On 7/8/2015 1:14 PM, 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.

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

On 7/8/2015 2:38 PM, John S wrote:
On 7/8/2015 10:48 AM, rickman wrote:
On 7/8/2015 10:09 AM, John S wrote:
On 7/2/2015 1:38 PM, rickman wrote:
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...

Try this - connect the output of an HF transmitter to an SWR bridge.
Now connect a piece of 75 ohm coax such as RG-59 to the output of the
SWR meter, and connect that to a 75 ohm resistive load. Do you think
the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1.



What you have described is a case of using the wrong swr bridge. You
are
trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm
bridge is
used it will show a 1:1 SWR.

The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no
reflected power.

My knowledge of antenna systems is limited, but I do know that this is
correct, there will be no reflection from the antenna.

If there is no reflections from the antenna, how can there be a loss in
the source end? There is NO power returned according to your own
statement.

I don't see any contradiction. The power comes from the source through
the source impedance. The source impedance will create a loss, no?


If the
transmitter output is 50 ohms there will be a loss in this matching
that
will result in less power being delivered to the feed line, but that
will not result in reflections in the feed line.

Why? What causes the loss? The transmitter output resistance? So that
would mean that one can never achieve more that 50% efficiency at the
transmitter's OUTPUT! And that would mean that a 1000W transmitter is
dissipating 500 watts under the BEST circumstances. Good luck on getting
that to work to your satisfaction.

Maybe "loss" isn't the right term then. The output of a 50 ohm source
driving a 75 ohm load will deliver 4% less power into the load than when
driving a 50 ohm load. That comes to -0.177 dB. Is there any part of
that you disagree with?

All of it. Let's say you have a 1A source and it has a 50 ohm impedance
in series with its output. With a 50 ohm load it will provide 50W to the
load. With a 75 ohm load it will provide 75W to the load. The only
difference is that the 50 ohm load will cause the source voltage (before
the series impedance) to be 100V while the 75 ohm load will require 112V
(before the series impedance). If the series impedance is 0 +/- j75
ohms, it will have no power loss. If the series impedance is 50 + j0 it
will have a 50W loss.

Oops! Source voltage will be 70.7V for 50 ohms and 90V for 75 ohms and
dissipation-less output impedance.

wrote in message
...

Once again, SWR is defined in terms of SOURCE impedance and LOAD
impedance. The normal LOAD for a transmitter is one end of a piece
of coax with an antenna on the other end.

The SWR at the near end of a piece of coax may or may not be the
same as the SWR at the far end of the coax.


--
Jim Pennino

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

I have never seen anything that mentions the Source impedance. Just the
ratio of the voltage or current going forward and reflected.

The SWR has to be the same at any point on the coax or transmission line
minus the loss in the line. A simple swr meter may show some differance
because of the way that kind of meter works. By changing the length of the
line , the apparent SWR may be differant at that point.

wrote:

Roger Hayter wrote:
Jeff wrote:


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


The laws are exactly the same for dc as ac, just at dc the frequency
dependant parts tend to the point that they are of no consequence.

Jeff

And, of course, they all do matter in the short time after switch-on
when things are settling.

To be pendatic again, there are 3 types of analysis: DC, AC, and transient.

In the real world, there are 3 sets of "laws" or equations for each case,
with DC being time invariant, AC in the frequency domain, and transient
in the time domain.

There is only one set of laws, but the maths is simpler for the simpler
cases. But the equations for the transient case will still give the
right answer for the DC case.

--
Roger Hayter

wrote:

John S wrote:
On 7/7/2015 1:58 PM, wrote:
John S wrote:
On 7/5/2015 7:21 PM, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately
50 Ohms as is trivially shown by reading the specifications for
the transmitter which was designed and manufactured to match a 50
Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything
after that false.


A quick google demonstrates dozens of specification sheets that
say the transmitter is designed for a 50 ohm load, and none that
mention its output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50
Ohm coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you
have EZNEC, you will not find a place to specify source impedance
but it will show the SWR.

A Smith chart is normalized to 1.

EZNEC allows you to set the impedance to anything you want and
assumes the transmission line matches the transmitter.


The EZNEC help file is very comprehensive. Please find any reference to
your assertion that there is an assumption of source impedance there and
provide information for us to verify your assertion.

Why don't you email the author and get his take on your assumptions?



Why don't YOU? You are the one in need of knowledge. If I do it and
report back here you will just doubt it or find something else to argue
about. Better you should do it first-hand.


EZNEC calculates the SWR presented to the SOURCE which is usually
placed at the antenna terminals.

EZNEC also calculates the SWR presented to the SOURCE which can be
modeled as a SOURCE at one end of a transmission line and the antenna
at the other end.

SWR is defined in terms of SOURCE impedance and LOAD impedance.

I am tired of typing in the same equations over and over again.


Zo is the characteristiic impedance of the transmission line and nothing
to do with the source impedance of whatever generator is supplying power
to the system.



--
Roger Hayter

Roger Hayter wrote:
wrote:

Roger Hayter wrote:
Jeff wrote:


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


The laws are exactly the same for dc as ac, just at dc the frequency
dependant parts tend to the point that they are of no consequence.

Jeff

And, of course, they all do matter in the short time after switch-on
when things are settling.

To be pendatic again, there are 3 types of analysis: DC, AC, and transient.

In the real world, there are 3 sets of "laws" or equations for each case,
with DC being time invariant, AC in the frequency domain, and transient
in the time domain.

There is only one set of laws, but the maths is simpler for the simpler
cases. But the equations for the transient case will still give the
right answer for the DC case.

Only for t = infinity.


--
Jim Pennino

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?


--
Jim Pennino

John S wrote:
On 7/8/2015 1:14 PM, 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.

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

The "base definition" can be whatever set of equations you pick that
are true.

BTW, the Vmax/Vmin DOES have a frequency dependant component that
determines WHERE Vmax and Vmin occur.

--
Jim Pennino

John S wrote:
On 7/8/2015 1:18 PM, wrote:
John S wrote:
On 7/7/2015 1:52 PM, wrote:
Brian Reay wrote:

Do the experiment.

Did it decades ago in electromagnetics lab with calibrated test equipmemnt,
not with amateur radio equipment.

Post the original lab notes, please. That way we cannot challenge the
accuracy of your memory.

Sorry, that was decades ago.

If you are so convinced, do the experiments yourself and post the results.

Or you could read an electromagnetics text on transmission lines and
show me the errors of my statements.

I did, decades ago. The results are that you are wrong. You surely trust
my memory as well as I trust yours, yes?

What I trust is what I can read in an electromagnetics text.


--
Jim Pennino

Ralph Mowery wrote:

wrote in message
...

Once again, SWR is defined in terms of SOURCE impedance and LOAD
impedance. The normal LOAD for a transmitter is one end of a piece
of coax with an antenna on the other end.

The SWR at the near end of a piece of coax may or may not be the
same as the SWR at the far end of the coax.


--
Jim Pennino

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

I have never seen anything that mentions the Source impedance. Just the
ratio of the voltage or current going forward and reflected.

It is generally not mentioned in Amateur publications.

The SWR has to be the same at any point on the coax or transmission line
minus the loss in the line. A simple swr meter may show some differance
because of the way that kind of meter works. By changing the length of the
line , the apparent SWR may be differant at that point.

There is no such thing as apparent SWR. It is what it is in a given
place.

Transmission line transformers.

http://highfrequencyelectronics.com/...TraskPart2.pdf

Impedance matching.

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


--
Jim Pennino