4
$\begingroup$

In radio communication, the distance traveling seems to be an important parameter before we choose the frequency band of communication.

There is also a path-loss model that describes a distance $\leftrightarrow$ frequency dependence of the attenuation with an isotropic antenna.

However, I fail to find any specific physical law that demonstrates the frequency dependence of the attenuation. Some even say that higher the frequency, higher the energy content. What are the factors for propagation distance?

$\endgroup$
7
  • 2
    $\begingroup$ Which kind of waves are you talking of? $\endgroup$
    – DarioP
    Nov 21, 2013 at 8:16
  • 1
    $\begingroup$ I'm afraid this is too broad a question at the moment. You need to be more precise e.g. are you talking about sound waves, light waves, water waves, or something else? The second paragraph is a separate question and should be posted separately. Google black body radiation for some background reading before you post. $\endgroup$ Nov 21, 2013 at 9:08
  • $\begingroup$ Correct me if I am wrong, but I believe this is related to the fact that an EM wave can only interact/be absorbed by objects which greater than of equal to its wavelength. $\endgroup$
    – fibonatic
    Nov 22, 2013 at 12:44
  • $\begingroup$ @DarioP and John I have rephrased my question, hopefully clarifies your points. $\endgroup$ Nov 22, 2013 at 13:57
  • $\begingroup$ @fibonatic sorry, I fail to get the link between your comment and my Q. I think we can safely assume that there are objects of all lengths in this world, required for the waves under consideration! $\endgroup$ Nov 22, 2013 at 13:58

1 Answer 1

1
$\begingroup$

If you search for "atmospheric attenuation vs wavelenghth" you will find many charts like this: They show that in our atmosphere there is not a simple relationship between wavelength and transmission. The various gases in the atmosphere contribute to the spectrum.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.