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Richard Clark wrote:
Frankly, I couldn't imagine how you could possibly continue to disagree - with yourself. I don't and you don't either. You just haven't realized it yet. Exactly what is it that you think you and I disagree about? -- 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! =----- |
wrote:
So when I use my voltmeter to measure a voltage, I actually have two voltages present which sum to the reading. Yes, that's why you need a directional voltmeter or directional wattmeter to separate the forward wave from the reflected wave when reflections exist. When I do this on a D-cell, what are these two voltages? Reflected voltage is zero during DC steady-state. It would seem that this belief will lead to some serious decidability issues. If standing waves exist, there is one forward wave component and one reflected wave component. If standing waves do not exist, there is no reflected wave component. Shirley, you can tell whether standing waves exist or not so you will know whether you are dealing with one or two waves. And don't ever trust your voltmeter again. Do you use a DC voltmeter to measure AC? Do you use a 60HZ AC voltmeter to measure 1 GHz RF voltages? Do you use a 100MHz o'scope to look at a light wave? If reflections exist and you are using a non-directional voltmeter, you will read the NET voltage. USE THE APPROPRIATE MEASURING INSTRUMENTS! -- 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! =----- |
W5DXP wrote:
wrote: So when I use my voltmeter to measure a voltage, I actually have two voltages present which sum to the reading. Yes, that's why you need a directional voltmeter or directional wattmeter to separate the forward wave from the reflected wave when reflections exist. When I do this on a D-cell, what are these two voltages? Reflected voltage is zero during DC steady-state. But how did you know this little cylinder with a bump on the end produced DC? And why can't DC circuits have net voltages derived from superposition? Seems that in your model you need extra information before you know what your voltmeter is telling you. I observe that you did not provide an answer for the other examples. What about that power plug on the wall? Too difficult? Feel free to assume that you own an abstract voltmeter which will tell you the voltage as a function of time (an oscilloscope maybe?). It would seem that this belief will lead to some serious decidability issues. If standing waves exist, there is one forward wave component and one reflected wave component. If standing waves do not exist, there is no reflected wave component. Shirley, you can tell whether standing waves exist or not so you will know whether you are dealing with one or two waves. And don't ever trust your voltmeter again. Do you use a DC voltmeter to measure AC? Do you use a 60HZ AC voltmeter to measure 1 GHz RF voltages? Do you use a 100MHz o'scope to look at a light wave? Since we are discussing the world of the ideal, we can assume the existence of ideal voltmeters. If reflections exist and you are using a non-directional voltmeter, you will read the NET voltage. USE THE APPROPRIATE MEASURING INSTRUMENTS! So I will rephrase my question for your benefit: When we measure a voltage using an ideal voltmeter, how do we know if this voltage is composed of components (I generalize, since similar difficulties arise in situations without reflections) so that we can do extra measurements and computations to learn the 'real' voltages and 'powers'? If you wish, the wall outlet provides an excellent opportunity for explanation by example. ....Keith |
wrote:
But how did you know this little cylinder with a bump on the end produced DC? And why can't DC circuits have net voltages derived from superposition? Seems that in your model you need extra information before you know what your voltmeter is telling you. The extra information is that you are dealing with an RF transmission line with reflections. The reflections are somewhat equivalent to a second source. And in a two-source DC circuit, you can indeed have net voltages derived from superposition of the two source voltages. Put them in series adding and you get a voltage maximum. Put them in series subtracting and you get a voltage minimum. What about that power plug on the wall? Too difficult? Too many sources, too many feedlines, too many loads. And the wavelength is really too long to illustrate RF transmission lines. I get a wavelength just over 3000 miles, which would stretch out straight from Texas to Alaska. When we measure a voltage using an ideal voltmeter, how do we know if this voltage is composed of components ... For an RF transmission line, use a directional wattmeter to ascertain if reflected power exists. If your 50 ohm SWR meter is in a 50 ohm Z0 environment, it will indicate the presence or absence of reflected power. If you wish, the wall outlet provides an excellent opportunity for explanation by example. Nope, it doesn't. That wall outlet doesn't have one source, one feedline, and one load. If you had one 60 Hz generator hooked up to 3100 miles of lossless transmission line, you could certainly observe reflections over that one wavelength of wire. -- 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 Thu, 28 Aug 2003 05:18:37 -0500, W5DXP
wrote: Richard Clark wrote: Frankly, I couldn't imagine how you could possibly continue to disagree - with yourself. I don't and you don't either. You just haven't realized it yet. Exactly what is it that you think you and I disagree about? Hi Cecil, Most amusing your disagreeing that you don't disagree. Isn't that enough? By the way, I will adjourn to my other posting to observe the quality of your work as this infinite regress (along with the tedium of other postings to this thread) serves no purpose. See ya there? 73's Richard Clark, KB7QHC |
W5DXP wrote:
wrote: And in a two-source DC circuit, you can indeed have net voltages derived from superposition of the two source voltages. Put them in series adding and you get a voltage maximum. Put them in series subtracting and you get a voltage minimum. But power can only be computed using the resultant voltage and current (or, if you prefer, NET voltage and current). What about that power plug on the wall? Too difficult? Too many sources, too many feedlines, too many loads. And the wavelength is really too long to illustrate RF transmission lines. I get a wavelength just over 3000 miles, which would stretch out straight from Texas to Alaska. Let me make it simpler then: the lamp cord from the wall to the table lamp; lamp off and lamp on. Most lamp cord looks like twin-lead; some have said it is approximately 100 Ohms but for the purposes of discussion feel free to assume any convenient Z0 for the cord. When we measure a voltage using an ideal voltmeter, how do we know if this voltage is composed of components ... For an RF transmission line, use a directional wattmeter to ascertain if reflected power exists. If your 50 ohm SWR meter is in a 50 ohm Z0 environment, it will indicate the presence or absence of reflected power. And if it is not an RF transmission line? If it is a DC circuit? Say your phone line when on hook? If it is a lamp cord? Do we really need directional wattmeters to understand what is happening? If you wish, the wall outlet provides an excellent opportunity for explanation by example. Nope, it doesn't. That wall outlet doesn't have one source, one feedline, and one load. If you had one 60 Hz generator hooked up to 3100 miles of lossless transmission line, you could certainly observe reflections over that one wavelength of wire. Please, replace wall outlet with lamp cord to table lamp. ....Keith |
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W5DXP wrote:
wrote: Let me make it simpler then: the lamp cord from the wall to the table lamp; lamp off and lamp on. Most lamp cord looks like twin-lead; some have said it is approximately 100 Ohms but for the purposes of discussion feel free to assume any convenient Z0 for the cord. I don't think you understand the problem. That lamp cord is about 0.0000005 wavelength at 60 Hz. Any reflected wave effects are completely negligible. It's like expecting reflections due to resistor leads at HF RF. They are theoretically there but usually too small to measure. While the standing wave will be difficulty to observe, none of the equations related to ideal transmission lines require a minimum length. The reflected voltage and power can be trivially calculated just as it is for RF. Let me do it for you. 120 Volt RMS line, open circuit, assume 100 Ohms for the lamp cord: Vf = 60V RMS, Vr = 60 V RMS Pf = 36 W, Pr = 36 W And if it is not an RF transmission line? If it is a DC circuit? Say your phone line when on hook? If it is a lamp cord? The transmission line has to be a non-negligible percentage of a wavelength for one to have to switch from circuit math to transmission line math. But you can switch early if you wish and should still get reasonable results. What's the wavelength of DC? What's the wavelength of voice audio? What is the wavelength of 60 Hz? If the phone line is an appreciable percentage of a wavelength, there will be reflections (echoes) as anyone doing a transcontinental phone call can attest. Echo cancellation is big business with the telephone companies. Your theory would make telephone line echoes impossible for long unterminated lines. It is known that unterminated telephone lines result in the largest magnitude of echoes. Of course, we were discussing sinusoidal steady state for which there will be no echoes. Echoes only happen when the line energy state is changing. It just occurred to me that you can revolutionize the telephone industry if you are right. You could patent your idea of no reflected energy on a long unterminated telephone line and sell it to the telephone companies for millions of dollars. Just have every intercontinental telephone line unterminated by using high impedance receivers on each end. Since no reflected energy could cross the voltage minimum point, echoes would be eliminated without the expensive echo cancellation equipment. Unfortunately, such a situation only occurs when the line is in steady-state which means no information could be sent, which would render the line economically unviable. Please, replace wall outlet with lamp cord to table lamp. Are you capable of measuring the 0.0001 degrees of phase shift between the forward wave and reflected wave? Your measuring instrument needs to be about plus or minus 0.0000001 degrees to get an accurate measurement. Don't need to measure phase shift to obtain forward and reflected voltages and powers. But I think I understand the technique. Whenever the theory would produce answers with which you might be uncomfortable, rather than producing those anwsers and attaining a better understanding of the viability of the theory, you prune the problem space to which the theory applies. Ergo, not for DC, not for 60 Hz, not for lamp cords, not for step functions. But the math from the theory works just fine in all these situations. Try it. ....Keith |
wrote:
... assume 100 Ohms for the lamp cord: Rash assumption. Please prove that six feet of zip cord exhibits a Z0 of 100 ohms to 60Hz signals. Of course, we were discussing sinusoidal steady state for which there will be no echoes. Echoes only happen when the line energy state is changing. In reality, there is no such thing as sinusoidal steady-state and noise in the system cannot be eliminated. The noise can be used to prove that echoes (reflections) are still happening. So nothing magic happens at sinusoidal steady-state. The wave reflection model does NOT magically become dysfunctional at steady-state. But the math from the theory works just fine in all these situations. Try it. I know the math works *IF* you can define the boundary conditions. Assuming that 50 ohm coax exhibits a Z0 of 50 ohms at 60 Hz is a rash assumption. -- 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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