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.

But, using the voltage ratio definition, you can work yourself back
around to various source and load impedances, reflection coefficients, etc.

Sure. In fact I once built a slotted line to measure load impedances
because I could not afford to buy a vector voltmeter at the time.

On 7/8/2015 4:52 PM, wrote:
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.

Fine. Then you should confine your arguments to those sources rather
than referring to decades old measurements to you may not remember
correctly.

On 7/8/2015 8:34 PM, wrote:
rickman wrote:
On 7/8/2015 3:38 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.

I was referring to a voltage source.


Instead of arguing about it, one can download QUCS for free which
will simulate the whole thing and one can see what really happens.

Download QUCS for your operating system:

http://qucs.sourceforge.net/

Generate a model consisting of a voltage source with a series resistance
of a few Ohms to simulate a solid state source or a much higher
resistance to simulate a vacuum tube source. Chose a convienient
frequency for the source.

Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html
to calculate an impedance matching network to match the resistance
you've chosen to 50 Ohms.

Put the matching circuit in the model.

Add a transmission line to the model.

Terminate the transmission line with a 50 Ohms resistor.

Add a fixed frequency AC simulation at the desired frequency.

Change various parameters to your heart's content to see what happens.

Change the matching network such that the output of your transmitter
is no longer 50 Ohms and see what happens.

When the QUCS output disagrees with your beliefs, you can argue with
the program.

Uh, I already did a simulation using LTspice. No one even commented on
the simulation as I recall. Besides, I don't really need a simulation
to measure the power delivered to a load. That is a *very* simple
circuit to calculate in a few seconds. But first we have to agree on
what we are discussing.

I'm not going to retrace this entire conversation, but someone said a
matching network which presents a complex impedance with a non-zero real
part and a zero imaginary part would reflect 100% of the wave which had
been reflected from the antenna back to the transmitter output. I was
trying to nail down this example so I could do some calculations on it.
If you want to do that I am happy to work on the problem. If not,
that's fine too.

--

Rick

On 7/8/2015 6:43 PM, wrote:
Ralph Mowery wrote:

wrote in message
news
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 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.

So, you are saying Zo is the source impedance while every one else
thinks it is the characteristic impedance of the line. Go back to your
books and look up the definition of Zo.

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

wrote in message
news
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".

--

Rick

rickman wrote:
On 7/8/2015 8:34 PM, wrote:
rickman wrote:
On 7/8/2015 3:38 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.

I was referring to a voltage source.


Instead of arguing about it, one can download QUCS for free which
will simulate the whole thing and one can see what really happens.

Download QUCS for your operating system:

http://qucs.sourceforge.net/

Generate a model consisting of a voltage source with a series resistance
of a few Ohms to simulate a solid state source or a much higher
resistance to simulate a vacuum tube source. Chose a convienient
frequency for the source.

Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html
to calculate an impedance matching network to match the resistance
you've chosen to 50 Ohms.

Put the matching circuit in the model.

Add a transmission line to the model.

Terminate the transmission line with a 50 Ohms resistor.

Add a fixed frequency AC simulation at the desired frequency.

Change various parameters to your heart's content to see what happens.

Change the matching network such that the output of your transmitter
is no longer 50 Ohms and see what happens.

When the QUCS output disagrees with your beliefs, you can argue with
the program.

Uh, I already did a simulation using LTspice. No one even commented on
the simulation as I recall. Besides, I don't really need a simulation
to measure the power delivered to a load. That is a *very* simple
circuit to calculate in a few seconds. But first we have to agree on
what we are discussing.

Does LTspice do transmission lines?

I'm not going to retrace this entire conversation, but someone said a
matching network which presents a complex impedance with a non-zero real
part and a zero imaginary part would reflect 100% of the wave which had
been reflected from the antenna back to the transmitter output. I was
trying to nail down this example so I could do some calculations on it.
If you want to do that I am happy to work on the problem. If not,
that's fine too.

Nope, I know what happens.

BTW, my response was not directed at any particular person other
than those that do not understand the conditions for maximum
power transfer given a fixed source.



--
Jim Pennino

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.

An ideal battery doesn't.


--
Jim Pennino

John S wrote:
On 7/8/2015 4:51 PM, wrote:
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.


You are just being argumentative. The WHERE doesn't matter in measuring
VSWR. You still measure correct VSWR wherever the locations.

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



--
Jim Pennino

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.


--
Jim Pennino

John S wrote:
On 7/8/2015 4:52 PM, wrote:
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.

Fine. Then you should confine your arguments to those sources rather
than referring to decades old measurements to you may not remember
correctly.

Instead of arguing about it, one can download QUCS for free which
will simulate the whole thing and one can see what really happens.

Download QUCS for your operating system:

http://qucs.sourceforge.net/

Generate a model consisting of a voltage source with a series resistance
of a few Ohms to simulate a solid state source or a much higher
resistance to simulate a vacuum tube source. Chose a convienient
frequency for the simulation.

Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html
to calculate an impedance matching network to match the resistance
you've chosen to 50 Ohms at the choosen frequency.

Put the matching circuit in the model.

Add a 50 Ohm transmission line to the model.

Terminate the transmission line with a 50 Ohm resistor.

Add a fixed frequency AC simulation at the choosen frequency.

Change various parameters to your heart's content to see what happens.

Change the matching network such that the output of your transmitter
is no longer 50 Ohms and see what happens.

When the QUCS output disagrees with your beliefs, you can argue with
the program.


--
Jim Pennino