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Thread: Building an CANTENNA

  1. #11
    Flight attendant Strix technica's Avatar
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    Quote Originally Posted by Strix technica View Post
    The core extends 1/4w in the other direction, also open impedance at its end. Radiation incident perpendicular to the cantenna will see two 1/4w conductors — which looks very much like a 1/2w dipole, albeit one where the feeder shield is somewhat decoupled from the active antenna. But the decoupling sleeve is not a balun because it's still connected to RF ground so, in that sense, it's more like a 1/2w dipole without a balun.
    Correction: apparently, a sleeve can act as a wavelength-specific symmetrising balun but can also act as a radiating (receiving) element, depending on which way it's oriented as described here.

    This explains, sort of, how sleeve baluns "symetrise" (for a specific wavelength), but it's unclear whether the shield side of the dipole is actually active or whether it merely increases output on the driven element, and therefore how it compares with a ferrite core balun in terms of radiating/receiving efficiency. NB: ferrite baluns have some insertion loss that sleeve baluns won't because of losses in the ferrite core, so it might might, er, balance out in the end.

    As applied to the cantenna, the decoupling sleeve is oriented to be a radiator/receiver, so my original point stands: the cantenna is a 1/2w dipole (largely balanced, apparently) for perpendicular incident radiation, ie for the most distant of signals.
    Last edited by Strix technica; 2017-05-09 at 12:43.

  2. #12
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    A 1/2w dipole has 2.15 dBi (0dBd) gain and a 1/4w GP has 2 dBi. Not sure how you got to 5dBi ?

    /M
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  3. #13
    Flight attendant Strix technica's Avatar
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    Quote Originally Posted by MrMac View Post
    A 1/2w dipole has 2.15 dBi (0dBd) gain and a 1/4w GP has 2 dBi. Not sure how you got to 5dBi ?
    I've read conflicting accounts about this but as far as I can make out, the principle goes as follows:

    On the one hand, a 1/4w mono has half the aperture of a 1/2w dipole and also half the impedance of a 1/2w dipole (36Ω v. 72Ω):

    P = V^2/Z. Substituting as factors of monopole/dipole, we get P = 0.5^2/0.5 = 0.5 = -3 dBd.

    This much is intuitive: half the length, half the gain.

    On the other hand, this considers only radiation incident on the whip and ignores the effect of the ground, the impedance of which can vary wildly from 0Ω to hundreds or thousands. On an essentially lossless metal ground plane that is at least 1/4w in radius, all of the power incident on the ground plane (along its azimuth) will be reflected onto the whip but still driven into half the impedance:

    So that's P = (0.5+0.5)^2/0.5 = +3 dBd or +5.15 dBi.

    But this only works for a whip directly on a perfect ground.

    Just as the vertical dipole's lateral gain comes at the expense of gain on axis, this extra gain has to come from somewhere; in this case, it comes from what would have been the pattern below the midpoint of the dipole. This gives it a more tightly confined and somewhat elevated pattern relative to a dipole as illustrated here. Note the caveats made in that description concerning the size of the ground plane in wavelengths.

    The extended pattern (gain) is welcome for ADS-B, but the fact that it is inclined is not because the weakest signals come from the horizon.

    One might conclude that, for our purposes, co-linear > 1/4w monopole > 1/2w dipole — certainly, that's how it worked out in practice for me (though my dipole didn't have a balun) — because though the 1/2w has a perpendicular pattern, it is roughly spherical in cross-section (convolved about the axis of the antenna) which means a lot of the pattern isn't where we want it, including reaching into the ground rather than toward the horizon. Co-linears (with sufficient elements) have a likewise perpendicular pattern that is flatter and elongated toward the horizon, which makes them good for ADS-B.

    How a 1/4w mono will behave depends entirely on its ground plane size and angle to horizon. For example, some pole-mounted monopoles have their RF ground radials declined to better match with 50Ω coax, however this induces some RF into the shield and so the signal will degrade by various means. That's not to say that an impedance matching unun would necessarily do any better instead because of the insertion loss they would introduce (if a 1:1.333 matching ratio were feasible anyway).
    Last edited by Strix technica; 2017-05-10 at 13:12.

  4. #14
    Captain abcd567's Avatar
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    Cantenna is neither a pure 1/2 WL Dipole, nor a pure 1/4 WL Monopole with ground-plane.
    It is a hybrid.


    IMAGE 1 of 2
    On right side is the old-timer Koaxial Antennen (Coaxial Antenna), aka Coaxial Dipole. It is a 1/2 WL Dipole as it has almost zero ground-plane due to very narrow copper tube/pipe used. The copper tube/pipe has diameter only slightly bigger than the diameter of the coaxial cable.

    On left is the wide-bodied variant of Coaxial Dipole. Due to very large diameter of the soda can (66mm, almost 1/4 WL), the bottom of soda can makes a substantial ground-plane. As a result, unlike Coaxial Dipole, the Cantenna does not behave as a pure 1/2 WL Dipole.

    (Translation in English in red color is added by me)




    IMAGE 2 of 2
    These are 3 Cantennas with different diameter can/pipe. All other dimensions are identical.
    The worst performer was the one with smallest dia (20mm dia copper water pipe with end cap), and best performer was the one with largest dia (66mm dia standard soda can).


  5. #15
    Captain abcd567's Avatar
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    Simulation of Cantenna
    Gain = 1.71 dBi
    SWR (50 ohms) = 1.35

    Can dia: 68mm
    Can length: 70mm
    Whip length: 68mm
    Environment: Free Space


  6. #16
    Flight attendant Strix technica's Avatar
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    Quote Originally Posted by abcd567 View Post
    Cantenna is neither a pure 1/2 WL Dipole, nor a pure 1/4 WL Monopole with ground-plane.
    It is a hybrid.
    I agree, and that's what I said — the cantenna's characteristics depend on the angle of incident radiation.

    However, the other day you insisted that it was a 1/4w monopole:

    Quote Originally Posted by abcd567 View Post
    Cantenna is NOT a 1/2w dipole.
    It is a 1/4w monopole, with a ground plane disc (bottom of the soda can), and a 1/4w long decoupling sleeve (cylindrical wall of the soda can).

    Quote Originally Posted by abcd567 View Post
    Due to very large diameter of the soda can (66mm, almost 1/4 WL), the bottom of soda can makes a substantial ground-plane.
    If the soda can has a diameter of 66mm, then the radius is 33mm. That makes the ground plane a bit under 1/8w.

    Quote Originally Posted by abcd567 View Post
    As a result, unlike Coaxial Dipole, the Cantenna does not behave as a pure 1/2 WL Dipole.
    Perpendicular incident radiation can't see the ground plane, so it'll look like a 1/2w dipole albeit, as I said before, with some degree of decoupling/symmetrisation that you don't get with a coaxial dipole. In fact, because of the lip and the concave shape of the ground plane, its 1/4w properties probably don't begin to kick in until the angle of incidence to the perpendicular is fairly substantial. Even when its 1/4w wave properties are apparent, you've got to consider that the ground plane isn't a full 1/4w given the geometry of where the active element is in relation to the edge of the disc.

    Quote Originally Posted by abcd567 View Post
    These are 3 Cantennas with different diameter can/pipe. All other dimensions are identical.
    The worst performer was the one with smallest dia (20mm dia copper water pipe with end cap), and best performer was the one with largest dia (66mm dia standard soda can).
    I believe it, especially for a/c overhead (where the gain is least). But even FL400 directly overhead is only 6.6 nm away, increasing to 9 nm at 45° (6.6/cos° 45). In fact, you have to get to 3.8° to the horizon before the plane at FL400 is 100 nm away (90 - arccos° 6.6/100) by which point you're well into dipole territory.

  7. #17
    Flight attendant Strix technica's Avatar
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    Quote Originally Posted by abcd567 View Post
    Simulation of Cantenna
    Gain = 1.71 dBi
    Vertical plane pattern looks like a 1/2w dipole to me, not remotely like a 1/4w mono, and yet its expected gain is less than that expected of a dipole.

    Perhaps this is not surprising, though, considering that this is a simulation of the cantenna transmitting rather than receiving. It would, indeed, look like a dipole if driven by a transmitter, and a tx simulator can't model different angles of incidence for rx.

    Also, the feedline doesn't appear to be modelled, but then I'm not familiar with 4nec2 so I'm unsure how important that is.

    ETA: the other cause of decreased gain might be that there is a 1/8w phase shift between radiation incident on the core and on the can. That's a quarter of a half wave so it's not insignificant (think of how the induced voltages will add up when the voltage maxima are at the ends of the dipole).
    Last edited by Strix technica; 2017-05-11 at 09:27. Reason: mention 1/8w phase shift

  8. #18
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    Quote Originally Posted by abcd567 View Post
    Since angle of tilt is not big in this one, the improvement by making another better cantenna is not likely to be big.

    It is easy to make a Cantenna. No harm making and trying another more accurate one.

    What is the length of coax between Cantenna and DVB-T? A long coax cable will result in reduction of reception.
    Hi sorry for the late reply, it is one meter long.

  9. #19
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    Quote Originally Posted by Strix technica View Post
    So that's P = (0.5+0.5)^2/0.5 = +3 dBd or +5.15 dBi.
    But this only works for a whip directly on a perfect ground.
    That is directivity, not actual gain.

    /M
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  10. #20
    Flight attendant Strix technica's Avatar
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    Quote Originally Posted by MrMac View Post
    That is directivity, not actual gain.
    Right. Antennae are passive devices, they don't amplify anything so antenna 'gain' is not like amplifier gain.

    Directivity, AIUI, is related to gain in antennae specifications. The tighter the pattern, the greater the gain relative to the ideal isotropic — in a given direction.

    If you think I've misapplied the theory and in particular the relationship between incident amplitude, impedance and transferred power, please do say how.
    Last edited by Strix technica; 2017-05-12 at 16:33.

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