VSWR doesn't matter?
 

Richard Fry wrote:

Maximizing the output power and efficiency of a broadcast r-f amplifier
dictates that its effective output Z must be greatly different than the
load Z it is expected to drive. In the case of broadcast transmitters,
that source impedance is low (a few ohms), compared to the typical 50 or
75 ohm Zo of the load it is driving.

And this it the reason that much of the voltage reflected from an
antenna/far-end mismatch returns from the tx back to the antenna to be
radiated...

This is not correct. It has nothing to do with the output impedance of
the drivers. Even a 50 ohm Thevenin source will reflect _all_ the power
back to the load as long as it is 'matched'.

Here is how to prove this to your self. Take a transmitter and tune it
up into a dummy load. The effective output impedance is now 50 ohms. Now
plug that transmitter into a trans-match that drives some coax to a long
wire. When the match is tuned there will be no energy flowing from the
match back to the transmitter. All the energy reflected from the long
wire gets reflected back from the match even if a 50 ohm source is used.

Best, Dan.

VSWR doesn't matter?
 

Cecil Moore wrote:
On Mar 14, 12:24 pm, Gene Fuller wrote:
Au contraire, mon frere. You continue to claim that a standing wave MUST
be made up of two traveling waves, but without proof.

On the contrary, I have presented at least three references as proof.
If I remember correctly, it was Ramo, Whinnery, Hecht, and Balanis.
You, OTOH, have presented none.

My contention is that this distinction is merely a matter of
mathematical preference. When standing waves occur, there is absolutely
no physical difference between the standing wave and its traveling wave
constituents.

Obviously false as proven by the different equations for the two types
of waves. We laid that one to rest long ago. In fact, it was you who
pointed out that standing wave phase is completely different from
traveling wave phase and cannot be used to measure phase shift through
a coil. If I remember correctly, it was the difference between
cos(x*wt) and cos(x)*cos(wt), i.e. *very* different.

Water is also a scalar. If you had one gallon per minute flowing into
a barrel and two gallons per minute flowing out of the barrel, would
you argue that there is no water flowing into the barrel and only one
gallon of water flowing out of the barrel? Or would you say the *net*
water flow is one barrel per minute out of the barrel?
This is totally irrelevant to the issue at hand. Try to keep on task.

No, it is virtually identical to your argument. Saying it is "totally
irrevelent" doesn't change anything. You are arguing that net energy
transfer is primary and the underlying energy components are
irrelevant if nonexistant.
--
73, Cecil, w5dxp.com


Cecil,

Until next time. I guess we will continue to disagree.

73,
Gene
W4SZ

VSWR doesn't matter?
 

Gene Fuller wrote:
Until next time. I guess we will continue to disagree.

Gene, how can you insist that standing waves are just
like traveling waves when their equations are so
different - different enough to make them virtually
opposites. The phase of a traveling wave varies with
distance - the phase of a standing wave doesn't. The
amplitude of a standing wave varies with distance -
the amplitude of a traveling wave doesn't (in a loss-
less transmission line). I can't think of a way that
those two types of waves are alike except for frequency.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

"Dan Bloomquist" wrote
This is not correct. It has nothing to do with the output impedance of the
drivers. Even a 50 ohm Thevenin source will reflect _all_ the power back
to the load as long as it is 'matched'.
_______________

Following is a quotation from a paper titled "A Study of RF Intermodulation
Between FM Broadcast Transmitters Sharing Filterplexed or Co-Located Antenna
Systems," by G. N. Mendenhall, P.E., who then was the VP of Engineering for
Broadcast Electronics, Inc, and now is VP of Product Engineering for Harris
Corporation, Broadcast Division. Broadcast Electronics and Harris
manufacture a full range of high power broadcast transmitters. Mendenhall
is highly regarded in the broadcast industry.

In the context of the quote, the "interfering signal" means a signal coupled
into the transmitter PA output circuits that was generated external to that
transmitter. But the interfering signal also could be a reflection of the
output signal of that transmitter from a mismatched load. Therefore the
information in the quote addresses the subject of these posts.

QUOTE
"Output Return Loss" is a measure of the amount.of interfering signal that
is coupled into the output circuit versus the amount that is reflected back
from
the output circuit without interacting with the nonlinear device.

To understand this concept more clearly, we must remember that although the
output circuit of the transmitter is designed to work into a fifty ohm load,
the
output source impedance of the transmitter is not fifty ohms. If the source
impedance were equal to the fifty ohm transmission line impedance, half
of the transmitter's output power would be dissipated in its internal output
source impedance.

The transmitter's output source impedance must be low compared to the load
impedance in order to achieve good efficiency. The transmitter therefore
looks
like a voltage source driving a fifty ohm resistive load. While the
transmission
line is correctly terminated looking toward the antenna (high return loss),
the
transmission line is greatly mismatched looking toward the output circuit of
the
transmitter (low return loss). This means that power coming out of the
transmitter
is (almost) completely absorbed by the load while interfering signals fed
into
the transmitter are almost completely reflected by the output circuit.
END QUOTE

If the terminating impedance for reflected/reverse power on a transmission
line looking back into the PA matched the impedance of that transmission
line, it is rather unclear why the PA would reflect _all_ that power back
toward the load, rather than dissipate it in that termination.

RF

VSWR doesn't matter?
 

Cecil Moore wrote:
Gene Fuller wrote:
Until next time. I guess we will continue to disagree.

Gene, how can you insist that standing waves are just
like traveling waves when their equations are so
different - different enough to make them virtually
opposites. The phase of a traveling wave varies with
distance - the phase of a standing wave doesn't. The
amplitude of a standing wave varies with distance -
the amplitude of a traveling wave doesn't (in a loss-
less transmission line). I can't think of a way that
those two types of waves are alike except for frequency.

Cecil,

OK, so you don't want to let this drop quite yet. I have dredged through
the muck of Google archives, and I found the following 5 exact quotes
from you. I believe these fairly represent your position, but if not,
please let me know about others. More on the other end . . .

******************

1) Quoting Balanis: "Standing wave antennas, such as the dipole, can be
analyzed as traveling wave antennas with waves propagating in opposite
directions (forward and backward) and represented by traveling wave
antenna currents I(f) and I(b)."

2) Kraus: "A sinusoidal current distribution may be regarded as the
standing wave produced by two uniform (unattenuated) traveling waves of
equal amplitude moving in opposite directions along the antenna."

3) From "Fields and Waves ...", by Ramo & Whinnery, in describing the
standing wave situation: "The total energy in any length of line a
multiple of a quarter wavelength long is constant, *merely interchanging
between energy in the electric field of the voltages and energy in the
magnetic field of the currents*." Again, proof that standing wave energy
doesn't flow. It just stands there being exchanged between the E-fields
and the H-fields. That is from page 40 of "Fields and Waves in
Communications Electronics", by Ramo, Whinnery, and Van Duzer.

4) I recognize that equation from "Optics", by Hecht. Pick any point,
'z', and see what you get. Hecht says, "It doesn't rotate at all, and
the resultant wave it represents *DOESN'T PROGRESS THROUGH SPACE* - it's
a standing wave." The RF equivalent of a standing wave of light that
doesn't progress through space is an RF standing wave that doesn't
progress through a wire. That's what I have been telling you guys.
Standing waves don't move. Standing wave current doesn't flow! Even in
empty space, a light standing wave doesn't progress through space, i.e.
IT DOESN'T MOVE! That is on page 289 of "Optics", by Hecht, 4th edition.

5) Here's a little help from Hecht of "Optics" fame. (quote)

E(x,t)=2E0t*sin(kx)*cos(wt)

This is the equation for a *standing wave*, as opposed to a traveling
wave. Its profile does not move through space; it is clearly not of the
(traveling wave) form f(x +/- vt) ...
Let the phasor E1 represent a (traveling) wave to the left, and E2 a
(traveling) wave to the right. ... (The sum) doesn't rotate at all, and
the resultant wave it represents doesn't progress through space - it's a
standing wave. (end quote)

******************

Quotes (1) and (2) do not use words that most people would associate
with "proof". Instead, the use of terms such as "can be analyzed" and
"may be regarded" completely support my position that the choice to use
standing waves or traveling waves is simply one of mathematical
convenience. When standing waves exist, there is no physical difference
between the standing waves and their constituent traveling wave components.

Quotes (3), (4), and (5) completely support my position again. When a
standing wave exists, there is no more hidden information buried in the
constituent traveling wave components. No flowing energy waves or other
such nonsense.

It is possible to have many mathematical descriptions of a physical
phenomenon. However, they all need to yield exactly the same physical
predictions or else one or more of the models are incomplete or wrong.

Of course there are traveling waves that are not simply mathematical
components of standing waves. All of the stuff about TDRs and ghosts
falls into that category. This message is not about those traveling
waves at all, so you can forget about bringing up all of your TV ghost
arguments.

73,
Gene
W4SZ

VSWR doesn't matter?
 

Gene Fuller wrote:

Quotes (1) and (2) do not use words that most people would associate
with "proof". Instead, the use of terms such as "can be analyzed" and
"may be regarded" completely support my position that the choice to use
standing waves or traveling waves is simply one of mathematical
convenience.

No technical author is going to be arrogant enough to use
the phrases "must be analyzed" or "must be regarded". Many
shortcuts do work from a mathematical standpoint but often
lose their ability to tell us anything about reality.

You were the first person to point that out - that when two
opposite direction traveling waves are superposed, the sum
of the superposition loses its changing phase. That's why
W7EL's and W8JI's phase measurements through a loading coil
on a standing wave antenna were of no value except to prove
that standing wave current doesn't change phase in a wire or
in a coil.

When standing waves exist, there is no physical difference
between the standing waves and their constituent traveling wave components.

Proven false by the previous quote from "Optics", by Hecht.

E(x,t)=2E0t*sin(kx)*cos(wt)

This is the equation for a *standing wave*, as opposed to a traveling
wave. Its profile does *NOT* move through space; it is clearly *NOT*
of the
(traveling wave) form f(x +/- vt) ...

Hecht apparently assumed the definition of the word "not" is
understood by the average reader and didn't need emphasis
so I added it. :-)

There is an obvious physical difference that can be seen
from the equations. Again, a standing wave has fixed phase
while a traveling wave has a variable phase. A standing
wave has a variable amplitude while a traveling wave has
a fixed amplitude. That's two ways they are entirely
different.

No flowing energy waves or other such nonsense.

Nothing like that assertion is supported in the quotes. Please
point out where any of those references assert that there is no
energy in a reflected wave or that reflected waves do not exist.

Ramo and Whinnery go so far as to vector sum the forward power
flow vector and the reflected power flow vector.

It is possible to have many mathematical descriptions of a physical
phenomenon. However, they all need to yield exactly the same physical
predictions or else one or more of the models are incomplete or wrong.

Plus they need to be linked to reality. Standing waves existing
without the component forward and reverse traveling waves is
divorced from reality. Neither you nor anyone else has been able
to provide even one real-world example of such.

Forward traveling wave + reflected traveling wave = standing wave

What happens to the standing wave when you take away the reflected
wave?

Forward traveling wave + nothing = forward traveling wave

i.e. there is no standing wave. So please tell us again how
you can build a standing wave from a single traveling wave.

... so you can forget about bringing up all of your TV ghost
arguments.

That rug of yours under which you try to sweep all the
reflected energy is going to explode one of these days. :-)
--
73, Cecil, w5dxp.com

VSWR doesn't matter?
 

Cecil Moore wrote:


Forward traveling wave + reflected traveling wave = standing wave

What happens to the standing wave when you take away the reflected
wave?

It's a different physical situation. None of this discussion has any
bearing on the new problem with only one traveling wave. When you when
finally understand the meaning of your own words, there may be hope for
progress. Until then, we are just boring everyone.

73,
Gene
W4SZ

VSWR doesn't matter?
 

Richard Fry wrote:
"Dan Bloomquist" wrote

This is not correct. It has nothing to do with the output impedance of
the drivers. Even a 50 ohm Thevenin source will reflect _all_ the
power back to the load as long as it is 'matched'.

_______________

Following is a quotation from a paper titled "A Study of RF
Intermodulation Between FM Broadcast Transmitters Sharing Filterplexed
or Co-Located Antenna Systems," by G. N. Mendenhall, P.E., who then was
the VP of Engineering for Broadcast Electronics, Inc, and now is VP of
Product Engineering for Harris Corporation, Broadcast Division.
Broadcast Electronics and Harris manufacture a full range of high power
broadcast transmitters. Mendenhall is highly regarded in the broadcast
industry.

In the context of the quote, the "interfering signal" means a signal
coupled into the transmitter PA output circuits that was generated
external to that transmitter. But the interfering signal also could be
a reflection of the output signal of that transmitter from a mismatched
load. Therefore the information in the quote addresses the subject of
these posts.

QUOTE
"Output Return Loss" is a measure of the amount.of interfering signal that
is coupled into the output circuit versus the amount that is reflected
back from
the output circuit without interacting with the nonlinear device.

To understand this concept more clearly, we must remember that although the
output circuit of the transmitter is designed to work into a fifty ohm
load, the
output source impedance of the transmitter is not fifty ohms. If the
source
impedance were equal to the fifty ohm transmission line impedance, half
of the transmitter's output power would be dissipated in its internal
output
source impedance.

The transmitter's output source impedance must be low compared to the load
impedance in order to achieve good efficiency. The transmitter therefore
looks
like a voltage source driving a fifty ohm resistive load. While the
transmission
line is correctly terminated looking toward the antenna (high return
loss), the
transmission line is greatly mismatched looking toward the output
circuit of the
transmitter (low return loss). This means that power coming out of the
transmitter
is (almost) completely absorbed by the load while interfering signals
fed into
the transmitter are almost completely reflected by the output circuit.
END QUOTE

If the terminating impedance for reflected/reverse power on a
transmission line looking back into the PA matched the impedance of that
transmission line, it is rather unclear why the PA would reflect _all_
that power back toward the load, rather than dissipate it in that
termination.

Hi Richard,
He says it in the last sentence. But here is an example. Take a 50 ohm
thevinin source. Power off, it looks like 50 ohms back into it. Take a
second thevinin source to represent a reflection and drive 5 volts into
the first source. Now set your first source 180 degrees to the
reflection and drive forward 5 volts.

(s)-----/\/\/\--------(c)-----/\/\/\--------(r)

(s)source (c)connection (r)reflection. With (s) 180 degrees out of phase
from (r), (r) will see a short at (c). It is because of the power
generated at the source that the impedance into it can look purely
reactive. And, you can use 5 ohms with 1 volt at the source, (c) will
still look like a short to (r). The source resistance doesn't matter as
long as a 'match' is made.

And for the same reason, why the 50 ohm line doesn't look like 50 ohms
is because of reflected power. Drive an open quarter wave line and it
looks like a short because the reflected voltage is 180 degrees out from
the source.

RF

Best, Dan. (DB :)

VSWR doesn't matter?
 

Cecil Moore wrote:


No technical author is going to be arrogant enough to use
the phrases "must be analyzed" or "must be regarded". Many
shortcuts do work from a mathematical standpoint but often
lose their ability to tell us anything about reality.


Cecil,

Utter rot. These experts are not careless. Textbooks are full of
examples of "must" and "may". The words are not chosen at random.

If you think a standing wave is a "shortcut", how about showing the
mathematical models that support your position? I, along with many
others, have shown the reverse many times.

73,
Gene
W4SZ

VSWR doesn't matter?
 

Gene Fuller wrote:
Cecil Moore wrote:
Forward traveling wave + reflected traveling wave = standing wave
What happens to the standing wave when you take away the reflected
wave?

It's a different physical situation.

The two components of the standing wave are the forward traveling
wave and the reverse traveling wave. I guess if the reverse
traveling wave disappears, you can't ignore it anymore, huh?
--
73, Cecil, w5dxp.com