# Propagation of electromagnetic waves in a lossy media

1. Apr 15, 2010

### Julle

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?

2. Apr 16, 2010

### Born2bwire

This is a good explaination: 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. Apr 16, 2010

### gnurf

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. Apr 16, 2010

### Julle

Thanks for the answers. The link does indeed give a good explanation. I think I have what I need now.

5. Apr 16, 2010

### Born2bwire

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. Apr 17, 2010

### gnurf

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?

7. Apr 17, 2010

### Born2bwire

\beta is not the wave number.

The spatial phase dependence term is something along the lines of
$$e^{i\mathbf{k}\cdot\mathbf{r}} = e^{-j\mathbf{k}\cdot\mathbf{r}}$$
The propagation term is related to the wave number by
$$\gamma = ik = -jk$$
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
$$k = \omega \sqrt{\epsilon\mu}$$
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. Dec 17, 2010

### ZunairaMaryam

plx Explain the difference between K and beta in physical definition?

9. Dec 17, 2010

### ZunairaMaryam

plz 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 travelling 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. Dec 17, 2010

### ZunairaMaryam

plus your explaination 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?