On Sun, 3 Oct 2004 09:16:53 +0100, "Ian White, G3SEK"
wrote:

The definition used by OFCOM, the UK licensing authority, contains the
same words: "The average power ... in one RF cycle at the crest of the
modulation envelope" - but it also contains two useful loopholes.

The full wording is: "The average power supplied to the antenna by a
transmitter during one RF cycle at the crest of the modulation envelope
taken under normal operating conditions."

That means UK amateurs are explicitly permitted to allow for feedline
loss (very handy at UHF and higher) and abnormal transients aren't
counted. Given our 400W PEP output limit, we need all the concessions we
can get.

Okay, gentlemen, I can see where you're coming from now.
Incidentally, the (UK) definition above could be construed to allow
for some really serious QRO if one takes "normal operating conditions"
to refer to *atmospheric* conditions rather than those of the station
set-up. When's the next sunspot minima? :-}
--

"What is now proved was once only imagin'd." - William Blake, 1793.

That means UK amateurs are explicitly permitted to allow for feedline
loss (very handy at UHF and higher) and abnormal transients aren't
counted. Given our 400W PEP output limit, we need all the concessions we
can get.

73 from Ian G3SEK
=======================================

It's also very handy on 160 meters.

For example, if the antenna is just a 15-feet length of wire and 400 watts
PEP are fed into it over 115 feet of 600-ohm single-wire transmission line,
who needs concessions?

A 15-feet length of wire, all by itself, is quite efficient on 160 meters.
Nearly all of the 400 watts fed into it will be radiated and the licensing
regulations are not violated.

Furthermore, because a transmission line of that particular length
accurately does the impedance matching, a tuner becomes redundent.
----
Reg.

On Sun, 3 Oct 2004 09:16:53 +0100, "Ian White, G3SEK"
wrote:

Gary Schafer wrote:

The definition of peak envelope power (PEP) is: "The average power
contained in one RF cycle at the crest of the modulation envelope".
(note that the definition says "AVERAGE power" not RMS power) This is
from the FCC definition.


The definition used by OFCOM, the UK licensing authority, contains the
same words: "The average power ... in one RF cycle at the crest of the
modulation envelope"

Hang on a minute, Ian!
I've just looked in your book and in section 6-5 you say in a passage
on Peak Envelope Power: "PEP is the RMS RF power level at the peak of
the modulating waveform." "RMS"? Which is it: RMS or AVERAGE??
Not getting confused, are you? ;-}
--

"What is now proved was once only imagin'd." - William Blake, 1793.

Paul Burridge wrote:
The definition used by OFCOM, the UK licensing authority, contains the
same words: "The average power ... in one RF cycle at the crest of the
modulation envelope"

Hang on a minute, Ian!
I've just looked in your book and in section 6-5 you say in a passage
on Peak Envelope Power: "PEP is the RMS RF power level at the peak of
the modulating waveform." "RMS"? Which is it: RMS or AVERAGE??

It's not a term I would use any more; but if the first sentence hadn't
tired you out, the rest of that sidebar would have told you exactly what
I meant by it.

Not getting confused, are you? ;-}

No, just getting more careful about writing things that can be
selectively misquoted :-(


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

On Sun, 3 Oct 2004 09:16:53 +0100, "Ian White, G3SEK"
wrote:


|
|That means UK amateurs are explicitly permitted to allow for feedline
|loss (very handy at UHF and higher) and abnormal transients aren't
|counted. Given our 400W PEP output limit, we need all the concessions we
|can get.


When our FCC changed the power limit from DC input to rf output I
assumed that the feedline was part of a "distributed" amplifier output
matching network and moved my Bird sensor to the antenna feedpoint.

Feedline loss? What feedline loss? [g]

"Reg Edwards" wrote in message
...

That means UK amateurs are explicitly permitted to allow for feedline
loss (very handy at UHF and higher) and abnormal transients aren't
counted. Given our 400W PEP output limit, we need all the concessions we
can get.

73 from Ian G3SEK
=======================================

It's also very handy on 160 meters.

For example, if the antenna is just a 15-feet length of wire and 400 watts
PEP are fed into it over 115 feet of 600-ohm single-wire transmission
line,
who needs concessions?

A 15-feet length of wire, all by itself, is quite efficient on 160 meters.
Nearly all of the 400 watts fed into it will be radiated and the licensing
regulations are not violated.

Furthermore, because a transmission line of that particular length
accurately does the impedance matching, a tuner becomes redundent.

=================================

But to keep things in proportion -

To radiate 400 watts from a 15-feet antenna wire, fed via a 115-feet,
600-ohm, single-wire, overhead transmission line, would require a
transmitter power of the order of 1.4 Megawatts.

So before ordering the materials to construct such an antenna and feedline
it would be better to forget all about it!

sorry smiley
----
Reg

On Sun, 03 Oct 2004 06:54:50 -0700, Bill Turner
wrote:

On Sun, 03 Oct 2004 03:29:17 GMT, Gary Schafer
wrote:

The definition of peak envelope power (PEP) is: "The average power
contained in one RF cycle at the crest of the modulation envelope".
(note that the definition says "AVERAGE power" not RMS power) This is
from the FCC definition.

_________________________________________________ ________

Right you are, but I'd like to know where the definition of average
power comes from. Is it the IEEE?

If the formal definition says VRMS x IRMS = average power, I suppose I
could live with that, but until then, I think it equals RMS power.

Sources, please.


Rms voltage and current are also called "the effective values".

If you have heard that average power is 1/2 of peak power we could
investigate why.

The rms voltage and rms current of a sine wave are found by
multiplying peak voltage by .707. Same for current.
If you multiply .707 by .707 that gives you .5 or 1/2.

1 volt peak Ac voltage times 1 volt peak / 1 ohm = 1 watt peak power.

1 volt peak times .707 = .707 rms volts. .707 rms volts times .707 rms
volts = .5 / 1 ohm = .5 watts average power.


"It takes twice the Ac peak power to provide the same amount of heat
as it does average DC power." Therefore 1/2 the peak Ac power is equal
to it's average power.

Note that rms voltage is defined as the amount of Ac voltage that will
cause the same amount of heating in a resistor as an amount of DC
voltage.


Also note that when describing rms voltage and its heating effects,
that it does not say amount of power required to do the same amount of
heating. It says the amount of rms voltage to do the same amount of
heating in a resistor. (thus, effective voltage) This is where many
get confused.

If you want to know the amount of power you must square the voltage
and divide by the resistance.
There is that .707 x .707 = .5 again for Ac power. (E squared /R)

2000 ARRL handbook 6.6 chapter 6, RMS VOLTAGES AND CURRENTS.

73
Gary K4FMX

On Mon, 04 Oct 2004 07:54:11 -0700, Bill Turner
wrote:

On Sun, 03 Oct 2004 16:33:46 GMT, Gary Schafer
wrote:

Rms voltage and current are also called "the effective values".

large snip

_________________________________________________ ________

You missed my question. Who defines these terms? I know that RMS,
average and effective are often used interchangeably, but who or what
organization says there is "no such thing as RMS power", and why?

No I didn't miss your question. I thought I gave you enough
information so that you could figure out the "why" yourself.

I don't know of any organization that says "there is no such thing as
rms power". What would be the point. That would be like formally
declaring "water doesn't run uphill", since there is no proof or
evidence that water runs uphill.

I also don't know of any organization that says there is such a thing
as rms power although you will often find it referred to in many
articles and advertisements. Even some of the older ARRL handbooks may
call it rms power but I don't think you will find that in the newer
versions.

Most any AC circuit theory will explain how values of a sine waves are
found and the relation of those values. The reference I quoted from
the ARRL handbook does very nicely at explaining it. It is worth a few
minutes to read. Only a couple of paragraphs will get you there. (you
will see the absence of rms power mentioned)

The thing to consider is that once you multiply an rms value by
another rms value the result is no longer an rms value.
If it were then you could use any of the rms factors to convert it to
peak etc.

Consider 1 volt peak x .707 = .707 volts rms. .707 volts x .707 volts
= .5. Divide that by 1 ohm and you have 1/2 watt.

Using rms factors, convert the rms values to peak. .707 volts x 1.414
= 1 volt peak. OK.

Now try and find peak power like you would find peak voltage from a
known rms value.
Do that with the power value that you want to call rms power and you
have .5 x 1.414 = .707 watts. Not the 1 watt peak you were looking
for.

73
Gary K4FMX

Yadda, yadda, yadda


The term "RMS POWER" while not technically correct for anything practical is
tossed about and I suspect it is 'meant' to mean true or average power as
generally understood by those schooled in the field. One of the reasons for
formal training (or understanding of that training) is so we have
terminology which we have in common. One word or phrase relates to the same
concept for everyone in the discussion.


"Peak-to-peak power" is quite meaningless. P-P voltage and current can be
measured, but power is a second order quantity requiring voltage and current
and an in-phase component as well...and multiplication of these quantities
(that's what makes it second order). This takes care of the phase
relationships and when the voltage goes negative, the current does (the
in-phase component does) and two negatives make a positive and you again
have positive power. There is NO amount of voltage or current which occurs
which is the P-P value. This is only in the eye of the beholder who chooses
to relate two parts of a waveform which occur at different times.


Question:

In a pulsed situation like a common bridge rectifier, capacitor input filter
DC supply, can the true power be determined by measuring the Irms and Vrms
(on the AC side) and doing P = Irms X Vrms ???

Anybody sure enough to say???


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

Ask a sensible question.
What is "true power"
Where in the circuit is it located.