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Reflection and Transmission of EM waves

  1. Apr 30, 2009 #1
    Hi all,

    I was reading "Geometrical Optics Reflected Fields" chapter from "Introduction to Uniform Geometrical Theory of Diffraction" textbook.

    The author assumed that the surface is perfectly conducting and therefore he only considered the reflected waves. He derived a formula for the reflected field, its polarization and the radii of curvature of its wavefront.

    I have two questions:

    1) If the incident wave front is spherical, the radii of curvature of the reflected waves will depend on the incident wave and the reflecting surface, but will the reflected wave front still be considered spherical?

    2) If the surface is not perfectly conducting, then a transmitted electric field exists. We can use fresnel coefficients to calculate its magnitude but how can we calculate its polarization and the radii of curvature of its wavefront?
     
  2. jcsd
  3. Apr 30, 2009 #2

    Born2bwire

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    1. If the scatterer is a large flat PEC plane, then yes the reflected waves would be spherical waves due to image theory. This is not a given when it is spherical surface. I don't know how it looks but you may be able to get an idea from Mie scattering series. The Mie series gives the reflection of a plane wave from a sphere but you can always decompose any point source (which gives rise to a spherical wave) as the supperposition of the product of three outgoing plane waves (each going along x, y, z or however you orient the local coordinate system). You might be able to then use the Mie series over the what I think is called the Weyl identity.

    2. It depends on how you model the surface. But you could treat it as a local infinite layered medium. If you have a sheet of dielectric, then the local interface will look like air-dielectric-air and you can calculate the generalized reflection and transmission coefficients for that interface and apply them to the incoming ray (a ray is localized plane wave). Balanis talks about the coefficients for a two and three layer system but I do not know if he works out the generalized solution. Weng Cho Chew's "Waves and Field in Inhomogeneos Media" does work his out. But there are other boundaries that are used like Impedance Boundary.

    If you do not have plane waves then what you have to do is decompose the incoming wave as a summation of plane waves. For a point source, this is done using the Weyl or Sommerfeld Identities. The reflection off of a layered medium from a point source is done using the Sommerfeld identity which decomposes the solution into the summation of horizontally traveling cylindrical waves multiplied by vertically traveling plane waves. This is done in Chew's book but it is a rather advanced topic.

    I'm not sure if UTD or GTD handles these kinds of situations. Usually they just have rules for certain geometries, i.e. flat, curved, wedges, spheres. Shooting and Bouncing Rays doesn't have these problems since the ray represents a plane wave and so you can use local approximations and use the Fresnel reflection coefficients. The SBR solvers can decompose a curved surface into a mesh of triangles so that each element is flat or I have seen solvers that use predefined shapes, like cylinders, but I do not know if they are handled differently when solved. I think the predefined shapes make it easy because then you do not need to worry about mesh densities and such.
     
  4. Apr 30, 2009 #3
    I can assume that I have only plane surfaces.
    Does that change the shape of the wave front of the transmitted waves?
     
  5. Apr 30, 2009 #4

    Born2bwire

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    Plane waves will always reflect as plane waves off of an infinite planar surface. Anything else and it gets complicated, sometimes there are closed form solutions, like with the Mie serie or scattering from a cylinder. Classical electromagnetics follows linear superposition. So you can theoretically decompose any wave into the summation of plane waves, or cylindrical or spherical. Sometimes that makes it easier to derive a solution (as in the case for cylinder and sphere scattering).

    So, if you have a plane wave coming in and reflecting from a scatterer, the reflections can always be represented as a summation of a finite or infinite number of plane waves, but rarely do you get a finite summation.
     
  6. Apr 30, 2009 #5
    Thanks a lot :)
     
  7. Apr 30, 2009 #6
    If I have a plane wave the radii of curvature of its wavefront are both infinity, the spreading factor is then 1.
    As a result the electric field would be E(s) = E(0) exp(-j ks).
    The electric field becomes independent of the distance from the reference point.
    Is this because the assumption here is a lossless medium and there is no spreading of waves so the field remains as it is?

    Can this be employed in SBR? What are the other alternatives?


    Can anybody tell me if the equation in this link is correct: http://en.wikipedia.org/wiki/Isotropic_radiator" [Broken]?

    Thanks in advance
     
    Last edited by a moderator: May 4, 2017
  8. Apr 30, 2009 #7

    Born2bwire

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    It doesn't spread,correct, but it also is infinite. When you have a source that is finite in two or three dimensions then you will have spreading due to the propagation of the wave itself. You can approximate a plane wave in SBR, just initialize a large plane of rays with the same polarization, amplitude, and wave vector. Just make sure that it is large enough to cover the entire scatterer and observation points of interest. If you are interested in the total field, don't forget that you want the reflected and initial rays that can reach, actually reach your observation point.

    As for the wikipedia page, I corrected a sign error in that equation a week or two ago. The exponential is supposed to cause a loss in the amplitude as r increases. With the j convention, a lossy medium has a negative imaginary part in the permittivity. With that correction I'm happy with it. Technically though, the sign convention is a matter of how they choose the time convention, whether time is [tex]e^{-i\omega t}[/tex] or [tex]e^{j\omega t}[/tex]. That's usually how it is defined but people can specify it willy nilly and it causes a lot of confusion I've found on Wikipedia.
     
    Last edited: Apr 30, 2009
  9. Apr 30, 2009 #8
    The other alternative to plane waves in the SBR is to consider a spherical wavefront that is distributed across the rays emitted from the point source. correct?

    As for the equation in the wikipedia .. shouldn't there be an E0 or something? and why the (4 PI r) factor?
     
  10. Apr 30, 2009 #9

    Born2bwire

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    Yeah, you could also use an isotropic radiator as your source too. It doesn't matter, it's whatever you want to simulate. Usually what you do is if you have a small antenna, then you have a point source that has the same radiation pattern as your desired antenna. Electrically large antennas can be estimated by either placing them far away or building part of the physical antenna. For example, a dish reflector antenna (satellite) is much larger than the wavelength. So one way to model it is to make a physical dish reflector, arm, and source box and then in the source place the necessary antenna source. It's whatever you want to model really.

    The 4 pi factor is convention. If you solve for the electric field due to a point source current of amplitude 1 A, then you get [tex]\frac{e^{ikr}}{4\pi r}[/tex]. Or, it is convenient because then the power density at 1 m away is 1 W/m^2. Either way, you can scale as you wish with the physical insight being that you are scaling it as if you were multiplying the 1 Amp point source current. The amplitude scales as 1/r because it is a spherical wave. The energy over a wavefront of the spherical wave is constant in a lossless medium. Thus, if the wave is expanding out as a spherical shell, then the surface area grows as [tex]4\pi r^2[/tex]. Thus, the power grows as r^2. Thus, to keep the power/energy along our closed surface constant, the amplitude of the field must drop as 1/r. This loss in power of 1/r^2 is referred to as the inverse square law.
     
  11. Apr 30, 2009 #10
    So basically I multiply the field in the wikipedia by the initial field as if I'm scaling the current right?

    The antenna measurements are made in the far field which lies beyond 2D^2/lambda. If I'm using a point source, it has no dimension. What is the far field boundary in this case?

    I want to take the reference point to be at the boundary of the far field instead of 1 m away from the antenna. This is because I'm simulating an indoor environment so it is not large compared to 1 m.
     
  12. Apr 30, 2009 #11

    Born2bwire

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    With a point source, the exact field is [tex]\frac{e^{ikr}}{{4\pi r}}[/tex], it has no near-field or far-field boundaries. This is not so if you are talking about a physical antenna like a wire dipole but no physical antenna can replicate an isotropic radiator (near-field or far-field).
     
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