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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 |
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. |
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. ;-) |
Yikes... this thread is still running.
As I explained before there is a "common" usage which has unfortunately been adopted (and printed in the almighty and "always correct" text), which uses the words "RMS power" to actually mean the *average power*. An earlier poster explained that this term was also adoppted (still non mathetically correct) in audio circles to indicate a specific test. This is not a correct usage because the term "RMS" means that it actually is a Root Mean Square value. [[square it, take the mean, then take the square root]] We just don't do this with power waveforms. Sidebar: I did extensive calculations of this type as a result of an article in QST earlier this year where it was proposed to use a VOM/DVM to measure line voltage and current(via a small series dropping resistor) to arrive at power draw of common ham equipment...OOPS! I was, however, a bit puzzeled, Roy, why you went to the trouble of calculating the the RMS value of a power earlier. I suspect it was just to show that the value is indeed different (didn't check your math). I will differ with Roy on one issue. The RMS value of voltage and current have, for many years, also been referred to as the "effective" values. This was to relate it to the DC heating effect (of resistance) we are all familiar with. It is, indeed just as "effective" as the same DC value, in producing power. This is another terminology issue I suspect some of you may wish to squabbling about, but is is not a 'what is correct technically' issue. It is cleat that probably all of you understand the math, but this is simply an nomenclature issue. I feel sorry for Bill because he seems to understand the math: understand about deriving the RMS power from the instantaneous power in the same way that RMS voltage or current is derived, ....yet he said: I just don't accept the definition. It's been taught the other way all my life, even though you say it's incorrect. I see your point, I just don't accept it. I don't understand when you say "I just don't accept the definition." Does this mean that you do not accept thst "RMS Power" implies (to some of us) that you have done the Root Mean Square calculation on the power waveform? Which deffinition is giving you grief? I am not not trying to prolong the pain (or this thread), it is just that I was born with a bone in my head that makes it hard for me to give up explaining some basic concept like this. (yep, it can be a curse) You're so close. -- Steve N, K,9;d, c. i My email has no u's. "Roy Lewallen" wrote in message ... Bill Turner wrote: On Tue, 19 Oct 2004 00:02:50 -0700, Roy Lewallen wrote: And now the real question: How much DC voltage would you have to apply to the resistor to get exactly the same power dissipation? 100 volts. The dissipation would be 100 watts. __________________________________________________ _______ Well. The answer you gave is exactly the answer I would have given, but you say my answer is wrong. Where have I said that's wrong? Of course it's not. I understand about deriving the RMS power from the instantaneous power in the same way that RMS voltage or current is derived, I just don't accept the definition. It's been taught the other way all my life, even though you say it's incorrect. I see your point, I just don't accept it. I have no idea who taught that to you, since the definition of RMS is in its very name (the square Root of the Mean of the Square of the function). It's your choice to ignore the accepted definition. I can only hope you don't teach your mistaken idea to others, who will then someday say the same thing. I will QRT for now, but thanks for taking the time to explain. I mean that sincerely and I do respect your point of view. You're welcome. The only reason I've taken the time for these postings is in the hope that it will help people understand and learn. Even if it hasn't worked for you, I hope some other readers have benefitted. Roy Lewallen, W7EL |
"K9SQG" wrote in message ... While the discussion is interesting, one needs to consider the purpose. For a pure sine wave, certain relationships are stable and hold true. For a complex waveform, like voice, the relationship between peak power and average/RMS power not a straightforward relationship. 73s, Evan Hi Evan, This is not the point of contention. I think, at times, some posters get the wrong idea about the subject being discussed. I think all thoses in the discussion will agree with your statement. The issue is whether or not the words "RMS Power" are a _correct_ term to use for what some of us are saying should be called _Average Power_. It is a discussion of terminology, not math or physics. a.k.a. what to _call_ this power, not what it _is_. -- Steve N, K,9;d, c. i My email has no u's. |
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. ;-) |
Steve Nosko wrote:
. . . I was, however, a bit puzzeled, Roy, why you went to the trouble of calculating the the RMS value of a power earlier. I suspect it was just to show that the value is indeed different (didn't check your math). Yes, that was the only reason. As I mentioned, I'd never before had the occasion to calculate the RMS value of a power waveform -- it's simply not useful for anything. Except to illustrate that it's different from the very useful value of average power. I will differ with Roy on one issue. The RMS value of voltage and current have, for many years, also been referred to as the "effective" values. This was to relate it to the DC heating effect (of resistance) we are all familiar with. It is, indeed just as "effective" as the same DC value, in producing power. This is another terminology issue I suspect some of you may wish to squabbling about, but is is not a 'what is correct technically' issue. It is cleat that probably all of you understand the math, but this is simply an nomenclature issue. I don't have any disagreement with this. I just have to keep cautioning people not to extrapolate it to power. That is, just because the RMS value of voltage or current is an "effective" or DC equivalent value, don't think it implies that the RMS value of power must be its "effective" or DC equivalent value. It's not. . . . I am not not trying to prolong the pain (or this thread), it is just that I was born with a bone in my head that makes it hard for me to give up explaining some basic concept like this. (yep, it can be a curse) . . . Egad, another person with the same genetic defect! Welcome! Roy Lewallen, W7EL |
I have some ideas, but that's all. I hope someone with more recent
direct experience with AM broadcasting than mine who really knows the answer will comment. I will go out on a limb and speculate that the carrier isn't being reduced during modulation. If it were, simple envelope detectors would produce serious distortion. And if the carrier isn't being reduced, then the power has to be greater when modulation is present. But let's see if an expert will comment -- if I'm wrong I'll gladly eat my words. Roy Lewallen, W7EL Bill Turner wrote: Last year I took a tour of the KFI transmitter site in Southern California and was fascinated by the 50kW transmitter. On the transmitter's front panel was a meter calibrated in output power. It read *steady* at 50kW, except when the operator dropped the power momentarily to 5kW, just to show he could. Ever since then, I've kicked myself for not asking why the power didn't rise with modulation. The transmitter was a Harris model DX50 (IIRC) which uses dozens of low power solid state modules which are switched on and off digitally to produce the RF output. Could it be that as they are switched on and off, they also are driven in such a way as to maintain constant power? In other words, when modulation is added the carrier power is reduced? It's the only thing that comes to mind, but there may be another reason. Ideas? -- Bill W6WRT |
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