Propagation of electromagnetic waves in a lossy media

In summary, the conversation discusses the concept of path loss for electromagnetic waves and how it is influenced by different mediums. The attenuation constant is determined by the propagation constant, which is a complex number with a real part representing attenuation and an imaginary part representing phase shift. The wave number, k, is also a complex number and is related to the propagation constant. It is not equal to (2*pi/lambda) in all cases and can be broken into axial components when the wave propagates in a specific direction. In a lossy medium, the wave number and attenuation are both affected.
  • #1
Julle
5
0
Hi

I've looking into path loss for electromagnetic waves and it's quite straight forward to figure out how it works in free space by looking at the free space path loss formula (http://en.wikipedia.org/wiki/Free-space_path_loss).

It has not been that easy figuring out how another medium would influence this loss. Given information about a medium, such as the complex dielectric constant, is it possible to find out the theoretical attenuation in the medium?

As I figure the loss will still be subjected to the inverse square law but with an added factor multiplied on to account for the loss in the medium. Is that correct?
 
Physics news on Phys.org
  • #2
This is a good explanation: http://www.amanogawa.com/archive/docs/EM7.pdf

In addition to the typical space loss factor, you also have an attenuation over the path of the wave. This is encompassed as a complex part of the wave number.
 
  • #3
Born2bwire said:
This is a good explanation: http://www.amanogawa.com/archive/docs/EM7.pdf

In addition to the typical space loss factor, you also have an attenuation over the path of the wave. This is encompassed as a complex part of the wave number.

According to your link, the attenuation constant, α, is the real part of the propagation constant, γ = α + jβ.

OP, starting with the material parameters σ, ε, and μ, you can apply your awesome complex algebra skills on the expression for the propagation constant:

γ = √(jωμ(σ + jωε))

Simplify this into its real and imaginary parts, and thus you will have your attenuation constant (in nepers per meter).
 
  • #4
Thanks for the answers. The link does indeed give a good explanation. I think I have what I need now.
 
  • #5
gnurf said:
According to your link, the attenuation constant, α, is the real part of the propagation constant, γ = α + jβ.

OP, starting with the material parameters σ, ε, and μ, you can apply your awesome complex algebra skills on the expression for the propagation constant:

γ = √(jωμ(σ + jωε))

Simplify this into its real and imaginary parts, and thus you will have your attenuation constant (in nepers per meter).

The source material implicitly includes a factor of j in those calculations, so that \gamma = -j*k where k is the wave number. The spatial phase dependence of a wave proceeds something like exp(ikr). The imaginary part of k creates a negative real part in the exponential which causes attenuation as the wave propagates.
 
  • #6
Born2bwire said:
\gamma = -j*k where k is the wave number

How can the wavenumber k (= ß on page 78 in your link?) be negative?
The spatial phase dependence of a wave proceeds something like exp(ikr). The imaginary part of k creates a negative real part in the exponential which causes attenuation as the wave propagates.

I still don't see how the wavenumber is a complex number?


The way I learned this was that the propagation constant

γ = √(jωμ(σ + jωε)) = α + jβ

determined the attenuation and phase shift of the wave. The real part of the propagation constant, α, was a positive quantity called the attenuation constant.

Expending the r.h.s of the wave equation solution (on p. 77) with γ = α + jβ, where the two terms represent waves propagating in positive and negative directions respectively, one ended up with the exponential factors exp(-αz) and exp(αz) for propagation in the +z and -z directions. The attenuation per unit length is equal to exp(α). And that was that.

Is this wrong or somehow not what you are saying?
 
  • #7
gnurf said:
How can the wavenumber k (= ß on page 78 in your link?) be negative?


I still don't see how the wavenumber is a complex number?


The way I learned this was that the propagation constant

γ = √(jωμ(σ + jωε)) = α + jβ

determined the attenuation and phase shift of the wave. The real part of the propagation constant, α, was a positive quantity called the attenuation constant.

Expending the r.h.s of the wave equation solution (on p. 77) with γ = α + jβ, where the two terms represent waves propagating in positive and negative directions respectively, one ended up with the exponential factors exp(-αz) and exp(αz) for propagation in the +z and -z directions. The attenuation per unit length is equal to exp(α). And that was that.

Is this wrong or somehow not what you are saying?
\beta is not the wave number.

The spatial phase dependence term is something along the lines of
[tex] e^{i\mathbf{k}\cdot\mathbf{r}} = e^{-j\mathbf{k}\cdot\mathbf{r}}[/tex]
The propagation term is related to the wave number by
[tex] \gamma = ik = -jk[/tex]
It's simply the product of the imaginary constant in the exponential and the wave number. You can see that this is true by the fact that the notes give the propagation as exp(\gamma r).

The wave number is given as
[tex]k = \omega \sqrt{\epsilon\mu}[/tex]
With a lossy permittivity, encapsulated as a conductivity giving rise to an imaginary part in the permittivity (see slide 76), the wave number is a complex number.
 
  • #8
plx Explain the difference between K and beta in physical definition?
 
  • #9
please explain the physical interpretation of complex wave number and how do we get the relation
exp(gamma) = exp(ik)?? (physically)

If wave is propagating in a random direction and k has only real part (that is beta) then we can write it as kx,ky,kz employing wave number along x direction y direction and z direction respectively,whose magnitude equals beta along the direction of wave propagation, but if k is having a complex part as well,and the wave is traveling in a random direction how could we break k into axial components ??

woulb in that case the wave numbers in particular directions be accompanied by attenuation along those directions and overall attenuation 'a' along the direction of wave propagation??
 
  • #10
plus your explanation employs k is not equal to (2*pi/lemda) in every case but beta dose equal to this for any media?K is equal to (2*pi/lemda) for lossless case only?
 

1. How do electromagnetic waves propagate in a lossy media?

Electromagnetic waves propagate in a lossy media through the absorption of energy by the medium, causing a decrease in the amplitude and intensity of the wave as it travels through the medium. This absorption is due to the conversion of the wave's energy into heat as it encounters resistive elements in the medium.

2. What is a lossy media?

A lossy media is a material or substance that can absorb or dissipate energy, resulting in a decrease in the amplitude and intensity of electromagnetic waves passing through it. Examples of lossy media include water, soil, and certain metals.

3. How does the dielectric constant of a lossy media affect the propagation of electromagnetic waves?

The dielectric constant of a lossy media, also known as the permittivity, is a measure of its ability to store electrical energy. In a lossy media, the presence of dielectric materials can cause the electromagnetic wave to be partially absorbed and converted into thermal energy, resulting in a decrease in the wave's amplitude and intensity.

4. What is the skin depth in a lossy media?

The skin depth in a lossy media is the distance at which the amplitude of an electromagnetic wave decreases to 1/e (approximately 37%) of its initial value. It is inversely proportional to the square root of the conductivity of the medium and can be used to characterize the penetration depth of the wave into the medium.

5. How does the frequency of an electromagnetic wave affect its propagation in a lossy media?

The frequency of an electromagnetic wave can affect its propagation in a lossy media by influencing the amount of energy that is absorbed by the medium. Generally, higher frequency waves are more easily absorbed by lossy media, resulting in a more rapid decrease in amplitude and intensity as the wave propagates through the medium.

Similar threads

  • Electromagnetism
Replies
4
Views
962
Replies
8
Views
1K
  • Electromagnetism
Replies
17
Views
3K
  • Electromagnetism
Replies
8
Views
2K
  • Classical Physics
2
Replies
65
Views
3K
  • Special and General Relativity
Replies
18
Views
1K
  • Quantum Physics
Replies
1
Views
981
  • Electromagnetism
Replies
22
Views
4K
Replies
6
Views
1K
Back
Top