On Tue, 24 Feb 2004 18:28:57 -0600, "Steve Nosko"
wrote:
I assumed Richard's intent here is that you only have to do the
calculation for one full period of the periodic component to derive the RMS
value - and if it is symmetrical, then only one half period will suffice.
This all assumes a symmetrical AC shape. .. I see no reason why this would
not be true.

And then you offer in contradiction:

However a DC biased periodic shape requires another squaring and root
operation if you capture all the components. It gets a bit more harry

Such is reality. There is no such thing as symmetry, except on the
academic page. However, I am not so pedantic as to suggest that
shortcuts don't abound; simply pedantic enough to point out you don't
make claims to accuracy (admittedly none were offered that I was
responding to) through fudge factors when so many alternatives remove
doubt.

The simple determination of RMS is the graphical integration of the
area under the curve. There are as many "correction factors" for RMS
as there are shapes, and they all derive from this simple concept.

Here I'll take issue with the ONE WORD "graphical". You can integrate
if you can describe the function of the wave shape mathematically.

Of course, but it is eminently doubtful if you can actually express it
mathematically. Far more here own o'scopes than works of multiple
regression. Graphical analysis is first year engineering stuff out of
drafting class.

When the computational horsepower requirement becomes enormous (there
are many here that give up too easily with complexity); it is the
provence of the "Old School" to suggest that since RMS is all based on
the notion of power, you simply measure the caloric result and ignore
shape altogether. This may be done with thermo-electric piles or
other measurable property transformers that perform the complexity of
integration through physics*. I can anticipate those who dearly
embrace the complexity that they shudder to face (such contradictions
of their love-hate relationships) when I hear Crest, or pulse/power
factor (or duty cycle) uttered. Clearly the problem will have
migrated from Power to some other consideration, but is dressed as an
RMS debate.

73's
Richard Clark, KB7QHC

Yikes! Not sure where you went on that last bit, Richard C...
Now, I ask. Do the power meters on the outside of our houses take all those
factors into consideration and REALLY show TRUE watt hours? I have one in
the basement and I think I figured out why I was seeing twice the reading I
should have (letting a light bulb sit on for awhile) ... I counted the
teeth to get the ratio of the gear train, just to find that it is printed
(somewhat cryptically) on the face) (I made a two wire / three wire
connection error)

Hi Steve,

Yikes? Look at your own response to shudder. ;-)

You offer the doubt and then correct your error in the space of three
sentences. From that I must suppose it was a rhetorical question, but
the "yikes" heading it promises more clarity in response to what you
apparently complain of. After all, a complex, compound sentence with
ellipses and nested parenthetical statements? You lose points for
unpaired braces and lack of punctuation throughout and at the end.

Do the power meters on the outside of our houses take all those
factors into consideration and REALLY show TRUE watt hours?
For at least a Century now.

The power companies offer as close to pure symmetry as you could buy.
They also offer long term time accuracy to far better than any source
short of WWVL.

73's
Richard Clark, KB7QHC

Steve Nosko wrote:
"I wonder if a "true RMS" DVM can handle this?"

So do I. The true rms meter seems a wonderful development to me.

Steve also wrote:

"Also ok, but not sure how it plays into the RMS discussion."

My speculation is that the effective value of a nonsinusoidal waveform
could be found by summation of its sinusoidal constituents.

But, it`s not difficult to find an effective value for not only
sinusoidal periodic waveforms but for
nonsinusoidal periodic waveforms as well. One can graphically take a
large number of equally spaced ordinates of the form, using at least one
complete alternation, Richard Clark. Both alternations are not needed
but could be used as a minus times a minus is a plus and each of the
ordinate values must be squared because power is a function of the
current squared, so both alternations when their ordinates are squared
produce positive values. Next, we sum the squared ordinate values and
divide by the number of ordinates. You get the average value of the
squared curve which is what we are looking for.

Or, you can construct the squared curve and integrate the area under the
squared curve by using a planimeter. Dividing the area by the baseline
length gives the same average value of the squared curve as iusing the
graphic ordinates above. Voila: rms.

Unless I`ve opened a new can of worms with this posting, I don`t know of
any difference of opinion I have with Steve.

Best regards, Richard Harrison, KB5WZI

On Tue, 24 Feb 2004 20:23:07 -0600 (CST),
(Richard Harrison) wrote:

Steve Nosko wrote:
"I wonder if a "true RMS" DVM can handle this?"

So do I. The true rms meter seems a wonderful development to me.


Hi Guys,

The first "True RMS" meter I worked on was from Fluke back in '75. It
was broke and I had to fix it, but most of the components had their
identification numbers scrubbed off and the schematic at that point
was just an outline with gozintas and comesouttas.

Well, Fluke was (is) just up the road from me, and I gave them a call
for details. Nocando came the reply. My first lesson in proprietary
design. Such technology was to kill for.

Today, I can buy far better from Radio Shack at 1/10th the cost.
Still and all, it only works through to voice frequencies (to lazy to
grab the spec sheet). If you want more precision/bandwidth, then
there are complex (non-digital) math chips that do it in real time
through analog multipliers and compressors (log function). One of my
tech gurus, Robert Pease of National Semi, challenges most digital
designs as a waste of time and energy. He successfully offers analog
designs that provide far better stability and accuracy.

73's
Richard Clark, KB7QHC

"Richard Harrison" wrote in message
...
[...]
Steve also wrote:
"Also ok, but not sure how it plays into the RMS discussion."

My speculation is that the effective value of a nonsinusoidal waveform
could be found by summation of its sinusoidal constituents.

That's what I thought you were going for. My gut feel is that this must
be valid, otherwise "Fourier ain't an exact solution after all". But it
sounds like a lot of work and you have to be able to extract all the
significant Fourier components (harmonics) and there is an element of
approximation here, just like my "mathematicall function" method, no?

....Of course doing the integration on the waveform takes some time to crank
through the math and get all the quantities collected correctly.


But, it`s not difficult to find an effective value for ...
nonsinusoidal periodic waveforms as well. One can graphically take

For what I wanted to do, the integration was easier. I may be wrong,
but Fourier also needs a function and you still need to integrate, no?

Now, to answer Richard Clark's comments about accuracy in another
post... I did assume that the wave shapes I was interested in were defined
(I think the word is) explicitly. I assumed a mathematical function. When
I compared the three waveforms which were appropriate (sine, triangle,
trapezoid) the results were very much the same (for ratio of RMS to
Average). (gotta remember to keep my sentences shorter).
From this I came to the conclusion that for wave shapes which differed
slightly from the ideal (assumed shapes) the values would be well within
acceptable bounds. Easily 5%. Yes, not exact, but much better than Bob
Shrader had assumed.


large number of equally spaced ordinates of the form, using at least one
complete alternation, Richard Clark.

I don't think they have to be equally spaced since the actual time
enters into the calculation (as you described later). This would make long
sections with a constant value easier.


Both alternations are not needed
but could be used as a minus times a minus is a plus and each of the

This assumes that the waveform is symmetrical around zero. I think in
general, one whole period is required.



or[...] using a planimeter.

Richard H. clearly prefers graphical solutions and that's ok.

Unless I`ve opened a new can of worms with this posting, I don`t know of
any difference of opinion I have with Steve.
Best regards, Richard Harrison, KB5WZI

Right now it appears that this half vs. full period is the only
difference I can see.

--
Steve N, K,9;d, c. i My email has no u's.

OOPS!

"Richard Clark" wrote in message
...
On Tue, 24 Feb 2004 18:28:57 -0600, "Steve Nosko"
wrote:
I assumed Richard's intent here is that you only have to do the
calculation for one full period ...symmetrical, then only one half
period...

And then you offer in contradiction:

However a DC biased periodic shape requires another squaring and root
operation if you capture all the components. It gets a bit more harry

This was a stupid thing to add at this point since it addresses
something which I did not explain, so I'm sure it looks weird. I didn't
even say what was in my mind. Namely, that with the DC + AC you still need
to do one full period of the periodic part.

What I was poorly referring to was this:
If you can break down the signal into component parts such as:
DC
Periodic part #1
Periodic part #2
etc
Then there is a formula for the total RMS which is the square root of
the sum of the squares. It is in one of the papers I linked previously.
That is one way to do an AC+DC situation. Second formula in the Intl Rect
paper:
http://www.irf.com/technical-info/an949/append.htm


[...] There is no such thing as symmetry, except on the
academic page. [...] you don't
make claims to accuracy (admittedly none were offered that I was
responding to) through fudge factors when so many alternatives remove
doubt.

I think is it safe to say that we each determine our own tolerance for
error. Five percent for power is ok for my purposes. SO approximating
waveform functins is just fine.


The simple determination of RMS is the graphical integration of the
area under the curve. There are as many "correction factors" for RMS
as there are shapes, and they all derive from this simple concept.

Here I'll take issue with the ONE WORD "graphical". You can
integrate
if you can describe the function of the wave shape mathematically.

Here, I'll take issue. I believe that the basis for the many
"correction factors" for RMS of various shapes is indeed the mathematical
integration of a function rather than the "Simple concept" of a graphical
solution (if that is what you meant). There is a mathematical integral of a
sine wave.

I think you'd be hard pressed to prove that the average of a sine wave is
2/pi exactly using graph paper. You could certainly say it sure looks close
to 2/pi, but is it exactly pi?... can't say for sure.

Case in point is the phase controlled sine wave made by SCR light
dimmers. I find it hard to believe you can graphicaly come up with the
formula:

Sqr-root[ D/2 + sin[pi(1-D)] cos[pi(1-D)]/2pi

I must also add that the graphical solution and mathematical integration
are different implementations of the same concept. I don't intend to say
that one is wrong and one is right.

Now, for Richard, C. Is the thing coming out of the AC outlet an exact
sine wave...no. It is very noticably flattened on the top by all the power
supplies drawing peak currents near the peak. There must be other
corruptins as well. However, if I assume it is a sine wave, will my
calculations come out very close, I believe Yes.

Now in all fairness, I won't dispute that you can use the graphical
method to find the RMS to any desired accuracy, just not exact...without
being able to integrate the waveform.


Of course, but it is eminently doubtful if you can actually express it
mathematically. Far more here own o'scopes than works of multiple
regression.

Graphical analysis is first year engineering stuff out of
drafting class.

Integration is first year engineering stuff out of calc class. At least
it was for me. I am still amazed that I remember doing tripple integrals
then, and thinking...Gee, this ani't so hard after all!. Couldn't do one to
save my life now.


[...] you simply measure the caloric result and ignore
shape altogether.

I always thought that the common method of measuring RF power was pretty
cool! The Thermistor or bolometer. Here you balance a bridge with DC or
low freq AC. It heats the thermistor to the correct resistance. Then, when
you add RF power, the thing heats up more and changes resistance. So, you
remove some DC power to get back to the correct resistance and that amount
is easy to figure. That is how much RF you put in. Cool. I think it is
correct to say that you absolutely cannot measure power *directly*. You
must measure something else which is affected/caused by the power...comment?


[..] migrated from Power to some other consideration

Not trying to migrate. I just think its a cool method.



Yikes! Not sure where you went on that last bit, Richard C...
Now, I ask. Do the power meters on the outside of our houses take all
those
factors into consideration
And, like me, Richard Clark makes a really long answer that says...Yes.

Got me!
I have NO idea how the watt-hour meter works...and don't want to try
to understand at this point.

73,
--
Steve N, K,9;d, c. i My email has no u's.

"Cecil Moore" wrote in message
...
Richard Harrison wrote:
If we have a Class C amplifier feeding power to the same antenna and
enjoying a conjugate match, we can have a source that takes less than
50% of the available energy. So, the transmitting antenna system can be
more efficient than the receiving antenna system, it seems to me.


Didn't see the original post...

I think the phrase "available energy" may draw discussion, but won't start
it. I suspect it is not the thing to focus on. However...

With a "conjugate match" the source dissipates 50% of the power and the load
the other 50%. This can't be changed. Draw the schematic to see. Rs = RL
therefore Ps = PL. (I'll use a big "L" so it looks like an "L")


------------------------
Then I go to something I have for a long time wondered about; the following
situation:
A (tube) amplifier is in conjugate match conditions. It is dissipating 10
watts in its plate. This is the limit of its plate dissipation. Model this
as a voltage source (Vs) with source resistance (Rs) and a load with load
resistance (RL). The conjugate match has removed all the reactance.

There is also 10 Watts to the load.

Now, assuming you can, increase the (plate) supply voltage by, say 20%.

This raises the source voltage (Vs) [may be the fatal flaw]. If the plate
resistance (Rs) stays the same (I don't know if it will - more flaw-fodder),
then the plate dissipation will also increase. SO... how about adjusting
the match so the plate sees a higher RL to get back to 10 watts plate
dissipation. RL must now increase 40% to do this. The source current must
stay the same if the source resistance stays the same and keep the same
source dissipation.

Now, what we have done is to increase RL and Load voltage (VL), and
therefore load power, but kept 10 watts at the plate. If I did the math
correctly the load now dissipates 40% more power. If, however, we
re-adjust the match back to conjugate we WILL get more power and 50% will be
dissipated in the stage...until it blows. SO...

I think this tells us that if the stage is dissipation limited, but not
breakdown limited, we can non-conjugate match for a higher power than
'before' by raising the supply voltage. It sorta looks like we are getting
more with a non-conjugate match. You don't have the 50-50 division of power
also. It is now 41.7% tube, 58.3% load

Snake oil? The math is correct, but there may be some practical limit not
considered.

--
Steve N, K,9;d, c. i My email has no u's.

Then there's the solid state power amplifier standard output resistance
formula.
Rs = Vcc^2 / (2*Po)
The implication should be obvious...

Dan Richardson wrote:

Let me try this one more time. You had posted earlier and I commented
on this:

"A receiving antenna must be resonant to enable full acceptance of
available energy, and it must be matched to avoid re-radiation of more
than 50% of the energy it is able to grab."

I commented on the first portion of your statement (above). My only
point is that it makes no difference if an antenna's is resonate or
not in determining how much energy it grabs.

That's it, nothing more.

73
Danny, K6MHE

SNIP

The effective aperture is the effective aperture is the effective
aperture ...

The antenna intercepts ALL the EM energy within it's effective aperture.

You, the listener, may be interested in only 2.8 KHz of that energy. :-)

Great, now to the issue.
\
How much power does the antenna pick up compared to the power enclosed
within the 2.4 Khz frequency spread that is considered useful ? Or can we
say how much energy is dumped to ground by a bandpass filter (plus insertion
losses). I know it may be absolutely quiet outside the bandwidth of choice
but I assume you get my drift with respect to efficiency of a simple dipole
to a radiator of high Q.
Regards
Art
"Dave Shrader" wrote in message
news:he8%b.120870$jk2.515312@attbi_s53...
Dan Richardson wrote:

Let me try this one more time. You had posted earlier and I commented
on this:

"A receiving antenna must be resonant to enable full acceptance of
available energy, and it must be matched to avoid re-radiation of more
than 50% of the energy it is able to grab."

I commented on the first portion of your statement (above). My only
point is that it makes no difference if an antenna's is resonate or
not in determining how much energy it grabs.

That's it, nothing more.

73
Danny, K6MHE

SNIP

The effective aperture is the effective aperture is the effective
aperture ...

The antenna intercepts ALL the EM energy within it's effective aperture.

You, the listener, may be interested in only 2.8 KHz of that energy. :-)

Dave Shrader wrote:
"The effective aperture is the effective aperture is the effective
aperture...."

And by any other name it would likely smell as sweet.

Terman has a definitionin of sorts in his 1955 edition on page 899:
Directive gain=(4 pi)(effective aperture)(k) divided by lambda squared.

Arnold B. Bailey is much more descriptive in "TV and Other Receiving
Antennas", starting around page 298. I`ve tried to condense and number
his points:
1. The radio field is expressed in volts per meter or in watts per
square meter.
2. Our standard dipole will have a capture area within the radio field,
broadside to the field, with the rod positioned in the direction of the
electric vector. (not cross-polarized)
3. Our 1/2-wave dipole has a capture area extending the 1/2-wavelength
of the rod, and 1/8-wavelength either side of the center of the rod.
4. The area thus calculates as lambda squared over 8.
5. This effective area representation is a simplification but it works
for most purposes.
6. If we can modify our 1/2-wave antenna to capture more signal than
normal, we say the antenna has gain.
7. "Capture area" is pure fiction but it gives the right answers.
8. Zero-dB gain antennas all have the same "capture area", and are
assumed to be resonant which is an essential requirement enabling full
performance.
9. Shape, size, power transfer capability, and directional response are
the four main factors which determine an antenna`s utility. (Bailey has
great elaboration on each of these four factors.)
10. Some antenna shapes operate well over a wide frequency bandwidth,
others don`t.
11. Fat antennas tend to be broadband and are shorter than a free-space
wavelength, indicating a reduced signal velocity along their lengths.
12. All parts of the receiving antenna rod intercept the radio wave.
13. Induced current is reflected by open-circuit ends of the antenna
rod.
14. Impedance at the center of the rod is the result of incident and
reflected waves interacting.
15. Resonance is a result of interaction of incident and reflected
signal components in the rod.
16. The antenna rod has a characteristic resistance independent of the
reflected signal, and this resistance is determined by the ratio of
voltage on the rod to current.. (This is found on page 306.)
17. From the size of an antenna rod we can determiner its capacitance
per unit length. This factor determines its characteristic resistance.
(voltage to current ratio)
18. Reflection coefficient at the ends of the rod depends on
configuration. Thicker antennas have smaller reflections at their ends.
19. Conical rods which are thicker toward their extremities produce weak
reflections, and tend to have broadband responses.
20. On thin dipoles at first resonance, there is substantial reflection
from its ends and the reflection is 180-degrees out of phase with the
incident signal.
21. 1/4-wave back from the open-circuit ends of the antenna, the voltage
is low and the current is high, so the impedance (a resistance) is low.
The resistance 1/4-wavelencth back from the open-circuit ends is much
lower than the characteristic resistance of the rod to the signal
traveling in either direction alonng the antenna rod.
22. With reflection----
R = Vincident - Vreflected / Iincident + Ireflected
(Whereas, the Ro = V/I
23. Fat antennas are less directional than thin antennas and keep their
broadside lobe together better at harmonic frequencies.

Don`t know what the particular value of effective aperture is. Terman
found it useful in describing the directional characteristics of a
matress antenna. Since effective aperture is related to directive gain
by constants, why not just stick with directive gain? Bailey, a leading
Bell Labs antenna researcher, said capture area is pure fiction anyway
as the grip on radiated energy does not start and stop abrubtly at the
imaginary boundaries used to describe it.

Obviously, effective aperture has enjoyed some popularity among some who
found it useful. The effective area of an antenna has a dictionary
definition: The square of the wavelength multiplied by the power gain in
a particular direction, divided by 4 pi. The "capture area" is defined
as: The area of the antenna elements that intercept radio signals.
I may have confused the definitions.

Best regards, Richard Harrison, KB5WZI

Steve Nosko wrote:
"I assumed Richard`s intent here is that you only have to do the
calculation for one full period...symmetrical, then only half
period..."

Maybe an advantage of graphical representation is manifestation of the
absurd. If the wave is not symmetrical around the zero volt axis, or if
one alternation has greater squared values, it may be apparent. You may
be able to see this in the mathematical expression for the waveform too.

The effective value of both alternations must be equal for the
determination of the value of only one alternation to suffice.

Steve also wrote:
"There ia a formula for the total rms which is the square root of the
sum of the squares.""

Sounds familiar. Sounds like Pythagoras.

Best regards, Richard Harrison, KB5WZI