Reading here and there that the signals of the on-going DX-expedition to
Glorioso Island are generally very low, I got the curiosity to simulate the
so-called "spiderbeam" antenna they are using (sized for the 10-meter band) on
EZ-NEC.

Doing that, I obtained an unexpected result. The simulated antenna shows a clear
SWR minimum at 29.0 MHz where impedance is 76 + j32 ohm.

I then checked SWR across the 24 - 34 MHz range with the following results:

- going up in range 29 - 34 MHz, the reactance steadily increases (+334 ohm at
34 MHz)

- going down in range 29 - 24 MHz, the reactance remains positive and steadily
increases up to 28.5 MHz, after which it starts to decrease, until it becomes 0
ohm at 27 MHz, and negative below that frequency. At 27 MHz impedance is 9 + j0
ohm (hence it is the resonant point).

I knew that the resonant point does not precisely coincide with the minimum SWR
point, but I would not have suspected such a big difference (2 MHz shift at 29
MHz!).

Any comment?

Tony I0JX
Rome, Italy

"Antonio Vernucci" wrote in message
.. .
Reading here and there that the signals of the on-going DX-expedition to
Glorioso Island are generally very low, I got the curiosity to simulate
the so-called "spiderbeam" antenna they are using (sized for the 10-meter
band) on EZ-NEC.

Doing that, I obtained an unexpected result. The simulated antenna shows a
clear SWR minimum at 29.0 MHz where impedance is 76 + j32 ohm.

I then checked SWR across the 24 - 34 MHz range with the following
results:

- going up in range 29 - 34 MHz, the reactance steadily increases (+334
ohm at 34 MHz)

- going down in range 29 - 24 MHz, the reactance remains positive and
steadily increases up to 28.5 MHz, after which it starts to decrease,
until it becomes 0 ohm at 27 MHz, and negative below that frequency. At 27
MHz impedance is 9 + j0 ohm (hence it is the resonant point).

I knew that the resonant point does not precisely coincide with the
minimum SWR point, but I would not have suspected such a big difference (2
MHz shift at 29 MHz!).

Any comment?

Tony I0JX
Rome, Italy


that is not surprising for an antenna that has a very low or very high
impedance at the resonant point. The SWR depends on the magnitude of the
impedances not the angle, so you could have a minimum SWR with a big
reactance and small real component.

Dave wrote:

. . .The SWR depends on the magnitude of
the impedances not the angle, so you could have a minimum SWR with a big
reactance and small real component.

That's not true. For example, impedances of 50 + j0, 35.36 + j35.36, and
0 + j50 ohms all have the same magnitude (50 ohms), but a 50 ohm cable
connected to loads of those impedances will have SWRs of 1, 2.41, and
infinity respectively.

Correct formulas for calculating SWR can be found in the ARRL Antenna
Book, the ARRL Handbook, or any respectable transmission line text.
Incorrect ones can, I'm sure, be found on the Web and elsewhere.

Roy Lewallen, W7EL

On Sat, 19 Sep 2009 18:03:25 +0200, "Antonio Vernucci"
wrote:

I knew that the resonant point does not precisely coincide with the minimum SWR
point, but I would not have suspected such a big difference (2 MHz shift at 29
MHz!).

Any comment?

Hi Tony,

What did you expect it to be?

73's
Richard Clark, KB7QHC

Antonio Vernucci wrote:
I knew that the resonant point does not precisely coincide with the
minimum SWR point, but I would not have suspected such a big difference
(2 MHz shift at 29 MHz!).

There's a thread over on eHam.net dealing with this same subject.
Many complex antennas exhibit this effect to a certain extent. The
reason is obvious. Our SWR meters are calibrated for 50 ohms and
an antenna may be resonant with a e.g. 9+j0 ohm feedpoint impedance.

That's a 50 ohm SWR of 5.6:1 where almost 1/2 of the RF is rejected
at the antenna when 50 ohm coax is being used. If the 50 ohm SWR
drops below 5.6:1 somewhere else it necessarily must exhibit a
higher resistance and reactance than exists at the 9 ohm antenna
feedpoint.

Moral: There is nothing magic about 50 ohms. If you were using
a transmission line with a Z0 of 9 ohms with a 9 ohm SWR meter,
you wouldn't notice anything worth reporting.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

"Cecil Moore" wrote in message
...
Antonio Vernucci wrote:
I knew that the resonant point does not precisely coincide with the
minimum SWR point, but I would not have suspected such a big difference
(2 MHz shift at 29 MHz!).

There's a thread over on eHam.net dealing with this same subject.
Many complex antennas exhibit this effect to a certain extent. The
reason is obvious. Our SWR meters are calibrated for 50 ohms and
an antenna may be resonant with a e.g. 9+j0 ohm feedpoint impedance.

That's a 50 ohm SWR of 5.6:1 where almost 1/2 of the RF is rejected
at the antenna when 50 ohm coax is being used. If the 50 ohm SWR
drops below 5.6:1 somewhere else it necessarily must exhibit a
higher resistance and reactance than exists at the 9 ohm antenna
feedpoint.

Moral: There is nothing magic about 50 ohms. If you were using
a transmission line with a Z0 of 9 ohms with a 9 ohm SWR meter,
you wouldn't notice anything worth reporting.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com


Actually, there is something 'magic' about 50 ohms. An air-dielectric
co-axial cable has minimum loss per metre when its characteristic impedance
is 76.7 ohms and the relative permittivity of polythene is 2.26 so a
polythene-dielectric co-axial cable has lowest loss when its characteristic
impedance is 76.7/SQRT(2.26) = 51 ohms, which is most often rounded down to
50. This is on the basis that the conductor loss greatly exceeds the
dielectric loss, which is true over most of the frequency range for which
solid polythene dielectric is appropriate.

Maximum power handling, for a polythene-dielectric cable, occurs at a much
lower impedance: 30/SQRT(2.26) = 20 ohms.

Chris

Actually, there is something 'magic' about 50 ohms. An air-dielectric
co-axial cable has minimum loss per metre when its characteristic impedance is
76.7 ohms

I presume that the 76.7-0hm figure comes from a trade-off beween RF current and
conductor resistance. In other words, increasing the impedance value, the RF
current would become lower (for a given RF power), but the inner conductor
resistance would become higher because of the lower diameter needed to obtain
the higher impedance value (for a given outer diameter cable). And viceversa.


and the relative permittivity of polythene is 2.26 so a polythene-dielectric
co-axial cable has lowest loss when its characteristic impedance is
76.7/SQRT(2.26) = 51 ohms, which is most often rounded down to 50.

Under the assumption that dielectric loss is negligible, a permittivity 2.26
time higher than that of air results in a lower inner conductor diameter, for a
given outer diameter cable and a given impedance. Probably, lowering impedance
from 75 to about 50 ohm, the loss advantage one experiences thanks to the higher
inner conductor diameter needed for the lower impedance value is higher than the
loss disadvantage caused by the higher RF current (for a given RF power).

Maximum power handling, for a polythene-dielectric cable, occurs at a much
lower impedance: 30/SQRT(2.26) = 20 ohms.

I do not succeed to understand that statement. Maximum power handling is bound
to maximum temperature which is in turn bound to dissipated power. If 50 ohm is
the impedance at which minimum loss occurs (for a given RF power), why lowering
impedance to 20 ohm should result in a loss reduction. In the equation
30/SQRT(2.26) = 20 ohms, which is meaning of the figure 30?

I wonder whether you could indicate us a reference where all those trade-offs
are mathematically discussed.

73

Tony I0JX

On Sun, 20 Sep 2009 17:48:51 +0200, "Antonio Vernucci"
wrote:

I wonder whether you could indicate us a reference where all those trade-offs
are mathematically discussed.

This should help:
http://www.microwaves101.com/encyclopedia/why50ohms.cfm


--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

This should help:
http://www.microwaves101.com/encyclopedia/why50ohms.cfm


Yes, very helpful. Thanks

Tony I0JX

"Jeff Liebermann" wrote in message
...
On Sun, 20 Sep 2009 17:48:51 +0200, "Antonio Vernucci"
wrote:

I wonder whether you could indicate us a reference where all those
trade-offs
are mathematically discussed.

This should help:
http://www.microwaves101.com/encyclopedia/why50ohms.cfm


--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558


Thanks Jeff, that reference does help but it gets a bit confused over
matters of relative permittivity, Er.

Some time ago (2005), in my work, I derived the whole lot from almost first
principles. It turns out that the series conductor loss (as opposed to the
shunt dielectric loss) is proportional to (1+p)/ln(p), where p is the ratio
of the inside diameter of the outer conductor (D) to the outside diameter of
the inner conductor, and to SQRT(Er). The minimum value of this loss is
found by differentiating the function of p with respect to p and that's what
gives the 76.7 ohms value for Er = 1 (it also involves a constant for copper
conductors, the root frequency and 1/D). The result scales with SQRT(Er)
for polythene.

I should have stated the _peak_ power handling because the 30 ohms (air)
value results from combination of the expression for the electric field
strength and the expression for the characteristic impedance (along the
lines of P = V^2/R). Minimising the field strength gives the greatest
resistance to dielectric breakdown, but a different value of p results when
the impedance is taken into account at the same time. Again, the result
scales with SQRT(Er).

The application for all this was analogue to digital terrestrial television
switch over - the digital signals have much greater peak-to-mean ratios than
the analogue ones, so flashover in air-spaced feeders is a potential power
limitation.

Chris