John S wrote:
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.

When a transmission line is connected to a load, the source for the load
IS the end of the transmission line.

Where else would you think the source is?

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

In message ,
writes


There are other kinds of "SWR", but you will never see them refered to
in Amateur literature.

What's wrong with good old-fashioned 'SWR'? Is the VSWR different from
the C*SWR?
*Current



--
Ian

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?

What you say is true in a literal sense. However there is certainly
such a thing as DC from the POV of what you want to know about a system
being calculated or measured with sufficient and relevant accuracy and
precision by treating it as DC. In real life this is remarkably common.


--
Roger Hayter

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 remains frequency dependent.

--
Roger Hayter

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.

--

Rick

"Jeff" wrote in message
...

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.


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.

On 7/9/2015 9:14 AM, Ralph Mowery wrote:
"Jeff" wrote in message
...

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.


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.

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?

--

Rick

On 7/9/2015 12:31 AM, wrote:
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?

Yes.

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.

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.

On 7/8/2015 8:25 PM, 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.


Did you include a transmission line? If not, see
TransmissionLineInverter.asc in the \Program Files
(x86)\LTC\LTspiceIV\examples\Educational folder to get you started.

Do you have EZNEC?

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.