On 7/8/2015 6:32 AM, 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

Thanks, Jeff. I agree, of course.

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.

--
Roger Hayter

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.

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.

On 7/7/2015 1:47 PM, Dave Platt wrote:
In article , rickman wrote:

Lol. You are a trip. I'm not going to spend $100 on a book just to see
if you are right. I was intrigued by the idea that a wire could carry a
signal without the resistance dissipating power according to P = I^2 R.
I guess there is some communication failure.

Yah.

It's a question of terminology. Unfortunately, one term has come to
be used for two (related but different) concepts.

There is "resistance", as in the E=I^2*R sort. If I recall correctly,
Maxwell refers to this as "dissipative" impedance. If you put current
through a dissipative resistance, a voltage drop develops across the
resistance, and power is dissipated.

There are plenty of examples of this, with which I'm sure you're
familiar.

There is also "resistance", as in "the 'real', non-reactive component
of a complex impedance, in which current is in phase with voltage."
This type of "resistance" is fundamentally non-dissipative - that is,
you can run power through it without dissipating the power as heat.

There are also good examples of this. One "textbook" example would be
a perfectly-lossless transmission line... say, one made out of a wire
and tube of a superconductor, cooled to below the superconducting
temperature.

I have analyzed this previously. Unless your transmission line is
infinitely long, eventually the wave reaches the other end and is either
dissipatively absorbed or is reflected back to the source where it
interacts. An infinitely long transmission line is not very
interesting. Dealing with the reflection from a finite transmission
line is what we are trying to analyze, so that model is not very useful.

So, no, I am not familiar with a non-dissipative load with current in
phase with the voltage.


You can (in principle) build such a superconducting coax to have
almost any convenient impedance... 50 or 75 ohms, for example. Since
we're theorizing, let's assume we can built one a few trillion miles
long... so long that the far end is light-years away.

If you hook a transmitter to one end of this and start transmitting,
it will "look" to the transmitter like a 50-ohm dummy load. The
transmitter itself won't be able to tell the difference. The
transmitter puts out an RF voltage, and the line "takes current"
exactly in phase with the voltage, in a ratio of one RF ampere per 50
RF volts.

But, there's a fundamental difference between this "resistance" and
that of a dummy load. A 50-ohm dummy load's resistance is
dissipative... all of the power going into it turns into heat, and is
dissipated in accordance with the fundamental laws of thermodynamics.

*None* of the power being fed into the superconducting coax, is
dissipated as heat in the coax. All of the power still exists, in its
original RF form. It's being stored/propagated down the coax without
loss.

When it hits a load at the other end, it may be dissipated as heat
there.

Or, perhaps not. What if what's at the other end of the
superconducting coax is a superconducting antenna, tweaked to present
an impedance of exactly 50 ohms? The RF will be radiated into space.

And, "free space" is another great example of a medium that has a
well-defined "resistance" (in the non-dissipated sense).

I believe this radiation *is* dissipative in the sense that the power is
removed from the system being analyzed. That is *exactly* why it is
considered to be due to a radiation "resistance". Note they do not
refer to it as a radiation "impedance".


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

One of the fundamental jobs of an antenna, is to match the impedance
of its feedline to the impedance of free space.

Now, any coax you can buy at the store has *both* types of
"resistance", of course. It has a dissipative component, and a
non-dissipative component. Typically, the more you spend and the more
you have to strain your back carrying it around, the lower the amount
of dissipative resistance (which is only good for keeping the pigeons'
feet warm) and the more predictable and precisely-defined the
non-dissipative part.

What you are calling "non-dissipative resistance" is only a way to
characterize the AC behavior. It has nothing to do with what we are
discussing and is in no way similar to resistance. Trying to analyze a
transmission line without considering the reflections from the other end
is only a transient solution which ignores the behavior of the
transmission line.

How about we construct an example circuit with a conjugate matching
network rather than deal with abstractions that have nothing to do with
the discussion?

If you want to continue to discuss the transmission line, then we need
to consider the reflection and find a steady state solution, not a
transient one, right?

--

Rick

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?

Nothing in this analysis addresses the theoretical maximum possible
efficiency of an arbitrary transmitter and an arbitrary load. In
particular I posted the results of a simulation that showed very clearly
that the loss in the transmitter output impedance can be well below 50%
of the total power drawn from the PSU. Just set the load impedance and
make your output impedance as low as you would like.

It is when you set the output impedance of the transmitter to a fixed
value that a matched load impedance will draw the maximum power from the
transmitter while the loss in the transmitter output will be 50%.

--

Rick

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.



--
Jim Pennino

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.



--
Jim Pennino

Jeff wrote:
On 07/07/2015 19:44, 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.



VSWR is defined as Vmax/Vmin on the transmission line and is independent
of phase or wavelength.

It can also be expressed in terms of the *magnitude* of the reflection
co-efficient. 1+|p|/1-|p|

The reflection coefficient can also be expresses as (Zl - Zo)/(Zl + Zo)
where Zl is the complex load impedance and Zo is the complex source
impedance.

The complex impedances are functions of wavelengths, i.e. frequency.

All complex numbers have a frequency dependant part.

In the real world of transmission lines and antennas, the source
impedance is usually 50 + j0 and thus the second part is ignored.

or in terms of forward and reflected power
1+sqtr(Pr/Pf)/1-sqrt(Pr/Pf))

VSWR is the same regardless, of phase, when measured at any point on a
lossless line. The phase of the reflection co-efficient will change but
not its magnitude.

Jeff



--
Jim Pennino

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.


--
Jim Pennino

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.


--
Jim Pennino