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Superposition of radio waves and energy conservation law?

  1. Apr 6, 2009 #1
    Hey all!

    Now I'm working on wireless communication, not a physician. In many courses in ECE, we learn about principle of superposition of radio waves. But I have a stupid question which has been considered no wonder so far.

    There are two radio senders(S1,S2) and one receiver(R). Let's suppoe that the distance between senders and receiver, the waveform frequency, phase and types are exactly same.

    By principle of superposition, at the position (R), the intensity is getting 4 times than there is only one sender because amplitude of two signals is getting double.

    What does that mean? It seriously makes me confusing.

    Does it mean that I can get double power if I use two different sources?

    There must not be free energy!! So I think that there might be something to lose for this free food.

    What is the cost I have to pay to get double intensity at the receiver in wireless communication?

    Thanks in advance!!
     
  2. jcsd
  3. Apr 7, 2009 #2
    I don't quite understand the question here.

    Radio waves combine 'in space' to double the power and the voltage goes up by root 2.

    It is similar to combining two signals in a combiner such as a wilkinson combiner (splitter) which involves impedance transformation and therefore voltage/currents are changed but power is always doubled (minus any loss in the combiner). Basically you step up the impedance of each power source to 100 Ohms (volts go up by root 2) and then you can parallel the two sources to get back to 50 Ohms.
     
  4. Apr 7, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hey ychang! Welcome to PF! :smile:

    (it's physicist! :biggrin: a physician is a medical doctor who isn't a surgeon :wink:)

    I take it you understand that the energy actually is doubled,

    but you want to know how that fits in with superposition, where two identical sources can interfere destructively, to give zero power, or constructively, to give 4 times the power?

    (and it applies equally to sound waves and water waves, so it isn't just a quantum thing)

    hmm :rolleyes: … I dunno … maybe it has something to do with energy being absorbed or generated "by" the boundary conditions? :redface:
     
  5. Apr 7, 2009 #4
    Suppose that the signals from both senders are exactly the same frequency and amplitude, and are in phase at the receiving antenna. Then the voltage from the (omni) antenna is twice the voltage of one signal (superposition) and the power then is quadrupled. because superposition depends on phase, if the receiver is moved about 1/2 wavelength closer to one sender, the two signals will cancel.
     
  6. Apr 7, 2009 #5

    tiny-tim

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    Yes … but where does the extra energy go (or come from)? :confused:
     
  7. Apr 7, 2009 #6
    Yes!! physicist!! Sorry about that.. :)

    I agree with Bob's answer. I calculated expected value of the power envelope at the receiver in terms of location. It gives me 2, not 4. Right... the maximum we can get is quadrupled depending on phase. On the otherhand, this superposition will make holes(the signals cancel) through the space. So the average power will stay in double.

    Here is another question. What happened if two sources are getting close to each other?
    Let's say much closer than wavelength... or zero!

    Can I have still chance to get quadrapled at certain point?
     
  8. May 29, 2009 #7
    Are the principles of superposition (the linear sum of partial differential equations) and conservation of energy compatible? They must be, right? But how.

    Energy is proportional to amplitude squared.

    http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000054000003000233000001&idtype=cvips&gifs=yes" [Broken]

    The abstract. Unfortunately the full article will cost you.
     
    Last edited by a moderator: May 4, 2017
  9. May 29, 2009 #8

    vk6kro

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    If you have two equally powered sources of signal, on the same frequency and in phase with each other, that are physically very close together in terms of wavelength, it is just the same as doubling the power in one antenna. The receiver will see a 3 dB increase in signal power.

    If they are well separated in terms of wavelength, say a quarter wavelength or more, there will be points in the surrounding landscape where the two waves will cancel each other out and other points where they will reinforce each other. And there will be other points that are only partly affected, either constructively or destructively.

    These points are not random and tend to occur in lines radiating from a point between the antennas, but depend on the spacing between the antennas.

    At these points, both waves still exist in their original strength.
    They have to, because the wave is moving out from the antenna and it is not affected by the presence of another signal on the same frequency.

    They don't cancel each other out in the air.
    If you put a directional antenna at one of these points, you could still receive both of them (one at a time) at normal strangth.
    It is just that when you place an antenna at that point, and it could receive both of them, the induced voltages in the antenna cancel each other out, or reinforce each other.
    .
     
  10. Jan 24, 2010 #9
    I guess the details depend on the antennas, but one important point should be noticed. The power radiated by an antenna is Prad = Rrad * I2 where Rrad is the radiation resistance of the antenna, which usually depends only on the geometry of the antenna (see http://en.wikipedia.org/wiki/Radiation_resistance ). Bringing two antennas close together will not always change the relevant part of the geometry. For example, in dipole antenna (see http://en.wikipedia.org/wiki/Dipole_antenna), the radiative resistance depends only on the length of the dipole. Therefore, the radiative resistance of two dipole antenna close together will be the same as each one separately, which is not what we expect with usual electrical components. For example, if we bring two cables in parallel, the total resistance is half the resistance in each cable, not the same. Assuming that the radiative resistance is small with respect to the total resistance of the antenna, we have that the resistance in the two antenna is half the resistance of one antenna. So, if the supplied voltage is kept constant, the current in the antenna is twice as large. So, the answer to the question "From where the energy comes from?" is simply that it comes from the I2 in the usual formula Prad = Rrad I2 and the fact that the current is twice as large because the total resistance is half the original value. In my opinion, the surprising part is that the radiative resistance Rrad for two antennas close together can be the same as for each antenna separately.
     
    Last edited by a moderator: Apr 24, 2017
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