Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?

Chris
"Roy Lewallen" wrote in message
...
| Paul, I apologize. My browser showed only the first of the two pages,
| and I didn't realize that the second was there. While looking for the
| quotation I found that the second page was simply scrolled off screen.
|
| Yes, there is one thing (on that second page) I do disagree with the
| author on, that the equivalent power, the product of Vrms and Irms, is
| "RMS" power. I did a brief web search to find out who the author was so
| I could contact him about that, and discovered that it's Joe Carr.
| Unfortunately, he died a short time ago.
|
| Maybe my suggestion about looking in non-mathematical texts for an
| explanation wasn't such a good idea. It appears that some of the authors
| of those texts don't fully understand the math either. I'll have to say
| that you certainly have provided some evidence as to how widespread the
| misconception is. Next time I'm downtown at Powell's Technical
| Bookstore, I'll leaf through a few volumes oriented toward technicians
| and see just how bad it is. All I have on my bookshelf in the way of
| basic circuit analysis texts is two (Pearson and Maler, and Van
| Valkenburg) which are intended for beginning engineering students, and
| they of course both have it right. A popular elementary physics text
| which I have, Weidner & Sells, _Elementary Classical Physics_, Vol. 2,
| succinctly summarizes (p. 913): "Thus, the average rate at which thermal
| energy is dissipated in the resistor is the product of the rms voltage
| across it and the rms current through it." This follows immediately
| below an equation showing the calculation of pav from the classical
| definition of average which I posted some time ago.
|
| In response to the question about Vrms, Irms, and equivalent power, I
| said that when Vrms = 100 volts and Irms = 1 amp, Pavg = 100 watts and
| Prms is about 122.5 watts. I haven't had any previous occasion to
| calculate RMS power, so I might have made a mistake. According to my
| calculation, for sinusoidal voltage and current, the RMS power equals
| the average power (which is Vrms X Irms when the load is resistive)
| times the square root of 1.5. Surely some of the readers of this group
| can handle the calculus involved in the calculation -- it's at the level
| taught to freshman engineering students, and now often taught in high
| school. The calculation isn't hard, but it's a little tedious, so
| there's ample opportunity to make a mistake. I'd very much appreciate if
| one of you would take a few minutes and double-check my calculation. Or
| check it with Mathcad or a similar program. I'll be glad to get you
| started if you'll email me. And I'll be glad to post a correction if I
| did make a mistake.
|
| Roy Lewallen, W7EL
|
| Paul Burridge wrote:
|
| On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
| wrote:
|
|
| I really appreciate the compliment, and will do my best to try and
| deserve it.
|
| Please look very carefully at the diagram at the URL you've posted, and
| notice that it's a voltage waveform (see the labeling of the vertical
| axis). Then read the text very carefully. Neither the diagram nor the
| text contradict what I've said. If you think it does, post the reason
| why, and I'll try to clear it up.
|
|
| Okay, here's the bit that you seem to take exception to (it's spread
| over both pages):
|
| "We can define the real power in an AC circuit as the equivalent DC
| power that would produce the same amout of heating in a resistive load
| as the applied AC waveform. [In a purely resistive load] we can use
| the root mean square (RMS) values (Vrms and Irms) to find this
| equivalent or RMS power."
|
| Then the equation "P = Vrms x Irms"
|
| Where "P" here explicitly refers to RMS power.
|
| This is something you have stated clearly that you disagree with.

On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:

Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?

AIUI the specified 4 Watts is the maximum*average* power allowed (in
the UK, anyway). When you modulate it 100% AM., it's still 4W average
power. If you fully modulate it with FM., it's *still* 4W average
power. But as you've seen here, for every assertion, there's a
contradiction. ;-)
--

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

Paul Burridge wrote:

[snip] But as you've seen here, for every assertion, there's a
contradiction. ;-)

Eh? I assert that 1+1=2, for 1 and 2 in the integers, and "+" defined
according to Peano's axioms. Got a valid contradiction?

--
Mike Andrews

Tired old sysadmin

On Wed, 20 Oct 2004 15:31:00 +0000 (UTC), (Mike
Andrews) wrote:

Paul Burridge wrote:

[snip] But as you've seen here, for every assertion, there's a
contradiction. ;-)

Eh? I assert that 1+1=2, for 1 and 2 in the integers, and "+" defined
according to Peano's axioms. Got a valid contradiction?

Post it to sci.math. Someone will - guaranteed! :-)
--

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

The average power of a 100% modulated 4 watt carrier is 6 watts, not 4.
(If you want to look at it in the frequency domain, where the total
power has to be the same as in the time domain, you've now got the
original carrier plus two sidebands. The power in the two sidebands
totals 2 watts.) And I'd give the answer to Chris' two questions as yes.

Roy Lewallen, W7EL

Paul Burridge wrote:

On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:


Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?


AIUI the specified 4 Watts is the maximum*average* power allowed (in
the UK, anyway). When you modulate it 100% AM., it's still 4W average
power. If you fully modulate it with FM., it's *still* 4W average
power. But as you've seen here, for every assertion, there's a
contradiction. ;-)

Pardon my saying so Roy, but I think you may be confusing the issue
here. We all know that you understand this stuff backward and forward
and most here have the highest regard for your expertise, including
me. I also agree that you are totally correct in what you say.

The question was not what is the average power of a 100% modulated 4
watt carrier.
It was "is the 4 watt maximum power of a CB radio actually average
power, not RMS right?"

He is trying to establish the meaning between so called (widely
misused) RMS power and average. And he is also trying to figure out
the relationship to pep.

Although you did acknowledge that he has the correct conclusion to his
question, at the same time I think that you have injected some doubt
in your answer. There are a lot of people that have trouble with some
of the basics of this stuff. Throwing a little twist like that in
often raises the confusion level with some. Again pardon me for
comments.

73
Gary K4FMX


On Wed, 20 Oct 2004 13:17:49 -0700, Roy Lewallen
wrote:

The average power of a 100% modulated 4 watt carrier is 6 watts, not 4.
(If you want to look at it in the frequency domain, where the total
power has to be the same as in the time domain, you've now got the
original carrier plus two sidebands. The power in the two sidebands
totals 2 watts.) And I'd give the answer to Chris' two questions as yes.

Roy Lewallen, W7EL

Paul Burridge wrote:

On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:


Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?


AIUI the specified 4 Watts is the maximum*average* power allowed (in
the UK, anyway). When you modulate it 100% AM., it's still 4W average
power. If you fully modulate it with FM., it's *still* 4W average
power. But as you've seen here, for every assertion, there's a
contradiction. ;-)

No pardon needed, Gary, and I appreciate your comments. I was only
correcting what Paul said:

AIUI the specified 4 Watts is the maximum*average* power allowed (in
the UK, anyway). When you modulate it 100% AM., it's still 4W average
power. . .

What's true is that the average *carrier* power is still 4 watts. But
that's not what the posting said.

I think it's important to be careful with our terminology. The problem
with being loose and free with it is that it causes us to keep having
breakdowns in communication. It also ends up giving people a mistaken
idea about how things work. A naive reader could easily take Paul's
statement to mean that the average power of a 100% modulated 4 watt
carrier is 4 watts -- that's what he said, after all (even though it
might not be what he meant). The lengthy discussion about "RMS power"
illustrates just how deeply rooted a misconception can get, simply from
being careless with terminology.

If anyone considers this to be just nit-picking, that's ok. If you
already understand it, just ignore my postings. But I hope it does serve
a positive purpose for some readers.

And I'm guilty, too! I should have said that the average power of a 100%
amplitude modulated 4 watt carrier is 6 watts *if the modulation is a
sine wave*. When modulated by voice, the average power over any given
interval can vary a great deal. That's why PEP is a more useful
measurement of the modulated signal. As Ian humbly pointed out, it's
really easy to be careless with terminology, and we all make mistakes.
But I think we should keep trying to do better.

Roy Lewallen, W7EL

Gary Schafer wrote:

Pardon my saying so Roy, but I think you may be confusing the issue
here. We all know that you understand this stuff backward and forward
and most here have the highest regard for your expertise, including
me. I also agree that you are totally correct in what you say.

The question was not what is the average power of a 100% modulated 4
watt carrier.
It was "is the 4 watt maximum power of a CB radio actually average
power, not RMS right?"

He is trying to establish the meaning between so called (widely
misused) RMS power and average. And he is also trying to figure out
the relationship to pep.

Although you did acknowledge that he has the correct conclusion to his
question, at the same time I think that you have injected some doubt
in your answer. There are a lot of people that have trouble with some
of the basics of this stuff. Throwing a little twist like that in
often raises the confusion level with some. Again pardon me for
comments.

73
Gary K4FMX


On Wed, 20 Oct 2004 13:17:49 -0700, Roy Lewallen
wrote:


The average power of a 100% modulated 4 watt carrier is 6 watts, not 4.
(If you want to look at it in the frequency domain, where the total
power has to be the same as in the time domain, you've now got the
original carrier plus two sidebands. The power in the two sidebands
totals 2 watts.) And I'd give the answer to Chris' two questions as yes.

Roy Lewallen, W7EL

Paul Burridge wrote:


On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:



Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?


AIUI the specified 4 Watts is the maximum*average* power allowed (in
the UK, anyway). When you modulate it 100% AM., it's still 4W average
power. If you fully modulate it with FM., it's *still* 4W average
power. But as you've seen here, for every assertion, there's a
contradiction. ;-)

Yes Chris, you are exactly right.

73
Gary K4FMX


On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:

Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?

Chris
"Roy Lewallen" wrote in message
...
| Paul, I apologize. My browser showed only the first of the two pages,
| and I didn't realize that the second was there. While looking for the
| quotation I found that the second page was simply scrolled off screen.
|
| Yes, there is one thing (on that second page) I do disagree with the
| author on, that the equivalent power, the product of Vrms and Irms, is
| "RMS" power. I did a brief web search to find out who the author was so
| I could contact him about that, and discovered that it's Joe Carr.
| Unfortunately, he died a short time ago.
|
| Maybe my suggestion about looking in non-mathematical texts for an
| explanation wasn't such a good idea. It appears that some of the authors
| of those texts don't fully understand the math either. I'll have to say
| that you certainly have provided some evidence as to how widespread the
| misconception is. Next time I'm downtown at Powell's Technical
| Bookstore, I'll leaf through a few volumes oriented toward technicians
| and see just how bad it is. All I have on my bookshelf in the way of
| basic circuit analysis texts is two (Pearson and Maler, and Van
| Valkenburg) which are intended for beginning engineering students, and
| they of course both have it right. A popular elementary physics text
| which I have, Weidner & Sells, _Elementary Classical Physics_, Vol. 2,
| succinctly summarizes (p. 913): "Thus, the average rate at which thermal
| energy is dissipated in the resistor is the product of the rms voltage
| across it and the rms current through it." This follows immediately
| below an equation showing the calculation of pav from the classical
| definition of average which I posted some time ago.
|
| In response to the question about Vrms, Irms, and equivalent power, I
| said that when Vrms = 100 volts and Irms = 1 amp, Pavg = 100 watts and
| Prms is about 122.5 watts. I haven't had any previous occasion to
| calculate RMS power, so I might have made a mistake. According to my
| calculation, for sinusoidal voltage and current, the RMS power equals
| the average power (which is Vrms X Irms when the load is resistive)
| times the square root of 1.5. Surely some of the readers of this group
| can handle the calculus involved in the calculation -- it's at the level
| taught to freshman engineering students, and now often taught in high
| school. The calculation isn't hard, but it's a little tedious, so
| there's ample opportunity to make a mistake. I'd very much appreciate if
| one of you would take a few minutes and double-check my calculation. Or
| check it with Mathcad or a similar program. I'll be glad to get you
| started if you'll email me. And I'll be glad to post a correction if I
| did make a mistake.
|
| Roy Lewallen, W7EL
|
| Paul Burridge wrote:
|
| On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
| wrote:
|
|
| I really appreciate the compliment, and will do my best to try and
| deserve it.
|
| Please look very carefully at the diagram at the URL you've posted, and
| notice that it's a voltage waveform (see the labeling of the vertical
| axis). Then read the text very carefully. Neither the diagram nor the
| text contradict what I've said. If you think it does, post the reason
| why, and I'll try to clear it up.
|
|
| Okay, here's the bit that you seem to take exception to (it's spread
| over both pages):
|
| "We can define the real power in an AC circuit as the equivalent DC
| power that would produce the same amout of heating in a resistive load
| as the applied AC waveform. [In a purely resistive load] we can use
| the root mean square (RMS) values (Vrms and Irms) to find this
| equivalent or RMS power."
|
| Then the equation "P = Vrms x Irms"
|
| Where "P" here explicitly refers to RMS power.
|
| This is something you have stated clearly that you disagree with.