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

VSWR doesn't matter?
 

Gene Fuller wrote:
Cecil Moore wrote:
If you think a standing wave is a "shortcut", how about showing the
mathematical models that support your position?

I already did, Gene, but here it is again:

forward traveling wave + reverse traveling wave = standing wave

just substitute the appropriate math symbols. There's a
trig identity that corresponds to the above equation.

Take away either the forward traveling wave or the reverse
traveling wave and the standing wave ceases to exist.
There's no valid mathematical model that supports the
position that standing waves can exist with the two
component waves.
--
73, Cecil, w5dxp.com

VSWR doesn't matter?
 

Cecil Moore wrote:
There's no valid mathematical model that supports the
position that standing waves can exist with the two
component waves.

Sorry, obviously should be "without", not "with".
I'm engaging in March Madness while I'm posting.
Gig 'Um, Aggies!
--
73, Cecil, w5dxp.com

VSWR doesn't matter?
 

Cecil Moore wrote:
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

Cecil,

If I cut off one leg I will probably fall over. So what?

Your desperate attempts at evasion are showing. 8-)

73,
Gene
W4SZ

VSWR doesn't matter?
 

Gene Fuller wrote:
Your desperate attempts at evasion are showing. 8-)

I'm not evading anything, Gene. I assert that a standing
wave cannot exist without its component traveling waves.
I can give any number of examples.

You are the one who refuses to provide just one example
of a standing wave existing without its component
traveling waves so exactly who is evading?
--
73, Cecil, w5dxp.com

VSWR doesn't matter?
 

Cecil Moore wrote:
Gene Fuller wrote:
Your desperate attempts at evasion are showing. 8-)

I'm not evading anything, Gene. I assert that a standing
wave cannot exist without its component traveling waves.
I can give any number of examples.

You are the one who refuses to provide just one example
of a standing wave existing without its component
traveling waves so exactly who is evading?
--
73, Cecil, w5dxp.com

And I say, "Big deal." You can set up any combination of components you
wish. The physical reality remains exactly the same as that (fully)
represented by the standing wave. You simply cannot derive new reality
by manipulating the math.

73,
Gene
W4SZ

VSWR doesn't matter?
 

Gene Fuller wrote:
You simply cannot derive new reality by manipulating the math.

But, Gene, that is exactly what you have done. Asserting
that two waves with different equations are the same
*IS* manipulating the math.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

Roy Lewallen wrote:

David G. Nagel wrote:

If you want a quick lesson in high vswr find a ham with an old tube
transmitter and see if he will hook it up to a mismatched load. The
cherry red plates are the reflected energy being absorbed. Transistors
will just turn to smoke under the same conditions.

Unfortunately, you'd be learning the wrong lesson.

The cherry color is due to the transmitter being loaded with an
impedance it's not designed for, causing the final to run at low
efficiency. You can disconnect the antenna and replace it with a lumped
RC or RL impedance of the same value and get exactly the same result.
Alternatively, you can attach any combination of load and transmission
line which give the same impedance, resulting in a wide variation of
"reflected energy", and get exactly the same result. All that counts is
the impedance seen by the transmitter, not the VSWR on the line or the
"reflected power".

The problem is that the idea of "reflected energy" turning the plates
hot is so easy to understand, that people aren't willing to abandon it
simply because it isn't true.

See http://eznec.com/misc/Food_for_thought.pdf for more.

Roy Lewallen, W7EL

Do us a favor, compute the S-vectors for an incandescent lamp with a linear
filament.
Then follow though with the same for a transmitter, transmission line and a
mismatched load.
You will find that is the reflected S-vector that adds heat to the plate.


--
JosephKK
Gegen dummheit kampfen die Gotter Selbst, vergebens.Â*Â*
--Schiller

VSWR doesn't matter?
 

Richard Fry wrote:

"Roy Lewallen" wrote
The problem is that the idea of "reflected energy" turning the plates hot
is so easy to understand, that people aren't willing to abandon it simply
because it isn't true.
_____________

But reflected energy/power does exist.

For an easy example, such reflections are evident in the picture seen on
an analog TV receiver when the match between the transmit antenna and the
transmission connected to it is bad enough.

In analog TV transmit systems with a typical 500+ foot length transmission
line from the tx to the antenna, a 5% reflection from a far-end mismatch
can be quite visible, showing as a "ghost" image that is offset from the
main image as related to the round-trip propagation time of the
transmission line.

RF

Poppycock, TV ghosting is caused by multipath length differences. Calculate
the position ratio and the horizontal scan frequency (15750 Hz is close
enough). That gives you the path length difference; it is generally on the
order of miles (= major terrain features).

--
JosephKK
Gegen dummheit kampfen die Gotter Selbst, vergebens.Â*Â*
--Schiller

VSWR doesn't matter?
 

VSWR doesn't matter?
 

"joseph2k" wrote
Poppycock, TV ghosting is caused by multipath length differences.
Calculate
the position ratio and the horizontal scan frequency (15750 Hz is close
enough). That gives you the path length difference; it is generally on
the
order of miles (= major terrain features).
________________

Analog TV ghosts can be produced within the TV transmit antenna system as
well as by reflections of the transmitted signal in the propagation
environment. I know this from my experience as an RCA Broadcast Field
Engineer, because I've evaluated and corrected many transmit antenna systems
that had been the source of such ghosts.

For example, a reflection from a mismatch between a 1,000 foot long,
air-dielectric transmission line and the TV transmit antenna connected to it
produces a ghost with ~ 2 µs delay from the main image. The active scan
width of an NTSC TV line is about 53 µs, so 2/53 = ~4% of the width of the
screen, or maybe 5% counting overscan. This ghost is easy to see in a
typical TV set/viewing setup.

RF

VSWR doesn't matter?
 

Dan Bloomquist wrote:

billcalley wrote:

snip

I've been reading the posts on this. One poster said this has been going
on for twenty years! (For the other groups, this thread has life on
rec.radio.amateur.antenna) It doesn't need to be so.

First, there should be no doubt that reflected power on a transmission
line is real. Sure, you can replace the line with a lump but that
doesn't clear up the question for others.

For the next two examples, see page 179:
http://cp.literature.agilent.com/lit...4753-97015.pdf
All examples assume the same impedance for source and line.

First example, step into an open line with a Thevenin source. The energy
is divided between the source and the line. Half the energy is moving
down the line and when it returns changes the impedance the source sees
to an open circuit. The energy does not flow back into the source, so,
where did it go? It is stored in the capacitance of the line.

Second example, step into a shorted line. When the energy returns the
source now sees a short. The energy does not flow back into the source,
so, where did it go? It is stored in the inductance of the line.

So here are two examples where the energy sent down the line do not
return to the source.

Wrong, wrong, wrong. Energy is not created or destroyed, but it can be
converted back and forth to mass. See mass defect and hydrogen fusion.


Third example. Send a pulse down the line. The Thevenin voltage source
will go to short, as it should, when the pulse falls. The pulse is
reflected from either an open or a short at the end of the line. All the
energy is dissipated in the source impedance when this pulse returns.
That is where the energy goes. And it is obviously the _same_ energy
created at the source.

Less wrong, line length makes no significant difference (unless it is really
lossy).


Sure, non of the cases above represent steady state AC. But they do show
that energy may or may not be returned to the real component of the
source.


Yes they do. Same model, same reflected power, same heating effect on the
last amplifier stage.

With the above in mind, it can be shown, (in some part II), that a real
accounting of energy from source to load and back is possible.
Equivalent circuits are just that, the trading of line for lump. But,
and this is really important, the only reason the effective impedance at
the input of a 50 ohm line is not 50 ohms is because of reflected energy.


Wrong, wrong, wrong again. Transmission line impedance is strictly a matter
of physical dimensions, and surrounding materials permitivity and
permeablity.

Best, Dan.

--
JosephKK
Gegen dummheit kampfen die Gotter Selbst, vergebens.Â*Â*
--Schiller

VSWR doesn't matter?
 

Richard Fry wrote:
For example, a reflection from a mismatch between a 1,000 foot long,
air-dielectric transmission line and the TV transmit antenna connected
to it
produces a ghost with ~ 2 µs delay from the main image. The active scan
width of an NTSC TV line is about 53 µs, so 2/53 = ~4% of the width of the
screen, or maybe 5% counting overscan. This ghost is easy to see in a
typical TV set/viewing setup.

We performed that exact experiment at Texas A&M
in the 50's. The ghosts were right where the
reflected waves predicted they would be. I
wonder how the modulation in a reflected wave
moves up and down the line without the reflected
wave also moving up and down the line? :-)
--
73, Cecil, w5dxp.com

VSWR doesn't matter? (Standing - travelling waves)
 

"Cecil Moore" wrote

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

Which brings us back to the great loading coil dispute.

In the resonant quarter wave monopole (say 80m), loaded with solenoid coil,
about 2/3 up the radiator, we experience significant (about 40%) current
drop at the top end of the coil. This is also demonstrated by bottom of the
coil getting warmer or hot proportionately, indicating that we have standing
wave circuit and some real current in the system, fortifying Cecil's
argument. RF is flowing along the radiator, "seeing" high impedance tip at
the end, being reflected, flowing back and being superimposed with the
forward wave. Reality that W8JI and other "defenders" had hard time to
swallow.

In the case of traveling wave antenna, like Beverage, terminated with
resistance, we can see the uniform current along the wire. Coil or slinky
inserted in such system will show the same current along the coil (minus
ohmic losses).

There is real life proof about what Cecil is saying above.

Relating to the standing wave circuit, I had question in my mind: how
important is to control the resistance and consider it in standing wave
antenna system. Example is that the current above the loading coil is
appreciably smaller than at the base, hinting that you perhaps do not need
low resistance (copper tubing vs. SS whip).

But...
if the standing wave is made of forward and reverse traveling waves, should
not we be trying to keep the resistance low in the system? Or is it
insignificant?
My pet peeve tells me that it would gain significance in the multi element
loaded arrays. Do the modeling programs capture that? They show slight
increase of current at the bottom (few turns) of the coil (when loading
inductance is properly modeled). Would that be due to the loss from that
point on, when forward and reflected wave is "meeting" the losses to
resistance and radiation and then with lesser amplitude superimposing with
forward wave?

Yuri, K3BU.us

VSWR doesn't matter? (Standing - travelling waves)
 

Yuri Blanarovich wrote:
But...
if the standing wave is made of forward and reverse traveling waves, should
not we be trying to keep the resistance low in the system? Or is it
insignificant?

The I^2*R losses in the conductors depend upon the net
current which is the phasor sum of the forward current
and reflected current. Since the net current is lower
above the coil in a mobile center-loaded antenna, the
resistance of the stinger is not as important as the
resistance of the bottom shaft which carries maximum
current. Whether the I^2*R losses in a stinger are
significant or not is a subjective call. If one is
looking for that last 0.1 dB that will win a mobile
antenna shootout, replacing the stainless steel
stinger with one-inch copper tubing might do the
trick.

Incidentally, in a thin-wire 1/2WL dipole, the reflected
current arriving back at the feedpoint is approximately
90% of the forward current, i.e. only about 10% of the
current is lost to radiation in its round trip to the
end of the antenna and back. The same is true of forward
and reflected voltage. The feedpoint impedance of a
1/2WL dipole results from the superposition of and
interference between the forward and reflected waves
on the standing-wave antenna at the feedpoint.

If we made a Z0=600 ohm 1/4WL open stub out of resistance
wire such that the feedpoint impedance is 73 ohms, we
would have a pretty good simulation of an antenna where
energy is converted to heat instead of being radiated.
The voltages and currents on the stub would correlate
closely with the voltages and currents on a 1/2WL dipole.
--
73, Cecil, w5dxp.com

VSWR doesn't matter?
 

On Mar 16, 10:22 am, Cecil Moore wrote:
Richard Fry wrote:
For example, a reflection from a mismatch between a 1,000 foot long,
air-dielectric transmission line and the TV transmit antenna connected
to it
produces a ghost with ~ 2 µs delay from the main image. The active scan
width of an NTSC TV line is about 53 µs, so 2/53 = ~4% of the width of the
screen, or maybe 5% counting overscan. This ghost is easy to see in a
typical TV set/viewing setup.

We performed that exact experiment at Texas A&M
in the 50's. The ghosts were right where the
reflected waves predicted they would be. I
wonder how the modulation in a reflected wave
moves up and down the line without the reflected
wave also moving up and down the line? :-)
--
73, Cecil, w5dxp.com

Awww come on Cecil - think it through...
First the standing wave is only standing by stroboscopic effect, i.e.
the instantaneous alignment (constructive and destructive) of the
forward and reflected wave fronts at a given point along the line...
The currents continue to stream past the standing point (in both
directions) carrying both the carrier and it's instantaneous
modulation products with it...
Also, the ghosts should be multiple... In the example cited of 2 uS
delay, there should be a ghost every 2 uS across the screen, albeit
each subsequent ghost a fixed number of dB weaker that the one before
due to the % radiated (plus line losses) for each round trip...

denny / k8do

VSWR doesn't matter?
 

Denny wrote:
Cecil Moore wrote:
I
wonder how the modulation in a reflected wave
moves up and down the line without the reflected
wave also moving up and down the line? :-)

Awww come on Cecil - think it through...
First the standing wave is only standing by stroboscopic effect, i.e.
the instantaneous alignment (constructive and destructive) of the
forward and reflected wave fronts at a given point along the line...
The currents continue to stream past the standing point (in both
directions) carrying both the carrier and it's instantaneous
modulation products with it...

OK, I'll change my wondering - I wonder how any rational
person can believe that the modulation in a reflected wave
moves up and down the line without the reflected wave also
moving up and down the line? :-)

Also, the ghosts should be multiple... In the example cited of 2 uS
delay, there should be a ghost every 2 uS across the screen, albeit
each subsequent ghost a fixed number of dB weaker that the one before
due to the % radiated (plus line losses) for each round trip...

Yes, that's exactly the way it appeared to be. Multiple
ghosting getting fainter across the screen.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

"Denny" wrote
Also, the ghosts should be multiple... In the example cited of 2 uS
delay, there should be a ghost every 2 uS across the screen, albeit
each subsequent ghost a fixed number of dB weaker that the one
before due to the % radiated (plus line losses) for each round trip...
_____________

The ghost image needs to be at least 1% of the peak amplitude possible for
the primary video signal before it becomes noticeable So in order for the
2nd and successive ghosts in my example to be visible to a TV viewer, the
reflection producing the first ghost would be so objectionable that a TV
station would not tolerate it, and fix the problem.

RF

VSWR doesn't matter?
 

Denny, K8DO wrote:
"The currents continue to stream past the standing point (in both
directions) carrying both the carrier and its instantaneous modulation
products with it."

Yes. Point is the a-c amplitudes of both forward and reflected currents
are varing at the r-f frequency but their phase relationship is fixed so
that the two currents may be represented by vectors (phasors) at the
point.

Amplitude modulation is a mixing process which produces sum and
difference frequencies that are not synchronized with either original
frequency. Stable frequency generators produce fixed locations for
maxima and minima*along the transmission line, however.

Cecil`s tests were designed to demonstrate atanding waves and phasor
addition. They likely worked as intended.

Best regards, Richard Harrison, KB5WZI

VSWR doesn't matter?
 

Richard Fry wrote:
The ghost image needs to be at least 1% of the peak amplitude possible
for the primary video signal before it becomes noticeable So in order
for the 2nd and successive ghosts in my example to be visible to a TV
viewer, the reflection producing the first ghost would be so
objectionable that a TV station would not tolerate it, and fix the problem.

As I remember, the Texas A&M experiment used a reflection
coefficient of 0.707, i.e. 50% of the signal energy was
reflected at the TV receiver. There also had to be considerable
reflections at the source but that fact was never discussed.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

Richard Harrison wrote:
Cecil`s tests were designed to demonstrate atanding waves and phasor
addition. They likely worked as intended.

One fact to note is that the virtual impedance changes
all up and down a transmission line yet no additional
reflections occur while the Z0 is constant. Reflections
occur only at *actual* impedance discontinuities, e.g. at
a junction of two different Z0s.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

Cecil Moore wrote in news:avVKh.3753$Qw.1263
@newssvr29.news.prodigy.net:

One fact to note is that the virtual impedance changes
all up and down a transmission line yet no additional
reflections occur while the Z0 is constant. Reflections
occur only at *actual* impedance discontinuities, e.g. at
a junction of two different Z0s.

Cecil, that is a simple statement of a scenario in which reflections
*may* occur, but *not always* occur.

Think about it and you will think of examples where a reflection does not
occur at the "junction of two Zos".

I am not quite sure what you mean by an "impedance discontinuity" beyond
the simple "junction of two different Zos" case.

The magnitude of a reflection (zero or otherwise) is *always* and *only*
related to whether the ratio of V to I for the "thing" (whether it is
another line, a lumped circuit or some combination) attached to the end
of the line is equal to Zo. The magnitude is calculated from V/I (Zl) and
Zo using a well known expression.

Should your "rule" be more correctly stated as Reflections
may occur only at *actual* impedance discontinuities, e.g. at a junction
of two different Z0s. Since it has "may" in there, it isn't a rule, is it
worth stating? It is just one of those "may"s that people like to parrot
until they become a Rule of Thumb (ROT).

Owen

VSWR doesn't matter?
 

Owen Duffy wrote:
The magnitude of a reflection (zero or otherwise) is *always* and *only*
related to whether the ratio of V to I for the "thing" (whether it is
another line, a lumped circuit or some combination) attached to the end
of the line is equal to Zo. The magnitude is calculated from V/I (Zl) and
Zo using a well known expression.

I assume you are talking about virtual reflection
coefficients based on virtual V/I impedances, something
that has gotten hams into conceptual trouble for any
number of years. Let's take a look at the S-Parameter
equations.

Port 1 | Port 2
---Z0---+---Z1---
a1-- | --a2
--b1 | b2--

b1 = s11*a1 + s12*a2
b2 = s21*a1 + s22*a2

As you probably know, s11 is a *physical* reflection
coefficient involving unequal impedances.
s11 = (Z1-Z0)/(Z1+Z0)
Z0 and Z1 are physical impedances, not virtual
impedances, i.e. *NOT* merely a V/I ratio.

a1 is the voltage wave incident upon Port 1.
s11*a1 is the reflection from Port 1.

If Z1 Z0, there exists an impedance discontinuity.
s11 0, and s11*a1 0, i.e. there exist reflections.
This is why I say: If there is a physical impedance
discontinuity, then reflections exist.

If reflections are to be canceled, then s12*a2
must be equal in magnitude and 180 degrees out of
phase with reflection s11*a1. Note that for
reflections to be canceled, they must first exist.

s12*a2 is the voltage not reflected from Port 2.

All this is covered in HP Application Note 95-1,
"S-Parameter Techniques" available from:

http://www.sss-mag.com/pdf/an-95-1.pdf
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

"Cecil Moore" wrote
...That's simply not true. When the load is connected directly to
the source, incident power is often still rejected, it just doesn't
have very far to "bounce". And since it is internal to the source,
the "bouncing" is difficult if not impossible to quantitize. etc
_______________

Does the lack of a technical response to Cecil's post (so far) mean
that his analysis and conclusions are understood and accepted?

Hopefully so.

RF

VSWR doesn't matter?
 

Richard Fry wrote:
"Cecil Moore" wrote
...That's simply not true. When the load is connected directly to
the source, incident power is often still rejected, it just doesn't
have very far to "bounce". And since it is internal to the source,
the "bouncing" is difficult if not impossible to quantitize. etc

Does the lack of a technical response to Cecil's post (so far) mean
that his analysis and conclusions are understood and accepted?

The "eliminate the transmission line" sword cuts both
ways. If the source cannot tell the difference between
driving a one wavelength transmission line and driving
a lumped circuit load directly, it follows that the load
cannot tell if it is being driven by a one-wavelength
transmission line or being driven directly by a source.
The incident signal looks the same in either case and the
load rejects (reflects) the same amount of forward power
either way. Except for the energy stored in the one-
wavelength transmission line, conditions are the same
in either case.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

"Cecil Moore" wrote in message
...
Richard Fry wrote:
"Cecil Moore" wrote
...That's simply not true. When the load is connected directly to
the source, incident power is often still rejected, it just doesn't
have very far to "bounce". And since it is internal to the source,
the "bouncing" is difficult if not impossible to quantitize. etc

Does the lack of a technical response to Cecil's post (so far) mean
that his analysis and conclusions are understood and accepted?

The "eliminate the transmission line" sword cuts both
ways. If the source cannot tell the difference between
driving a one wavelength transmission line and driving
a lumped circuit load directly, it follows that the load
cannot tell if it is being driven by a one-wavelength
transmission line or being driven directly by a source.
The incident signal looks the same in either case and the
load rejects (reflects) the same amount of forward power
either way. Except for the energy stored in the one-
wavelength transmission line, conditions are the same
in either case.
--
73, Cecil http://www.w5dxp.com

in steady state... where your favorite s equations hold. this is true. it
is not true in the general case where you account for startup transients.

VSWR doesn't matter?
 

Dave wrote:
in steady state... where your favorite s equations hold. this is true. it
is not true in the general case where you account for startup transients.

Thanks Dave, since I was talking about steady-state,
I should have said so.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

Richard Fry wrote:

Does the lack of a technical response to Cecil's post (so far) mean
that his analysis and conclusions are understood and accepted?

Hopefully so.

In my case, it's because I plonked him long ago.

Roy Lewallen, W7EL

VSWR doesn't matter?
 

Roy Lewallen wrote:
Richard Fry wrote:
Does the lack of a technical response to Cecil's post (so far) mean
that his analysis and conclusions are understood and accepted?

Hopefully so.

In my case, it's because I plonked him long ago.

For pointing out that an antenna is a distributed
network, not a lumped circuit.
--
73, Cecil http://www.w5dxp.com

VSWR doesn't matter?
 

Points well taken.

In my "Food for thought" essay
(http://eznec.com/misc/Food_for_thought.pdf) I use a voltage source in
series with a resistor for most examples. Among the calculations are
those showing the power dissipation in the resistor. I've used this
simple circuit a number of times to illustrate various points regarding
transmission line operation and the effects of traveling voltage and
current waves.

People who aren't willing to accept the points being made borrow from
the politicians' play book and immediately declare the source to be a
"Thevenin equivalent" and therefore any calculation of source power to
be invalid and meaningless. This handily diverts the discussion from the
fundamental topic to something more to the attacker's liking. It can
then proceed to endless arguments about the magnitude and linearity of a
transmitter's output impedance, and whether or not it constitutes a
"dissipationless resistance". The discussion has followed this path many
times, and I'm sure will do so many times more.

The essay shows that "reflected power" is NOT absorbed or dissipated by
the source resistor in my simple circuit -- which is NOT a Thevenin
equivalent of a transmitter or anything else (although, as I point out,
it is a reasonable model for some signal generators). What remains for
the people promoting the notion of waves of average energy propagating
like voltage and current waves to show is how their theories can explain
the resistor dissipation in the very simple circuit I used. (How about a
single equation showing the resistor dissipation as a function of
"reflected power"?) Only after that is done is it necessary to begin the
argument about what the output of a transmitter looks like.

Roy Lewallen, W7EL

J. Mc Laughlin wrote:
I teach my students that prior to analysis of an electrical, electronic, or
EM network/system one must ask and answer a critical question. The question
is: Is the network/system linear, close enough to linear for engineering
purposes, or not linear?

If linear, or essentially linear, one brings into play linear analysis.
Thevenin equivalents, which are only equivalent as far as what they do to
the outside world, are a part of linear analysis.

Most RF power amplifiers that deliver more than one or two watts are
non-linear circuits. Typically, the active device conducts for only a
fraction of each cycle. How else could one get DC power to RF power
efficiencies of over 50 %? Great care must be taken in modeling such
circuits.

A simple example: Consider a transformer fed bridge rectifier (very
non-linear) that is connected to an (old fashion) series L, shunt C filter.
In steady state, if L is large enough, one may model the rectifier as a
series of series connected voltage sources with harmonically related
frequencies (and a DC source). It is left as an exercise for the student to
decide on the sizes, frequencies, and phases of the sources. (Because of
the LPF properties of the LC network, one does not need many harmonics.)
Then one may apply superposition (the essence of a linear process) to
estimate the ripple on the load. However, the model just described is
invalid if L is too small or if L is non-linear. The model is insufficient
to predict the losses in the rectifier. This example is not likely to be
found in current electronic texts, but we all know for whom they are
written.

Techniques exist for dealing with many non-linear networks. They must be
used with great care. If one holds one's nose, one might find an
"equivalent" for a transmitter that suffices for describing what happens
outside of the transmitter, but not inside of the transmitter. Please do
not make conclusions about the "equivalent" itself. Please discriminate
between linear and non-linear networks.

Thus ends the lecture. 73 Mac N8TT

--
J. Mc Laughlin; Michigan U.S.A.
Home:

VSWR doesn't matter?
 

I teach my students that prior to analysis of an electrical, electronic, or
EM network/system one must ask and answer a critical question. The question
is: Is the network/system linear, close enough to linear for engineering
purposes, or not linear?

If linear, or essentially linear, one brings into play linear analysis.
Thevenin equivalents, which are only equivalent as far as what they do to
the outside world, are a part of linear analysis.

Most RF power amplifiers that deliver more than one or two watts are
non-linear circuits. Typically, the active device conducts for only a
fraction of each cycle. How else could one get DC power to RF power
efficiencies of over 50 %? Great care must be taken in modeling such
circuits.

A simple example: Consider a transformer fed bridge rectifier (very
non-linear) that is connected to an (old fashion) series L, shunt C filter.
In steady state, if L is large enough, one may model the rectifier as a
series of series connected voltage sources with harmonically related
frequencies (and a DC source). It is left as an exercise for the student to
decide on the sizes, frequencies, and phases of the sources. (Because of
the LPF properties of the LC network, one does not need many harmonics.)
Then one may apply superposition (the essence of a linear process) to
estimate the ripple on the load. However, the model just described is
invalid if L is too small or if L is non-linear. The model is insufficient
to predict the losses in the rectifier. This example is not likely to be
found in current electronic texts, but we all know for whom they are
written.

Techniques exist for dealing with many non-linear networks. They must be
used with great care. If one holds one's nose, one might find an
"equivalent" for a transmitter that suffices for describing what happens
outside of the transmitter, but not inside of the transmitter. Please do
not make conclusions about the "equivalent" itself. Please discriminate
between linear and non-linear networks.

Thus ends the lecture. 73 Mac N8TT

--
J. Mc Laughlin; Michigan U.S.A.
Home: