen)
revealed the situation - elliptical is a type of time-varying polarization in
which the E-field magnitude is time-changing as well as the E-field angle.
-- 
%  Randy Yates                  % "I met someone who looks alot like you,
%% Fuquay-Varina, NC            %             she does the things you do, 
%%% 919-577-9882                %                     but she is an IBM."
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO   
http://home.earthlink.net/~yatescr

"Howard Long" <howard@howardlongxxx.com> wrote in message 
news:duig5b$6br$1@nwrdmz03.dmz.ncs.ea.ibs-infra.bt.com...
>
> Nope - it's somewhere between circular and linear. 78 degrees would be 
> slant polarisation (although normally slant polarisation is +/-45 
> degrees).

OK I'll add a bit here having seen Clay's notes. I would imagine that for 
your purposes, you would only use linearly polarised antennas, and so for 
your purposes you are only seeing the linear component.

To generate circular polarisation, imagine to linear antennas mounted 
orthogonally to each other. You feed one 90 degrees after the other using a 
bit of coax acting as a quarter wavelength delay line. That's circular 
polarisation.

If you don't feed the second antenna exactly 90 degrees later, or the 
antennas are not precisely orthogonal, then you now have elliptical 
polarisation.

Regards, Howard

Dang, Clay, how about a thorough answer?   ;)

On Mon, 6 Mar 2006 17:17:33 -0500, "Clay S. Turner"
<Physics@Bellsouth.net> wrote:

>Hello Randy,
>
>In a nutshell, the answer is yes - polarization gets messed with. In a 
>single path environment, the polarization may be either rotated by 
>reflection off of an obliquely oriented non-metallic object or reflection 
>off of a metallic object. In the 1st case, the reflection is still linearly 
>polarized. The 2nd case is more complicated. The plane parallel and plane 
>perpendicular components will each reflect with a different reflection 
>coefficients and phase shifts. So in general, metallic reflection creates 
>elliptical polarization. This happens all across the EM spectrum including 
>radio and light waves. For the nonmetallic case, the phase shifts are the 
>same although the reflectivies are still different - so linear stays linear.
>
>You probably recall from your Physics studies about the index of refraction 
>which simply is the ratio of the speed of light in a vacuum to the speed of 
>light in matter. Also you will recall that this usually is frequency 
>dependent, hence this is why a prism decomposes white light into a 
>spectrum - each frequency has a different velocity in the glass and because 
>of the oblique incidence is refracted by a different amount.
>
>If one applies Maxwell's equations and the continuity relations to the 
>problem of wave propagation across a matter boundary, one arrives at 
>Fresnel's equations for the reflection and transmission coefs based on the 
>indices of refraction. It turns out that metals can be properly described as 
>having complex indices of refraction where the imaginary part corresponds to 
>the absorptivity of the material. For example, aluminum has an index of 
>refraction of 1+4.45i at the wavelength of 632.8 nm (the orange-red line 
>produced by a He-Ne laser.) And from Fresnel's equation one finds that 
>linearly polarized light when incident at 78 degrees and oriented so both 
>the plane parallel and plane perpendicular components have equal power upon 
>incidence, you will create circularly polarized light. Back when I used to 
>make holograms, I showed how to match the polarization rotation in each of 
>the light beams so the light's interference was maximized. A wrote a CAD 
>program to track the effects of metallic polarization caused by the aluminum 
>mirrors used in the optical setup. I was able to design polarization 
>independent optical setups where a less expensive non polarized laser may be 
>used.  A science based on metallic polarization used to measure thickness of 
>thin films is called ellipsometry.
>
>Now back to the RF. Since any arbitrary polarization can be decomposed into 
>the sum of other polarizations, you can now see the a multipath situation 
>involving metallic objects can result in some pretty screwed up 
>polarizations. But even without the metal components, the polarization can 
>get rotated around if your received signal ends up being composed of mostly 
>reflected energy. Even with vertical reflecting structures, if you are near 
>the transmitting antenna and its height is different from your receiving 
>antenna's height, you have a certain amount of obliqueness, and can see a 
>rotated polarization.
>
>IHTH,
>
>Clay
>
>
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org