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

Antonio Vernucci wrote:
. . .
Another interesting observation is that, at 29 MHz (i.e. where the
antenna impedance is 76 + j32 ohm and the SWR on a 75-ohm cable shows
the minimum value of 1.95) one can find a cable length at which the
impedance appears to be purely resistive and equal to 1.95*75 = 146 ohm
(or 75/1.95 = 38.5 ohm). This fact is deceiving as, seeing a purely
resistive impedance, one could be led to concluding that the real
antenna resonant frequency is 29 MHz, whilst in reality it resonates at
27 MHz (although knowing what is the real antenna resonant frequency may
not be so important).
. . .

No one with a basic understanding of transmission lines would think that
the frequency at which resonance occurs (X = 0) at the input end is the
same frequency at which the load is resonant, except for two special
cases -- if the line Z0 equals the load resistance at the load's
resonant frequency, or the line is an integral number of quarter
wavelengths long at the load's resonant frequency. And, as you imply,
the resonant frequency of the antenna itself has no significance.
Transmission lines have been used for over a hundred years for impedance
matching, transforming a load of complex impedance into a purely
resistive impedance of a desired value.

I raised the above arguments just as a confirmation of the fact that
understanding what to do before attempting to adjust antennas is not
that easy.

The way to begin is to gain a basic understanding of how transmission
lines transform impedances. The ARRL Antenna Book is a good resource. If
a person's knowledge is limited to only vague understandings of SWR and
resonance, antennas and transmission lines will be a constant source of
mysterious and unexpected results.

Roy Lewallen, W7EL

Antonio Vernucci wrote:
. . .
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. . .

Yes, and this is why foamed dielectric cable has lower loss than solid
dielectric cable. Not because of lower dielectric loss (at least below a
few GHz), but because it has a larger center conductor for the same
impedance and outside diameter.

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
Antonio Vernucci wrote:
. . .
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. . .

Yes, and this is why foamed dielectric cable has lower loss than solid
dielectric cable. Not because of lower dielectric loss (at least below a
few GHz), but because it has a larger center conductor for the same
impedance and outside diameter.

Roy Lewallen, W7EL


You've got it ... spread the word to all those amateurs who are hung up on
(negligible) dielectric loss!

Chris

"Antonio Vernucci" wrote in
:

....
I raised the above arguments just as a confirmation of the fact that
understanding what to do before attempting to adjust antennas is not
that easy.

Well, it was easier until people that don't understand the fundamentals
of transmission lines got access to instruments that measure R and X, and
used their new found capability to prop up the "resonant antennas work
better" myth.

For many common ham antenna *systems* (eg a length coax feed to a centre
fed, approximately half wave dipole using an effective balun), system
efficiency is best when transmission line losses are least, and
minimising line VSWR is a good first cut for best efficiency. Having done
that, an ATU at the tx to transform the load to that required by the tx
so that it can deliver its rated power with specification linearity may
be needed.

If you drill down on the resonance myth, its greatest validity is that
for some types of antenna systems (including the one described above),
resonance delivers a low VSWR, approximately the minimum VSWR, and in
those systems leads to approximately lowest line loss, resulting in best
efficiency. Nothing to do with the 'technical' explanation that I heard
the other day that a "resonance antenna fairly sucks the energy out of
the transmitter". It is a course a fallacy that resonant antennas
naturally "work better", or that resonance is a necessary condition for
high efficiency.

It is pointed out to me from time to time that the article that I
referred you to earlier is way above the head of the average MFJ259B
user, but it is my contention that you cannot realise much of the
potential of the MFJ259B or the like without understanding transmission
lines. VNAs are the new wave of instruments with potential exceeding
typical user's desire for understanding.

Owen

christofire wrote:

You've got it ... spread the word to all those amateurs who are hung up on
(negligible) dielectric loss!

Chris

I've been doing what I can. I pointed it out on this newsgroup on Sept.
12, 1998, and repeat it whenever the opportunity arises.

Roy Lewallen, W7EL

christofire wrote:
"Cecil Moore" wrote in message
Moral: There is nothing magic about 50 ohms.

Actually, there is something 'magic' about 50 ohms.

It appears that you are using a different definition of
magic from the one I was using soI'll say the same thing
in different words:

There is nothing supernatural about 50 ohms.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

My post below is not exactly on target for the thread, but I believe
useful. It's Sec 11.3 from Chapter 11 of Reflections, the whole of
which is available on my web page at www. w2du.com.
The title of the Sec is "A Reader Self-test and Minimum-SWR
Resistance."

Sec 11.3 A Reader Self-Test and Minimum-SWR Resistance

" Everyone knows that when a 50-ohm transmission line is terminated
with a pure resistance of 50 ohms, the magnitude of the reflection
coefficient,, rho , is 0, and the SWR is 1:1. Right? Of course! With
that in mind, here is a little exercise to test your intuitive skill.
If we insert a reactance of 50 ohm in series with the 50-ohm
resistance, the load becomes Z = 50 + j50. The SWR will be 2.618:1.
Now for the question. With this 50-ohm reactance in the load, is the
SWR already at its minimum value with the 50-ohm resistance, or will
some other value of resistance in the load reduce the SWR below
2.618:1? You say the SWR is already the lowest with the 50-ohm
resistance, because, after all, the line impedance, ZC, is 50 ohms?
Sorry, wrong. With reactance in the load, the minimum SWR always
occurs when the resistance component of the load is greater than ZC.
In fact, the more the reactance, the higher the resistance required
for to obtain minimum SWR. For any specific value of reactance in the
load there is one specific value of resistance that produces the
lowest SWR. I call this resistance the "minimum-SWR resistance."
Finding the value of this resistance is easy. First you normalize the
reactance, X, by dividing it by the line impedance, ZC. The normalized
value of X is represented by the lower case x. Thus x = XC / ZC. Then
we solve for the normalized value of resistance r, from Eq 5-1, which
is repeated here.

r = sqrt (x^2 - 1) Eq 5-1

Let's try it on the example above. The normalized value of 50 ohms
of reactance X, is x = 1. Substituting in Eq 5-1, r = sqrt 2 = 1.414.
So the true value of the minimum-SWR resistance is 1.414 x 50 =
70.7ohms. While the 50-ohm resistance yields a 2.618:1 SWR, the
70.7-ohm resistance in series with the 50-ohm reactance yields an SWR
of 2.414:1. Not a great deal smaller, but still smaller than with the
50-ohm resistance.

So let's try a more dramatic example, this time with a 100-ohm
reactance, which has a normalized value x = 2.0. With a 50-ohm
resistance, the SWR is now 5.828:1. However, with the normalized
minimum-SWR resistance, r = sqrt 5 = 2.236. Multiplying by 50, we get
R = 111.8 ohms. With this larger resistance in series with the 100-ohm
reactance, the SWR is reduced from 5.828:1 to 4.236:1. The results of
this exercise didn't turn out quite the way you expected, did it?"

For further proof of this concept I suggest reviewing the remainder of
this Sec using the Smith Chart, available from my web page.

Walt, W2DU