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Using two connected Yagi antennas to "bend" a mobile signal

  1. May 9, 2017 #1
    I have a typical problem. I have two bars mobile signal in my attic - none at ground floor level. The reason is a roadside pole transmitter less than 500m away installed down in a dip behind houses (duh!) An electrically powered signal booster is illegal in the UK which atracts a £5000 fine!
    Thus: could I have a roof top yagi antenna pointed at the transmitter connected via the BNC cable to another yagi pointing ground floor wards? The idea is that the two yagis will collect and re-radiate any signals going back and forwards..
    What do you think?
    Here's a diagram of what I have in mind:
    yagi.gif
     
  2. jcsd
  3. May 9, 2017 #2

    davenn

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    .

    hi there
    welcome to PF :smile:

    yes it works, bit it is quite inefficient. it is essential to keep the linking cable VERY short ( less than a couple of metres and low loss cable - minimum RG213,) as the signal levels going between the Yagi's is a very low level

    As I said low loss cable RG213 or better, eg CFD400 or similar. ALSO DONT use connectors .... just another source of loss ... coax directly between feedpoints of the two Yagi's


    Dave
     
    Last edited: May 9, 2017
  4. May 9, 2017 #3

    Baluncore

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    Pairs of Yagi antennas do work well as VHF passive reflectors. But UHF is more difficult since the aperture of the smaller antennas is significantly lower. A periscope antenna would be more applicable to your situation. https://en.wikipedia.org/wiki/Periscope_antenna
    So for UHF and mobile phone, a flat sheet metal reflector makes a better mirror. Depending on the polarisation you might be able to use a mesh like a cake tray that will be less visible and have less windage.
     
  5. May 9, 2017 #4
    Hi guys,

    Thanks for the input so far. It sounds like its worth a little experimentation. :-)

    So I take it the tv antenna co-ax I have would be useless? The Yagis you can buy on ebay have a short length of coax already attached and if I mount in the attic space I could keep the cable to a few centimeters.

    I had looked at a yagi as I can pick them up on Fleabay for circa £12GBP (~$18) each. I would need to build a periscope antenna. I am assuming this is just a flat metal panel, or would an attempt at a parabolic curve be useful? If I install in the attic space I could play with cradboard and alluminium sheet initially plus an antenna to test the idea.

    I can also pick up a couple of these for a similar price to the Yagis, but I don't think this is what you mean:
    http://www.ebay.co.uk/itm/UK-GSM-CD...epeater-Booster-Indoor-360sq-yd-/311750108630
     
  6. May 9, 2017 #5
    Ahhh ... finally found an image of what you mean. Literally a flat plate designed to bounce the signal downwards, periscope-antenna.png
    where me and the mobile are the dish in this diagram. Is that correct? Would it be that simple? What size of plate for 900mhz, or will any do?
     
    Last edited by a moderator: May 9, 2017
  7. May 9, 2017 #6
    In a typical case, assume 10m from the rooftop down to the mobile. This has a path loss of about 32dB.
    The two Yagis will have a combined gain of, say, 2 x 12dBi, so that the overall loss of the system you are adding is 32 - 24 = 8dB.
    This is not too bad, considering that the mobile signal above the roof might well be 8dB or more above threshold.
     
  8. May 9, 2017 #7

    Baluncore

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    That is right.
    900 MHz has wavelength = 333 mm so a half wave will be 166 mm. I would try odd multiples of half a wavelength. Say 3λ/2 = 500 mm square = 20". If you know polarisation you might just use an aluminium rod of that length.
     
  9. May 10, 2017 #8
    Yes, about 3 half waves square would give a gain as a reflector of about 24dBi I think. But the polarization may be is uncertain for mobile, so a square mesh would be best.
     
  10. May 10, 2017 #9
    I feel a weekend project coming on ...

    I'll let you know the results.
    Cheers :-)
     
  11. May 10, 2017 #10

    davenn

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    a plane reflector wont give any gain .... it is a just reflector, period. :wink:
     
  12. May 10, 2017 #11

    Baluncore

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    True. A very big mirror, hundreds of wavelengths across, has a gain close to 0 dB. But a small plane reflector is a scatterer with a real divergence loss. The scatterer directivity is controlled by the aperture area of the plate. So the incident energy that arrives almost parallel at the scatterer is then spread out over a wider angle and area on the ground below.
    For a 3/2 λ dimension, the beamwidth of the scatterer will be 57° / (3/2) = 38°, which decides the accuracy that you must aim the mirror and the area you will have below that will be reasonably well illuminated.
     
  13. May 10, 2017 #12

    berkeman

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    Where does the gain come from? I'm having trouble seeing that...
     
  14. May 10, 2017 #13
    The easiest way of seeing this is if you imagine the alternative arrangement consisting of two dishes connected back-to-back. If each has a gain of, say, 12 dBi, then the pair give 24dBi. The gain comes form the collecting area on receive and because the energy is concentrated into a beam on transmit. In the case of a plane reflector, the same happens - a large receive collecting area and on transmit, a large aperture with a uniform phase, to radiate a narrow beam. However, remember that the second path down to the handset has a spreading loss which is additional to that from the base station to the reflector.
    In some cases the gain of the reflector can even be a little more than expected, because the size can be chosen so it excludes the second Fresnel Zone, which is cancelling.
     
  15. May 10, 2017 #14

    berkeman

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    Interesting. I see gain with parabolic reflectors concentrating the energy to the receive point, but with a flat reflector, I would think you would get the same energy density reflected downward as you would get in the volume behind the reflector if the reflector were not there. With no reflector you would have 0dB gain right behind where the reflector would be positioned, and normal spreading losses behind that, no?
     
  16. May 10, 2017 #15

    davenn

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    again .... there is NO GAIN with a plane (flat) reflector
     
  17. May 10, 2017 #16
    May I suggest another parallel? If, instead of the two dishes we have a huge sheet of metal with a hole cut in it the diameter of the dishes, the illumination of the aperture will be the same as for the case of the transmitting dish. And the reflector is nothing more electrically than the hole in the metal sheet.
    I am not suggesting something for nothing, as there is always the second path involved with its spreading loss.
    An alternative calculation is to use the Radar Equation, but we need care in correctly stating the radar cross section of the plate, as radar cross sections are based on spheres.
     
  18. May 10, 2017 #17

    davenn

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    and what is the point of that ? it's still not a dish that has gain
    you are just attenuating the signal severely by only allowing what makes it through the aperture to continue on to its intended destination
     
  19. May 10, 2017 #18

    Baluncore

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    An aperture can be an infinite conductive sheet with an arbitrary hole in it, or by swapping the arbitrary hole with conductor and the conductor with space, it makes the 'complementary' aperture, that is the small reflective sheet as used in a periscope antenna. In general, an antenna obeys the same rules, and has the same characteristics, as it's complement.

    The waves that pass through, or reflect from, an aperture do not continue to propagate in straight lines. They spread out because the aperture has a dimension of only a few wavelengths. The radar equation is applicable. There is a 1/r2 path loss while propagating to the aperture, then another 1/r2 path loss due to divergence while propagating away from the aperture.

    Although the energy selected is proportional to the area of the aperture, and the divergence following the aperture is also less for larger apertures, no matter how you work it, the net result is always a greater divergence and therefore a negative gain in dB. That is the tyranny of the radar equation.
     
  20. May 11, 2017 #19
    I agree about the overall loss for the line of sight portion of the path being increased by the use of a reflector.
    I think my numbers using the concept of passive reflector gain will be the same as using the Radar Equation.
    A description of passive reflector gain is given in the following book by the Microflect Company, page 38:-
    http://az276019.vo.msecnd.net/valmo...roflect-passive-repeater-catalog.pdf?sfvrsn=6
     
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