Diffraction of RF - how is it affected by frequency?

In summary: Increasing the wavelength will reduce the fractional size of the slot so the slot will tend to a line ('point') source which will become an Omnidirectional Source. Decreasing the wavelength will make the fractional slot width will be bigger and the majority of the energy will go straight through (θ approaches zero).The simplest example of diffraction is the diffraction by two very thin sources. The Young's Slits experiment is usually the first approach to diffraction; when there are two or more very small sources, the term 'interference' is used. (I chose this link because it's much more basic and uncluttered than the Wiki pages) Cancellation and enhancement of the waves
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Dickie
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I'm currently trying to get my head around the properties of UHF vs VHF propagation, specifically to do with diffraction around terrain and obstacles of various sizes (i.e buildings, vegetation etc).

I understand from experience that UHF is generally more suitable in built up, urban environments and VHF produces slightly longer ranges but I've yet to find a straightforward and convincing explanation why. I've asked my current lecturer who has given me some information about back scattering and that the higher frequency RF is back scattered less, thus propagating further around angular, man-made structures but I still can't rationalise this into an intuitive, logical understanding to understand what it is about the frequency that changes its diffraction properties.

Is there a mathematical or other logical explanation at all that can explain this?
 
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Over smooth, flat unobstructed Earth we find that the ground reflection tends to cancel the direct path, and above 30 MHz we usually find that attenuation increases rapidly with distance. Power falls at 12dB when the distance is doubled.
In an urban environment, the signal from a high transmitter reaches a mobile at street level via multiple reflections and also by diffraction over rooftops etc. Ths is very hard to calculate, but tests were carried out in Japan by Okumura give reliable empirical data on th statistics of path loss in these circumstances. See Wiki https://en.wikipedia.org/wiki/Okumura_model
Diffraction is only one part of the propagation mechanism in the urban environment. If the wavelength is large compared to the obstruction then diffraction is greater. But notice that free space path loss increases with frequency.
There are nomograms available giving the approximate diffraction loss over hills and knife edges for cases where we have a clean path. I can find a link if required.
 
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Dickie said:
Is there a mathematical or other logical explanation at all that can explain this?
Modelling can be on a number of levels. Undergraduate Physics of EE is quite adequate for getting familiar with the basics of diffraction. The simplest things are ideal slits and round holes and you can get an idea of how you might expect the signals of various wavelengths to behave in typical topography. There is a massive rang of wavelengths used for 'Radio Communications' and each octave has its own characteristics in practice. The maths soon becomes inadequate because the nature of the surfaces is as relevant as the sizes and shapes.
Dickie said:
but I've yet to find a straightforward and convincing explanation why.
You need to read around a lot before you will find something at a level that suits you. https://www.eetimes.com/rf-basics-radio-propagation/# but no single source will be enough for you, I'm sure. Searching with terms including RF Propagation will yield results.
Sorry not to be any more helpful about specifics but your OP is very open ended. That's always the problem with a new topic.
 
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Thank you both for your replies. After reading the info and links above, I managed to find the following equation related to single slit diffraction experiments:

d sin(theta) = n * lambda

Re-arranging to find theta, there is a clear mathematical link between reduced wavelength and a smaller diffraction angle, however I still can't understand why that is the case logically. That said, this is sufficient for the level of knowledge I require at the moment in my systems engineering course so will do for now!

Thanks again.
 
  • #5
Dickie said:
Thank you both for your replies. After reading the info and links above, I managed to find the following equation related to single slit diffraction experiments:

d sin(theta) = n * lambda

Re-arranging to find theta, there is a clear mathematical link between reduced wavelength and a smaller diffraction angle, however I still can't understand why that is the case logically. That said, this is sufficient for the level of knowledge I require at the moment in my systems engineering course so will do for now!

Thanks again.
Increasing the wavelength will reduce the fractional size of the slot so the slot will tend to a line ('point') source which will become an Omnidirectional Source. Decreasing the wavelength will make the fractional slot width will be bigger and the majority of the energy will go straight through (θ approaches zero).
 
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Dickie said:
I still can't understand why that is the case logically.
The simplest example of diffraction is the diffraction by two very thin sources. The Young's Slits experiment is usually the first approach to diffraction; when there are two or more very small sources, the term 'interference' is used. (I chose this link because it's much more basic and uncluttered than the Wiki pages) Cancellation and enhancement of the waves from the two slits depends on the relative phases of the waves (the path difference). The formula shows that a half wavelength path difference is greater as the wavelength decreases and so the angle for cancellation must be smaller. (That's your "logical" argument in a nutshell).

Diffraction, in general, is the result of an infinite number of contributions over a source of finite width but it's only a step further in the same direction as for interference from just two sources.
 

Related to Diffraction of RF - how is it affected by frequency?

1. What is diffraction of RF?

Diffraction of RF (radio frequency) is a phenomenon that occurs when an electromagnetic wave encounters an obstacle or passes through an opening. It causes the wave to bend and spread out, resulting in a change in its direction and intensity.

2. How is frequency related to diffraction of RF?

The frequency of an electromagnetic wave is directly related to its wavelength. As the frequency increases, the wavelength decreases, and vice versa. This means that the higher the frequency of an RF wave, the smaller its wavelength, and the more it will diffract when encountering an obstacle or opening.

3. How does diffraction of RF affect signal strength?

Diffraction of RF can cause a decrease in signal strength as the wave spreads out and loses intensity. This is particularly true for high frequency RF waves, which are more susceptible to diffraction. However, in some cases, diffraction can also improve signal strength by allowing the wave to reach areas that would otherwise be blocked by obstacles.

4. Can diffraction of RF be controlled or minimized?

Diffraction of RF cannot be completely eliminated, but it can be controlled or minimized through various methods. These include using directional antennas, which focus the RF wave in a specific direction, and using higher frequency waves, which have smaller wavelengths and are less susceptible to diffraction.

5. How does the environment affect diffraction of RF?

The environment can have a significant impact on the diffraction of RF. Obstacles such as buildings, trees, and mountains can cause the wave to diffract and change direction, potentially affecting signal strength. Additionally, atmospheric conditions such as temperature, humidity, and air pressure can also influence the diffraction of RF waves.

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