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In the Friis transmission equation, can you please explain the significance of the factor

$$\left(\frac{\lambda}{4\pi R}\right)$$

and what it helps us achieve? Specifically in terms of path loss.

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  • $\begingroup$ That part is the (inverse of the) path loss. Could you clarify your question? Do you want know what path loss is? Do you want to know why it has that value? $\endgroup$
    – Kevin Reid AG6YO
    Apr 7, 2015 at 21:48
  • $\begingroup$ @KevinReidAG6YO Thank you. I understand that it is the inverse of path loss. What does it help us achieve. In the sense, how can the formula be used to help us in research, etc. and maybe how it links with the other gain and power parameters. Thank you. $\endgroup$
    – Max
    Apr 8, 2015 at 17:14
  • $\begingroup$ What it helps us achieve is computing the received power, or the maximum distance, as part of the equation — it's not all that meaningful by itself, other than "the part that isn't dependent on the antennas”. If you want to understand why it is what it is, this question might help: What does the Friis transmission equation represent and how is it derived?. If that's not what you're looking for, please try to ask a more concrete question. $\endgroup$
    – Kevin Reid AG6YO
    Apr 8, 2015 at 20:43
  • $\begingroup$ @KevinReidAG6YO Thank you for your answer, i'll use to deploy further research. Can i ask you another question? How does the design of a satellite communications system allow for the effect of potential losses in transmission? $\endgroup$
    – Max
    Apr 8, 2015 at 22:30
  • $\begingroup$ Don't ask further questions in comments; post them separately. $\endgroup$
    – Kevin Reid AG6YO
    Apr 9, 2015 at 0:40

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