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Friis Path Loss on a Segmented Path Derivation

  1. Apr 6, 2017 #1
    Let's say we have a path Length AC. Point A transmits a signal at Ptransmit.
    From Friis' Path Loss formula: (assumming Gain for receiver and transmitter antennae are 1.)
    ##P_{received} = P_{transmit} (\frac{λ}{4π* LengthAC})^2##
    ,when point C contains the receiver.

    What if I divided this into two segments. They should still have the same answer right?
    Let's say point B is somewhere between A and C.
    Theoretically, the path loss in AB plus the path loss in BC should equal the path loss in AC. If I do this using the formula.

    ##P_{received at B} = P_{transmit} (\frac{λ}{4π* LengthAB})^2##
    ##P_{received at C} = P_{received at B} (\frac{λ}{4π* LengthBC})^2##

    Combining these two should get me to the total of the equation before. How should I combine them? I couldn't just multiply or add them together. The math would be messy. Am I missing some assumptions here?
  2. jcsd
  3. Apr 6, 2017 #2


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    If you have two paths in tandem, they should be treated as two paths, each having two antennas and a free space attenuation. The losses in Decibels of the two paths should be added.
    The reason that the answer is different to treating it as a single path is that most attenuation occurs at the start of the path, and for two paths this happens twice. For example, suppose a given path has an attenuation of 60dB. If you double the path length, it becomes only 66dB. But if the two paths are used in tandem, say by using a passive repeater, it will give 120dB.
  4. Apr 6, 2017 #3
    So, how do you exactly define the start of a path when something like this involves an interaction at B?
    let's say there's a reflection at point B. Will it "start" a new path? or is it safe to treat the total reflected path distance and treat it as a single path loss?
  5. Apr 6, 2017 #4


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    For the two paths in tandem to be calculated as one, the receiving antenna on the first hop must be large enough to catch all the radiated energy. So it might need to be a kilometre in diameter!
    For cases where the intermediate point is back-to-back dishes, passive reflectors, reflections from buildings etc etc, treat it as two paths. The repeater has its gain twice - once for receiving and again for transmitting.
  6. Apr 6, 2017 #5
    Hmm. I think I'm getting the problem with my equation. As you said, I should be getting all energy. that just means the Friis' Path Loss Formula estimates a certain propagation area and a finite point where it is received. From the wikipedia page, I see it has assumed an isotropic case.

    In the case of having reflections, I have to treat it as two paths. since it "starts" again as another wavefront.

    I just clarified what you wrote in this text. Thank you!
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