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CB "Swing"? (proper AM modulation)
I will probably get some flames from this but here it goes. I have been into
CB radio for a number of years but don't agree with most of what I hear. That's just one of the reasons I'm looking to amateur radio. One of the things I often hear in CB circles is that one should turn a 4 watt AM radio down to 1 1/2 watts and let it "SWING". How is this possible? What really happens when you do this? I think I know. So, how much carrier should you have for an amp or final stage with a known max output. In other words, if it can produce 8 watts max unmodulated carrier, is a 4 watt carrier ideal? If it produces 100 watts, is 50 watts ideal? How much "space"does it need for proper modulation? Is there a website that explains this well? I'm an electronics tech thirsting for knowledge. Chris |
There is no limit to AM modulation, it is not limited to 100 percent in the positive direction. It is limited to 100 percent in the negative direction, to prevent cutoff. Positive limits are set by the mdoulation linearity of the output stage with increasing positive voltage. Most AM BC broadcasters in this country use assemtrical modulation for this reason. Peter |
It sounds like a technique I remember calling "controlled carrier". The
carrier power was reduced when you weren't talking, then was increased with the audio in an AGC-like manner. Sounded a little weird, but not badly distorted. The objective was to reduce the average dissipation of the final stage, so smaller tubes and a lighter duty power supply could be used. But I don't see why you'd use a method like this with a low power transmitter, since it's trivial to make one that easily handles the power requirements of standard AM. So I don't really think that's what is meant by "swing". I'd bet good money that whatever "swing" is, it doesn't improve quality or signal strength, and very likely introduces distortion that causes splatter. If the transmitter was designed for 100% modulation of a 4 watt carrier, and you reduce the carrier without a proportional reduction of the audio, you'll be overmodulating and consequently distorting and splattering. What are the supposed benefits of this "swing"? Roy Lewallen, W7EL |
"Roy Lewallen" wrote in message ... | It sounds like a technique I remember calling "controlled carrier". The | carrier power was reduced when you weren't talking, then was increased | with the audio in an AGC-like manner. Sounded a little weird, but not | badly distorted. The objective was to reduce the average dissipation of | the final stage, so smaller tubes and a lighter duty power supply could | be used. | | But I don't see why you'd use a method like this with a low power | transmitter, since it's trivial to make one that easily handles the | power requirements of standard AM. So I don't really think that's what | is meant by "swing". I'd bet good money that whatever "swing" is, it | doesn't improve quality or signal strength, and very likely introduces | distortion that causes splatter. If the transmitter was designed for | 100% modulation of a 4 watt carrier, and you reduce the carrier without | a proportional reduction of the audio, you'll be overmodulating and | consequently distorting and splattering. | | What are the supposed benefits of this "swing"? | | Roy Lewallen, W7EL To reduce the drive power to an amplifier and make the modulation louder. Chris |
Chris wrote:
To reduce the drive power to an amplifier and make the modulation louder. Chris I see. But reducing the carrier won't make the modulation louder, only more distorted. Well, let me back up a little. What I said is true if the modulation is 100%. But let's suppose that the transmitter is capable of only 50% modulation. In that case, you *can* make the modulation louder by increasing the amount of audio applied to the carrier. If the transmitter is fundamentally designed to handle 100% modulation, this would require only more audio gain or a "hotter" microphone. That would be the best way to make your modulation louder. But let's say that instead, you reduce the carrier from 4 watts to 1. Then the 1 watt carrier would be 100% modulated. (100% modulation of a 4 watt carrier takes 2 watts. 50% modulation takes only 1/2 watt, which will modulate a 1 watt carrier 100%.) Now you have 100% modulation of the 1 watt carrier. There's the same amount of transmitted audio power as before -- 1/2 watt --, so you're really not making the audio any stronger, and no one will be able to copy you any better than before. (In fact, your weaker signal will have more trouble getting through in the presence of noise or interference.) But if you're the only signal being heard, the receiver's AGC (automatic gain control) will react to your weaker carrier by turning up the receiver's gain, making the audio sound louder. The person receiving your signal can make your audio just as loud with a 4 watt carrier by manually turning up the volume. So I'll relent and say that reducing the carrier might make your audio sound louder -- but only if your transmitter is undermodulated in the first place, there's no stronger signal to control the receiver AGC, and if you don't reduce the carrier so much that it makes the modulation exceed 100%. But your ability to get through interference and noise will probably be reduced. Roy Lewallen, W7EL |
Chris wrote:
I will probably get some flames from this but here it goes. I have been into CB radio for a number of years but don't agree with most of what I hear. That's just one of the reasons I'm looking to amateur radio. One of the things I often hear in CB circles is that one should turn a 4 watt AM radio down to 1 1/2 watts and let it "SWING". How is this possible? What really happens when you do this? I think I know. So, how much carrier should you have for an amp or final stage with a known max output. In other words, if it can produce 8 watts max unmodulated carrier, is a 4 watt carrier ideal? If it produces 100 watts, is 50 watts ideal? How much "space"does it need for proper modulation? Is there a website that explains this well? I'm an electronics tech thirsting for knowledge. In amplitude modulation, the strength - "amplitude" - of the carrier is adjusted by the audio you wish to transmit. The degree to which this strength is adjusted is the "modulation percentage". If the carrier is cut completely at negative voice peaks, and strengthened to twice its normal level at positive peaks, then the signal is said to be "100% modulated". The receiver at the other end can only detect the *changes* in carrier strength - not the carrier itself. If you reduce the modulation percentage, you reduce the strength of the changes - the strength of the signal the other guy can hear. On the other hand, the laws of physics prohibit negative power. Once you've modulated 100% - and reduced the carrier to zero at negative peaks - you CAN'T go any further. It's physically impossible. If you try, you'll generate sharp cutoffs that result in "splatter" - strong interfering noises in adjacent channels. (and your signal on the channel you're meaning to transmit on will become seriously distorted and difficult to understand) [0] So the point is, you want to modulate as close to 100% as practical while ensuring you never *exceed* 100%. For normal "high-level" modulation the amount of audio power required to achieve 100% modulation is half the RF power. A 4-watt carrier requires two watts of audio to modulate it 100%. -- Doug Smith W9WI Pleasant View (Nashville), TN EM66 http://www.w9wi.com [0] It is possible to exceed 100% in the *positive* direction - increasing the carrier beyond twice its normal level - without causing distortion and interference. Such schemes are common at AM broadcast stations. I'm not aware of any CB radio that contains such a circuit. |
Roy Lewallen wrote:
It sounds like a technique I remember calling "controlled carrier". The carrier power was reduced when you weren't talking, then was increased with the audio in an AGC-like manner. Sounded a little weird, but not badly distorted. The objective was to reduce the average dissipation of the final stage, so smaller tubes and a lighter duty power supply could be used. No, "controlled carrier" was something else. In 100% positive modulation, the carrier amplitude is constant regardless of program material. You (somehow!) maintain proper absolute phase through the chain so that you know a positive-going audio signal at the transmitter audio input terminals will result in increasing power in the modulated carrier. You then allow the positive-going signal peaks to exceed 100% while limiting negative-going peaks to less than 100%. Broadcast modulation monitors are able to display negative-going and positive-going modulation peaks independently. (at a broadcast station, the FCC requires that carrier power be between -- IIRC, my copy of Part 73 is missing -- 80 and 110% of the authorized figure. DX-60B-style controlled carrier wouldn't comply, though I suppose you could develop a system that didn't swing the carrier quite as far.) -- Doug Smith W9WI Pleasant View (Nashville), TN EM66 http://www.w9wi.com |
Doug Smith W9WI wrote:
Roy Lewallen wrote: It sounds like a technique I remember calling "controlled carrier". The carrier power was reduced when you weren't talking, then was increased with the audio in an AGC-like manner. Sounded a little weird, but not badly distorted. The objective was to reduce the average dissipation of the final stage, so smaller tubes and a lighter duty power supply could be used. No, "controlled carrier" was something else. A quick web search shows that what I described is properly called "dynamic carrier control". My mistake. I only recall having seen one such amateur transmitter, and it was over 40 years ago. . . Roy Lewallen, W7EL |
Roy Lewallen wrote:
A quick web search shows that what I described is properly called "dynamic carrier control". My mistake. I only recall having seen one such amateur transmitter, and it was over 40 years ago. . . I owned one, a Heath DX-60B. Got my General in 1974, DSB-carrier AM phone was already essentially obsolete. When I got a 33 signal report from a 40-meter station four miles away, I decided to stick to CWgrin... Seems to me there was a circuit for homebrewing dynamic carrier control in the ARRL Handbook for awhile. -- Doug Smith W9WI Pleasant View (Nashville), TN EM66 http://www.w9wi.com |
So let's use a typical amp rated at 100 watts AM/CW/FM and 200 watts PEP on
SSB. Assuming that the transmitter is modulated at 100%, how many watts should the carrier be? Chris "Roy Lewallen" wrote in message ... | Chris wrote: | | | To reduce the drive power to an amplifier and make the modulation louder. | | Chris | | I see. But reducing the carrier won't make the modulation louder, only | more distorted. | | Well, let me back up a little. What I said is true if the modulation is | 100%. | | But let's suppose that the transmitter is capable of only 50% | modulation. In that case, you *can* make the modulation louder by | increasing the amount of audio applied to the carrier. If the | transmitter is fundamentally designed to handle 100% modulation, this | would require only more audio gain or a "hotter" microphone. That would | be the best way to make your modulation louder. | | But let's say that instead, you reduce the carrier from 4 watts to 1. | Then the 1 watt carrier would be 100% modulated. (100% modulation of a 4 | watt carrier takes 2 watts. 50% modulation takes only 1/2 watt, which | will modulate a 1 watt carrier 100%.) Now you have 100% modulation of | the 1 watt carrier. There's the same amount of transmitted audio power | as before -- 1/2 watt --, so you're really not making the audio any | stronger, and no one will be able to copy you any better than before. | (In fact, your weaker signal will have more trouble getting through in | the presence of noise or interference.) But if you're the only signal | being heard, the receiver's AGC (automatic gain control) will react to | your weaker carrier by turning up the receiver's gain, making the audio | sound louder. The person receiving your signal can make your audio just | as loud with a 4 watt carrier by manually turning up the volume. | | So I'll relent and say that reducing the carrier might make your audio | sound louder -- but only if your transmitter is undermodulated in the | first place, there's no stronger signal to control the receiver AGC, and | if you don't reduce the carrier so much that it makes the modulation | exceed 100%. But your ability to get through interference and noise will | probably be reduced. | | Roy Lewallen, W7EL |
Oh, how do you convert from WPEP to WRMS? Does it have to be converted to
voltage and multiplied by .707? Chris "Chris" wrote in message k.net... | So let's use a typical amp rated at 100 watts AM/CW/FM and 200 watts PEP on | SSB. Assuming that the transmitter is modulated at 100%, how many watts | should the carrier be? | | Chris | "Roy Lewallen" wrote in message | ... | | Chris wrote: | | | | | | To reduce the drive power to an amplifier and make the modulation | louder. | | | | Chris | | | | I see. But reducing the carrier won't make the modulation louder, only | | more distorted. | | | | Well, let me back up a little. What I said is true if the modulation is | | 100%. | | | | But let's suppose that the transmitter is capable of only 50% | | modulation. In that case, you *can* make the modulation louder by | | increasing the amount of audio applied to the carrier. If the | | transmitter is fundamentally designed to handle 100% modulation, this | | would require only more audio gain or a "hotter" microphone. That would | | be the best way to make your modulation louder. | | | | But let's say that instead, you reduce the carrier from 4 watts to 1. | | Then the 1 watt carrier would be 100% modulated. (100% modulation of a 4 | | watt carrier takes 2 watts. 50% modulation takes only 1/2 watt, which | | will modulate a 1 watt carrier 100%.) Now you have 100% modulation of | | the 1 watt carrier. There's the same amount of transmitted audio power | | as before -- 1/2 watt --, so you're really not making the audio any | | stronger, and no one will be able to copy you any better than before. | | (In fact, your weaker signal will have more trouble getting through in | | the presence of noise or interference.) But if you're the only signal | | being heard, the receiver's AGC (automatic gain control) will react to | | your weaker carrier by turning up the receiver's gain, making the audio | | sound louder. The person receiving your signal can make your audio just | | as loud with a 4 watt carrier by manually turning up the volume. | | | | So I'll relent and say that reducing the carrier might make your audio | | sound louder -- but only if your transmitter is undermodulated in the | | first place, there's no stronger signal to control the receiver AGC, and | | if you don't reduce the carrier so much that it makes the modulation | | exceed 100%. But your ability to get through interference and noise will | | probably be reduced. | | | | Roy Lewallen, W7EL | | |
Chris wrote:
So let's use a typical amp rated at 100 watts AM/CW/FM and 200 watts PEP on SSB. Assuming that the transmitter is modulated at 100%, how many watts should the carrier be? Chris Zero. As universally used, SSB means "single sideband suppressed carrier". There is no transmitted carrier in this mode. And, modulation percentage has no meaning when talking about SSB, since it refers to the relationship between the modulation and the carrier. Roy Lewallen, W7EL |
Chris wrote:
Oh, how do you convert from WPEP to WRMS? Does it have to be converted to voltage and multiplied by .707? Chris While the conversion from PEP to RMS is simple for a sine wave or other simple waveform, it's not simple when dealing with a real voice waveform. It depends heavily on the characteristics of the voice, and any audio processing (such as compression or RF clipping) that might be taking place. Typically, the PEP value of an unprocessed voice waveform is many times the RMS value. Roy Lewallen, W7EL |
There is no limit to AM modulation, it is not limited to 100 percent in the positive direction. It is limited to 100 percent in the negative direction, to prevent cutoff. Positive limits are set by the mdoulation linearity of the output stage with increasing positive voltage. Most AM BC broadcasters in this country use assemtrical modulation for this reason. You're dead right of course Peter. i believe you're AM broadcasters are limited to 125% positive while here in the UK we have 100% positive peak limit. FYI the late model Nautel PDM broadcast transmitters are capable of around 200% positive peak and up to 50KHz in frequency response! Broadcast transmitters are limited for obvious reasons. -- Philip de Cadenet G4ZOW Transmitters 'R' Us http://www.transmittersrus.com |
"Roy Lewallen" wrote in message ... Doug Smith W9WI wrote: Roy Lewallen wrote: It sounds like a technique I remember calling "controlled carrier". The carrier power was reduced when you weren't talking, then was increased with the audio in an AGC-like manner. Sounded a little weird, but not badly distorted. The objective was to reduce the average dissipation of the final stage, so smaller tubes and a lighter duty power supply could be used. No, "controlled carrier" was something else. A quick web search shows that what I described is properly called "dynamic carrier control". My mistake. I only recall having seen one such amateur transmitter, and it was over 40 years ago. . . Roy Lewallen, W7EL The Heath DX-60 used "controlled carrier" modulation. It was a form of screen modulation. Pete |
"Chris" wrote in message k.net... I think I've got it figured out. For AM, the carrier should be 1/8 of the PEP maximum or 1/4 of the max carrier. So an amp rated 100W AM/FM/CW, 200 WPEP SSB should run with a 25 W carrier on AM. Does that sound right to everyone? The peak power on AM is 4 times the carrier power. But, don't forget, AM has TWO sidebands, not one. Pete |
"Roy Lewallen" wrote in message ... It sounds like a technique I remember calling "controlled carrier". The carrier power was reduced when you weren't talking, then was increased with the audio in an AGC-like manner. Sounded a little weird, but not badly distorted. The objective was to reduce the average dissipation of the final stage, so smaller tubes and a lighter duty power supply could be used. But I don't see why you'd use a method like this with a low power transmitter, since it's trivial to make one that easily handles the power requirements of standard AM. So I don't really think that's what is meant by "swing". I'd bet good money that whatever "swing" is, it doesn't improve quality or signal strength, and very likely introduces distortion that causes splatter. If the transmitter was designed for 100% modulation of a 4 watt carrier, and you reduce the carrier without a proportional reduction of the audio, you'll be overmodulating and consequently distorting and splattering. What are the supposed benefits of this "swing"? Roy Lewallen, W7EL CBers are very susceptible to urban myth and legend. Not understanding the "normal" nature of an AM carrier, often they won't see the needle move very much (assuming some cowboy jockey hasn't been in the rig snippin' and clippin'), they get kind of excited. "HEY! I ain't got no "swang", so therefore, they think they aren't "gittin'" out. Voodoo techs have been able to take advantage of this by monkeying around with the sets and/or using funky, cheap meters to show the unknowing how much their radio is "swangin'". "LOOK! Ya got 8 watts o' carrier and 40 watts of 'swang'"! The CB guys eat it up! AND willingly part with $$$ to get this "Swang". Swing is fully embedded in CB psyche and, like the "coax length" (18 FEET! 18 FEET! Ya gots to have 18 feet of coax!!!!!!) bullsh--, it is part of the "holy" grail of CB radio! LMAO! J |
Such as believing that an M3/CB Fools' Licence makes
then into a Radio Ham overnight. "Jerry" wrote in message . .. CBers are very susceptible to urban myth and legend. |
Jerry, you can trust me when I say that hams have their full share of
beliefs in voodoo physics and misunderstandings about how even the simplest phenomena take place. A quick scan of the rec.radio.amateur.antenna archives provides ample evidence in itself, but there's plenty of other evidence scattered about. So let's not be too hasty at calling the kettle black. Roy Lewallen, W7EL Jerry wrote: CBers are very susceptible to urban myth and legend. Not understanding the "normal" nature of an AM carrier, often they won't see the needle move very much (assuming some cowboy jockey hasn't been in the rig snippin' and clippin'), they get kind of excited. "HEY! I ain't got no "swang", so therefore, they think they aren't "gittin'" out. Voodoo techs have been able to take advantage of this by monkeying around with the sets and/or using funky, cheap meters to show the unknowing how much their radio is "swangin'". "LOOK! Ya got 8 watts o' carrier and 40 watts of 'swang'"! The CB guys eat it up! AND willingly part with $$$ to get this "Swang". Swing is fully embedded in CB psyche and, like the "coax length" (18 FEET! 18 FEET! Ya gots to have 18 feet of coax!!!!!!) bullsh--, it is part of the "holy" grail of CB radio! LMAO! J |
In article ,
Gary Schafer writes: Well Chris, you got a lot of good advice and some completely wrong advice. Funny thing is nobody really answered your question! [lots of useful information snipped...] This is how you convert average power to PEP. I say average power because RMS power is really a misnomer. There is no such thing as RMS power as so many commonly refer to. It is really AVERAGE power. It is derived from RMS current and RMS voltage but what you get when you multiply RMS voltage by RMS current or resistance is an AVERAGE value, not an RMS value. I completely agree about RMS 'power'. RMS applies to a voltage or current waveform. Caveat: you cannot simply multiply RMS voltage by RMS current unless you have a purely resistive load. Oddly enough, if the load is a fixed resistance (measured as a conductance G) in parallel with a reactance then the E_rms^2 * G (E_rms^2/R when purely resistive) value continues to apply. When it is a fixed resistance in series with a reactance, I_rms^2 * R continues to apply. And these extend to the nonsinusoidal case as well, again under the restrictions given above (fixed G in the first case, fixed R in the second). This may seem trivial but it becomes important when trying to convert from one form to another. Using the wrong notation can give you wrong answers. As to the above, the average voltage or current of a sine wave is 50% of the peak. Actually, it is .6366 times the peak. The averaging is done continuously over the positive half cycle of the sine wave, and is essentially the area under the sine curve divided by the length of the half-cycle (2/3.14159, or, if you calculated in degrees, 114.592/180). The true average value of a sine curve is to average over a full cycle, but is not very interesting: it's zero. The .6366 number is useful because it is what a VOM typically measures - average DC voltage after passing through a full-wave rectifier and before applying a scale factor of .7071/.6366 to produce E_rms - all based on the assumption of a sinewave input. Average value of voltage or current waveform is of limited use; but RMS value is directly related to the ability to deliver power to a resistive load. The rms value of a sine wave is .707 of the peak voltage or current. If you multiply .707 (rms voltage) by .707 (rms current) you get .5 or 50%. This is AVERAGE. It is no longer an rms value. By the way, 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) The average power is in fact all that is of interest in AC circuits, and is the average over time of "instantaneous power" E*I. If you do not use average power you have to deal with the fact that reactances accept power during half of the cycle and then feed it back during the other half; and for nonsinusoidal waveforms do the same but in not so regular a pattern. PEP is just averaging over the shortest reasonable interval (one cycle of the carrier frequency), then keeping only the largest such value seen. Useful to regulators because it is a measure of your maximum capacity to interfere with other signals. [...] 73 Gary K4FMX |
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I'm not sure who the "we" is in your statement, but it certainly doesn't
include people who understand the concepts and mathematics involved. The problem with your approach is that there *is* such as thing as RMS power (as I described in an earlier posting), and it is *not* the amount that gives the same heating effect as the same amount of DC power. So while you may think of RMS power in this way, your thinking is wrong. I can calculate my RMS speed in going from point A to point B. But it isn't equal to the distance I went divided by the time it took to get there (unless, of course, I went the whole way at a constant speed) -- for exactly the same reason that the RMS power isn't the equivalent heating power. You can think of the RMS speed as being the distance divided by the time, but it isn't, and your thinking doesn't make it so. And that also doesn't make the RMS speed a useful value. Roy Lewallen, W7EL Paul Burridge wrote: I guess you're right, but when we speak of RMS power (sic) what we actually envisage is the amount of AC power dissipated in a resistive load that gives rise to the same heating effect as the equivalent amount of DC power. Sorry if that's not well put, but you no doubt get my drift... So it may be a misnomer, but that doesn't make it useless. |
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 |
Bill Turner wrote:
On Mon, 18 Oct 2004 11:15:43 -0700, Roy Lewallen wrote: The problem with your approach is that there *is* such as thing as RMS power (as I described in an earlier posting), and it is *not* the amount that gives the same heating effect as the same amount of DC power. So while you may think of RMS power in this way, your thinking is wrong. __________________________________________________ _______ Let me ask one more question and then I'll shut up. Actually, three. Suppose you have a 100 ohm resistor and you apply a 100 VRMS sine wave to it. Will 1 amp of RMS current flow? I would think yes, but after the previous discussions, I'm not taking anything for granted. :-) Yes. So if you have 100 VRMS and 1 ARMS, how much power is the resistor dissipating? Just pure power in terms of heat generation, forget whether it's RMS or anything else. How many watts? 100 watts. This is the average power. (The RMS power is about 122.5 watts, if I did the integration right -- corrections are welcome. RMS power is calculated from the instantaneous power exactly like RMS voltage or current is calculated from instantaneous voltage or current.) 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. Breathlessly awaiting, I would hope that any Technician or higher class amateur could immediately answer those questions, no waiting required. The problem people seem to be having is the feeling (or, I'm afraid in some cases, certainty) that an RMS value of a quantity times an RMS value of another quantity HAS TO BE the RMS value of the product of the quantities. But it isn't. Here's something you can check on your calculator: Take sin(5 degrees) times sin(10 degrees). Is the product of the two equal to sin(50 degrees)? While the product assumption works for some functions, such as square and square root for example, it doesn't work for others, like RMS and sine. Unfortunately, if you (any reader, not Bill personally) aren't able to follow the mathematics involved in calculating RMS and average values (which I posted in brief form a short while ago), this is bound to remain something of a mystery to you, and it looks like you're left with four choices: 1. Hang tightly to a mistaken idea about the meaning of RMS and average, and ignore or resist any explanation that contradicts them. This will include explanations in any textbook. 2. Learn enough math to understand the definitions of average and RMS, which should make the explanations understandable. 3. Assume that I and the textbook authors know what we're talking about and believe us, without really understanding why. 4. Find a textbook oriented toward people without a very strong math background, which hopefully can explain it in terms you can understand. The fourth choice is probably the best for people who are really interested in learning and can't devote the time necessary to learn the math. The main problem with the first choice is that you'll be likely to pass the misconceptions on to others. But the choice is yours. Roy Lewallen, W7EL |
Let me add one more comment that will hopefully help in understanding this.
I think I might have identified another misconception that might be contributing to the confusion. That misconception is that the RMS value of a waveform is the "equivalent" or "heating" value. This leads to the mistaken notion that the RMS value of the power must be the "equivalent" or "heating" value. Voltage and current, by themselves, don't do any heating. Heating takes power. And the amount of heating is the determined by the average power. If two different power waveforms have the same average value, they'll create the same amount of heat and otherwise do the same amount of work, even if their RMS values are different. So what's the whole thing about RMS? The only importance of RMS is that when you multiply the RMS values of two waveforms together (such as V * V, I * I, or V * I), you get the average of the product of the two. That is, RMS(i) * RMS(v) = Avg(i * v), where i and v are the instantaneous values of the current and voltage, and i * v is the instantaneous power. Likewise, RMS(v) * RMS(v) / R = Avg(v*v/R) and RMS(i) * RMS(i) * R = Avg(i*i*R). In fact, RMS(x) * RMS(y) = Avg(x * y) where x and y can be any periodic waveforms or quantities. (I derived these mathematically in an earlier posting.) So we calculate the RMS values of voltage and current only so we can use them to calculate the average power -- not because the RMS value of any waveform is its "equivalent" or "heating" value -- which it isn't. Roy Lewallen, W7EL |
Bill Turner wrote:
On Tue, 19 Oct 2004 01:39:57 -0700, Roy Lewallen wrote: Voltage and current, by themselves, don't do any heating. __________________________________________________ _______ I don't follow you on this. Voltage can be present without any current, but current can only flow if voltage is present, so what is meant by "current by itself"? -- Bill W6WRT An electron beam in a vacuum can flow current without any voltage drop, as can a superconductor. Granted, these are not everyday occurrences, but they _do_ occur. The real point that Roy is talking about is that whenever a voltage drop occurs simultaneously with a current there is power being consumed -- and if it isn't coming out of the system as some other form of power then it's being dissipated as heat. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com |
On Tue, 19 Oct 2004 06:32:55 -0700, Bill Turner
wrote: Well. The answer you gave is exactly the answer I would have given, but you say my answer is wrong. 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 agree. On the one had we've had a pillar we'd always accepted knocked away by Roy; on the other hand, it was Roy who knocked it away. Had it been anyone else I'd have dismissed them as a nutter. 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. As I'm sure we all do. If this is a misconception it must be an extremely widespread one. I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) and have posted the relevant page he http://www.burridge8333.fsbusiness.co.uk/cvcvcv.gif If we are wrong, it appears we're not alone... -- "What is now proved was once only imagin'd." - William Blake, 1793. |
Sure, current can be present without voltage (in a short circuit) just
as easily as voltage can be present without current (in an open circuit). But what I meant was that the value of voltage doesn't determine heating, nor does the value of current. It's the value of power that does. Roy Lewallen, W7EL Bill Turner wrote: On Tue, 19 Oct 2004 01:39:57 -0700, Roy Lewallen wrote: Voltage and current, by themselves, don't do any heating. __________________________________________________ _______ I don't follow you on this. Voltage can be present without any current, but current can only flow if voltage is present, so what is meant by "current by itself"? -- Bill W6WRT |
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 |
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. Roy Lewallen, W7EL Paul Burridge wrote: On Tue, 19 Oct 2004 06:32:55 -0700, Bill Turner wrote: Well. The answer you gave is exactly the answer I would have given, but you say my answer is wrong. 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 agree. On the one had we've had a pillar we'd always accepted knocked away by Roy; on the other hand, it was Roy who knocked it away. Had it been anyone else I'd have dismissed them as a nutter. 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. As I'm sure we all do. If this is a misconception it must be an extremely widespread one. I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) and have posted the relevant page he http://www.burridge8333.fsbusiness.co.uk/cvcvcv.gif If we are wrong, it appears we're not alone... |
My statements about voltage and current referred to ideal circuits,
which are immensely useful in gaining understanding of fundamental circuit operation. One can argue for a long time about how perfect a real circuit can be made, and I won't get into that. I hope I made my point without the need to. Roy Lewallen, W7EL Bill Turner wrote: On Tue, 19 Oct 2004 06:49:01 -0700, Tim Wescott wrote: An electron beam in a vacuum can flow current without any voltage drop, as can a superconductor. __________________________________________________ _______ I can't argue the superconductor, but in a vacuum tube there must be a source of voltage to cause the current to flow. Even if you are talking about the "contact potential" caused by a heated cathode, there is still voltage present, isn't there? And of course, the anode voltage causes current to flow there too. -- Bill W6WRT |
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. -- "What is now proved was once only imagin'd." - William Blake, 1793. |
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. |
There is a nice simple explanation in the ARRL handbook that I posted
earlier in the thread on "power output formula". Guess nobody read it. 2000 ARRL handbook 6.6 chapter 6, RMS VOLTAGES AND CURRENTS. 73 Gary K4FMX On Tue, 19 Oct 2004 16:33:53 -0700, Roy Lewallen wrote: 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. |
Paul Burridge wrote:
If this is a misconception it must be an extremely widespread one. I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) and have posted the relevant page he http://www.burridge8333.fsbusiness.co.uk/cvcvcv.gif If we are wrong, it appears we're not alone... Any fool can write a book that includes the words "RMS power". I've done it myself. I have known since age 15 how to calculate the average power or an arbitrary AC waveform: "Take the root-mean-square averages of the voltage and current first, and then multiply those results together. Doing it the other way around gives the wrong answer, stupid boy!" My particular mistake was to try to use "RMS power" as a shorthand label. I naively assumed that everybody would understand that it was short for: "This is power that has been calculated by a *correct* use of RMS averaging, which we all know means taking the RMS average of the voltage or current first, and obviously it does not mean taking the RMS average of the power, because that would *not* be correct, and now that we've wasted half a paragraph on this point, and completely derailed the original train of thought, which was mostly about something else, I'd rather like to get back to our main topic." Just writing "RMS power" seemed so much more appealing... BIG mistake! It fell right into the gap between people who don't know how to calculate power correctly; and people who do know, but were all too eager to assume they'd found a fault. I learned from that mistake... but the mistake itself is immortalised 6000 times over, in cold type. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) ============================= Yet another Beta-tested bible is discredited! ;o) ;o) |
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. |
Reg Edwards wrote:
I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) ============================= Yet another Beta-tested bible is discredited! ;o) ;o) I wouldn't trust anything from that publisher. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
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. |
On Wed, 20 Oct 2004 13:17:28 +0100, "Ian White, G3SEK"
wrote: Reg Edwards wrote: I've found a reference to RMS power and how it's calculated in 'Practical Radio Frequency Test & Measurement' (Newnes) ============================= Yet another Beta-tested bible is discredited! ;o) ;o) I wouldn't trust anything from that publisher. Who was *your* publisher, Ian? -- "What is now proved was once only imagin'd." - William Blake, 1793. |
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