John S wrote:
On 7/8/2015 1:39 PM, 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.

No. It is defined as Vmax/Vmin on the line. Show an equation that
defines SWR as the matching of the source to the line.

OK, since you insist, one more time:

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


--
Jim Pennino

Roger Hayter wrote:
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.

No, Zo is the source impedance and it is irrelevant what the source
physically is and has nothing to do with power.

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

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



--
Jim Pennino

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.

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

"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."

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.



--
Jim Pennino

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.

--

Rick

"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.

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

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.



--
Jim Pennino

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.

That is one definition.

In fact since SWR is defined as the maximum to minimum voltage ratio, the
"V" in VSWR is redundant.

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

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

Or you can start from source and load impedances.

The whole Vmax/Vmin definition is a leftover from the early days when
RF measurments were done on Lecher lines or slotted lines and you
actually looked for the min and max points on the line.

I doubt few under about 50 have heard of slotted lines or Lecher lines
much less ever used one.


--
Jim Pennino

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?

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.