On Mon, 07 Jan 2008 12:43:24 -0800
Roy Lewallen wrote:

Roger Sparks wrote:

On Tue, 01 Jan 2008 23:13:18 -0800
Roy Lewallen wrote:

Second analysis: +0.5 input reflection coefficient


clip.......

so that the initial voltage at the line input is simply sin(wt). At any
point that the forward wave has reached, then,

vf1(t, x) = sin(wt - x)

The open far end of the line has a reflection coefficient of +1, so as
before the returning wave is:

vr1(t, x) = sin(wt + x)

(The change from -x to +x is due to the reversal of direction of
propagation. You'll see this with every reflection.) At any time between
when vf1 reflects from the output end (t = 2*pi/w) and when it reaches
the input end of the line (t = 4*pi/w), the total voltage at any point
the returning wave has reached is

vtot = vf1(t, x) + vr1(t, x) = sin(wt - x) + sin(wt + x) [Eq. 1]


Roy's work is only provided as an example of the use of mathematical notation.

The notation "vf1(t, x)" is not fully explained in the text, so any
reader who is unfamiliar with the notation must do some additional
study. This describes my situation, so after some study, this is what
I think these terms mean, focusing on the two coordinates, wt and x.

Roy is using phasor math here, a subject comprehensively covered
elsewhere.

vf1(t, x) means that vf1 is a function of (varies with) both t and x.
Change t and/or x, and vf1 changes. This is not phasor notation, which I
specifically avoided. It is a simple description of the voltage at any
time and position. Plug t and x (and w, which stays fixed for the whole
analysis) into the argument of the sine function, poke it into your
pocket calculator, hit the sin key, and you have the voltage at that
time and place.

(Link http://campus.murraystate.edu/tsm/ts...h6_5/ch6_5.htm)

There are only two brief illustrations of a phasor (figs. 6-25 and 6-26)
in that document, which is otherwise just fine. The remainder of it is a
discussion of time functions like I've used.

The sum of "wt" and "x" denote the phase angle of the wave, measured in
degrees, both referenced to zero. Two angles are needed, one to represent the physical location of each wave as a function of time, the second to locate the point of examination of the wave. "wt" locates the wave in time and physical space . "x" locates the examination point in physical space (on a transmission line in this case).

That's correct except for the description of wt. wt does not describe
the location of the wave in any way - it only tells how the value of the
wave changes with time. w converts the time, in seconds, to an angle in
radians which the sine function can be applied to.

How is "x" located on the transmission line?

x is the distance from the input end of the line. It should be in the
same units as wt. I rather carelessly specified that it be in electrical
degrees, but that would be only if wt is also converted to degrees.

My automatic assumption was that "x" was based at the input to the
line. Thus, if the input was on the left side of the page, the line
would extend to the right, and x would become larger when moving
right.

Yes, that's correct.

This led to a contradiction when I considered how the wave
reflects from the right side upon encountering a discontinuity.
Suddenly, a positive reflected wave represented by sine(-x) was made
sine(x) at an open circuit. At position sin(90), the wave reversed and
became sine(-90). It was if a factor of -1 was applied to the
analysis, which was not the case here.

The change from -x to +x is due to the change in direction of
propagation. The input end of the line is at x = 0. In the examples, the
line is an integral number of wavelengths long, so the far end of the
line is x = n*2*pi radians (or n*360 degrees) where n is an integer. So
at those points, sin(wt - x) and sin(wt + x) are the same value.

After considerable thought and time, I realized that the origin of "x"
was not at the input point, but was at the reflection point. Then the
notation made perfect sense.

Uh, oh.

Forward wave-
input-_____________________________________|Reflectio n point
++x - increasing x - Reflected wave 0

Scale both the sine wave and physical line in degrees. The wave will repeat every 360 degrees (2pi radians). The x scale can be as long (in degrees or radians) as desired

Think of the sine wave as an curve traced on a MOVING sheet of paper. The paper/curve moves in the direction of the forward wave. As the sine wave trace passes the zero point, the wave has a(unit) magnitude, described as sin(wt). Pick any wt and match wt to 0 on the physical scale. Then go to the x location of interest to find the magnitude of the sine wave at that location for one instant of time (wt).

If a reflected wave is present, think of the same wave picture but moving in the opposite direction. When the reflection point is an open circuit, the reflected wave is generated by the forward wave so the two waves will always have a 1 to 1 corespondence at the zero point (point of reflection). The two waves are related by

For the forward wave, vf(t, x) = sin(wt - x)
For the reflected wave, vr(t, x) = sin(wt + x)

For the total voltage at any point (open circuit case).
vtot = vf(t, x) + vr(t, x)
= sin(wt - x) + sin(wt + x)

This looks ok so far, but with x = 0 at the input end of the line, as I
described. Let me show you -- start TLVis1, demo 1, and pause it just as
the initial wave reaches the end. (The line is two wavelengths, by the
way, not 3 as I incorrectly said in my introductory posting.) You've
stopped it at wt = 4*pi radians, or 720 degrees, using a frequency of 1
Hz for simplicity. (I'll use degrees since most of us think better in
degrees than radians.) What is the value of the voltage 1/4 wave from
the input end of the line (x = 90 degrees)? sin(wt - x) = sin(720 - 90
deg.) = sin(-90 deg) = -1, which you can see is correct. What's the
value 1/4 wave from the far end of the line (x = 630 deg.)? sin(720 -
630) = sin(90) = +1, which you can see is also correct.

Now start the analysis again and run it until the reflected wave reaches
the source, and pause it again. You're now at wt = 1440 degrees. What's
the value of the reflected wave at x = 90 degrees, that is, 90 degrees
from the input end of the line? sin(wt + x) = sin(1440 + 90) = sin(90) =
+1. 90 degrees from the far end (x = 630 deg), the value is sin(1440 +
630) = -1. These are what what the graph shows. You'll find that at
every point, at every time, the value of the forward wave is k * sin(wt
- x) and the reverse wave k * sin(wt + x) with x = 0 at the line input.
(k changes at each reflection). Please let me know if you see any time
or position for which this isn't true.

When the reflection point is a short circuit, the voltage becomes zero at the short. The sum of forward and reflected wave voltages (at the short) will always be zero.

That's correct.

Mathematically, this is accomplished by multiplying either the forward or reflective part of the equation a factor of -1.

It's done by applying a reflection coefficient of -1 to the impinging
wave to result in a reflected wave which is equal and opposite in phase.

The effect is to shift the phase of both forward and reflected waves by 90 degrees for a total of 180 degrees.

No. The analysis using superposition treats the forward wave and reverse
waves separately. Reflection has no effect on the forward wave, and it
creates a reflected wave. The reflected wave is the inverse of the
forward wave if the reflection coefficient is -1.

We will apply the factor of -1 to the forward wave to maintain
polarity consistancy with the open circuit case.

vfS(t, x) = -sine(wt - x) For short circuit reflection.

I'm not quite sure what you're doing here. If the impinging wave is
sin(wt - x), the reflected wave is -sin(wt + x) upon hitting a short
circuit. If the impinging wave is sin(wt + x), its reflection is -sin(wt
- x).

Notice that if x = 0,

vtotS(t, x) = -vfS(t, x) + vrS(t, x)
= -sine(wt) + sine(wt)
= 0

For practice, let's analyze a transmission line fed at one end, open at the other. The relfections will have a voltage equal to the forward voltage at all times so we will use the equation
vtot = sin(wt - x) + sin(wt + x)

Assuming one volt peak for the sine wave amplitude (a unit amplitude), find the voltage at points vtot(wt, x),(0,0),(90,0),(180,0),(270,0),(360,0). Notice that every voltage will be computed at the reflection point, x = 0.

I'm not sure, but it looks like you're analyzing the sum of one forward
and one reflected wave, at the far end of the line. At wt = 0, there is
no voltage at the far end of the line. Perhaps you're also referencing
the time to the reflection point, that is, t = 0 is when the wave reflects?

For 0,0, vtot(wt,x) = sine(0 - 0) + sine(0 + 0) = 0
For 90,0 vtot(wt,x) = sine(90 - 0) + sine(90 + 0) = 2v
For 180,0, vtot(wt,x) = sine(180 - 0) + sine(180 + 0) = 0
For 270,0, vtot(wt,x) = sine(270 - 0) + sine(270 + 0) = -2v
For 360,0, vtot(wt,x) = sine(360 - 0) + sine(360 + 0) = 0

Those are correct values. You get the same result when you measure x
from the input end of the line and time from when the source is turned on.

In this example, one complete sine wave cycle has passed the x origin.

By which I believe you mean the far end of the line.

For the next example, let's assume we want to compute the voltage at the point located 45 degrees from the reflection point at the same time that the voltage at the end is at a maximum. The maximum points for voltage at the end occurs when the 90 and 270 degree phases reach the reflection point. Lets look at the 90 degree case.

For 90,45, vtot(wt,x) = sine(90 - 45) + sine(90 + 45)
= sine(45) + sine(135)
= 0.707 + 0.707
= 1.414v
For 90,135, vtot(wt,x) = sine(90 - 135) + sine(90 + 135) = 0
= sine(-45) + sine(225)
= sine(225) + sine(225)
= -0.707 + (-0.707)
= -1.414v

We can continue this process to find the voltage envelope for any location/phase relationship desired. It works equally well for the short circuit case.

Where this would run into trouble would be on a line which is not an
integral number of wavelengths long.

No, it would work for any line length. The line length is the largest x value picked. For instance, to see the voltages at 597 degrees for a line 597 degrees long, make the x line at least 597 degrees long. Because the phase completely rotates every 360 degrees, a line 597-360 = 237 degrees long will have the same voltage relationship.

Assume we want to compute the total voltage at the input when the input is at peak applied voltage (as if from in ideal source) at +90 degrees. Using the 597 degree long line, wt at the end of the line will be 90 + 597 = 687 degrees. We are examining the line at 597 degrees so x would be 597 degrees.

For 687,597, vtot(wt,x) = sine(687 - 597) + sine(687 + 597)
= sine(327 - 237) + sine(327 + 237)
= sine(90) + sine(564)
= sine(90) + sine(204)
= 1 + (-.406)
= 0.594v

I hope you come up with the same value.


Certainly, you could use any point on or off the line as the x
reference, and any time t as the t reference, as long as you modify the
equations appropriately, as I believe you've shown. Your equations have
reversed the sign of x and the roles of forward and reflected waves, to
arrive at the same result when the two are summed.

The equations I presented use the input end of the line as the x
reference and the time of source turn-on as the t reference. The
equations were simpler because of the chosen line lengths, but apply to
all line lengths when the appropriate phase terms are added.


Let's look at the case for short circuited 0 degree case, where the phase is 0 degrees, and examination point is 90 degrees.

For 0,90 (short circuit),
vtotS(wt,x) = - sine(0 - 90) + sine(0 + 90)
= - sine(-90) + sine(90)
= - sine(270) + sine(90)
= -(-1) + 1
= 2

Now advance the phase at 45 degrees, and examine x = 90 degrees

For 45,90 (short circuit)
vtotS(wt,x) = - sine(45 - 90) + sine(45 + 90)
= - sine(-45) + sine(135)
= - sine(315) + sine(135)
= -(-0.707) + (0.707)
= 1.414

Now advance the phase to 90 degrees, examine at x = 90 degrees.
For 90,90, (short circuit),
vtotS(wr,x)= - sine(90 - 90) + sine(90 + 90)
= - sine(0) + sine(180)
= - 0 + 0
= 0

This discussion focused on voltage. Similar notation is used in equations for current and power.

This is an interesting alternative analysis. Is there any point on the
line at any time for which it predicts a different result than my
analysis?

Not that I know of. I think you are not really using x, but x plus line length in your equations. This continues to have me confused.

Does it correctly show the voltage of each forward and
reflected wave individually as well as its sum? If so, then it's an
equally valid analysis for this condition of integral wavelength lines.

So far as I have tested, yes. There is no reason to use the reflection point as the reference unless it simplifies the equations or improves understanding.


If I get the time I'll try to extend either the analysis or program, or
both, to non-integral-wavelength line lengths.

Roy Lewallen, W7EL

Using the reflection point as the zero reference seems to correspond with an observation you made about the end of the line controlling the SWR.

73, Roger, W7WKB

On Mon, 07 Jan 2008 18:12:59 GMT, Cecil Moore
wrote:

feedpoint SWR is 1.032:1.

Which, of course, has nothing to do with Standing Waves ON THE
ANTENNA.

On Mon, 7 Jan 2008 18:26:16 -0500, "AI4QJ" wrote:


"Richard Clark" wrote in message

Now, if you would simply take my advice to heart: strip away the
static and ask the question that is plaguing you.

OK. Thank you for your patience. You said "The SWR on the wire is load-
based and is equal to 8:1 as evidenced by CURRENT on the wire. :-)

Hi Dan,

Well, in fact I don't recall having said anything about "Load-Based,"
but it is immaterial to your question(s).

You cannot make a SWR measurement on a receive antenna any other way."

You are still trapped in a loose thread of associations.

We will get back to that in a moment (I hate doing this, trying to
intuit what you are after). The specific comment you recite above is
wholly out of context, and is being joined to a new one. This is why
I ask you to strip out the noise and ask the question that is
bothering you.

My very simple question: How do I measure that ISWR current?

Specifically, as you have already suggested, with a loop pickup.
However, in my career I have never encountered its use as a precision
device when measuring voltage is so much easier. EZNEC does not
report voltage (although it could be computed, I will leave that to
others as there is no significant information obtained in that
exercise).

Please describe possible meters and probes and where on the receive antenna
do I measure this current?

I've commented on that already. But this tied back into receive and
you will not find a satisfactory, bench answer for that which is
practical (in the sense of $).

At the input?

For waves on a line you go to the line. EZNEC provides this quite
clearly and even provides a graphic solution if you are not into the
actual numbers.

At the nodes? At the zero crosses?
How do you calculate ISWR? Is it Imax/Imin?

That is the conventional way to observe Standing Waves.

I am only trying to
conceptualize what you mean. Do you have to block the reverse current
somehow and measure forward current and vice versa? What does this do to the
standing wave? What does any direct current measurement do to the ISWR
itself?

Dan, you've got yourself wrapped around the axle and here I have to
force a solution to your problem that you only vaguely offer through a
chain of disconnected references above.

First, you spend an inordinate amount of effort quoting to the point
of receiving, and yet none of your questions appear to be aware of the
significance of receiving with these breed of (Traveling Wave)
antennas.

Cecil's wire, one foot off the ground, is strongly influenced by its
proximity. I demonstrated that. However, Cecil divorces this effect
while embracing his goal of explaining away the confusion about
Traveling Wave antennas. This one foot high monstrosity is most
closely allied with the Beverage antenna which was developed with
ground in mind and even here, Cecil corrupts this concept to turn his
model into a Traveling Wave transmission line. In effect, his is a
bait and switch argument as he abandoned the "explanation" 350 posts
ago and never returned until I forced the argument.

Cecil, under protest to my proofs of Standing Wave on these Traveling
Wave, then supplied a citation that heads this very thread:
"Because the Beverage is a
traveling wave, terminated antenna, it has no standing
waves resulting from radio signals."

It contains FROM radio signals. FROM is an externality. In other
words the antenna must be externally excited to test the validity of
this specific statement. Further, this statement has its own internal
logic in that the antenna (the Beverage) is a RECEIVE antenna.
Further, this statement has its own internal logic in that a Beverage
antenna employs ground as an active element. Hence, all internal
logical characteristics are consistent with the statement about
externality. Any reference to source SWR is wholly inappropriate.

EZNEC is perfectly capable of measuring a receive antenna, if you can
supply an antenna to excite it. I did so explicitly. The reports
offered by EZNEC will give you a source SWR reading like Cecil quoted,
but that is entirely unrelated to the receive antenna performance. I
did not report that reading for that reason, Cecil did - for whatever
reason, but a reason that bore no relation to Standing/Traveling waves
on the test antenna.

EZNEC will report all currents on all wires everywhere. When the
Beverage is excited by a source 100 kilometers away, those currents in
the wires defining the Beverage exhibit classic Standing Waves. This
is quite simple.

Now as to your enquiry in how to measure those currents. That is not
practical at the nanoAmpere levels, but software does it quite easily.
If I were tasked to do it, I would immediately transform this into the
voltage model, modulate the source with an unique pattern, and perform
a synchronous detection of the levels involved along the length of the
line. A trivial concept that is difficult in practice.

Now, I have pounded out a lot of words in an effort to eke out your
question. Was it answered?

73's
Richard Clark, KB7QHC

On 6 Jan, 18:39, "AI4QJ" wrote:
"art" wrote in message

...

"Did you ever think that your post would last this long? Obviously the
regular contributors
in this group cannot handle the truth and thus will not consult
anything.Now the experts are argueing over the term SWR a very, very,
deep discussion revealing things unknown to the amateur community at
this time. [...]"

Hello Art, the concept of SWR is extremely misunderstood even by people with
degrees in electronics engineering. It is assumed to be simple, yet many
people get it wrong. Indeed, a good understanding is essential for antenna
development. I do not fault anyone for not understanding the concept because
standing waves, simple as they may seem, are actually expressed as the
product of a cos wave over distance and a sine wave over time. Many things
are happening over the length of the antenna as the function is operating.
If you think of it, it is the essence of space-time and it may be productive
for you to consider it even more broadly in your own hypotheses, more
broadly that is by possibly incorporating the mechancial SWR analogues to
voltage/current SWR's and who knows what new ideas may come to mind with
your model.

I don't understand a lot of this talk about waves bouncing which is
fortunate. Icould not possibly stay on a thread where everybody is
talking past each other and then changing the subject as they didn't
understand the subject in the first place. In ham radio nothing is
believed if it is contrary to the norm.This bouncing wave thing will
never come to closure as all participants are deaf. As far as me
getting involved all the answers involved in my description of radio
are known facts in the scientific world and fully coroberated. Heck
they are even corroberated by existing antenna computor programs and
actual tests. I can't see how these waves fit in with classical
science so it must be another invented science that it referes to. Now
if the trend changed to debate the voracity of existing accepted data
is proved to be incorrect then they would have my attention but the
group is not competant enough for that trail.
Regards
Art

Cecil Moore wrote:
Jim Kelley wrote:

Cecil Moore wrote:

Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.


Evidently, you can't recognize a trig identity when you see one.


One really should take a look at the math before waving one's
hands and opening one's mouth in ignorance.

Please enlighten us as to exactly what trig "identity" will make
the following terms equal.

E1*e^j(wt-kx) ?=? E2*e^j(wt-kx) + E2*e^j(wt+kx)

Seems to me the only condition for which they are equal is when
E2=0, i.e. when reflections (and therefore standing waves) don't
exist.

The purpose of pointing out the trigonometric relationship between the
sum of sines and the product of sine and cosine was to illustrate
that, contrary to your assertion, there isn't a difference in the
waves. The traveling waves can either be written mathematically as
two separate traveling waves, or as one standing wave. It makes no
difference; the waves are the same in either case irrespective of how
you choose to describe them mathematically. Do you grasp the meaning
here, or not?

Thanks,

ac6xg

Richard Clark wrote:
. . .
Specifically, as you have already suggested, with a loop pickup.
However, in my career I have never encountered its use as a precision
device when measuring voltage is so much easier. EZNEC does not
report voltage (although it could be computed, I will leave that to
others as there is no significant information obtained in that
exercise).
. . .

As a general rule, the voltage can't be computed because of the spacing
between the wire and whatever reference point you're measuring voltage
to. The voltage depends on the path taken between the two points
(conceptually, how you arrange the voltmeter leads), so there's no
single answer. That's why EZNEC computes only current.

This is also one of the several problems with treating an antenna like a
transmission line. The similarity between the two is just enough to
tempt people to take the analogy farther than it holds.

Roy Lewallen, W7EL

On Mon, 7 Jan 2008 20:13:32 -0500
"AI4QJ" wrote:


"Roger Sparks" wrote in message
...
On Mon, 07 Jan 2008 12:34:03 -0600

After considerable thought and time, I realized that the origin of "x"
was not at the input point, but was at the reflection point. Then the
notation made perfect sense.

Forward wave-
input-_____________________________________|Reflectio n point
++x - increasing x - Reflected wave 0

Scale both the sine wave and physical line in degrees. The wave will
repeat every 360 degrees (2pi radians). The x scale can be as long (in
degrees or radians) as desired

Think of the sine wave as an curve traced on a MOVING sheet of paper. The
paper/curve moves in the direction of the forward wave.

Try this:

Pick a distance x = pi/3. Now take the sheet of paper perpendicular to the
dipole wire. Move it at a constant velocity in the perpendicular direction
with a pen recorder tracing the magnitude of the voltage on the antenna to
the paper as it moves. There will be a cosine function drawn on the paper.

Given: I = Io*cos(kx)*cos(wt)

Thanks for the examples. You use the term "I" which is usually the current, but the math makes sense for voltage.

To find the distance x = pi/3, I assume you mean from one end of the dipole, back toward the dipole center?


At pi/3 radians, cos(pi/3) = 0.5

The cosine function drawn on the paper moving at right angles to the antenna
will be:

I = 0.5*Io*cos(kt).

Now move the paper to x = pi/2.

cos(pi/2) = 0

The function drawn on the paper will be I = 0.

Now move the paper to x = pi

cos(pi) = -1

The function drawn on the paper, at RIGHT ANGLES tio the antenna will be:

I = -Io*cos(wt)

As cos(wt) rotates between 0 and 2pi, cos(wt) moves between +1 and -1. The voltage will always be the negative of the initial voltage times the cosine of the instantaneous angle wt.

That is how a standing wave operates.

OK, I think I understand.

Thanks for sending the examples.


73, Roger W7WKB

AI4QJ wrote:
"Richard Clark" wrote in message

Now, if you would simply take my advice to heart: strip away the
static and ask the question that is plaguing you.

OK. Thank you for your patience. You said "The SWR on the wire is load-
based and is equal to 8:1 as evidenced by CURRENT on the wire. :-)

That was an ignorant mistake. Like a rank novice, Richard assumed
a 50 ohm SWR but the antenna is *NOT* a 50 ohm antenna. The
acutal SWR is barely above 1.0:1.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
Cecil, under protest to my proofs of Standing Wave on these Traveling
Wave, then supplied a citation that heads this very thread:
"Because the Beverage is a
traveling wave, terminated antenna, it has no standing
waves resulting from radio signals."

It contains FROM radio signals.

That was a quote from one of my references. But I suspect
that a Beverage antenna must be excited in its directional
dimension for that to be true. It probably would have been
more accurate to say it has no reflections rather than to
say it has no standing waves.
--
73, Cecil http://www.w5dxp.com

AI4QJ wrote:
"Cecil Moore" wrote in message
There's no confusion. Terminated rhombics and Beverages are
traveling-wave antennas. Their currents are primarily:
I(x,t) = Io*cos(kx-wt)
A 1/2WL dipole is a standing-wave antenna. It's current is
primarily:
I(x,t) = Io*cos(kx)*cos(wt)

People need to understand the huge difference in the 2 waveforms that you
present.

I agree with you but what do you do with people who choose
to remain ignorant? Allow their ignorance to dominate the
newsgroup simply because they think they already know
everything?
--
73, Cecil http://www.w5dxp.com