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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 |
On Wed, 20 Oct 2004 16:34:05 -0700, Roy Lewallen
wrote: 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. If *you* don't know, Roy, WTF does?? -- "What is now proved was once only imagin'd." - William Blake, 1793. |
Paul, why don't you simply say -
Total power = Carrier power + Power in the two sidebands. When no sidebands, just the carrier power. ---- Reg, G4FGQ |
On Thu, 21 Oct 2004 13:41:13 +0000 (UTC), "Reg Edwards"
wrote: Paul, why don't you simply say - Total power = Carrier power + Power in the two sidebands. When no sidebands, just the carrier power. "Total power"?? I dear... I feel another lengthy, definition spin-off thread coming on... ;-) -- "What is now proved was once only imagin'd." - William Blake, 1793. |
"Roy Lewallen" wrote in message ... Steve Nosko wrote: . . . 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 So it's genetic!... Yea. And I do this in front of a class or 5-15 green students. It SURE is rewarding when someone says :"Oooh! NOW I get it!" 73, -- Steve N, K,9;d, c. i My email has no u's. |
Top post...
Well, friends, rather than speculate as we most certainly love to do, I just got off the phone with John, the KFI AM Engineer. (11:15am CDT Oct 21) He said that althought that meter is a pretty standard toroid type power sampler (very like the common ham units, but it has no reflected output), it is not designed to read PEP, but couldn't say what they did in the design. He went on to speculate that the meter ballistics(how fast the meter responds to changes) may be a factor. He also said that due to the level of compression used, you won't see much dynamic range (variation of power with modulation level changes) He said that "it *does* vary if you look closely." He advised that you *should have* looked at the final current meter. It *does* vary widely. He also said that Harris is doing "some pretty interesting things in that transmitter." He described the transmitter as a "50kw D to A converter" with hot redundancy (extra amps on line) and if one of those smaller amps dies, you can see a "notch" in the output waveform, but that it doesn't cause any high frequency sidebands--due in part to the filtering effect of the antenna bandwidth, but he couldn't elaborate if this was the only bandwidth limiting factor preventing splatter from said notches. Thinking about it I wonder if "notch" was not the best word, but that "flat spot" might be mbtter...but then, I didn't think to ask...Oh well. So there ya go... One comment I feel compelled to make here. It is all too common for us to ASSUME that EVERY measurement made is absolutely 100% correct. I see Engineers do this all the time (and though 99.99% of the time this probably is right, there are those times when some corrupting factor gives a strange reading and the experienced Engineer is the one who figures out that something is amiss IN THE SHORTEST TIME-- cuz' we all get confused by strange observations). It is sort of a "It says so on the HP digital meter, so it has to be 100%" assumption. Easy trap to fall into. Meter says "power, therefore it must be reading MY interpretation of power." Heep the faith AND keep asking... -- Steve N, K,9;d, c. i My email has no u's. "Roy Lewallen" wrote in message ... 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 |
On Wed, 20 Oct 2004 15:54:12 -0700, Bill Turner
wrote: 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.) _________________________________________________ ________ I have a question about AM sideband power. I have always believed what Roy says above, that the sidebands add to the total radiated power. You can see this when watching an S-meter. However. 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? I'll take a stab at this. It is probably because of how the power is being measured. An actual sample of output power is rarely used for monitoring. They may have just a simple rf voltmeter measuring the antenna line voltage, calibrated in watts. Many watt meters work in this fashion. When the carrier is modulated, the composite signal, carrier and side band voltage swings up to twice the voltage and down to zero volts with 100% positive and negative modulation. So the average voltage is the voltage that the carrier produces itself. The meter can't follow the swings fast enough so it stays at the average which is equal to the carrier. The plate current meter is in the same situation. It stands still also with modulation for the same reason even though the plate voltage may swing between 2 times and zero. If the modulation is not symmetrical, positive peaks greater than negative peaks, then you will see a slight upward kick with modulation as the average is no longer equal to the carrier voltage. An antenna current meter on the other hand will show a definite upward kick with modulation. The current meter is usually the thermal couple type of meter that actually requires power for it to operate. Heating of a resistor in it is what makes it work. Heating that thermal couple requires power. Since power output only increases with modulation, it never goes below carrier level, the antenna ammeter sees an average power increase with modulation. 73 Gary K4FMX |
"Gary Schafer" wrote in message ... On Wed, 20 Oct 2004 15:54:12 -0700, Bill Turner wrote: 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.) _________________________________________________ ________ I have a question about AM sideband power. I have always believed what Roy says above, that the sidebands add to the total radiated power. You can see this when watching an S-meter. However. 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? I'll take a stab at this. It is probably because of how the power is being measured. An actual sample of output power is rarely used for monitoring. They may have just a simple rf voltmeter measuring the antenna line voltage, calibrated in watts. Many watt meters work in this fashion. When the carrier is modulated, the composite signal, carrier and side band voltage swings up to twice the voltage and down to zero volts with 100% positive and negative modulation. So the average voltage is the voltage that the carrier produces itself. The meter can't follow the swings fast enough so it stays at the average which is equal to the carrier. The plate current meter is in the same situation. It stands still also with modulation for the same reason even though the plate voltage may swing between 2 times and zero. If the modulation is not symmetrical, positive peaks greater than negative peaks, then you will see a slight upward kick with modulation as the average is no longer equal to the carrier voltage. An antenna current meter on the other hand will show a definite upward kick with modulation. The current meter is usually the thermal couple type of meter that actually requires power for it to operate. Heating of a resistor in it is what makes it work. Heating that thermal couple requires power. Since power output only increases with modulation, it never goes below carrier level, the antenna ammeter sees an average power increase with modulation. ============== Yet 'total' RF power can be neatly measured by means of a calibrated oscilloscope. Frank GM0CSZ / KN6WH |
Gary Schafer wrote in message . ..
On Wed, 20 Oct 2004 15:54:12 -0700, Bill Turner wrote: I have a question about AM sideband power. I have always believed what Roy says above, that the sidebands add to the total radiated power. You can see this when watching an S-meter. This is true. Consider a classic plate modulated Class C AM transmitter running 200 watts input. It requires 100 watts of audio for modulation. Assume it has plate efficiency of 75%. With no input to the modulator, the transmitter produces 150 watts of RF from the 200 watts input. With sine-wave audio modulation, the input rises to 300 watts (200 watts DC plus 100 watts audio) and the total RF output rises to 225 watts. That's not a lot - only 75 watts more (75% of 100). However. 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? I'll take a stab at this. It is probably because of how the power is being measured. True - but not for the reasons you state. An actual sample of output power is rarely used for monitoring. They may have just a simple rf voltmeter measuring the antenna line voltage, calibrated in watts. Many watt meters work in this fashion. When the carrier is modulated, the composite signal, carrier and side band voltage swings up to twice the voltage and down to zero volts with 100% positive and negative modulation. So the average voltage is the voltage that the carrier produces itself. No, it isn't. The meter can't follow the swings fast enough so it stays at the average which is equal to the carrier. The plate current meter is in the same situation. It stands still also with modulation for the same reason even though the plate voltage may swing between 2 times and zero. True, but that's not the problem. If the modulation is not symmetrical, positive peaks greater than negative peaks, then you will see a slight upward kick with modulation as the average is no longer equal to the carrier voltage. An antenna current meter on the other hand will show a definite upward kick with modulation. The current meter is usually the thermal couple type of meter that actually requires power for it to operate. Heating of a resistor in it is what makes it work. Heating that thermal couple requires power. Since power output only increases with modulation, it never goes below carrier level, the antenna ammeter sees an average power increase with modulation. Not the issue. Both an RF voltmeter and a thermocouple ammeter require some power to work. If the antenna is a linear load, the voltage and current must retain a linear relationship, so that increasing one increases the other. IOW, voltmeter vs. ammeter makes no difference. Sorry. The reason the power meter didn't move was probably due to averaging. The measeurement system probably averages the power over several seconds, rather than trying to follow the audio. If they had transmitted just an unmodulated carrier for a long-enough time, you'd see the power output drop. But BC stations don't do that intentionally - "dead air" is a real no-no. 73 de Jim, N2EY |
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In article , Gary Schafer
writes: I set the carrier level out at 20 watts. Modulated the transmitter with a 1000 hz tone up to 100% modulation as seen on the scope monitoring the output of the transmitter. What sort of rig? Using a drake w4 watt meter (same as a bird) the watt meter held steady at 20 watts. It read the same at full 100% modulation as it did with no modulation, just the carrier. No ALC involved here either. Watching the scope the rf output voltage did double with 100% modulation as would be expected. Now had I put a thermocouple rf ammeter in the coax line I would have seen an increase in the line current with modulation. But you didn't put an RF ammeter in the line, nor an RF voltmeter. What did I do wrong? I don't think you did anything wrong. What sort of rig was it? How modulated? 73 de Jim, N2EY |
The explanations make sense to me.
The average *voltage* of an amplitude modulated signal equals the carrier voltage, provided that the modulation has no DC offset (that is, its average value is zero). This would be true for any waveform, provided that it's AC coupled and it's not DC level shifted, clipped, or otherwise distorted after the AC coupling. (It wouldn't be true of a classically overmodulated carrier, for example.) The average of an AC coupled waveform is always zero, if averaged over a time period that's long compared to the time constant of the coupling network. So a power meter that's really reading the voltage should stay at the carrier level, provided that its time constant is comparable to or longer than the time constant of the AC coupling of the modulation signal. I'd expect this to be the case for a typical meter and typical audio modulation. Even if you're monitoring the true power level, the average typically wouldn't get much greater than the carrier if you were modulating it with a normal voice. Compression would tend to raise the average some, though. As I understand the Bird wattmeter, it basically takes current and voltage samples and adds them (rather than vectorially multiplying them, as a true power detector would have to do). What this would do to its response to average power I'm not sure, but I wouldn't expect it to give an accurate indication of either the carrier power or total power of a modulated signal. We're confronted all the time with measurements that seem to contradict established theory. Some people regretfully are quick to embrace these as evidence that established theory is wrong. It does take some effort and knowledge to dig a bit to find out why there's a disagreement. But with a miniscule frequency of exceptions, the digging always reveals that we're not measuring what we think we are, we're using the wrong theory, or we're applying it wrong. Roy Lewallen, W7EL |
Bill Turner wrote:
Isn't the S-meter in a typical receiver voltage driven? They do kick upwards with modulation. -- Bill W6WRT S-meters are driven by the AGC voltage, which is usually derived from the peak voltage of the signal. The peak value of an AM signal does increase with modulation, so it wouldn't surprise me to see one kick up a little once in a while on voice peaks when listening to AM. Of course, they always kick upwards when listening to SSB -- just about none of what I've written applies to that mode. Roy Lewallen, W7EL |
In message , Bill Turner
writes On Fri, 22 Oct 2004 19:34:26 -0700, Roy Lewallen wrote: So a power meter that's really reading the voltage should stay at the carrier level, provided that its time constant is comparable to or longer than the time constant of the AC coupling of the modulation signal. I'd expect this to be the case for a typical meter and typical audio modulation. _________________________________________________ ________ Isn't the S-meter in a typical receiver voltage driven? They do kick upwards with modulation. -- Bill W6WRT Try making a simple RF voltmeter consisting of a diode detector driving a sensitive moving coil meter (say 100microamp fsd) or a multimeter reading volts. If the RF source doesn't already have a DC path to ground (ie it is AC coupled), you will also need to add a shunt resistor or RF choke. Make the resistor as low as possible without killing the RF too much. Apply sufficient RF to give (say) a half-scale reading. Now modulate the signal with audio (to 100% if possible). The meter reading will stay the same. Now add a fairly large capacitor across the meter (say 1uF). The reading will increase. If the RF source impedance is low compared with the resistance of the meter, the capacitor will discharge very little when the diode is 'off', and the cap will charge up to the peak voltage. With a good modulation waveform, 100% mod will double the reading. (Note: Any capacitor value between 'far too small' and 'more than enough' will give a reading somewhere between. I did this about 30 years ago to add a simple mod depth indicator to and old signal generator which already had an RF level meter. Anyway, all this talk about power and PA efficiency tends to complicate the explanation. Just think of the spectrum of the signal. No mod = 0dB reference = the carrier. Add mod. Sidebands appear each side of the carrier. The carrier level stays at 0dB. Set mod at 100%. Each sideband should be 6dB below the carrier level. Total power = 0dB + -6dB + -6dB = 1 + 1/4 + 1/4 = 1.5 (Note: This is regardless how the modulation was applied). Peak power can be got from the voltage waveform. The peak RF voltage doubles with 100% mod. The peak power is proportional to the square of the voltage, so it quadruples. So, to sum up... for AM: 100% mod, power increases x1.5 wrt 0% mod. 100% mod, peak power is x4 wrt carrier. Ian. -- |
My friend, N7TCY wrote this bit regarding RMSpower a few years ago. We work with inverters so there was some call for it. http://www.eskimo.com/~bgudgel/cgi-bin/rms/rms.html boB K7IQ On Thu, 21 Oct 2004 10:49:39 -0500, "Steve Nosko" wrote: "Roy Lewallen" wrote in message ... Steve Nosko wrote: . . . 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 So it's genetic!... Yea. And I do this in front of a class or 5-15 green students. It SURE is rewarding when someone says :"Oooh! NOW I get it!" 73, |
boB_K7IQ wrote:
My friend, N7TCY wrote this bit regarding RMSpower a few years ago. We work with inverters so there was some call for it. http://www.eskimo.com/~bgudgel/cgi-bin/rms/rms.html That looks like a good analysis. I've just put the finishing touches on a non-mathematical analysis which I hope will be helpful for folks who can't or don't want to wade through the math. It uses simple square waves to illustrate the concepts. The link is http://eznec.com/Amateur/'RMS Power'.pdf Comments and corrections are welcome, either posted here or emailed to me. Roy Lewallen, W7EL |
Hm, my browser doesn't like the name I gave the file. So I've uploaded a
second copy with a different name. You can also get it as http://eznec.com/Amateur/RMS_Power.pdf Roy Lewallen, W7EL Roy Lewallen wrote: That looks like a good analysis. I've just put the finishing touches on a non-mathematical analysis which I hope will be helpful for folks who can't or don't want to wade through the math. It uses simple square waves to illustrate the concepts. The link is http://eznec.com/Amateur/'RMS Power'.pdf Comments and corrections are welcome, either posted here or emailed to me. Roy Lewallen, W7EL |
Both the Rosenbaum and Lewallen papers look very good and well written.
I would, however, Roy, make a more explicit statement that while RMS power can be calculated, is has no practical value for the normal considerations of power...and add a note indicating something like that: "The phrase "RMS Power" has been used in some circles not as an exact use of the term RMS, but rather as an informal "standard" that actually means "average power under specific text conditions". (I refer to an earlier post telling of the single channel, steady state audio power amp measurement) This use of the term "RMS" was originally initiated to call attention to the specific test. Unfortunately, this use has caused some confusion in the use of this terminology and is some of the motivation for the paper." 73, Steve, K9DCI -- Steve N, K,9;d, c. i My email has no u's. "Roy Lewallen" wrote in message ... Hm, my browser doesn't like the name I gave the file. So I've uploaded a second copy with a different name. You can also get it as http://eznec.com/Amateur/RMS_Power.pdf Roy Lewallen, W7EL Roy Lewallen wrote: That looks like a good analysis. I've just put the finishing touches on a non-mathematical analysis which I hope will be helpful for folks who can't or don't want to wade through the math. It uses simple square waves to illustrate the concepts. The link is http://eznec.com/Amateur/'RMS Power'.pdf Comments and corrections are welcome, either posted here or emailed to me. Roy Lewallen, W7EL |
Thanks for the suggestions, Steve. I've updated the file to incorporate
them. The new file is at http://eznec.com/Amateur/RMS_Power.pdf. During recent trips to Powell's Technical Bookstore and the library, I looked through a number of books about electrical circuits which are oriented toward hobbyists and technicians (that is, ones lacking the math of a college level circuits text). I'm glad to say I didn't find any which were plainly wrong about average and RMS power (like the Joe Carr book quoted earlier here). But what nearly all of them do is to introduce RMS voltage and current pretty early on in the text, and explain that the RMS values of voltage and current are important because they represent equivalent heating values (which is correct). From then on, they simply use E and I with the assumption that they represent RMS values of voltage and current. At some point, they introduce the equation P = E * I or, in the more advanced ones, E * I * cos(phase angle), and maybe at that point mention that P is the equivalent heating power (which is also correct). What I didn't see in any of them was the fact that the product of the RMS values of E and I is the *average*, and *not* the RMS value of P. It's easy to understand, then, why a lot of people, like a number of the folks who posted comments and questions here, naturally (and incorrectly) assume that the product of Erms and Irms is RMS power. The books simply don't contain the information you'd need in order to discover that Erms * Irms = Pavg. Hopefully the paper posted by Rosenbaum and the one I did will help fill the void. Roy Lewallen, W7EL Steve Nosko wrote: Both the Rosenbaum and Lewallen papers look very good and well written. I would, however, Roy, make a more explicit statement that while RMS power can be calculated, is has no practical value for the normal considerations of power...and add a note indicating something like that: "The phrase "RMS Power" has been used in some circles not as an exact use of the term RMS, but rather as an informal "standard" that actually means "average power under specific text conditions". (I refer to an earlier post telling of the single channel, steady state audio power amp measurement) This use of the term "RMS" was originally initiated to call attention to the specific test. Unfortunately, this use has caused some confusion in the use of this terminology and is some of the motivation for the paper." 73, Steve, K9DCI |
On Thu, 18 Nov 2004 15:52:41 -0800, Roy Lewallen
wrote: Thanks for the suggestions, Steve. I've updated the file to incorporate them. The new file is at http://eznec.com/Amateur/RMS_Power.pdf. During recent trips to Powell's Technical Bookstore and the library, I looked through a number of books about electrical circuits which are oriented toward hobbyists and technicians (that is, ones lacking the math of a college level circuits text). I'm glad to say I didn't find any which were plainly wrong about average and RMS power (like the Joe Carr book quoted earlier here). But what nearly all of them do is to introduce RMS voltage and current pretty early on in the text, and explain that the RMS values of voltage and current are important because they represent equivalent heating values (which is correct). From then on, they simply use E and I with the assumption that they represent RMS values of voltage and current. At some point, they introduce the equation P = E * I or, in the more advanced ones, E * I * cos(phase angle), and maybe at that point mention that P is the equivalent heating power (which is also correct). What I didn't see in any of them was the fact that the product of the RMS values of E and I is the *average*, and *not* the RMS value of P. It's easy to understand, then, why a lot of people, like a number of the folks who posted comments and questions here, naturally (and incorrectly) assume that the product of Erms and Irms is RMS power. The books simply don't contain the information you'd need in order to discover that Erms * Irms = Pavg. Hopefully the paper posted by Rosenbaum and the one I did will help fill the void. Hi Roy, there does seem to be an unsettling amount of misinformation, errors and poor explanations in the majority of text books I've encountered, I'm sorry to say. No wonder there's such a huge amount of confusion surrounding these aspects of our hobby. :-( -- "What is now proved was once only imagin'd." - William Blake, 1793. |
Ok Chris...here's the deal with "swing"...it's pretty much BS. By turning the power down to 2 watts and letting the xmtr go to 10 watts is just going to cause splatter, TVI and a whole host of other problems just so you can show off to your CB buddies how much you can make the s-meter swing on thier end. Anything in excess of 100 percent modulation is a waste of power and will make you sound like ca-ca. Unless you want to be a smartass and a prick, follow the guide below.
100 percent AM modulation as far as us hams are concerned is unmodulated carrier times 4. So if you have a 375 watt "dead carrier" with 100 percent modulation you will show a peak reading of 1500 watts with modulation (ham legal limit, BTW)...so a 4 watt dead carrier...well 100 percent modulation will be 16 watts. That's with a linear response out of the modulator and final circuitry...FCC rules say 4 modulated 5 as far as wattage for CB. The AMC (automatic modulation control) does this (4 watts dead key, 5 max "swing") by providing a non-linear response out of the modulator and final circuitry combination, accomplished usually with a diode or diode and potentiometer. Removal or adjustment of these devices, along with realignment, (peaking out to coin CB terms) can acheive the desired modulation result described here. Type in CB mods into google and you'll find a bunch of stuff if you're so inclined. To also pick another statement out from your post, "That's just one of the reasons I'm looking to amateur radio." That's your best bet...get your ham license. Blows 11 meter away. Take it from an ex freebander who now is a licensed ham. john Hope that answers your question Quote:
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