An antenna question--43 ft vertical
 

"Jeff" wrote in message
...
By 'apparent SWR' he means as indicated SWR on the meter, and yes it can
change at various point on the line due to inadequacies in the meter;
the
'real' VSWR will of course remain the same at any point on a lossless
line.

Jeff

That is what I mean Jeff. If there is any SWR, by changing the length of
the line, the voltage/current changes in such a maner that at certain
points
you may get a 50 ohm match at that point.


Absolutely NOT. By changing the length of a transmission you will NEVER
create the situation where you get a 50 ohm match from an initial
mismatch.

This is clearly demonstrable on a Smith chart. Take any starting point
other than a pure 50 ohms and add a length of transmission line. What you
will find is that as you increase the length of line your point will
merely rotate around the chart at a fixed radius (known as a constant VSWR
circle), it will never spiral into the centre which is 50 ohms and where
it must be for a perfect match.

The only time that it will start to spiral inwards is if the line is
lossy, but you will need a very long length, and the spiralling inwards is
due to the loss in the coax NOT any matching characteristics due to the
length of line.

If such an effect as you are talking about is observed it is merely due to
the finite, and often poor, directivity of the SWR meter giving you a
false reading.

Also it is worth noting that achieving 50 ohms as a magnitude |Z| of the
complex impedance (Sqrt(R^2+X^2)) is not the same as getting a good match
with 50 ohms resistive. Even if |Z| = 50 ohms it will have a VSWR greater
than 1 if Z0. Again, plot the point on a Smith chart and you will see
that it can never be in the centre of the chart.

Jeff

That is easy to disprove Jeff.

If I have a 50 ohm load and use a 1/2 wave of any impedance line other than
50 ohms, the swr will be greater than 1:1, except at 1/2 wave multiplies of
the line. At this point there will be a 50 ohm match. The swr of the
line will not actually be 1:1 but some greater value.

An antenna question--43 ft vertical
 

On 7/9/2015 11:40 AM, Jeff wrote:
you may get a 50 ohm match at that point.

https://en.wikipedia.org/wiki/Standi...dance_matching

"if there is a perfect match between the load impedance Zload and the
source impedance Zsource=Z*load, that perfect match will remain if the
source and load are connected through a transmission line with an
electrical length of one half wavelength (or a multiple of one half
wavelengths) using a transmission line of any characteristic impedance
Z0."

This wiki article has a lot of good info in it. I have seen a lot of
stuff posted here that this article directly contradicts.... I wonder
who is right?


That is a very specific case where the source is not at the system
impedance and happens to be equal to the load impedance, there will also
be standing waves on the transmission line and associated losses as the
VSWR on the line will be equal to the magnitude of the mismatch between
the transmission line impedance and the load impedance.

Jeff

Yes, this is a specific case just as you indicated in the section you
both wrote and snipped...

By 'apparent SWR' he means as indicated SWR on the meter, and yes it can
change at various point on the line due to inadequacies in the meter; the
'real' VSWR will of course remain the same at any point on a lossless
line.

Jeff

"It can change at various points on the line".

That's all I was attempting to indicate.

--

Rick

An antenna question--43 ft vertical
 

rickman wrote:
On 7/8/2015 9:07 PM, John S wrote:
On 7/8/2015 4:48 PM, wrote:
John S wrote:
On 7/8/2015 12:47 PM, wrote:
John S wrote:

So, at 1Hz the law has changed, eh? What new law do I need to use?

To be pendatic, there is only one set of physical laws that govern
electromagnetics.

However for DC all the complex parts of those laws have no effect and
all the equations can be simplified to remove the complex parts.

In the real, practical world people look upon this as two sets of
laws, one for AC and one for DC.

A good example of this is the transmission line which does not exist
at DC; at DC a transmission line is nothing more than two wires with
some resistance that is totally and only due to the ohmic resistance
of the material that makes up the wires.

So, is .01Hz AC or DC, Jim? How about 1Hz? 10Hz? Where does AC begin and
DC end?

It is called a limit.

If there is NO time varying component, it is DC, otherwise it is AC.

Are you playing devil's advocate or are you really that ignorant?

Then there is no such thing as DC because even a battery looses voltage
over a period of time. DC voltage sources have noise.

Are just being argumentative or are you really that ignorant?

Even if you have a theoretical voltage source, there are no circuits
(other than imaginary) that have been on since before the big bang and
will be on for all time in the future.

So what?

Is there some point to all this other than to be argumentative?

How long before someone brings up the fact that a resistor generates
AC signals as some kind of straw man objection to DC theory?


--
Jim Pennino

An antenna question--43 ft vertical
 

Jeff wrote:

Can you measure VSWR on a 1 meter long Lecher line at 1 MHz?

VSWR is not meaningful in such a situation, however, you can measure
return loss and Reflection Coefficient etc.. Of course that in not to
say that VSWR is not used in situations where it is not appropriate in
order to indicate how good a match is, when RL or Reflection Coefficient
would be more appropriate.

Jeff

Jeff

Are you trying to say that VSWR is not meaningfull at 160M (to put it
in an Amateur context)?

For those that don't know, a Lecher wire is just a carefully contructed,
rigid parallel transmission line upon which one would slide a high
impedance sensor to find voltage minimum, maximum, and where they
occured. That and a Smith chart were used to solve transmission line
and impedance matching problems and were often home built by Amateurs
in the early VHF days.

Today you would use a VNA (Vector Network Analyzer).

--
Jim Pennino

An antenna question--43 ft vertical
 

Jeff wrote:
Actually, VSWR can be defined several ways, one of which is:

(1 + |r|)/(1 - |r|)

Where r is the reflection coefficient which can be defined a:

(Zl - Zo)/(Zl + Zo)

Where Zl is the complex load impedance and Zo is the complex source
impedance.

Note that a complex impedance has a frequency dependant part.

So, since Vmax/Vmin (the base definition) has no frequency dependent
part, does that invalidate it?


No, of course not, the other equations are NOT a definition of VSWR,
they are formulas the link other quantities to VSWR.

So who exactly declared which set of definitions is the one and
true definition of VSWR?

Is P=EI or P=E^2R?

Taking Reflection Coefficient for example, it requires the phase
information to be removed before conversion to VSWR by using only its
magnitude.

So what?

To emphasise this VSRW is phase independent and on a lossless
transmission line; its value does not change anywhere along that line;
that is equivalent to rotating around a constant VSWR circle on a Smith
Chart.

Jeff

--
Jim Pennino

An antenna question--43 ft vertical
 

Jeff wrote:
On 08/07/2015 19:14, wrote:
John S wrote:
On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

Actually, VSWR can be defined several ways, one of which is:

(1 + |r|)/(1 - |r|)

Where r is the reflection coefficient which can be defined a:

(Zl - Zo)/(Zl + Zo)

Where Zl is the complex load impedance and Zo is the complex source
impedance.

Note that a complex impedance has a frequency dependant part.



Note the the definition of VSWR uses the magnitude of the reflection
coefficient, |r|, which removes the phase and frequency dependant parts.

Jeff

The magnitude DEPENDS on the frequency dependant parts.

--
Jim Pennino

An antenna question--43 ft vertical
 

Jeff wrote:

Note the the definition of VSWR uses the magnitude of the reflection
coefficient, |r|, which removes the phase and frequency dependant parts.

Jeff

The magnitude remains frequency dependent.


It may or may not be depending on the load and how the phase of the
reflection coefficient changes with frequency.

Ergo it DOES remain frequency dependent in the general case.


--
Jim Pennino

An antenna question--43 ft vertical
 

Wayne wrote:


wrote in message ...

John S wrote:
On 7/8/2015 7:27 PM, Wayne wrote:


"John S" wrote in message ...

On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

That's also my understanding of the definition.
In fact since SWR is defined as the maximum to minimum voltage ratio,
the "V" in VSWR is redundant.

Sort of. There is also ISWR but it is not used frequently.

# Not sort of, but is.

# There is also PSWR.

And both go back to the Vmax/Vmin definition.

The PSWR is a tricky one because you can end up with a power ratio instead
of a voltage ratio.

Actually, no, PSWR has nothing to do with power ratios as in RF power,
rather it has to do with power ratios as in values raised to the second
power.


--
Jim Pennino

An antenna question--43 ft vertical
 

rickman wrote:
On 7/8/2015 7:43 PM, wrote:
Ralph Mowery wrote:

wrote in message
...
Ralph Mowery wrote:


Can you show any place where the SWR definition mentions the Source
impedance ?

I have several times now, but once again:

SWR = (1 + |r|)/(1 - |r|)

Where r = reflection coefficient.

r = (Zl - Zo)/(Zl + Zo)

Where Zl = complex load impedance and Zo = complex source impedance.

https://en.wikipedia.org/wiki/Reflection_coefficient

http://www.antenna-theory.com/tutori...nsmission3.php

You might check that again. I don't see Zo being defined as the complex
source impedance, but rather as the transmission line characteristic
impedance... not the same thing at all.


YOu have just proven my point. Read carefully from your refernce to
Wikipedia :

"The reflection coefficient of a load is determined by its impedance and
the impedance toward the source."

Notice it says TOWARD and not THE SOURCE.

Notice it actually says "the impedance toward the source".

From the second referaence notice that it says load impedance and impedance
of the transmission line. Nothing mentions the source at all:

What the hell do you think the transmission line is in this case if
not the source?

"The reflection coefficient is usually denoted by the symbol gamma. Note
that the magnitude of the reflection coefficient does not depend on the
length of the line, only the load impedance and the impedance of the
transmission line. Also, note that if ZL=Z0, then the line is "matched". In
this case, there is no mismatch loss and all power is transferred to the
load."

Perhaps you would like the second link better as it has pictures.

Of maybe this one that explains it all starting with lumped equivelant
circuits.

http://www.maximintegrated.com/en/ap...dex.mvp/id/742

Notice that ALL the links talk about the source impedance.

How about this one?

https://en.wikipedia.org/wiki/Standi...dance_matching

I think this has some very interesting analysis, very specifically
referring to "purely resistive load impedance".

So what?

A purely resistive anything is a special case of the general problem.


--
Jim Pennino

An antenna question--43 ft vertical
 

On 7/9/2015 1:27 PM, wrote:
rickman wrote:
On 7/8/2015 9:07 PM, John S wrote:
On 7/8/2015 4:48 PM, wrote:
John S wrote:
On 7/8/2015 12:47 PM, wrote:
John S wrote:

So, at 1Hz the law has changed, eh? What new law do I need to use?

To be pendatic, there is only one set of physical laws that govern
electromagnetics.

However for DC all the complex parts of those laws have no effect and
all the equations can be simplified to remove the complex parts.

In the real, practical world people look upon this as two sets of
laws, one for AC and one for DC.

A good example of this is the transmission line which does not exist
at DC; at DC a transmission line is nothing more than two wires with
some resistance that is totally and only due to the ohmic resistance
of the material that makes up the wires.

So, is .01Hz AC or DC, Jim? How about 1Hz? 10Hz? Where does AC begin and
DC end?

It is called a limit.

If there is NO time varying component, it is DC, otherwise it is AC.

Are you playing devil's advocate or are you really that ignorant?

Then there is no such thing as DC because even a battery looses voltage
over a period of time. DC voltage sources have noise.

Are just being argumentative or are you really that ignorant?

Even if you have a theoretical voltage source, there are no circuits
(other than imaginary) that have been on since before the big bang and
will be on for all time in the future.

So what?

Is there some point to all this other than to be argumentative?

How long before someone brings up the fact that a resistor generates
AC signals as some kind of straw man objection to DC theory?

The point is that separating DC and AC as being ruled by separate "laws"
is pointless. Just discuss the topic of interest rather than digressing
onto pointless diversions.

--

Rick