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On Tue, 24 Feb 2004 18:28:57 -0600, "Steve Nosko"
wrote: I assumed Richard's intent here is that you only have to do the calculation for one full period of the periodic component to derive the RMS value - and if it is symmetrical, then only one half period will suffice. This all assumes a symmetrical AC shape. .. I see no reason why this would not be true. And then you offer in contradiction: However a DC biased periodic shape requires another squaring and root operation if you capture all the components. It gets a bit more harry Such is reality. There is no such thing as symmetry, except on the academic page. However, I am not so pedantic as to suggest that shortcuts don't abound; simply pedantic enough to point out you don't make claims to accuracy (admittedly none were offered that I was responding to) through fudge factors when so many alternatives remove doubt. The simple determination of RMS is the graphical integration of the area under the curve. There are as many "correction factors" for RMS as there are shapes, and they all derive from this simple concept. Here I'll take issue with the ONE WORD "graphical". You can integrate if you can describe the function of the wave shape mathematically. Of course, but it is eminently doubtful if you can actually express it mathematically. Far more here own o'scopes than works of multiple regression. Graphical analysis is first year engineering stuff out of drafting class. When the computational horsepower requirement becomes enormous (there are many here that give up too easily with complexity); it is the provence of the "Old School" to suggest that since RMS is all based on the notion of power, you simply measure the caloric result and ignore shape altogether. This may be done with thermo-electric piles or other measurable property transformers that perform the complexity of integration through physics*. I can anticipate those who dearly embrace the complexity that they shudder to face (such contradictions of their love-hate relationships) when I hear Crest, or pulse/power factor (or duty cycle) uttered. Clearly the problem will have migrated from Power to some other consideration, but is dressed as an RMS debate. 73's Richard Clark, KB7QHC Yikes! Not sure where you went on that last bit, Richard C... Now, I ask. Do the power meters on the outside of our houses take all those factors into consideration and REALLY show TRUE watt hours? I have one in the basement and I think I figured out why I was seeing twice the reading I should have (letting a light bulb sit on for awhile) ... I counted the teeth to get the ratio of the gear train, just to find that it is printed (somewhat cryptically) on the face) (I made a two wire / three wire connection error) Hi Steve, Yikes? Look at your own response to shudder. ;-) You offer the doubt and then correct your error in the space of three sentences. From that I must suppose it was a rhetorical question, but the "yikes" heading it promises more clarity in response to what you apparently complain of. After all, a complex, compound sentence with ellipses and nested parenthetical statements? You lose points for unpaired braces and lack of punctuation throughout and at the end. Do the power meters on the outside of our houses take all those factors into consideration and REALLY show TRUE watt hours? For at least a Century now. The power companies offer as close to pure symmetry as you could buy. They also offer long term time accuracy to far better than any source short of WWVL. 73's Richard Clark, KB7QHC |
Steve Nosko wrote:
"I wonder if a "true RMS" DVM can handle this?" So do I. The true rms meter seems a wonderful development to me. Steve also wrote: "Also ok, but not sure how it plays into the RMS discussion." My speculation is that the effective value of a nonsinusoidal waveform could be found by summation of its sinusoidal constituents. But, it`s not difficult to find an effective value for not only sinusoidal periodic waveforms but for nonsinusoidal periodic waveforms as well. One can graphically take a large number of equally spaced ordinates of the form, using at least one complete alternation, Richard Clark. Both alternations are not needed but could be used as a minus times a minus is a plus and each of the ordinate values must be squared because power is a function of the current squared, so both alternations when their ordinates are squared produce positive values. Next, we sum the squared ordinate values and divide by the number of ordinates. You get the average value of the squared curve which is what we are looking for. Or, you can construct the squared curve and integrate the area under the squared curve by using a planimeter. Dividing the area by the baseline length gives the same average value of the squared curve as iusing the graphic ordinates above. Voila: rms. Unless I`ve opened a new can of worms with this posting, I don`t know of any difference of opinion I have with Steve. Best regards, Richard Harrison, KB5WZI |
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"Richard Harrison" wrote in message
... [...] Steve also wrote: "Also ok, but not sure how it plays into the RMS discussion." My speculation is that the effective value of a nonsinusoidal waveform could be found by summation of its sinusoidal constituents. That's what I thought you were going for. My gut feel is that this must be valid, otherwise "Fourier ain't an exact solution after all". But it sounds like a lot of work and you have to be able to extract all the significant Fourier components (harmonics) and there is an element of approximation here, just like my "mathematicall function" method, no? ....Of course doing the integration on the waveform takes some time to crank through the math and get all the quantities collected correctly. But, it`s not difficult to find an effective value for ... nonsinusoidal periodic waveforms as well. One can graphically take For what I wanted to do, the integration was easier. I may be wrong, but Fourier also needs a function and you still need to integrate, no? Now, to answer Richard Clark's comments about accuracy in another post... I did assume that the wave shapes I was interested in were defined (I think the word is) explicitly. I assumed a mathematical function. When I compared the three waveforms which were appropriate (sine, triangle, trapezoid) the results were very much the same (for ratio of RMS to Average). (gotta remember to keep my sentences shorter). From this I came to the conclusion that for wave shapes which differed slightly from the ideal (assumed shapes) the values would be well within acceptable bounds. Easily 5%. Yes, not exact, but much better than Bob Shrader had assumed. large number of equally spaced ordinates of the form, using at least one complete alternation, Richard Clark. I don't think they have to be equally spaced since the actual time enters into the calculation (as you described later). This would make long sections with a constant value easier. Both alternations are not needed but could be used as a minus times a minus is a plus and each of the This assumes that the waveform is symmetrical around zero. I think in general, one whole period is required. or[...] using a planimeter. Richard H. clearly prefers graphical solutions and that's ok. Unless I`ve opened a new can of worms with this posting, I don`t know of any difference of opinion I have with Steve. Best regards, Richard Harrison, KB5WZI Right now it appears that this half vs. full period is the only difference I can see. -- Steve N, K,9;d, c. i My email has no u's. |
OOPS!
"Richard Clark" wrote in message ... On Tue, 24 Feb 2004 18:28:57 -0600, "Steve Nosko" wrote: I assumed Richard's intent here is that you only have to do the calculation for one full period ...symmetrical, then only one half period... And then you offer in contradiction: However a DC biased periodic shape requires another squaring and root operation if you capture all the components. It gets a bit more harry This was a stupid thing to add at this point since it addresses something which I did not explain, so I'm sure it looks weird. I didn't even say what was in my mind. Namely, that with the DC + AC you still need to do one full period of the periodic part. What I was poorly referring to was this: If you can break down the signal into component parts such as: DC Periodic part #1 Periodic part #2 etc Then there is a formula for the total RMS which is the square root of the sum of the squares. It is in one of the papers I linked previously. That is one way to do an AC+DC situation. Second formula in the Intl Rect paper: http://www.irf.com/technical-info/an949/append.htm [...] There is no such thing as symmetry, except on the academic page. [...] you don't make claims to accuracy (admittedly none were offered that I was responding to) through fudge factors when so many alternatives remove doubt. I think is it safe to say that we each determine our own tolerance for error. Five percent for power is ok for my purposes. SO approximating waveform functins is just fine. The simple determination of RMS is the graphical integration of the area under the curve. There are as many "correction factors" for RMS as there are shapes, and they all derive from this simple concept. Here I'll take issue with the ONE WORD "graphical". You can integrate if you can describe the function of the wave shape mathematically. Here, I'll take issue. I believe that the basis for the many "correction factors" for RMS of various shapes is indeed the mathematical integration of a function rather than the "Simple concept" of a graphical solution (if that is what you meant). There is a mathematical integral of a sine wave. I think you'd be hard pressed to prove that the average of a sine wave is 2/pi exactly using graph paper. You could certainly say it sure looks close to 2/pi, but is it exactly pi?... can't say for sure. Case in point is the phase controlled sine wave made by SCR light dimmers. I find it hard to believe you can graphicaly come up with the formula: Sqr-root[ D/2 + sin[pi(1-D)] cos[pi(1-D)]/2pi I must also add that the graphical solution and mathematical integration are different implementations of the same concept. I don't intend to say that one is wrong and one is right. Now, for Richard, C. Is the thing coming out of the AC outlet an exact sine wave...no. It is very noticably flattened on the top by all the power supplies drawing peak currents near the peak. There must be other corruptins as well. However, if I assume it is a sine wave, will my calculations come out very close, I believe Yes. Now in all fairness, I won't dispute that you can use the graphical method to find the RMS to any desired accuracy, just not exact...without being able to integrate the waveform. Of course, but it is eminently doubtful if you can actually express it mathematically. Far more here own o'scopes than works of multiple regression. Graphical analysis is first year engineering stuff out of drafting class. Integration is first year engineering stuff out of calc class. At least it was for me. I am still amazed that I remember doing tripple integrals then, and thinking...Gee, this ani't so hard after all!. Couldn't do one to save my life now. [...] you simply measure the caloric result and ignore shape altogether. I always thought that the common method of measuring RF power was pretty cool! The Thermistor or bolometer. Here you balance a bridge with DC or low freq AC. It heats the thermistor to the correct resistance. Then, when you add RF power, the thing heats up more and changes resistance. So, you remove some DC power to get back to the correct resistance and that amount is easy to figure. That is how much RF you put in. Cool. I think it is correct to say that you absolutely cannot measure power *directly*. You must measure something else which is affected/caused by the power...comment? [..] migrated from Power to some other consideration Not trying to migrate. I just think its a cool method. Yikes! Not sure where you went on that last bit, Richard C... Now, I ask. Do the power meters on the outside of our houses take all those factors into consideration And, like me, Richard Clark makes a really long answer that says...Yes. Got me! I have NO idea how the watt-hour meter works...and don't want to try to understand at this point. 73, -- Steve N, K,9;d, c. i My email has no u's. |
"Cecil Moore" wrote in message ... Richard Harrison wrote: If we have a Class C amplifier feeding power to the same antenna and enjoying a conjugate match, we can have a source that takes less than 50% of the available energy. So, the transmitting antenna system can be more efficient than the receiving antenna system, it seems to me. Didn't see the original post... I think the phrase "available energy" may draw discussion, but won't start it. I suspect it is not the thing to focus on. However... With a "conjugate match" the source dissipates 50% of the power and the load the other 50%. This can't be changed. Draw the schematic to see. Rs = RL therefore Ps = PL. (I'll use a big "L" so it looks like an "L") ------------------------ Then I go to something I have for a long time wondered about; the following situation: A (tube) amplifier is in conjugate match conditions. It is dissipating 10 watts in its plate. This is the limit of its plate dissipation. Model this as a voltage source (Vs) with source resistance (Rs) and a load with load resistance (RL). The conjugate match has removed all the reactance. There is also 10 Watts to the load. Now, assuming you can, increase the (plate) supply voltage by, say 20%. This raises the source voltage (Vs) [may be the fatal flaw]. If the plate resistance (Rs) stays the same (I don't know if it will - more flaw-fodder), then the plate dissipation will also increase. SO... how about adjusting the match so the plate sees a higher RL to get back to 10 watts plate dissipation. RL must now increase 40% to do this. The source current must stay the same if the source resistance stays the same and keep the same source dissipation. Now, what we have done is to increase RL and Load voltage (VL), and therefore load power, but kept 10 watts at the plate. If I did the math correctly the load now dissipates 40% more power. If, however, we re-adjust the match back to conjugate we WILL get more power and 50% will be dissipated in the stage...until it blows. SO... I think this tells us that if the stage is dissipation limited, but not breakdown limited, we can non-conjugate match for a higher power than 'before' by raising the supply voltage. It sorta looks like we are getting more with a non-conjugate match. You don't have the 50-50 division of power also. It is now 41.7% tube, 58.3% load Snake oil? The math is correct, but there may be some practical limit not considered. -- Steve N, K,9;d, c. i My email has no u's. Then there's the solid state power amplifier standard output resistance formula. Rs = Vcc^2 / (2*Po) The implication should be obvious... |
Dan Richardson wrote:
Let me try this one more time. You had posted earlier and I commented on this: "A receiving antenna must be resonant to enable full acceptance of available energy, and it must be matched to avoid re-radiation of more than 50% of the energy it is able to grab." I commented on the first portion of your statement (above). My only point is that it makes no difference if an antenna's is resonate or not in determining how much energy it grabs. That's it, nothing more. 73 Danny, K6MHE SNIP The effective aperture is the effective aperture is the effective aperture ... The antenna intercepts ALL the EM energy within it's effective aperture. You, the listener, may be interested in only 2.8 KHz of that energy. :-) |
Great, now to the issue.
\ How much power does the antenna pick up compared to the power enclosed within the 2.4 Khz frequency spread that is considered useful ? Or can we say how much energy is dumped to ground by a bandpass filter (plus insertion losses). I know it may be absolutely quiet outside the bandwidth of choice but I assume you get my drift with respect to efficiency of a simple dipole to a radiator of high Q. Regards Art "Dave Shrader" wrote in message news:he8%b.120870$jk2.515312@attbi_s53... Dan Richardson wrote: Let me try this one more time. You had posted earlier and I commented on this: "A receiving antenna must be resonant to enable full acceptance of available energy, and it must be matched to avoid re-radiation of more than 50% of the energy it is able to grab." I commented on the first portion of your statement (above). My only point is that it makes no difference if an antenna's is resonate or not in determining how much energy it grabs. That's it, nothing more. 73 Danny, K6MHE SNIP The effective aperture is the effective aperture is the effective aperture ... The antenna intercepts ALL the EM energy within it's effective aperture. You, the listener, may be interested in only 2.8 KHz of that energy. :-) |
Dave Shrader wrote:
"The effective aperture is the effective aperture is the effective aperture...." And by any other name it would likely smell as sweet. Terman has a definitionin of sorts in his 1955 edition on page 899: Directive gain=(4 pi)(effective aperture)(k) divided by lambda squared. Arnold B. Bailey is much more descriptive in "TV and Other Receiving Antennas", starting around page 298. I`ve tried to condense and number his points: 1. The radio field is expressed in volts per meter or in watts per square meter. 2. Our standard dipole will have a capture area within the radio field, broadside to the field, with the rod positioned in the direction of the electric vector. (not cross-polarized) 3. Our 1/2-wave dipole has a capture area extending the 1/2-wavelength of the rod, and 1/8-wavelength either side of the center of the rod. 4. The area thus calculates as lambda squared over 8. 5. This effective area representation is a simplification but it works for most purposes. 6. If we can modify our 1/2-wave antenna to capture more signal than normal, we say the antenna has gain. 7. "Capture area" is pure fiction but it gives the right answers. 8. Zero-dB gain antennas all have the same "capture area", and are assumed to be resonant which is an essential requirement enabling full performance. 9. Shape, size, power transfer capability, and directional response are the four main factors which determine an antenna`s utility. (Bailey has great elaboration on each of these four factors.) 10. Some antenna shapes operate well over a wide frequency bandwidth, others don`t. 11. Fat antennas tend to be broadband and are shorter than a free-space wavelength, indicating a reduced signal velocity along their lengths. 12. All parts of the receiving antenna rod intercept the radio wave. 13. Induced current is reflected by open-circuit ends of the antenna rod. 14. Impedance at the center of the rod is the result of incident and reflected waves interacting. 15. Resonance is a result of interaction of incident and reflected signal components in the rod. 16. The antenna rod has a characteristic resistance independent of the reflected signal, and this resistance is determined by the ratio of voltage on the rod to current.. (This is found on page 306.) 17. From the size of an antenna rod we can determiner its capacitance per unit length. This factor determines its characteristic resistance. (voltage to current ratio) 18. Reflection coefficient at the ends of the rod depends on configuration. Thicker antennas have smaller reflections at their ends. 19. Conical rods which are thicker toward their extremities produce weak reflections, and tend to have broadband responses. 20. On thin dipoles at first resonance, there is substantial reflection from its ends and the reflection is 180-degrees out of phase with the incident signal. 21. 1/4-wave back from the open-circuit ends of the antenna, the voltage is low and the current is high, so the impedance (a resistance) is low. The resistance 1/4-wavelencth back from the open-circuit ends is much lower than the characteristic resistance of the rod to the signal traveling in either direction alonng the antenna rod. 22. With reflection---- R = Vincident - Vreflected / Iincident + Ireflected (Whereas, the Ro = V/I 23. Fat antennas are less directional than thin antennas and keep their broadside lobe together better at harmonic frequencies. Don`t know what the particular value of effective aperture is. Terman found it useful in describing the directional characteristics of a matress antenna. Since effective aperture is related to directive gain by constants, why not just stick with directive gain? Bailey, a leading Bell Labs antenna researcher, said capture area is pure fiction anyway as the grip on radiated energy does not start and stop abrubtly at the imaginary boundaries used to describe it. Obviously, effective aperture has enjoyed some popularity among some who found it useful. The effective area of an antenna has a dictionary definition: The square of the wavelength multiplied by the power gain in a particular direction, divided by 4 pi. The "capture area" is defined as: The area of the antenna elements that intercept radio signals. I may have confused the definitions. Best regards, Richard Harrison, KB5WZI |
Steve Nosko wrote:
"I assumed Richard`s intent here is that you only have to do the calculation for one full period...symmetrical, then only half period..." Maybe an advantage of graphical representation is manifestation of the absurd. If the wave is not symmetrical around the zero volt axis, or if one alternation has greater squared values, it may be apparent. You may be able to see this in the mathematical expression for the waveform too. The effective value of both alternations must be equal for the determination of the value of only one alternation to suffice. Steve also wrote: "There ia a formula for the total rms which is the square root of the sum of the squares."" Sounds familiar. Sounds like Pythagoras. Best regards, Richard Harrison, KB5WZI |
Steve Nosko wrote:
"I think the phrase "available energy" may draw discussion, but I won`t start it." I plead guilty. I knew when I wrote it that the choice was poor but couldn`t think of a better phrase at the moment. Terman says on page 76 of his 1955 edition: "Alternatively, a load impedance may be matched to a source of power in such a way as to make the power delivered to the load a maximum. (see footnote) The power delivered to the load under these conditions is termed the "available power" of the power source." Walter Maxwell straightened me out on my carelessness on this long ago, and I repeated anyway. Dang me! Best regards, Richard Harrison, KB5WZI |
Steve Nosko wrote:
"With a "conjugate match" the source dissipates 50% of the power and the load the other 50%." This is true only if all the source resistance is the type that converts electrical energy to heat. There is a non-dissipative resistance. Switched-off time in the Class-C amplifier is part of its internal resistance. You can deliver all the available power of the Class-C amplifier to the load and dissipate less than 50% in the source. Best regards, Richard Harrison, KB5WZI |
Steve Nosko wrote:
"A (tube) amplifier is in conjugate match conditions. It is dissipating 10 watts in its plate. This is the limit of its plate dissipation----There is also 10 watts in the load. Now assuming you can, increase the (plate) supply voltage by , say 20%---(may be the fatal flaw)." Likely so. If everything remains linear, 20% more voltage increases power by 1.2 squared, or 1.44 times. If the tube was already dissipating its maximum sustainable power, expect an early failure due to the overload. If you were not already in a maximum power transfer condition, and a condition which might provoke flashover within the tube, a readjustment of the match might shift more of the available power to the load and thus relieve the tube of some of the dissipation. Best regards, Richard Harrison, KB5WZI |
"Richard Harrison" wrote in message ... Steve Nosko wrote: "With a "conjugate match" the source dissipates 50% of the power and the load the other 50%." This is true only if all the source resistance is the type that converts electrical energy to heat. There is a non-dissipative resistance. Switched-off time in the Class-C amplifier is part of its internal resistance. You can deliver all the available power of the Class-C amplifier to the load and dissipate less than 50% in the source. Best regards, Richard Harrison, KB5WZI I think this becomes very academic (or perhaps a better word is awkward) because originally you framed this as a "conjugate match" situation. Now, you are in terms of time varying parameters. I think the analysis must stay in one realm or the other. My mental models have trouble switching back and forth. I can't speak to tubes, but I do know that to get the most out of a transistor power amp (transmitter type) up to about 200 MHz you design the output match to be for a collector resistance of Vcc^2/(2Po). IF I recall, this can be derived easily if you assume the transistor pulls all the way to ground and the output tank swings up to 2 x Vcc. I was shown, and understood it way back then, but can't recall the path to the solution off the top of my head. This, then, always started the discussion of whether this "matched" the transistors output impedance or some other more esoteric concept. Then talk of "average resistance" came in and eyes would glaze over.... The designer would then go back to the bench, work to optimize the design to his requirements, test it over temp, etc. and ship it. "There comes a time to shoot the Engineer and ship the product." is the title of a famous editorial from the early 70's (IIR). Got it around here somewhere... -- 73 Steve N, K,9;d, c. i My email has no u's. |
Steve Nosko wrote:
"Then there`s the solid-state power amplifier standard output resistance formula. Rs = Vcc^2 / (2*Po) The implication should be obvious." It looks like Ohm`s law to me, P=Vsq / R. The implication of (2*Po) is that 50% of the power is in the source and 50% of the power is in the load. If so, it`s a Class-A amplifier formula, but the semiconductors could be biased to cut-off (Class-B) to reduce dissipation in the transistors when they are idle. The best collector load resistance is often not that which produces maximum output, but that which produces maximum "undistorted output". Best regards, Richard Harrison, KB5WZI |
"Richard Harrison" wrote in message ... Steve Nosko wrote: "I think the phrase "available energy" may draw discussion, but I won`t start it." I plead guilty. I knew when I wrote it that the choice was poor but couldn`t think of a better phrase at the moment......... Best regards, Richard Harrison, KB5WZI I think this is what causes much of the discussion here. Terms are not always as clear cut as things like: voltage or resistance. We know those, but some of the others are more vague...or just what concept they point to is not obvious. It is also sometines difficult to determine what the KEY point or word is of a question or statement. When I read yours, I made the assumption that "available power" was not where your question pointed. I went along the; "Can we get a non 50-50 division? and not waste so much in the stage" path. It is clear that different responders "Hear" different question. Cryminy crum! I think my brain has forgotten how to hit the keys "on" in the correct order. I keep getting things like "questino" for "question"....or is it a keyboard timing issue??? As my fingers fly along... -- Steve N, K,9;d, c. i My email has no u's. |
I wrote a long boring post on "effective aperture". I apologize. I
regret not becoming familiar with Kraus`s "Antennas" long ago. Kraus makes many ideas clear with few words. On page 43 of his 1950 edition of "Antennas" Kraus writes: "The ratio of the power W in the terminating impedance to the power density of the incident wave will be defined as the effective aperture Ae. Thus Ae = W/P If W is in watts and P is in watts per square meter, the Ae is in watts per square meters." Best regards, Richard Harrison, KB5WZI |
Steve, have we now moved to an antenna that has a preamplifier on it for
listening ? If so I need not continue to struggle to follow the thread day by day to determine its implications to the subject at hand. The half power thingy I presume is understood by all so IS something very exciting to be revealed that shows that the dipole is really an efficient radiator after all, but only if you put a class C amplifier on it ? There are stacks of information in books and lots of words in a dictionary, but sooner or later one has to give a reason as to why they are reading out loud if it is meant to be instructive or explanaatory with respect to the thread, i.e. dipole and its impedance. You can start a different thread which would help out for when one checks out the archives unless the intent is meant to be destructive Regards Art .. "Steve Nosko" wrote in message ... Hi Richard... "Richard Harrison" wrote in message ... [...] "----Second, it is the RMS current through the tube which will waste power, so that is what we must be concerned with." I don`t believe current through a Class C amplifier consists of an ordinary sine wave. And I didn't say that it does nor do I believe it does. I'm inclined to take my 100MHz storage scope to to the 6146's of my TS830s and see for myself. Your words imply (at least I infer) you are thinking that only a sine wave has an RMS value. Every wave of any shape has an effective or RMS value - its heating or "power causing" value. [...] I think it consists of short unidirectional pulses. The tuned "tank circuit" is the source of sine waves. This certainly has to be correct. The tank will most likely cause some sine-like VOLTAGE waveform, but the tube current has to be pulses of some shape. This is a very timely discussion in view of the AC power meter QST article and the extensive investigation I just completed on several pulse shapes.. RMS is the effective value, not the average value, of an a-c ampere. I will differ here. The RMS value is more appropriately described as the power producing value of ANY wave form. Pulses can produce heat just as well as sine wave AC. We all know this from a practical view since tubes can only conduct in one direction and the plates DO get hot. ...as the heating value of an ampere is proportional to the current squared. This is actually a simplification. P=ExI Power is the product of voltage and current *only*. Because this is a second order effect, in a resistance it can be related to either voltage squared or current squared... because that captures the second order character. Maybe there's a better way to say it mathematically, but I don't know it. When we get to non sine shapes, then we have to fall back on the actual definition. root [avg of square] ...with the integral and all. http://www.ultracad.com/rms.pdf [...snip...] Ordinarily, with nonsinusoidal currents, the ratio of maximum to effective value is not the square root of 2. Best regards, Richard Harrison, KB5WZI Doing the math for pulses with the shape of sine, triangle (a single slope with sudden end) and trapezoid (a sudden start to one level then a slope to a peak and a sudden end), I decided to look at the RMS to AVERAGE ratio since average is what a common meter will measure in Bob Shrader's article (AC watt meter Jan 04 QST). I was particularly interested in the sine-shaped pulses of various duty cycle because the current of common power supplies occurs in short pulses with a sine-like shape that are near the peak of the voltage waveform. It was interesting that for all these shapes, this ratio was very similar. One relatively simple thing to understand which came out of the analysis was that the average value is directly proportional to the duty cycle as you might reasonably postulate. Where duty cycle is the ratio of "on" time to off time. Where "on" time is the time that ANY current flows. Whereas the RMS is proportional to the Square root of the duty cycle. e.g. drop the duty cycle to half and the RMS drops to .707. I have to do some verification, but it sure looks as though Bob's numbers can be as much as three times what he quoted, depending on the waveshape and some measurements I made. http://www.irf.com/technical-info/an949/append.htm Trapezoid=rectangular. Also for the phase controlled sine, the things that look like tau and a small n are both pi i.e. sin [pi x (1-D)] cos [pi x (1-D)] and denominator of 2 x pi Some average & RMS values here. http://home.san.rr.com/nessengr/techdata/rms/rms.html More (better) average formulas: http://www.st.com/stonline/books/pdf/docs/3715.pdf NOW I know where the average value of a sine wave comes from = (2/pi) The Greek delta = d. A calculator for RMS: http://www.geocities.com/CapeCanaveral/Lab/9643/rms.htm |
Richard Harrison wrote:
Thus Ae = W/P If W is in watts and P is in watts per square meter, the Ae is in watts per square meters." Seems to me that Ae would be in square meters. :-) -- 73, Cecil, W5DXP |
"Steve Nosko" wrote With a "conjugate match" the source dissipates 50% of the power and the load the other 50%. This can't be changed. ============================= Without disagreeing with what you say - A conjugate match is not relevant in the present discussion because there is seldom, if ever, a conjugate match between a PA and its antenna system. The tuning-up process is NOT intended to produce such a match. Tuning up is just the simple process of adjusting the transmitter load resistance to be equal to its designed-for load resistance, usually an arbitrary 50 ohms. The internal resistance of a transmitter is NOT 50 ohms. It is not a design feature. It is whatever happens to appear after the designer has met a series of other requirements. The designer himself does not know what the internal resistance is unless, out of curiosity, he bothers to measure or calculate it. ---- Reg, G4FGQ |
aunwin wrote:
The half power thingy I presume is understood by all so IS something very exciting to be revealed that shows that the dipole is really an efficient radiator after all, but only if you put a class C amplifier on it? Art, I assume you know that Class-C amplifiers are not usually used for SSB since they are not linear for SSB. The basic confusion is between linear systems and non-linear systems. If the amplifying device (singular) conducts over the entire 360 degrees of an RF cycle and the output waveform is a reasonable copy of the input waveform, then that device is said to be linear. If you have two amplifying devices operating in anything except Class-A operation, the output of each individual device is not linear. That's the kicker. The "two non-linear device" option is not available at the antenna for receiving purposes. A normal dipole cannot receive Class-C (non-linear) signals. -- 73, Cecil, W5DXP |
Reg Edwards wrote:
The internal resistance of a transmitter is NOT 50 ohms. It is not a design feature. It is whatever happens to appear after the designer has met a series of other requirements. The designer himself does not know what the internal resistance is unless, out of curiosity, he bothers to measure or calculate it. Proving that, like 50/60 Hz AC power, efficiency may be more important than maximum power transfer. Edison actually thought that AC generators would burn up while trying to deliver maximum power. That's why he backed DC. -- 73, Cecil, W5DXP |
Cecil, W5DXP wrote:
"Seems to me that Ae would be in square meters." Cecil is wide awake. If you divide watts by watts per square meter, you end up with square meters. Appropriate for for an aperture, and dimensionally correct. I must have been half asleep and typed something from the wrong line. Kraus made no mistake. I did. I apologize. Best regards, Richard Harrison, KB5WZI |
On Wed, 25 Feb 2004 12:22:32 -0600, "Steve Nosko"
wrote: [...] you simply measure the caloric result and ignore shape altogether. I always thought that the common method of measuring RF power was pretty cool! The Thermistor or bolometer. Here you balance a bridge with DC or low freq AC. It heats the thermistor to the correct resistance. Then, when you add RF power, the thing heats up more and changes resistance. So, you remove some DC power to get back to the correct resistance and that amount is easy to figure. That is how much RF you put in. Cool. I think it is correct to say that you absolutely cannot measure power *directly*. You must measure something else which is affected/caused by the power...comment? Hi Steve, There are many classes of caloric devices, two of which you identify that are common within the Metrologist's art, and wholly absent from amateur activities. So here I must make a slight correction of your description. Power meters contain two (2) such devices which form the balanced halves of a bridge. One side is exposed to the RF, the other side is exposed to the simpler DC or AC power that is known to a high degree of accuracy. What you describe is the detector implementation of the same devices (which exhibit non-linearity to perform detection). They would, in the fashion you describe, offer good "relative" power indication, but not absolute power (except through substitution methods). As such, they are fairly common in precision VSWR instrumentation especially when they are driven by 1KHz modulated power sources, and in turn drive special AC VTVM's scaled to present dB and VSWR to very high resolution. A list of the methods: The Crystal: 1N21/23/25/26... The Bolometer (low power caloric) The Barretter (a Bolometer): Sperry 821, PRD 630A The Wollaston wire (a Barretter): actually a 0.01A glass fuse The Carbon filament (a Barretter) The Thermistor (a Bolometer): Western Electric 28A The Thermocouple The Thermopile (lotsa Thermocouples) 73's Richard Clark, KB7QHC |
Cecil
I am talking about the half power thingy with respect to a series circuit of which a antenna is designed around. Reg politely made a separate thread on a specific part of that thread which referred to impedance, presumably because Q was being bandied about where he thought probably it was irelevant. Am I wrong to think that because a different thread was not made the impedance question and amplifiers was relavent and somebody was not being destructive/impolite? Now to your pont of what is the kicker...a normal dipole cannot receive Class-C (non linear signals)"? Having being told about this thing that somebody read somewhere, where is it leading to? Seems like I entered a class on engineering and after 5 minuites I am wondering what sort of professor I had that not only just read books out loud for his money but thought it was O.K. to read from a wood working book. Can't we assume that a antenna is a closed series circuit containing only passive items ? If you have in mind that we must we consider an antena as a Class something or other amplifier when determining its impedance then I am hopelessly lost in a thread that can only end up nasty like some others did and drive some more people away because somebody wanted to play games of obstruction with the intent to annoy. Now I see that somebody decided to change this thread heading instead of starting a new thread . Now we are talking about RMS meters and how they can be used ? Are we talking digital or analogue, hand held or otherwise, high enough accuracy to satisfy all ( Nah that is asking to much) Well it is an antenna newsgroup so it must radiate , somebody read it in a book and assumes that all are unaware of it so he wishes to describe it so others can make sense of what he read and why he read it! Regards Art I am not pointing the finger at you Cecil, I have no idea who the culprit is or what his intent is. ?Cecil Moore" wrote in message ... aunwin wrote: The half power thingy I presume is understood by all so IS something very exciting to be revealed that shows that the dipole is really an efficient radiator after all, but only if you put a class C amplifier on it? Art, I assume you know that Class-C amplifiers are not usually used for SSB since they are not linear for SSB. The basic confusion is between linear systems and non-linear systems. If the amplifying device (singular) conducts over the entire 360 degrees of an RF cycle and the output waveform is a reasonable copy of the input waveform, then that device is said to be linear. If you have two amplifying devices operating in anything except Class-A operation, the output of each individual device is not linear. That's the kicker. The "two non-linear device" option is not available at the antenna for receiving purposes. A normal dipole cannot receive Class-C (non-linear) signals. -- 73, Cecil, W5DXP |
Art, KB9MZ wrote:
"The half power thingy I presume is understood by all so IS something very exciting to be revealed that shows that the dipole is really an efficient radiator after all, but only if you put a class C amplifier on it?" Art was the introducer of efficiency into this impedance thread. Something from Art about radiation per unit length of an antenna, if I recall. We noted that the antenna itself is usually so efficient there`s not much to talk about, but there are differences in the effectiveness of getting a signal on and off the air via an antenna. There`s coupling the antenna to the radio. This has been argued here since before the "47 KW CB" thread, and that was years ago. If Art can get more signal out of an antenna which is as small or smaller than ordinary without putting more current into that antenna, assuming orientation, polarization, and the other usual conditions are fair for the competition, I`m excited. Best regards, Richard Harrison, KB5WZI |
Art, KB9MZ wrote:
"Can`t we assume that an antenna is a closed series circuit containing only passivee items?" Sure, but you have an incomplete representation. An antenna is coupled more or less to the entire universe. It is usually tightly coupled to a radio or similar apparatus by some mutual impedance which makes a codependency. Best regards, Richard Harrison, KB5WZI |
"Richard Harrison" wrote in message ... Steve Nosko wrote: "A (tube) amplifier is in conjugate match conditions. ... Now assuming you can, increase the (plate) supply voltage by , say 20%---(may be the fatal flaw)." Likely so. If everything remains linear, 20% more voltage increases power by 1.2 squared, or 1.44 times. If the tube was already dissipating its maximum sustainable power, expect an early failure due to the overload. That's what I went on to say (get more power and 50% will be dissipated in the stage...until it blows), but there was more to the story which is relavant. -- Steve N, K,9;d, c. i My email has no u's. |
"Richard Harrison" wrote in message ... Steve Nosko wrote: "Then there`s the solid-state power amplifier standard output resistance formula. Rs = Vcc^2 / (2*Po) The implication should be obvious." It looks like Ohm`s law to me, P=Vsq / R. Well... I don't call THAT ohm's law, but rather, oh, I suppose, the power formula, but that's symmantics. It is a transposition of (whatever you call) that formula. The implication of (2*Po) is that 50% of the power is in the source and 50% of the power is in the load. Again, I don't recall teh derivation, but it works. I don't believe it related to a mathematical constraint that the power be equally split. Can't speculate further withoug working it out. This is the "maximum output" load. Don't know off-hand what the limiting factor is, but this is what they desigend for and I don't thing you could get more out without killing the part or its lifetime. Don't recall anyone blowing parts with the wrong load...pretty robust parts. You just couldn't get any more out. If so, it`s a Class-A amplifier They are class C. VHF FM PAs. formula, but the semiconductors could be biased to cut-off (Class-B) to reduce dissipation in the transistors when they are idle. The best collector load resistance is often not that which produces maximum output, but that which produces maximum "undistorted output". Best regards, Richard Harrison, KB5WZI |
aunwin wrote:
Can't we assume that a antenna is a closed series circuit containing only passive items ? A passive antenna, when it is receiving a signal, is a pretty good approximation to a Thevenin Equivalent circuit. The received signal is the generator. The generator impedance is the radiation resistance (and the lossy R's). We've got a transmission line and usually a 50 ohm load in the receiver. If you have in mind that we must we consider an antena as a Class something or other amplifier when determining its impedance then I am hopelessly lost in a thread that can only end up nasty like some others did Why not, for the purposes of limiting the discussion, consider only a Thevenin Equivalent 50 ohm source for the transmitter? That way, the entire system will be linear and easy to discuss. I am not pointing the finger at you Cecil, ... If you were, which one would it be? :-) -- 73, Cecil, W5DXP |
"Reg Edwards" wrote in message ... "Steve Nosko" wrote With a "conjugate match" the source dissipates 50% of the power and the load the other 50%. This can't be changed. ============================= Without disagreeing with what you say - A conjugate match is not relevant in the present discussion because there is seldom, if ever, a conjugate match between a PA and its antenna system. The tuning-up process is NOT intended to produce such a match. Tuning up is just the simple process of adjusting the transmitter load resistance to be equal to its designed-for load resistance, usually an arbitrary 50 ohms. The internal resistance of a transmitter is NOT 50 ohms. It is not a design feature. It is whatever happens to appear after the designer has met a series of other requirements. The designer himself does not know what the internal resistance is unless, out of curiosity, he bothers to measure or calculate it. ---- Reg, G4FGQ Hi Reg, This is another track, but one I've pondered, but probably shouldn't now... And one that I believe has seen MANY words on NetNews, no? So I'll put this out anyway and see what I get. As we adjust the classical pi output network, I think we are starting high and gradually lowering the load impedance seen by the tube, no? Start with max C at the load cap and decrease it for max Po (keep in resonance with the plate cap) I believe in the pi config. the BIG load "C" transforms the load (50ohms) up to a "hi" Z seen by the plate. This may be wrong, though I believe not, so will continue with the concept. As we do this the power to the load keeps increasing up to a point. If we keep going, the Po drops. So, we have reached the optimum Z as seen by the tube. I think the question at this point is "why / what" is this the "optimum" for? I suspect the arguments are FOR and CON conjugate match. I think it must be agreed that it is clearly the optimum for Po, no? -- Steve N, K,9;d, c. i My email has no u's. |
"Cecil Moore" wrote in message ... Reg Edwards wrote: The internal resistance of a transmitter is NOT 50 ohms. It is not a design feature. It is whatever happens to appear after the designer has met a series of other requirements. The designer himself does not know what the internal resistance is unless, out of curiosity, he bothers to measure or calculate it. Proving that, like 50/60 Hz AC power, efficiency may be more important than maximum power transfer. ... 73, Cecil, W5DXP That's also where I was going with my "raise the plate voltage" concept. -- Steve N, K,9;d, c. i My email has no u's. |
I'll never go there in the RF arena even though another hobby of mine is laser light shows. Placing "photon" and "RF" in the same paragraph is ... what shall I call it, flame-fodder? "Richard Clark" wrote in message ... On Wed, 25 Feb 2004 14:46:42 -0600, "Steve Nosko" wrote: I think the phrase "available energy" may draw discussion, but won't start it. Uh-Oh I suspect it is not the thing to focus on. However... This has in the past lead to a cataract of 600 posting threads With a "conjugate match" the source dissipates 50% of the power and the load the other 50%. This can't be changed. This is the 1mm fuse that starts that torrent. Once lit, no one escapes. ... Snake oil? The math is correct, but there may be some practical limit not considered. Hi Steve, Practical or otherwise will be dismissed to the discussion of photons and general relativity by the end of two days, if this doesn't fall flat on its face with mine being the only comment. ;-) 73's Richard Clark, KB7QHC |
"Richard Clark" wrote in message ... On Wed, 25 Feb 2004 12:22:32 -0600, "Steve Nosko" wrote: [...] you simply measure the caloric result and ignore shape altogether. I always thought that the common method of measuring RF power was pretty cool! The Thermistor or bolometer. Here you balance a bridge with DC or low freq AC. It heats the thermistor to the correct resistance. Then, Yea... Then I give only a partial description since trying to give it completely would take pages, I decided to "overview" it. However... ... and wholly absent from amateur activities. OOPS! "wholly"?? I've got one. A really nice (but un temp compensated) thermistor mount) (but don't tell anyone that I have nothing to drive it with -- the rest of the bridge.) So here I must make a slight correction of your description. Power meters contain two (2) such devices which form the balanced halves of a bridge. I gotta go back & look, (this was from the 60's) but there are two and they are in series for the low freq and parallel for the RF. On the low freq side it is a 200ohm mount. Now I hafta' remember how the RF was handled .... musta' been to only one...let me think about this... One side is exposed to the RF, the other side is exposed to the simpler DC or AC power that is known to a high degree of accuracy. What you describe is the detector implementation of the same devices (which exhibit non-linearity to perform detection). They would, in the fashion you describe, offer good "relative" power indication, but not absolute power (except through substitution methods). OOPS again! You DO get absolute because you know how much low freq power you remove. Isn't that the principle of the HP437 & 438's? or are they doing something else...diode ... er...uh crystal if you're across the p ond? I always thought they were, caloric, as you say. -- Steve N, K,9;d, c. i My email has no u's. As such, they are fairly common in precision VSWR instrumentation especially when they are driven by 1KHz modulated power sources, and in turn drive special AC VTVM's scaled to present dB and VSWR to very high resolution. A list of the methods: The Crystal: 1N21/23/25/26... The Bolometer (low power caloric) The Barretter (a Bolometer): Sperry 821, PRD 630A The Wollaston wire (a Barretter): actually a 0.01A glass fuse The Carbon filament (a Barretter) The Thermistor (a Bolometer): Western Electric 28A The Thermocouple The Thermopile (lotsa Thermocouples) 73's Richard Clark, KB7QHC |
Consider me burned out too, but I did change the subject line to allign
w/the change in topic. -- Steve N, K,9;d, c. i My email has no u's. "aunwin" wrote in message news:zmq%b.127262$jk2.539846@attbi_s53... Steve, have we now moved to an antenna that has a preamplifier on it for listening ? If so I need not continue to struggle to follow the thread day by day to determine its implications to the subject at hand. The half power thingy I presume is understood by all so IS something very exciting to be revealed that shows that the dipole is really an efficient radiator after all, but only if you put a class C amplifier on it ? There are stacks of information in books and lots of words in a dictionary, but sooner or later one has to give a reason as to why they are reading out loud if it is meant to be instructive or explanaatory with respect to the thread, i.e. dipole and its impedance. You can start a different thread which would help out for when one checks out the archives unless the intent is meant to be destructive Regards Art . "Steve Nosko" wrote in message ... Hi Richard... "Richard Harrison" wrote in message ... [...] "----Second, it is the RMS current through the tube which will waste power, so that is what we must be concerned with." I don`t believe current through a Class C amplifier consists of an ordinary sine wave. And I didn't say that it does nor do I believe it does. I'm inclined to take my 100MHz storage scope to to the 6146's of my TS830s and see for myself. Your words imply (at least I infer) you are thinking that only a sine wave has an RMS value. Every wave of any shape has an effective or RMS value - its heating or "power causing" value. [...] I think it consists of short unidirectional pulses. The tuned "tank circuit" is the source of sine waves. This certainly has to be correct. The tank will most likely cause some sine-like VOLTAGE waveform, but the tube current has to be pulses of some shape. This is a very timely discussion in view of the AC power meter QST article and the extensive investigation I just completed on several pulse shapes.. RMS is the effective value, not the average value, of an a-c ampere. I will differ here. The RMS value is more appropriately described as the power producing value of ANY wave form. Pulses can produce heat just as well as sine wave AC. We all know this from a practical view since tubes can only conduct in one direction and the plates DO get hot. ...as the heating value of an ampere is proportional to the current squared. This is actually a simplification. P=ExI Power is the product of voltage and current *only*. Because this is a second order effect, in a resistance it can be related to either voltage squared or current squared... because that captures the second order character. Maybe there's a better way to say it mathematically, but I don't know it. When we get to non sine shapes, then we have to fall back on the actual definition. root [avg of square] ...with the integral and all. http://www.ultracad.com/rms.pdf [...snip...] Ordinarily, with nonsinusoidal currents, the ratio of maximum to effective value is not the square root of 2. Best regards, Richard Harrison, KB5WZI Doing the math for pulses with the shape of sine, triangle (a single slope with sudden end) and trapezoid (a sudden start to one level then a slope to a peak and a sudden end), I decided to look at the RMS to AVERAGE ratio since average is what a common meter will measure in Bob Shrader's article (AC watt meter Jan 04 QST). I was particularly interested in the sine-shaped pulses of various duty cycle because the current of common power supplies occurs in short pulses with a sine-like shape that are near the peak of the voltage waveform. It was interesting that for all these shapes, this ratio was very similar. One relatively simple thing to understand which came out of the analysis was that the average value is directly proportional to the duty cycle as you might reasonably postulate. Where duty cycle is the ratio of "on" time to off time. Where "on" time is the time that ANY current flows. Whereas the RMS is proportional to the Square root of the duty cycle. e.g. drop the duty cycle to half and the RMS drops to .707. I have to do some verification, but it sure looks as though Bob's numbers can be as much as three times what he quoted, depending on the waveshape and some measurements I made. http://www.irf.com/technical-info/an949/append.htm Trapezoid=rectangular. Also for the phase controlled sine, the things that look like tau and a small n are both pi i.e. sin [pi x (1-D)] cos [pi x (1-D)] and denominator of 2 x pi Some average & RMS values here. http://home.san.rr.com/nessengr/techdata/rms/rms.html More (better) average formulas: http://www.st.com/stonline/books/pdf/docs/3715.pdf NOW I know where the average value of a sine wave comes from = (2/pi) The Greek delta = d. A calculator for RMS: http://www.geocities.com/CapeCanaveral/Lab/9643/rms.htm |
There you go again wandering off and bull****ting about something that you
read some where and no body else knows it. If you think as you say that all has been said that needs to be said about antennas then open another thread where you can deposite your volumous verbal diarrea. The subject was antennas and impedance but you seem to think that every thing is known so you can turn to any book on any subject and empty yourself on us as if you are on a bathroom stool which is your county seat. Now if you want to attack me regarding my antenna then fine start a thread and we will go at it but remember it is not in a book so you will have to think for yourself or resort to reading any old book and regurgitating. Now you could find something about say loop antennas that you are sure nobody else knows about or are we not upto that part in Krause's book that you are currently reading ? And while we are at it, now you are getting absent minded please put a marker in the book where you left off. And another thing I will echo what Cecil says to the like of you....an antenna is an antenna and not to be regarded as a system. My antenna does not cry out aloud that it has had its amplifier stolen. Another point..you mention "we" since when do you think you talk for everyone here, only a few days ago your excuse was that you was not a professor,! Today you screwed up with dimension units that you copied incorrectly from YOUR book and frankly you have to post more than others because people do not agree with you most of the time as you seem unable to stay on target and get confused about what the subject is or how to interpret something that you read and screw around trying to find a way to make it relavent. If you remembered what ever you learned in school you would be able to talk for yourself and then debate for yourself now that Terman is dead, really dead . Now you also said something about what I said on this thread regarding current or something ( I think I said 'I believe' which is a long way from acting as a know all tho I must admit I haven't been up lots of towers,put them up in lots of countries and all those other stories that you tell ) to which you replied bull**** or balloney which from you should be enough so that you can walk away like a messiah you doesn't have to explain himself based on some fabulous light around his head that he thinks he has earned thru his fabulous teachings. Another thing are you the only one that was correct on Reggies question.....no I didn't see a response in the early days that I read that thread... probably you knew all along as you read something like it in a book so dumped that on the thread anyway. Now if you want to change the heading of this thread then be my guest since you and you alone knows everything there is to know about antennas as you read this book on a tower in Del Fuego or was it in the bathroom? This is not a personal attack by the way it is just a disagreement like saying baloney or bull****, justification is not required. I am sure you can identify with that with you not being a professor and thus allowed to spout off. Art Looking forward to hearing you hit my antenna and the mathematics that go with it regarding the amplifier hidden in the elements. Last time you said ' it' breaks all the laws of physics, the laws that you read in a book somewhere and that was sight unseen, the antenna that is, the laws were chiselled in a rock , graphical form that you saw the last time you climbed a mountain where a bush once burned and you became a prophet "Richard Harrison" wrote in message ... Art, KB9MZ wrote: "The half power thingy I presume is understood by all so IS something very exciting to be revealed that shows that the dipole is really an efficient radiator after all, but only if you put a class C amplifier on it?" Art was the introducer of efficiency into this impedance thread. Something from Art about radiation per unit length of an antenna, if I recall. We noted that the antenna itself is usually so efficient there`s not much to talk about, but there are differences in the effectiveness of getting a signal on and off the air via an antenna. There`s coupling the antenna to the radio. This has been argued here since before the "47 KW CB" thread, and that was years ago. If Art can get more signal out of an antenna which is as small or smaller than ordinary without putting more current into that antenna, assuming orientation, polarization, and the other usual conditions are fair for the competition, I`m excited. Best regards, Richard Harrison, KB5WZI |
No I can't remember ever putting a finger/s up at anybody.
But I have this feeling that this thread is now long enough and has included so many subjects of discussion that we have the beginnings of a full scale augument again. However if it means that this is the time a certain person leaves the group in discussed then no harm done. Forget antennas and impedance and the subject change you are advocating, let the row start now, say something, anything so we get a repeat of the last one. Can't you say balonney or bull**** at the next guru and tell something different to another guru and blame it on the first guru? Throw a fishing pole with a carrot on it! Art Art Art "Cecil Moore" wrote in message ... aunwin wrote: Can't we assume that a antenna is a closed series circuit containing only passive items ? A passive antenna, when it is receiving a signal, is a pretty good approximation to a Thevenin Equivalent circuit. The received signal is the generator. The generator impedance is the radiation resistance (and the lossy R's). We've got a transmission line and usually a 50 ohm load in the receiver. If you have in mind that we must we consider an antena as a Class something or other amplifier when determining its impedance then I am hopelessly lost in a thread that can only end up nasty like some others did Why not, for the purposes of limiting the discussion, consider only a Thevenin Equivalent 50 ohm source for the transmitter? That way, the entire system will be linear and easy to discuss. I am not pointing the finger at you Cecil, ... If you were, which one would it be? :-) -- 73, Cecil, W5DXP |
On Thu, 26 Feb 2004 15:20:06 -0600, "Steve Nosko"
wrote: OOPS again! You DO get absolute because you know how much low freq power you remove. Isn't that the principle of the HP437 & 438's? or are they doing something else...diode ... er...uh crystal if you're across the p ond? I always thought they were, caloric, as you say. Hi Steve, They are Bolos (caloric) using PRD type elements. From your description (barring the equipment name), it sounded like the typical detector application. I can see you originally described a self balancing bridge: The HP bridge is powered three ways. It has a DC source to offer temperature compensation and range changes (zero setting). It has an AC Oscillator (~10 KHz) source in quadrature (topographically orthogonal, across the bridge instead of over it). Basically, the bridge temperature components (the Bolometer) act as an oscillator level stabilizing feedback loop (much like the famous HP 200 which used a lamp's negative resistance). The third power input is the unknown RF which unbalances the heat causing the feedback to re-regulate where the drive level changes (which is monitored by the meter as the indication of the RF power applied). The sum of the operation was that the bridge always operates at null with the same heat burden for any applied RF power (within constraints of course). And yes, they are characterized as 200 Ohm mounts. 73's Richard Clark, KB7QHC |
On Wed, 25 Feb 2004 14:46:42 -0600, "Steve Nosko"
wrote: snip.... With a "conjugate match" the source dissipates 50% of the power and the load the other 50%. This can't be changed. Draw the schematic to see. Rs = RL therefore Ps = PL. (I'll use a big "L" so it looks like an "L") Dear Steve, You may find the information in this article to be of interest relative to the efficiency in a conjugate match situation. http://www.qsl.net/w9dmk/ObservationsOnMPTT.htm 73, Bob |
I just get an advertisement when I click on your link. Mac N8TT
-- J. Mc Laughlin - Michigan USA "Robert Lay W9DMK" wrote in message ... On Wed, 25 Feb 2004 14:46:42 -0600, "Steve Nosko" wrote: snip.... Dear Steve, You may find the information in this article to be of interest relative to the efficiency in a conjugate match situation. http://www.qsl.net/w9dmk/ObservationsOnMPTT.htm 73, Bob |
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