On Sun, 20 Sep 2009 22:41:29 -0400, Walter Maxwell
wrote:

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

On Sun, 20 Sep 2009 22:41:29 -0400, Walter Maxwell
wrote:

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

Sorry, I goofed on Eq. 5-1. The corrected eq is r = sqrt (x^2 + 1).

Walt, W2DU

Walter Maxwell wrote in
:

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!

Well, it for a distortionless 50 ohm line.

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?

Continuing on the distortionless example, if you visualise this on a
Smith chart, for any constant X and R independently variable, the value
of R for minimum VSWR will be such that the tangent to the reactance
circle is also a tangent to the VSWR circle at that point (R,X), and R
for minimum VSWR will always be greater than Ro for Xl0.

However, Zo for practical cables is not real, not quite.

Owen

christofire wrote:

"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!

It isn't the amateurs so much as the advertising. Marketing departments
highlight the foam dielectric because it's more obvious, and pretty soon
even the manufacturers are believing their own publicity.

As for the 50-ohm impedance, the reasons why it became an industry
standard are interesting but purely historical. The reason for using it
now is almost exclusively because it *is* an industry standard.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

On Sep 21, 1:32*am, Ian White GM3SEK wrote:
christofire wrote:

"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!

It isn't the amateurs so much as the advertising. Marketing departments
highlight the foam dielectric because it's more obvious, and pretty soon
even the manufacturers are believing their own publicity.

As for the 50-ohm impedance, the reasons why it became an industry
standard are interesting but purely historical. The reason for using it
now is almost exclusively because it *is* an industry standard.

--

73 from Ian GM3SEK * * * * 'In Practice' columnist for RadCom (RSGB)http://www.ifwtech.co.uk/g3sek

Resonance means little. It is like reverse engineering where it is
assumed or understood that the transmission line will be 50 ohms! The
point to remember is that the less the resistive component that one
measures at the antenna the more the power is shifting over from the
resistive loss to the radiative resistance and nothing more. For
matching there is an advantage when the load is totally resistive
because the matching becomes less complicated. Obviously as more
energy is shifted over to radiative purposes it is more difficult to
feed as we do not know how to switch power transmission from a
parallel line to a singular line. But the fact remains, the less the
resistive losses the more power goes to radiation which is exactly
what you are trying to achieve.

Art Unwin wrote:
Resonance means little.

snip a bit

The point to remember is that the less the resistive component that one
measures at the antenna the more the power is shifting over from the
resistive loss to the radiative resistance and nothing more.

It is truly amazing the things that come from his keyboard. This one
statement proves he understands nothing.

And he contradicted his previous stand on resonance, too! 2 in one blow.

tom
K0TAR