Normal incidence over conductors

The following image shows the different wave components (incidence, reflection and transmission) that we are going to study in this post:

Reflection and Transmission Coefficients

The reflection coefficient from medium n1 to medium n2 (refraction indexes) can be expressed as the ration between the electric field reflection component and the electric field incidence component.

 

The transmission coefficient also depends on the refraction indexes and it is the ration between the transmission field component and the incidence field component:

If there is normal incidence, the following relation is met:

 

The standing wave ratio is the relation between the maximum and the minimum values of the electric field:

Characteristics of a Conductor Medium

The following expressions are characteristics of a conductor medium:

Intrinsic impedance (Ohms):

Phase velocity (meter/second): Propagation constant (radian/meter): 

In this case, the attenuation constant and the phase constant are equivalent (Neper/meter):

 

Wavelength (meters):

Perfect Conductor

In a perfect conductor, we have:

The field components of the incident wave are:

The field components of the reflected wave are:

 

Therefore, the total components of the standing wave are: 

 

Incidence Exercise:

A plane wave with linear polarization arrives to a real conductor plane with a finite high conductivity. The current density which is induced in the conductor will stop being strictly superficial and will penetrate in some way in the conductor.

1. Is there any electric field in the conductor?
2. What is the value of the wave impedance in the conductor? Use the following values:  μ =μ0 = 4π x 10-7 H/m, σ = 5,0 x 105 1/Ωm, and calculate it for f = 90 MHz and for f = 10GHz.

1.

The field inside a perfect conductor is zero. In our case, σ is finite (the conductor is real as supposed to ideal), so the field won’t be zero.

2.

The impedance for an uniform plane wave is expressed as:

where

Therefore, the permittivity won’t be always a real number. By using the previous formulas, we have:

For f=10 GHz For f=90 GHz

Conclusions

Why did we study polarization in the previous post and now we study the incidence? There are important applications where these two concepts are developed:

• The electric field of the signals from AM radio stations is linearly polarized, with the electric field perpendicular to the Earth. Thus,the receiver antenna needs to be in vertical position to achieve maximum reception.
• The electric field of TV signals is linearly polarized with the horizontal direction. Therefore, the receiver antennas need to be in horizontal position in the roof of the houses and buildings.
• The wave transmitted by the FM radio stations are usually circularly polarized.

And these are just some of the simplest applications. Do you know any other applications where polarization and incidence is important? Leave a comment below! 🙂