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Dave Shrader wrote:
So, are you pursuing Physics or Applications type knowledge? And apparently, never the twain shall meet. :-) -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
What happens if the twisted short is grounded as in lightning protection?
Is there any succinct difference? -- 73 Hank WD5JFR "Cecil Moore" wrote in message ... Reg Edwards wrote: By how much does the length of the short-circuiting bar at the end of the 1/4-wave stub affect its resonant length? I just twist the ends together. I don't use a bar. What open-circuit resonant frequency does the stub change to when the short circuiting bar is removed? Same resonant frequency - virtual impedance at the input is reversed. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Dave wrote:
you can look at it various ways. As I mentioned above; engineers, technicians, wave mechanics, physicists, experimenters, etc., ALL live on this list grin. Mea Culpa, I'm a retired engineer. Just remember, stubs WORK very well. Thank the owner of your local repeater for tuning the stubs [AKA CAVITIES] to reduce the desense below receiver response levels, typically -120 dB or more attenuation. 1. the wave traveling wave analysis... the harmonic goes down the stub, reflects back, and when it gets back to the transmission line it is 180 degress out of phase with the next cycle so it cancels it at the junction... therefore there is no harmonic to propagate down the line past the junction. this of course will raise the hackles of the anti-reflectionists who will then say there is also no harmonic left to propagate down the stub which means there is a virtual short at the junction, but no way to generate it since nothing can be there to go down the stub. 2. the power analysis. power goes in, power comes out, it all reflects back and forth until the energy becomes infinite and the amp blows up... but of course energy is conserved and momentum must go somewhere so the stub probably walks across the table with each wave reflection. 3. the sinusoidal steady state analysis. this takes the stub and transforms the shorted impedance at the far end back to the junction and then does all calculations as if the real short existed at the junction.... this will of course annoy the reflectionists who will point out in never ending detail how you can't explain tv ghosts, radar, or other transient phenomena this way... of course by assuming the sinusoidal steady state at the start you exclude those systems from this type of analysis, but that won't stop the protests. 4. the s analysis... who know what this will say except cecil who will find some way to steer the discussion over to it. 5. the optical layer analysis... see above. "Henry Kolesnik" wrote in message ... I know they work! One of my reasons for asking the question is I've not found any mention in the literature of where the "attenuation/rejection/reflection/filter residue" goes. -- 73 Hank WD5JFR "Dave" wrote in message ... the harmonic is 'attenuated' in that the magnitude of it is reduced when the stub is in line. i look at it like you could replace the stub with a lumped filter at the same point so the term attenuation makes more sense than reflections or rejections... i don't really care where the harmonic goes, i want to know how much it is attenuated by so i can compare with other types of filters. "Henry Kolesnik" wrote in message .com... Dave Nice site, I like the "white paper" approach as I prefer the info without the glitter. I've only read a few items and I quote: "This is a plot of the attenuation provided by the stub. You can see that it provides about 32db of attenuation at 28.25Mhz. " I've noticed that the literature I've purused indicates that stubs either attenuate or reject. None say reflect! I don't want to get into a discussion of word definitions becasue reflect and feject are close but attenuate is not in the same class. Comments... -- 73 Hank WD5JFR "Dave" wrote in message ... "Henry Kolesnik" wrote in message y.com... I know that a shorted 1/4 wave stub exhibits a very high impedance. But for the 2nd harmonic it's a 1/2 wave stub and exhibits a very low impedance or a short. There are claims that this can be used to filter the even harmonics. Shorts can't diisipate power and must reflect, so how does a stub work? stubs work very nicely. you can get practical stub information at my web site, including how to build a 40m to 15m 3rd harmonic stub filter: http://www.k1ttt.net/technote/techref.html#filters as you may have noticed by now you have kicked the proverbial hornets nest. reflections are a touchy word in this group, usually attracting the endless argument that travels from thread to thread. in time this will deteriorate into name calling and endless argument over reflections, interference, virtual impedances, and a few other topics. |
On Fri, 11 Jun 2004 18:32:14 GMT, "Henry Kolesnik"
wrote: What if the source doesn't or can't dissipate? Hi Hank, Simple enough to find out. Disconnect the output of your transmitter. Set the drive level of your transmitter for full power. Defeat the ALC. Key down. By the clock or watch, take notes at 1 minute intervals for, say, 10 minutes. Prepare notes into report. Submit report here. Total time to perform: 30 seconds - 30 minutes depending mostly on access to the ALC defeat. This answers all equivocations about does/doesn't; will/won't; might/mightn't; could/couldn't; or can/can't. Think about it, you have probably spent more time than that writing to and reading from this thread to no certain conclusion. 73's Richard Clark, KB7QHC |
Cecil Moore wrote:
SNIP Heh, heh, you're on your own for that discussion. I personally believe it is possible for reflected energy to wind up decreasing the power consumption from the DC source that is supplying the finals, which winds up decreasing the load on the 60 Hz power grid, which decreases the oil required from the Middle East, which tends to decrease the possibility of world war, which bodes well for the human race, but that is another thread for another time. -- 73, Cecil http://www.qsl.net/w5dxp Good engineering CAN solve the world's problems!!! Just use enough 1/4 wavelength stubs and we can totally eliminate the #1 or #2 cause of international tension!! Now, the Oil exporting states will face economic ruin! We will have to give them foreign aid!! Damn ... taxes will have to go up!! That means we have to elect a Demo Krapp as president. Cecil, I didn't know you were campaigning for what's his name!! I better be careful or my tongue may slip out of my cheek. |
Sometimes, it's useful to think about problems such as this in terms of
analogous systems that behave in similar ways. For example, imagine a gasoline-powered engine driving a water pump though a friction belt and pulley. The pump is attempting to pump water into a short plugged pipe (shorted transmission line). After the first couple of seconds the reflected water from the plugged end creates a back pressure on the pump, a situation in which the pump is then not capable of delivering additional water to that which is stored in the pipe. Now, a number of things can occur, short of mechanical failure of some part of the system: 1) the belt between the motor and pump can start to slip, dissipating the energy generated by the motor as heat lost to friction in the belt and pulley 2) the motor can slow or cease to function, incapable of working against the increased torque (impedance) created by the belt that has been constrained by the pump that can't pump water into the plugged pipe. In case 1), the power is being generated by the motor, but it is not being delivered to the load. It is being dissipated (converted to heat) `in between. In case 2), the power simply cannot be created. The motor can't work against such a load. The power is not being dissipated anywhere -- it's simply not being generated. As Cecil has suggested, in this case, the engine stops, which stops using gasoline, which reduces the demand on the Middle East oil supply, which lowers prices, which ..... Case 2) is not one in which reflections "cancel" the energy being generated, although it can be thought of that way. Energy can't be "canceled" ( a fundamental law). If a generator isn't delivering energy to a load, it's because the energy either doesn't exist (isn't being generated), or it's being rerouted, stored, or being converted to another form. Although this analogy could be complicated by making it a pumping system that generates sinusoidal water pressure waves with wavelength-related pipe lengths, it doesn't need to be for it to work. Al WA4GKQ "Cecil Moore" wrote in message ... Henry Kolesnik wrote: What if the source doesn't or can't dissipate? Heh, heh, you're on your own for that discussion. I personally believe it is possible for reflected energy to wind up decreasing the power consumption from the DC source that is supplying the finals, which winds up decreasing the load on the 60 Hz power grid, which decreases the oil required from the Middle East, which tends to decrease the possibility of world war, which bodes well for the human race, but that is another thread for another time. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Better specify the kind of pump because a centrifugal will have a different
outcome versus a positive displacment! -- 73 Hank WD5JFR "Al" wrote in message ... Sometimes, it's useful to think about problems such as this in terms of analogous systems that behave in similar ways. For example, imagine a gasoline-powered engine driving a water pump though a friction belt and pulley. The pump is attempting to pump water into a short plugged pipe (shorted transmission line). After the first couple of seconds the reflected water from the plugged end creates a back pressure on the pump, a situation in which the pump is then not capable of delivering additional water to that which is stored in the pipe. Now, a number of things can occur, short of mechanical failure of some part of the system: 1) the belt between the motor and pump can start to slip, dissipating the energy generated by the motor as heat lost to friction in the belt and pulley 2) the motor can slow or cease to function, incapable of working against the increased torque (impedance) created by the belt that has been constrained by the pump that can't pump water into the plugged pipe. In case 1), the power is being generated by the motor, but it is not being delivered to the load. It is being dissipated (converted to heat) `in between. In case 2), the power simply cannot be created. The motor can't work against such a load. The power is not being dissipated anywhere -- it's simply not being generated. As Cecil has suggested, in this case, the engine stops, which stops using gasoline, which reduces the demand on the Middle East oil supply, which lowers prices, which ..... Case 2) is not one in which reflections "cancel" the energy being generated, although it can be thought of that way. Energy can't be "canceled" ( a fundamental law). If a generator isn't delivering energy to a load, it's because the energy either doesn't exist (isn't being generated), or it's being rerouted, stored, or being converted to another form. Although this analogy could be complicated by making it a pumping system that generates sinusoidal water pressure waves with wavelength-related pipe lengths, it doesn't need to be for it to work. Al WA4GKQ "Cecil Moore" wrote in message ... Henry Kolesnik wrote: What if the source doesn't or can't dissipate? Heh, heh, you're on your own for that discussion. I personally believe it is possible for reflected energy to wind up decreasing the power consumption from the DC source that is supplying the finals, which winds up decreasing the load on the 60 Hz power grid, which decreases the oil required from the Middle East, which tends to decrease the possibility of world war, which bodes well for the human race, but that is another thread for another time. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
This place is where one can just about learn anything or nothing or learn
incorreclty. AKA the politics of reflections. -- 73 Hank WD5JFR "Richard Clark" wrote in message ... On Fri, 11 Jun 2004 18:32:14 GMT, "Henry Kolesnik" wrote: What if the source doesn't or can't dissipate? Hi Hank, Simple enough to find out. Disconnect the output of your transmitter. Set the drive level of your transmitter for full power. Defeat the ALC. Key down. By the clock or watch, take notes at 1 minute intervals for, say, 10 minutes. Prepare notes into report. Submit report here. Total time to perform: 30 seconds - 30 minutes depending mostly on access to the ALC defeat. This answers all equivocations about does/doesn't; will/won't; might/mightn't; could/couldn't; or can/can't. Think about it, you have probably spent more time than that writing to and reading from this thread to no certain conclusion. 73's Richard Clark, KB7QHC |
Is there any objective reality when it comes to reflections? How do we
really know? Henry Kolesnik wrote: This place is where one can just about learn anything or nothing or learn incorreclty. AKA the politics of reflections. |
Well perhaps I should reword my comment. It seems like many responders are
getting into the politics of reflections, some into the theory but I believe it's a science and should be explainable. Theories are used to explain what goes on inside a black box that we can't open, just an input and output. But a shorted 1/4 wave stub is about as far as you can get from a black box.. I'm curious! There are no stupid questions, just stupid people asking and that's me. -- 73 Hank WD5JFR "Dave Shrader" wrote in message news:Mp1zc.42427$HG.38082@attbi_s53... Is there any objective reality when it comes to reflections? How do we really know? Henry Kolesnik wrote: This place is where one can just about learn anything or nothing or learn incorreclty. AKA the politics of reflections. |
On Sun, 13 Jun 2004 18:54:36 GMT, Dave Shrader
wrote: Is there any objective reality when it comes to reflections? How do we really know? Hi Dave, I offered a very simple test. There is the path of an objective result, or the path of a subjective and ponderous appeal. You do have a rig, do you not? You could perform the several steps to come to a conclusion I presume - otherwise disabuse me of this talent I inferred in your behalf and state which political party you are affiliated with. 73's Richard Clark, KB7QHC |
Dave Shrader wrote:
Is there any objective reality when it comes to reflections? How do we really know? Use a TDR. Observe TV ghosting. The reflection model is perfectly valid and consistent and the outcomes using the reflection model are the same as any other valid consistent model. An S-parameter analysis uses reflection/transmission coefficients, i.e. the wave reflection model. If an S-parameter analysis is not valid, a whole @#$%-load of electronics should not be functioning as they do. I'm not entirely certain, but I believe the H-parameter analysis, the Y-parameter analysis, and the Z-parameter analysis all use the reflection model and are valid for impedance discontinuities in transmission lines. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Originally you asked:
so how does a stub work? On Sun, 13 Jun 2004 19:21:19 GMT, "Henry Kolesnik" wrote: Well perhaps I should reword my comment. snip complaint of politics But a shorted 1/4 wave stub is about as far as you can get from a black box.. I'm curious! Hi Hank, Seems like several many have offered the simple mechanics to satisfy the question. Would you like to comment why they did not? If you find the reflection based argument tedious (and it can be that in spades); then perhaps you should offer an outline of the terms to be employed or the constraints of the stub's application to reduce shot-gun answers that bloat the thread. 73's Richard Clark, KB7QHC |
Henry Kolesnik wrote:
Well perhaps I should reword my comment. It seems like many responders are getting into the politics of reflections, some into the theory but I believe it's a science and should be explainable. Theories are used to explain what goes on inside a black box that we can't open, just an input and output. But a shorted 1/4 wave stub is about as far as you can get from a black box.. I'm curious! There are no stupid questions, just stupid people asking and that's me. The solid-state political creed seeks to discredit reflections by using only the solid-state shortcut physics. These shortcuts certainly work, but they do not dictate reality. Man, not Mother Nature, takes those shortcuts. Witness the inability of anyone to present an example of the existence of standing-waves in a single source, single feedline, single mismatched load system, without the existence of a forward- traveling wave and a rearward-traveling (reflected) wave. The same solid-state political creed dictates the definition of source power. It says that if reflected power is taken in the source, then it was never generated in the first place. Never mind that energy can be proven to have made a round-trip to the load and back to the source. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote:
Dave Shrader wrote: Is there any objective reality when it comes to reflections? How do we really know? I offered a very simple test. First, you must prove that you are a member of objective reality. That's a very difficult assignment, given most of your postings. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sun, 13 Jun 2004 14:57:11 -0500, Cecil Moore
wrote: I offered a very simple test. First, you must prove that you are a member of objective reality. too simple to perform, hmm? |
Richard Clark wrote:
SNIP Hi Dave, I offered a very simple test. There is the path of an objective result, or the path of a subjective and ponderous appeal. You do have a rig, do you not? You could perform the several steps to come to a conclusion I presume - otherwise disabuse me of this talent I inferred in your behalf and state which political party you are affiliated with. 73's Richard Clark, KB7QHC Richard, I'm a retired EE and have used stubs extensively in USAF Missile systems C-band and S-band design. They work, they get warm, they increase the insertion loss in the transmission line system, they work, they work, ... We use the word 'THEORY' too loosely in ham discussions. Theory is the state of experimental verification between Hypothesis and Law. Many times we use the word 'Theory' when we should say that Physics/Mathematics explain the observation as follows: etc. In a shorted 1/4 wavelength transmission line 'stub' we have a current maximum at the physical short circuit. We have a high impedance 1/4 wavelength from the physical short circuit. The voltage and current have a sinusoidal relationship along the length of the 'stub'. There is a forward and reflected wave within the 'stub'. There are several loss components within the 'stub' including I^2*R and V^2/Rl [Capacitive dielectric losses] and 1/2L*I^2 [Leakage Inductance losses]. Therefore the presence of a stub increase the insertion loss in the transmission line sub-system; conversely, the removal of the stub reduces the insertion loss in the transmission line sub-system. In EM Physics, circa 1958-59, the equations of state for the stub included both a transient response and a steady state response. I have not solved these equations in more than 20 years [I probably forgot how to solve them in any event! ] With all this said and done I find it interesting to follow the discussions and learn many things, some of which are correct grin. |
In the absence of an observer there is no objective reality.
In the absence of an observer any reality, if it exists, is unknowable. I observe, therefore ... + + + Cecil Moore wrote: First, you must prove that you are a member of objective reality. That's a very difficult assignment, given most of your postings. :-) -- 73, Cecil http://www.qsl.net/w5dxp |
With all this said and done I find it interesting to follow the
discussions and learn many things, some of which are correct grin. ================================= How do you know you are learning anything? |
Dave Shrader wrote:
In EM Physics, circa 1958-59, the equations of state for the stub included both a transient response and a steady state response. Hey Dave, looks like we both studied the same physics. But did you know that Mother Nature changed her mind during the last 50 years? :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
I'm not entirely certain, but I believe the H-parameter analysis, the Y-parameter analysis, and the Z-parameter analysis all use the reflection model and are valid for impedance discontinuities in transmission lines. -- 73, Cecil http://www.qsl.net/w5dxp Those are all examples of two port parameters, good for your basic linear, time invariant system. (While you're at it, don't forget the transmission parameters, Cecil.) They'll work with any old lumped or distributed system as long as it's linear, has two ports, and is time invariant. You can't really use S-parameters to prove reflections exist because S-parameter theory assumes reflections exist even when they don't, as in the case of transistor characterization - much like calculating the VSWR in a zero length transmission line. They're good for analysis, though, as long as you don't get carried away and look on them as proving something about God, the Universe and Everything. 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
You can't really use S-parameters to prove reflections exist because S-parameter theory assumes reflections exist even when they don't, ... On the contrary, if one assumes same-cycle reflections, the reflection model works perfectly. Why wouldn't a 50 ohm source looking into a 100 ohm resistor suffer an immediate reflection (energy rejection)? :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
Message-id: Tdonaly wrote: You can't really use S-parameters to prove reflections exist because S-parameter theory assumes reflections exist even when they don't, ... On the contrary, if one assumes same-cycle reflections, the reflection model works perfectly. Why wouldn't a 50 ohm source looking into a 100 ohm resistor suffer an immediate reflection (energy rejection)? :-) -- 73, Cecil http://www.qsl.net/w5dxp Why wouldn't it, indeed. 73, Tom Donaly, KA6RUH |
"Tdonaly" wrote in message
... Cecil wrote, Message-id: Tdonaly wrote: You can't really use S-parameters to prove reflections exist because S-parameter theory assumes reflections exist even when they don't, .... On the contrary, if one assumes same-cycle reflections, the reflection model works perfectly. Why wouldn't a 50 ohm source looking into a 100 ohm resistor suffer an immediate reflection (energy rejection)? :-) -- 73, Cecil http://www.qsl.net/w5dxp Why wouldn't it, indeed. 73, Tom Donaly, KA6RUH if there are no components that introduce a time delay it would. so no transmission lines allowed... hence no reflections. and you are back to a simple lumped resistor problem. |
Dave wrote:
"Tdonaly" wrote: Cecil wrote, On the contrary, if one assumes same-cycle reflections, the reflection model works perfectly. Why wouldn't a 50 ohm source looking into a 100 ohm resistor suffer an immediate reflection (energy rejection)? :-) Why wouldn't it, indeed. if there are no components that introduce a time delay it would. All real-world components introduce a time delay. so no transmission lines allowed... hence no reflections. and you are back to a simple lumped resistor problem. RF energy travels at the speed of light. For the length of time it takes for the energy to reach the resistor and reflect back to the source, the source sees 50 ohms. That length of time is not zero. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Dave wrote, if there are no components that introduce a time delay it would. so no transmission lines allowed... hence no reflections. and you are back to a simple lumped resistor problem. Maybe, but don't tell that to the analytical types. Zero length transmission lines are part of the stock in trade of some people who find the concept useful. 73, Tom Donaly, KA6RUH |
Cecil wrote,
RF energy travels at the speed of light. For the length of time it takes for the energy to reach the resistor and reflect back to the source, the source sees 50 ohms. That length of time is not zero. -- 73, Cecil http://www.qsl.net/w5dxp Even though the length of time is not zero, it's close enough so that the theorems of network analysis work (at least at the lower frequencies). Of course, if you want to do a reflection analysis on a couple of resistors, each of which is only a minute fraction of a wavelength long, well, it's a free country. S-parameters assume reflections and such, and are very useful to the people who use them, but, as an intellectual tool, it doesn't matter whether there are any real reflections or not as long as the answers come out right. They don't really prove reflections in and of themselves, anyway, since it's hard to prove something based on the assumption that it's true. (I assume it, therefore it's true.) 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
Dave wrote, if there are no components that introduce a time delay it would. so no transmission lines allowed... hence no reflections. and you are back to a simple lumped resistor problem. Maybe, but don't tell that to the analytical types. Zero length transmission lines are part of the stock in trade of some people who find the concept useful. At 3 GHz, one inch of wire is close to 1/4WL. :-) 12*984'/3000 = 3.936" -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Tue, 15 Jun 2004 06:15:16 -0500, Cecil Moore
wrote: At 3 GHz, one inch of wire is close to 1/4WL At 3 GHz nobody uses wire. |
"Cecil Moore" wrote in message ... Tdonaly wrote: Dave wrote, if there are no components that introduce a time delay it would. so no transmission lines allowed... hence no reflections. and you are back to a simple lumped resistor problem. Maybe, but don't tell that to the analytical types. Zero length transmission lines are part of the stock in trade of some people who find the concept useful. At 3 GHz, one inch of wire is close to 1/4WL. :-) 12*984'/3000 = 3.936" -- 73, Cecil http://www.qsl.net/w5dxp Now I get it! (Cecil, that is) LOL Just like the Civil Defense days... "This is a test. Had it been an actual response, Cecil would have made an intelligent comment about the real question". What a "tweeker", is Cecil. What a little devil. (:-) I'm thinkn' it proves the old adage about the idle mind and a workshop.... 73, Steve P.S. Remember that sea of entropy we were supposed to drown in? Well, upon closer examination, it turns out to be ignorance. BTW... This is my comment on the general population, not any specifics here... |
Richard Clark wrote:
Cecil Moore wrote: At 3 GHz, one inch of wire is close to 1/4WL At 3 GHz nobody uses wire. The "why" of that statement proves my point. -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Tue, 15 Jun 2004 21:10:54 -0500, Cecil Moore
wrote: At 3 GHz, one inch of wire is close to 1/4WL At 3 GHz nobody uses wire. The "why" of that statement proves my point. It is Tom's match point. |
Richard Clark wrote:
Cecil Moore wrote: At 3 GHz, one inch of wire is close to 1/4WL At 3 GHz nobody uses wire. The "why" of that statement proves my point. It is Tom's match point. Doubt it. EM waves travel the same speed at 1 MHz or 3 GHZ. If you get a reflection at 3 GHz, you certainly get a reflection at 1 MHz. All the steady-state shortcuts in the world won't change that fact of physics. What's sad is you don't even realize it. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Tue, 15 Jun 2004 23:05:23 -0500, Cecil Moore
wrote: don't even realize it ;-) |
Henry, WD5JFR wrote:
"Shorts can`t dissipate power, so how does a stub work?" Henry also wrote: "I know that a shorted 1/4 wave stub exhibits a very high impedance. But for the 2nd harmonic it`s a 1/2 wave stub and exhibits a very low impedance or a short." Henry is correct. Connect a resistance directly to a transmitter. How much energy is absorbed by the resistance? Ohm`s law is the first approximation. Current is directly proportional to the applied voltage if the transmitter`s internal reistance is negligible. The shorted 1/4-wave stub exhibits an open circuit at its mouth and accepts only enough current to supply its losss which are none in the perfect stub. So, it takes no power from the transmitter afer its circulating current is etablished. At 2X the 1/4-wave freqency, the stub is 1/2 wavelengh and does not transform the short at one end to an open circuit at its other end. Instead, the 1/2-wave directly presents the short circuit at its far end. How much curent can the transmitter supply to a short circuit? It depends on the internal impedance of the transmitter. "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and Wing explains how "Even Harmonics" are suppressed by the 1/4-wave short-circuited stub on page 29. The gist is that the stub is imperfect and its resistance saps harmonic energy which is allowed into the stub by its low impedance. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
The shorted 1/4-wave stub exhibits an open circuit at its mouth and accepts only enough current to supply its losss which are none in the perfect stub. Richard, I think you would be surprised if you measured the RF current through the short at the shorted end. It will be the in-phase sum of the forward current and reflected current and is quite high. I once melted the insulation at the end of a shorted 1/4WL piece of RG8X. The heat came from high I^2*R losses at the short. The SWR inside a perfect stub is infinite. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil, W5DXP wrote:
"Richard, I think you would be surprised if ypu measured RF current through the short at the shorted end." I expect high circulating current but without loss not much current is required of the power source. In his 1955 edition on page 106 Terman says: "Thus, if the line is short-circuited at the load, then at frequencies in the vicinity of a frequency for which the length is an odd number of quarter wavelengths long, the impedance will be high and will vary with frequency in the vicinity of resonance (i.e., frequency corresponding to quarter wavelength) in exactly the same manner as does the impedance of an ordinary parallel resonant circuit. It is therefore possible to describe resonance on a transmission line in terms of impedance at resonance and the equivalent Q of the resonance curve. On page 107, Terman gives a 200 MHz example. 2-inch air-dielectric coax is used for 1/4-wave short-circiuited stubs about 15 inches long. The resonant impedance is more than 250,000 ohms with a Q of 3000. How much current flows into an impedance of more than 1/4-million ohms? Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
How much current flows into an impedance of more than 1/4-million ohms? As you know, transmission lines transform impedances. That stub likely has an SWR around 5000:1. Assuming a driving voltage of 250V, the current at the 1/4-million ohm point is about 0.001 amps. However, 1/4WL away, at the short in the 1/4WL stub, the current will be ~5 amps with a voltage of ~0.05V. That's an impedance of ~0.01 ohm. The shorted stub has transformed the impedance from 1/4-million ohms at the mouth to ~0.01 ohm at the short. That's what transmission lines do. The question is not, "How much current flows into an impedance of more than 1/4-million ohms?" The question is: How much current flows 1/4WL away from that point at the shorted end of the stub? The answer is the sum of the forward current and reflected current. The voltage at the shorted end of a 1/4WL stub is the difference between the forward voltage and the reflected voltage. The voltage at the mouth of the stub is the sum of the forward voltage and reflected voltage. The current at the mouth of the stub is the difference between the forward current and reflected current. In the example above, that difference is ~0.001 amp. The forward current flowing inside the stub is ~2.505 amps and the reflected current flowing inside the stub is ~2.495 amps. If you don't believe the above, simply measure the RF current at the shorted end of the stub. Someone modeled it the other day and even the modeling program indicated that the current was sky high at the shorted end of a 1/4WL stub. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil, W5DXP wrote:
"Someone modeled it the other day and even the modeling program indicated that the current was sky high at the shorted end of a 1/4 WL stub." Agreed that the resistance of the short is small and allows a large current even when the volts are low. The small load resistance, ZL is transformed into another resistance, ZS that is inversely proportional to ZL. Terman expresses this in equation (4-31) as: ZS = Zo squared / ZL So, the smaller ZL, the larger ZS. As ZL goes to zero, ZS goes to infinity. A high impedance means: accepts little current for a given voltage. The open end of a good 1/4-wave short-circuited stub is defined as a high impedance. So much so that it is also called a "metallic insulator". Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
A high impedance means: accepts little current for a given voltage. The open end of a good 1/4-wave short-circuited stub is defined as a high impedance. So much so that it is also called a "metallic insulator". If it was indeed a high physical impedance, like a 250K resistor, it could be removed and no much would happen at its resonant frequency. Unfortunately, it is not a physical impedance and is merely a V/I ratio, a virtual impedance. There is nothing at the mouth of the stub capable of causing reflections. All the reflections occur at the shorted end of the stub. The forward and reflected voltages add in phase and the forward and reflected currents add 180 degrees out of phase at the mouth of the stub. The ratio of forward voltage to forward current is the Z0 of the stub as is the ratio of reflected voltage to reflected current. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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