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Modeling EM transmission

  1. Oct 13, 2008 #1
    Hi all,

    I have brave hopes of modeling the following problem:

    Imagine an antenna transmitting an ~50V 100kHz sine wave. Imagine a receiving antenna placed three or four feet apart, with an amplification and detection circuit.

    First, measure the amplitude of the received wave. Next, put your hand between the antennas. This will "shunt" away some of the current, causing the received wave's amplitude to decrease.

    We've built this little thing and done some preliminary testing. Now I want to model it.

    I have COMSOL and the RF module, but I'm unsure how I can simulate the voltage source, or the antennas. If anyone has any insight, or has done work on a similar problem, let me know.

    Thanks,

    Luke
     
  2. jcsd
  3. Oct 13, 2008 #2
    Loops or ferrite rods? 50 V means nothing unless into a specific impedance.

    First thoughts are that a loop or ferrite rod will generate a large wavefront and a hand will have little effect esp at such a low frequency.
     
  4. Oct 13, 2008 #3
    Hey Pumblechook, thanks for responding!

    The antennas are currently 100x100x1mm copper plates, fed by solid wire to a breadboard. The signal is generated by a PIC IC using PWM to create a square wave, whos frequency is matched to an LC tank circuit, generating the signal on the antenna.

    I'm not sure actually how to calculate the impedance of the antenna, and we hope to do some real antenna design in the future.

    However, we do see a 1.5V 100kHz sine wave at the receive end, which changes by about 500mv when a hand is in-between the plates.

    I have faith that it works a little because I've seen it. But I am completely at a loss as to how to model it.

    Thanks for your help.

    -Luke
     
  5. Oct 13, 2008 #4
    I don't think they are antennas.. Far far too small.. A half wavelength dipole would be 1500 metres long - nearly a mile. 100 mm is 1/30,000 of a wavelength.

    I think they are working like the plates of a capacitor and your hand will be a lossy dielectric. How a far apart are they? You are not radiating as such.. It is capacitive coupling... all electric field..no magnetic field. This may well be far more sensitive than trying to block an EM wave which as I said will have a large wavefront at 100 kHz.

    Portable radios use coils wound on ferrite rods at that sort of frequency. Loops at that frequency will be 1 - 2 metres in diameter.

    In any case there is no real radiation between closely coupled antennas. Loops or ferrite rods will tend to coupling inductively when close together.. all magnetic field..

    http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/RadCom/part7/page5.html


    It may be if you have a hi-q tank circuit and your hand is detuning the tank circuit.

    You are well in the near field which is up to around one wavelength from the source... 3 km.


    """The near-field is remarkable for reproducing classical electromagnetic induction and electric charge effects on the EM field, which effects "die-out" with increasing distance from the antenna (proportional to the cube of the distance), far more rapidly than do the classical radiated EM far-field (proportional to the distance). ...."

    http://en.wikipedia.org/wiki/Far_field
     
    Last edited: Oct 13, 2008
  6. Oct 16, 2008 #5
    hi!

    all i can think of is modelling you antenna as a patch antenna with air as dielectric substrate.

    that way you can easily calculate the input impedance of your antenna. (many ieee transaction papers have it - also, il take a look in my backup - i think i have the papers with the equations there - dont worry, they're not tough ones, you can calculate them on a sc.calculator in about 5 min.)
    that should give you the input impedance, (assume that you're using infinite substrate in your model, its quicker and requires less system power, and the loss of generality is not too significant unless you're doing a thesis..)
    then you're good to go :)
     
  7. Oct 16, 2008 #6
    A path antenna at 100 kHz only 100 mm x 100 mm.

    I dont think so some how.

    A patch typically has half wavelength sides. That is 1500 metres..... 2.25 sq km.
     
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