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Roy Lewallen, W7EL wrote:
"Either Cecil is misquoting me, or Richard is misquoting Cecil."

I apologize if I misquoted anyone. It was unintentional,

On page 239 of Kraus` 1950 edition of "Antennas" he wrote:
"It is generally assumed that the current distribution of an
infinitesimally thin antenna (l/a=infinity) is sinusoidal, and that the
phase is constant over a 1/2-wavelength interval, changing abruptly by
180-degrees between intervals."

Cosine has the same shape as sine. They are identical except for a
90-degree phase displacement. Sine starts at zero. Cosine starts at one.

Sinusoidal covers both sine and cosine shapes.

Best regards, Richard Harrison, KB5WZI

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Cecil Moore wrote:
Roy Lewallen wrote:

Richard Harrison wrote:

Cecil, W5DXP wrote:
"Assuming the source signal is a pure sine wave, if the standing wave
current "isn`t in general sinusoidally shaped (as Roy said)", then the
antenna would have to be introducing harmonic radiation that doesn`t
exist in the source signal."


Either Cecil is misquoting me, or Richard is misquoting Cecil.


Richard quoted me correctly. I did *NOT* quote you. I merely
stated what I thought you said.

Cecil calls the total current the "standing wave current".


That is absolutely false. Total current equals standing wave
current plus traveling wave current. Or standing wave current
equals total current minus traveling wave current. Subtract
out the traveling wave component and a pure standing wave
component is left.


I thought I was following this somewhat, but now it appears that I am not.

I thought there were two traveling waves, forward and reverse, with the
reverse wave being a reflection of the forward wave from the end of the
antenna. I thought that the total current was the sum of the two waves,
could be measured with an ammeter, and that it could be referred to as the
standing wave.

Are you now saying that the standing wave and the reverse wave are the same
wave?

Confusedly,
John

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John - KD5YI wrote:
I thought there were two traveling waves, forward and reverse, with the
reverse wave being a reflection of the forward wave from the end of the
antenna. I thought that the total current was the sum of the two waves,
could be measured with an ammeter, and that it could be referred to as
the standing wave.

If the antenna were lossless, that would be true. However, the
antenna loses RF energy to radiation, ground, I^2*R, etc. Since
the same thing is true of a transmission line, let's first look
at a transmission line.

Let's say we measure (at the tuner)
a forward power of 100 watts and a reflected
power of 80 watts. We can assume that the losses in the line
and the power delivered to the load add up to 20 watts. The
20 watts can be associated with a pure forward traveling wave.
The other 80 watts of the forward power and the 80 watts of
reflected power can be associated with a pure standing wave.
The equations for such would look something like this.

Itot = Ifor*cos(kx+wt) + Iref*cos(kx-wt) =

Itot = I1*cos(kx+wt) + I2*cos(kx)*cos(wt)

I1 would be considered to be the part of the traveling wave that
is delivering net power to the load. I2 would be considered to be
the standing wave current delivering no net power to the load.
It's just another way of mathematically partitioning the currents.

Let's assume for the sake of discussion that the forward wave at
the antenna feedpoint is 100 watts and the reflected wave at the
antenna feedpoint is 80 watts. That means there are 20 watts of
total losses. We can partition the currents in the same way that
we did in the transmission line.

Incidentally, if we assume the antenna is 1/2 wavelength with a
Z0 of 600 ohms, we can calculate the feedpoint impedance just as
if it were a piece of transmission line with a resistive termination.
Anybody want to try that exercise? What impedance is seen looking
into a 600 ohm 1/4WL open stub when the forward power is 100 watts
and the reflected power is 80 watts?

Are you now saying that the standing wave and the reverse wave are the
same wave?

The reverse wave is half of the standing wave. The other half of
the standing wave is part of the forward wave. One might say that
the energy in the reverse wave neutralizes the ability of the
forward wave to deliver that same amount of energy to the load.
And indeed, we know the power delivered to the load is the forward
power minus the reflected power.
--
73, Cecil http://www.qsl.net/w5dxp

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On Tue, 23 May 2006 09:33:44 -0500, (Richard
Harrison) wrote:

Sinusoidal covers both sine and cosine shapes.

Hi Richard,

Given two things:

1.) the frequency of this allusion to shape, sin/cos;
B.] the tolerance of errors that are sometimes as great as ±59%;

it then follows that the shape of current distribution along the
quarterwave element can also be described as a quarter circle arc.

Hello to those avidly seeking to participate,

The question of the new millennium must be: "What significance does
this shape bring to us?" For those with misty-eyed philosophies,
please adorn your response with stars and moon beams in the Subject
line as a courtesy to serious correspondents.

73's
Richard Clark, KB7QHC

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On Tue, 23 May 2006 08:55:42 -0500, (Richard
Harrison) wrote:

f(x) = y = mx + b is called linear because its graph is a straight line.

Hi Richard,

Try listening to Mozart with a speaker driven with such a "linear"
gain device. [Some things are so simple, they defy further analysis.]

This, again, goes to the matter of a finely staged distortion trial
with tone-deaf judges.

73's
Richard Clark, KB7QHC

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Cecil Moore wrote:
Roy Lewallen wrote:
. . .
Richard quoted me correctly. I did *NOT* quote you. I merely
stated what I thought you said.

Cecil calls the total current the "standing wave current".


That is absolutely false. Total current equals standing wave
current plus traveling wave current. Or standing wave current
equals total current minus traveling wave current. Subtract
out the traveling wave component and a pure standing wave
component is left.

I could have sworn Cecil had used it in that sense before. But Ok, I'll
go along with the new definition. Any basic analysis of transmission
line operation shows that the total current is simply the sum of the
traveling wave currents. Consequently the "standing wave current" as
defined by Cecil is zero. So by "standing wave current", Cecil means zero.

This is getting stranger by the day.

Roy Lewallen, W7EL

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I'm sure that somewhere in one of your texts you can find the definition
of linear as applied to networks. Once you do, though, a little thought
is required to discover that y = mx + b doesn't satisfy the criteria for
network linearity.

To be linear, a network has to satisfy superposition. This means that:

If y1 is the response to excitation x1 and y2 is the response to
excitation x2, then the response to x1 + x2 must be y1 + y2.

Let's try that with your function.

The response to x1 is:

y(x1) = mx1 + b

The response to x2 is:

y(x2) = mx2 + b

The sum of y(x1) and y(x2) is:

y(x1) + y(x2) = m(x1 + x2) + 2b

But response to x1 + x2 is:

y(x1 + x2) = m(x1 + x2) + b

These are not equal as they must be to satisfy superposition and
therefore the requirements for linearity.

Roy Lewallen, W7EL

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"But of course you realize that the function y = mx + b doesn`t meet the
requirements of a linear function when applied to network theory."

Works for me.

Linear means the graph of the function is a straight line.

f(x) = y = mx + b is called linear because its graph is a straight line.

A straight line is the shortest distance between two points.

In y = mx + b, m is a constant determining the slope of the line. x is
is the independent variable. b is the offset or point along the x-axis
where the line crosses.

y then is a linear function of x because its slope is always mx, but
displaced in the x-direction by a constant value, namely b.

y is linear the same as IR is linear, or by substitution, E is linear in
Ohm`s law where E=IR. For any value of I, voltage = IR and the graph of
I versus E is a straight line with a slope equal to R.

Resistance is a common factor in network theory.

Best regards, Richard Harrison, KB5WZI

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Roy Lewallen wrote:
Any basic analysis of transmission
line operation shows that the total current is simply the sum of the
traveling wave currents. Consequently the "standing wave current" as
defined by Cecil is zero. So by "standing wave current", Cecil means zero.

Your last two statements are false. Please stop trying
to tell others what I mean when it is obvious that you
don't know what I mean.

This is getting stranger by the day.

Why can't you and I just have a reasonable technical
discussion? Sooner or later, the readers who care at
all about the truth are going to see through your
attempts to falsify what I have said.

Since you obviously have not understood what I said in
the past I guess I need to simplify it for you so here goes:

Let A be the forward traveling current.

Let B be the rearward traveling current.

Let C be the part of the forward traveling current
that flows through the load.

Let D be the part of the forward traveling current
that doesn't flow through the load and becomes B
after being reflected.

A = C + D The total forward current is divided
into two components, one accepted by the load and
one rejected by the load.

The following is phasor addition

Total Current = A + B = (C + D) + B = C + (D + B)

(D + B) is the pure standing wave current component,
a part of the forward current equal to the reflected
current.

The total current is indeed the phasor sum of the
forward and reflected current.

The total current is also the sum of the load current
plus the pure standing wave current. The pure standing
wave current envelope is sinusoidal.
--
73, Cecil http://www.qsl.net/w5dxp
The day a Guru starts believing that He already knows
everything is the day He becomes hopelessly ignorant.

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Roy Lewallen wrote:
I'm sure that somewhere in one of your texts you can find the definition
of linear as applied to networks. Once you do, though, a little thought
is required to discover that y = mx + b doesn't satisfy the criteria for
network linearity.

You have missed the point, Roy. It is the relationship between
functions that has to be linear. If f(x) is the input and y
is the output where y = h(x) = k*f(x), a plot of the output
Vs input will yield a straight line of the form mx + b.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
John - KD5YI wrote:

I thought there were two traveling waves, forward and reverse, with
the reverse wave being a reflection of the forward wave from the end
of the antenna. I thought that the total current was the sum of the
two waves, could be measured with an ammeter, and that it could be
referred to as the standing wave.


If the antenna were lossless, that would be true. However, the
antenna loses RF energy to radiation, ground, I^2*R, etc. Since
the same thing is true of a transmission line, let's first look
at a transmission line.

Let's say we measure (at the tuner)
a forward power of 100 watts and a reflected
power of 80 watts. We can assume that the losses in the line
and the power delivered to the load add up to 20 watts. The
20 watts can be associated with a pure forward traveling wave.
The other 80 watts of the forward power and the 80 watts of
reflected power can be associated with a pure standing wave.
The equations for such would look something like this.

Itot = Ifor*cos(kx+wt) + Iref*cos(kx-wt) =

Itot = I1*cos(kx+wt) + I2*cos(kx)*cos(wt)

I1 would be considered to be the part of the traveling wave that
is delivering net power to the load. I2 would be considered to be
the standing wave current delivering no net power to the load.
It's just another way of mathematically partitioning the currents.

Let's assume for the sake of discussion that the forward wave at
the antenna feedpoint is 100 watts and the reflected wave at the
antenna feedpoint is 80 watts. That means there are 20 watts of
total losses. We can partition the currents in the same way that
we did in the transmission line.

Incidentally, if we assume the antenna is 1/2 wavelength with a
Z0 of 600 ohms, we can calculate the feedpoint impedance just as
if it were a piece of transmission line with a resistive termination.
Anybody want to try that exercise? What impedance is seen looking
into a 600 ohm 1/4WL open stub when the forward power is 100 watts
and the reflected power is 80 watts?

Are you now saying that the standing wave and the reverse wave are the
same wave?


The reverse wave is half of the standing wave. The other half of
the standing wave is part of the forward wave. One might say that
the energy in the reverse wave neutralizes the ability of the
forward wave to deliver that same amount of energy to the load.
And indeed, we know the power delivered to the load is the forward
power minus the reflected power.



Never mind.