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Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
"Denny" wrote in message ... Nice graphic, Cecil.. But the thread has drifted beyond recognition.. Part of the original dispute across a couple of threads as I remember it, was the contention that there is no energy contained within the reflected wave and therefore no energy contained within the standing wave, i.e. a mere artifact... I simply wanted to point out that the standing wave on a line does contain energy and it is a childishly simple exercise to prove it, therefore the reflected wave must contain energy... As far as the questioner, where does the energy go between the standing wave peaks - oy vey.... If it is a real question - as opposed to a rhetorical device which I hope was the intent - then the profound ignorance of basic physics is vastly beyond the limited space I have to go over it... See ANY introductory level, physics textbook for details... cheers ... denny the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Actually, the people who thought about waves 100 years ago knew quite well about superposition, standing waves, etc. Where did you get the idea that they didn't? 73, Tom Donaly, KA6RUH |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
Dave wrote: ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage ... Interesting statement, experiments are rather easy to do, such as this URL demonstrates: http://www.scribd.com/doc/188483/Standing-Waves-Lab Regards, JS |
Standing-Wave Current vs Traveling-Wave Current
the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri |
Standing-Wave Current vs Traveling-Wave Current
Cecil Moore wrote:
... Why is the ignorance level about traveling waves so high on this newsgroup? It's the result of those inadequate lumped circuit models. In Einsteins' spirit, let's have a real look at waves (basically the KISS rule): http://www.colorado.edu/physics/2000...ing_wave1.html you must go to the bottom of each page and click to view the next of the series. The standing wave is "driven" by the forward & reverse traveling waves, yet best thought of as being "separate in existence" (there are a total of 3 waves!) ... and can only/really exist within strict confines of design--or, resonance ... But then, this is nothing new, or, you already knew that ... I just like the way this is all presented--on those pages, or, even newbies are introduced to the depth of the argument ... Regards, JS |
Standing-Wave Current vs Traveling-Wave Current
"Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
... They are the real waves, the 'standing' ones are just figments of your imagination. ... Or, EXACTLY, in the way resonance is a figment of the imagination ... ROFLOL! JS |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
"Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH |
Standing-Wave Current vs Traveling-Wave Current
Tom Donaly wrote:
... Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH Better yet, in a mixer, where two frequencies are combined/mixed to come up with a unique third freq which is then amplified, say to a KW, and used to xmit a signal carrying data--well, this third signal is only a figment of our imagination! strange ... the power of imagination. JS |
Standing-Wave Current vs Traveling-Wave Current
"Dave" wrote in message news:O5ibj.1026$Pt6.868@trndny07... "Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. ....and Kraus, Terman, Jasik who imagined standing wave antennas? So what is the superposition of forward and reflected waves? Laying down waves? :-) Cecil, I admire your patience! Yuri |
Standing-Wave Current vs Traveling-Wave Current
John Smith wrote:
Tom Donaly wrote: ... Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH Better yet, in a mixer, where two frequencies are combined/mixed to come up with a unique third freq which is then amplified, say to a KW, and used to xmit a signal carrying data--well, this third signal is only a figment of our imagination! strange ... the power of imagination. JS Better yet, a true white noise generator where all frequencies exist simultaneously, just set your filter to the one you want ... JS |
Standing-Wave Current vs Traveling-Wave Current
Denny wrote:
Nice graphic, Cecil.. But the thread has drifted beyond recognition.. Part of the original dispute across a couple of threads as I remember it, was the contention that there is no energy contained within the reflected wave and therefore no energy contained within the standing wave, i.e. a mere artifact... Unfortunately, that goes against the distributed network reflection model that I learned at Texas A&M during the 50s. The joules/sec in any EM wave are ExB. The total joules stored in the transmission line are exactly the number of joules necessary to support the forward energy and reflected energy. I simply wanted to point out that the standing wave on a line does contain energy and it is a childishly simple exercise to prove it, therefore the reflected wave must contain energy... As far as the questioner, where does the energy go between the standing wave peaks - oy vey.... If it is a real question - as opposed to a rhetorical device which I hope was the intent - then the profound ignorance of basic physics is vastly beyond the limited space I have to go over it... See ANY introductory level, physics textbook for details... The standing wave current is the phasor sum of the forward current and reflected current. The standing wave voltage is the phasor sum of the forward voltage and reflected voltage. At the point where the standing wave current is equal to zero, the standing wave voltage is at a maximum indicating that all of the EM energy at that point is contained in the E-field. At the point where the standing wave voltage is equal to zero, the standing wave current is at a maximum indicating that all of the EM energy at that point is contained in the H-field. Standing waves are an artifact of the superposition of forward and reflected waves. Where the two E-fields cancel, the total H-field will at a maximum. Where the two H-fields cancel, the total E-field will be at a maximum. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Tom Donaly wrote:
Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. Tom, I know that superposition is a linear process. Please don't try to imply otherwise. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Yuri Blanarovich wrote:
Cecil, I admire your patience! Every time there is a new iteration on the subject of standing wave current being used to perform meaningless delay measurements through a loading coil, a few more individuals comprehend what I am saying. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
"Tom Donaly" wrote in message t... Dave wrote: "Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH ARGH! i was too nice saying that the ancient guys that started the name 'standing' waves didn't understand superposition, neither does everyone in this group! YES, superposition works in this case, why would it not work??? |
Standing-Wave Current vs Traveling-Wave Current
On Dec 22, 8:43 am, Denny wrote:
Nice graphic, Cecil.. But the thread has drifted beyond recognition.. Part of the original dispute across a couple of threads as I remember it, was the contention that there is no energy contained within the reflected wave and therefore no energy contained within the standing wave, i.e. a mere artifact... I'd suggest this is a mischaracterization of the contention. I have seen no disagreement with the notion that the line contains energy. Assertions about a lack of energy in reflected waves is not inconsistent with the line containing energy. I simply wanted to point out that the standing wave on a line does contain energy and it is a childishly simple exercise to prove it, therefore the reflected wave must contain energy... Prepare yourself to rethink this connection. As far as the questioner, where does the energy go between the standing wave peaks - oy vey.... If it is a real question - as opposed to a rhetorical device which I hope was the intent - It was not rhetorical, but an educational question that followed from the claim. With the claim that a lit flourescent bulb demonstrates the presence of energy, it is entirely reasonable to question what a dark lamp means and the original post did not suggest this understanding. then the profound ignorance There is no need to descend to the level of insult commonly used by some of the more prolific posters. of basic physics is vastly beyond the limited space I have to go over it... See ANY introductory level, physics textbook for details... -------------------------------------------------------------- Let us consider a transmission line.... There IS a voltage and current distribution on this line. For the moment attempt to forget standing waves, travelling waves, forward waves, reflected waves, .... Just that: There IS a voltage and current distribution on this line. These distributions can be expressed as functions of distance along the line and time: V(x,t) I(x,t) These are the instantaneous real voltage and current at a particular location (x) and time (t). They can be measured with a voltmeter and ammeter, though this gets more challenging at higher frequencies. Now we know from basic electricity that Power is Volts times Amps, so we have: P(x,t) = V(x,t) * I(x,t) P(x,t) is the instantaneous power at any point and time on the line. Power being the rate of energy flow, P(x,t) is the instantaneous energy flow at that point and time on the line. If you disagree with any of the above please read no further and post any objections now. Good! Agreement. So let's consider the specific example of sinusoidal signal applied to a transmission line that is open at the end. After settling, there is a voltage and current distribution on this line, but how can we describe it? Now some of you are immediately thinking "standing wave", and you'd be right. Its an excellent description, but we need to look at the details. So V(x,t) = A cos(x) cos(wt) where w is radians/second and x is measured in degrees back from the open end. Consider t=0. The spatial voltage distribution is a sinusoid with a maximum at the open end. As time advances, this spatial sinusoid drops in amplitude until the voltage everywhere on the line is 0, then the amplitude heads towards minus max. Noting that the zero crossings are always in the same place and the shape is sinusoidal leading to the name "standing wave". From a time perspective, every point on the line has a sinusoidal voltage, but the amplitude changes with position. The peaks and zero crossings occur at the same time everywhere, thus the claim that there is no phase shift as one moves down the line. The current is also a sinusoid, but shifted 90 degrees from the voltage sinusoid, thus there is a current zero where-ever there is a voltage maximum. Now power is really interesting. Recall that P(x,t) = V(x,t) * I(x,t) At certain values of t, the voltage everywhere on the line is 0, so at these times, no energy is flowing anywhere on the line. Similarly for current. And at certain positions on the line (n*180+90) the voltage is always 0, so the power is always 0 at these points. No energy is ever flowing at these points. Similarly for current at points (n*180). The sumary, being as pedantic as I can is that "standing wave" is "merely" a description of the voltage and the current on the line. And "merely" is in quotes, because it is a very powerful and useful description, but be careful not to ascribe too much to it. Always keep in mind that the "standing wave" is a description of the conditions on the line, not the creator of those conditions. Now some of you will be saying "Yes!", it IS the forward and reverse waves that create the conditions on the line. Not so fast. These too are "merely" partial descriptions that when summed, describe the conditions on the line. Very powerful and useful, but again descriptions, not creators. Some readers overuse forward and reverse waves and start ascribing power to them. These readers think that the forward wave transports energy to the end of the line which is reflected back in the reverse wave. To those readers I offer the following counter-proof.... In the setup above used for "standing waves" it can be seen that there is zero power in the line every 90 degrees back from the open end. At a zero power point, no energy is being transferred. Therefore, the forward and reverse waves can not be transferring energy across these points. Conclusion: forward and reverse waves do not always transport energy. ....Keith |
Standing-Wave Current vs Traveling-Wave Current
"Keith Dysart" wrote in message ... On Dec 22, 8:43 am, Denny wrote: Nice graphic, Cecil.. But the thread has drifted beyond recognition.. Part of the original dispute across a couple of threads as I remember it, was the contention that there is no energy contained within the reflected wave and therefore no energy contained within the standing wave, i.e. a mere artifact... I'd suggest this is a mischaracterization of the contention. I have seen no disagreement with the notion that the line contains energy. Assertions about a lack of energy in reflected waves is not inconsistent with the line containing energy. I simply wanted to point out that the standing wave on a line does contain energy and it is a childishly simple exercise to prove it, therefore the reflected wave must contain energy... Prepare yourself to rethink this connection. As far as the questioner, where does the energy go between the standing wave peaks - oy vey.... If it is a real question - as opposed to a rhetorical device which I hope was the intent - It was not rhetorical, but an educational question that followed from the claim. With the claim that a lit flourescent bulb demonstrates the presence of energy, it is entirely reasonable to question what a dark lamp means and the original post did not suggest this understanding. then the profound ignorance There is no need to descend to the level of insult commonly used by some of the more prolific posters. of basic physics is vastly beyond the limited space I have to go over it... See ANY introductory level, physics textbook for details... -------------------------------------------------------------- Let us consider a transmission line.... There IS a voltage and current distribution on this line. For the moment attempt to forget standing waves, travelling waves, forward waves, reflected waves, .... Just that: There IS a voltage and current distribution on this line. These distributions can be expressed as functions of distance along the line and time: V(x,t) I(x,t) These are the instantaneous real voltage and current at a particular location (x) and time (t). They can be measured with a voltmeter and ammeter, though this gets more challenging at higher frequencies. Now we know from basic electricity that Power is Volts times Amps, so we have: P(x,t) = V(x,t) * I(x,t) P(x,t) is the instantaneous power at any point and time on the line. Power being the rate of energy flow, P(x,t) is the instantaneous energy flow at that point and time on the line. If you disagree with any of the above please read no further and post any objections now. Good! Agreement. So let's consider the specific example of sinusoidal signal applied to a transmission line that is open at the end. After settling, there is a voltage and current distribution on this line, but how can we describe it? Now some of you are immediately thinking "standing wave", and you'd be right. Its an excellent description, but we need to look at the details. So V(x,t) = A cos(x) cos(wt) where w is radians/second and x is measured in degrees back from the open end. Consider t=0. The spatial voltage distribution is a sinusoid with a maximum at the open end. As time advances, this spatial sinusoid drops in amplitude until the voltage everywhere on the line is 0, then the amplitude heads towards minus max. Noting that the zero crossings are always in the same place and the shape is sinusoidal leading to the name "standing wave". From a time perspective, every point on the line has a sinusoidal voltage, but the amplitude changes with position. The peaks and zero crossings occur at the same time everywhere, thus the claim that there is no phase shift as one moves down the line. The current is also a sinusoid, but shifted 90 degrees from the voltage sinusoid, thus there is a current zero where-ever there is a voltage maximum. Now power is really interesting. Recall that P(x,t) = V(x,t) * I(x,t) At certain values of t, the voltage everywhere on the line is 0, so at these times, no energy is flowing anywhere on the line. Similarly for current. And at certain positions on the line (n*180+90) the voltage is always 0, so the power is always 0 at these points. No energy is ever flowing at these points. Similarly for current at points (n*180). This is where your argument falls apart. you can not apply superposition to power as it is a non-linear relationship with the voltage and current. this is where lots of the arguments on this group begin and get stuck forever. the argument you state in the above paragraph is a contradiction on the most basic level... take a step back and consider this: V(x,t)=Z0*I(x,t) which also must be true at every point on the line for the forward and reflected waves, each taken separately. so you can write equations like: Vf=Z0*If and Vr=Z0*Ir (i'll leave off the (x,t) for now) and by superposition you can also do: V=Vf+Vr I=If+Ir Which work just fine if you do fancy graphs and animations to create the illusion of 'standing' waves along the line. And as long as you keep the two equations separate everything is simple. BUT, lets look at the a place where the standing current wave is always zero and the voltage standing wave is of course a maximum. if you plug those into ohm's law to find the impedance at that point you get: Z=V/I (where V is large, and I is zero) so you get an infinite impedance. does this surprise anyone? it shouldn't, this is what the smith chart shows you should happen every half wave along the line. the result of superimposing the waves results in changes in the measured impedance as you move along the line. note this is NOT a change in Z0, that is forever a constant and property of the physical line independent of the waves imposed on it. So what does this really mean? on the surface it would tend to support the assertion that there is no power flow at the points where the current is zero, and the impedance measures is infinite. BUT there is a catch. P=VI is a basic representation for instantaneous power at a point given voltage and current, again (x,t) left off but that doesn't matter. but also you have to keep the relationship: V=Z0*I so you can rewrite the P equation 2 different ways: P=V^2/Z0 = I^2*Z0 now, obviously these have to hold for any wave that exists in the line. so for the forward wave they hold up just fine, and for the reflected wave they hold up just fine.. BUT if you look at the 'standing' wave they fall apart. i.e. for the spot where the current is zero and the voltage is a maximum you get: P=V^2/Z0 which is a large number AND P=I^2*Z0 which is a small number you can't have it both ways at one point at the same time! now, what is really happening? Given: 1. you have 2 waves. 2. these 2 waves are traveling in opposite directions. 3. each of these waves obeys ohms law. 4. the voltage and/or current of these waves obeys the superposition principle. 5. you only need to look at voltage OR current since they are linearly related to each other at every point in each wave. If you dispute any of the above, do not pass go, do not collect 200$, go back to school and take fields and waves 101 all over again. Lets think voltage waves for now, this is an arbitrary decision as noted above. and lets consider a lossless line with a short or open end so there is 100% reflection. So, as these two waves travel along the line they periodically go in and out of phase with each other, there are animations that show this very nicely. lets think about the time where the two waves completely cancel each other so the voltage along the line is zero... did the 2 waves dissappear? no, because obviously an instant later they are back out of phase and the voltage doesn't cancel on the line. is the power zero?? no, since energy can neither be created nor destroyed, it didn't just dissappear, so neither can the flow of energy that is called power. and since we know that when this voltage minimum occurs there is a current maximum in the superimposed waves the power represented by that superposition would contradict the power represented by the voltage minimum... they can't both be correct at the same time! So what do you take away from this? 1. Standing waves have no physical significance, they do not represent power or energy, they do not obey ohms law, they are ONLY a result of superposition of the voltage and/or current waves in the line. 2. can you measure standing waves? Yes, of course. that is how they got their name, you could measure them and they didn't seem to move on the line. but this is only because simple measurement tools can't distinguish the forward and reflected components that make them up. 3. if you want to talk about power and energy you MUST use the individual traveling waves. now what does that mean for this lossless line that has 100% reflection?? 1. the reflected wave magnitude is equal to the forward wave just traveling in the opposite direction. (voltage and/or current) 2. the power in the forward wave is equal to the power in the reflected wave but traveling in the opposite direction. 3. there is energy stored in the line equal to the sum of the integrated power in the two waves each taken separately. DO NOT sum them first then integrate, that is NOT a legal operation since as has been shown the superposition principle does not apply to power as it is a non-linear relationship! 4. in steady state the net power flowing past any point in the line is zero... that is, there is as much power flowing one direction as the other. ok, i'm done with my lecture... you guys can now ignore me if you want and go back to your regularly scheduled misconceptions and circular arguments. 4. |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
Let's consider your and my bank accounts. If we generated a net bank account of yours plus mine on a daily basis and reported the results at the end of the month, even though perfectly accurate, what relationship would that report bear to reality. Let us consider a transmission line.... Keith, that was the best example I have ever seen on how an over-simplified math model can confound and confuse. As long as one understands the underlying principles, a simplification can be useful. But when the simplification is presented as an example of the laws of physics when it actually violates the laws of physics, it's time to object. As a useful shortcut, there's nothing wrong with your standing- wave analysis. It's the conclusions you draw about the underlying laws of physics that violate the laws of physics. EM waves must follow certain laws of physics. 1. Being photonic in nature, they must move at the speed of light at the VF of whatever medium they are in. 2. The power density of an EM wave is ExH in joules/sec/unit-area. The voltage and current are the *result* of the E-field and the H-field. In phasor notation: V dot I = E cross H, e.g. adjusted for the unit-area of the transmission line. 3. The joules in any volume must be conserved. In a lossless transmission line, joules that have been put into the transmission line and have not left the transmission line are still in there. 4. EM waves cannot exist without momentum. The momentum in any volume must be conserved. If EM waves are confined to a volume, they are moving at the speed of light and reflecting off the physical boundaries containing the volume. Do standing waves meet the definition of EM waves? 1. Standing waves do not move at the speed of light. Therefore, standing waves are technically not EM waves. Standing waves are a simplified mathematical construct and have no stand-alone existence outside of the human mind, i.e. standing waves simply cannot exist without the underlying forward and rearward traveling waves. 2. Standing waves contain exactly the sum of the energy components of the forward and reverse traveling waves. The conservation of energy principle will not allow anything else. 3. The joules contained in the standing waves are exactly the number of joules needed to support valid measurements of forward power and reflected power. These joules are supplied by the source during the transient power-up state. 4. EM waves cannot stand still in a transmission line. Any EM energy confined to a transmission line must necessarily be moving at the speed of light in the medium. In a Z0-matched system, where no reflected energy is allowed to be incident upon the source, all EM energy confined to a lossless transmission line and contained in the standing waves must be reflecting back and forth between two reflection points (physical impedance discontinuities). About standing waves: 1. The standing wave voltage is not moving. It is oscillating in place. It is not an EM wave. It's phasor could just as easily be rotating in the opposite direction and nothing would change. 2. The standing wave current is not moving. It is oscillating in place. It is not an EM wave. It's phasor could just as easily be rotating in the opposite direction and nothing would change. 3. There is no net average energy flow in either direction. Therefore, the net power is zero. Pfor - Pref = 0 However, the forward joules passing a point in one second are still there and so are the reflected joules/sec. The *only* energy in a transmission line with standing waves is EM energy. Standing wave energy does NOT meet the definition of EM energy. Therefore, standing waves may be a useful math model but that model has a built in technical contradiction when forced upon reality. For instance, in a one-second long lossless transmission line with forward power = 200w and reflected power = 100w, the total energy contained in that line is 200+100 = 300 joules and the EM wave nature of those 300 joules has not changed because they are allocated to the standing waves. They are still moving at the speed of light in the medium, 200 joules in the forward direction and 100 joules in the rearward direction. Here is an EXCEL spread sheet of the transient build-up to steady- state when the steady-state forward power is 200 watts and the steady-state reflected power is 100 watts in a one-second long lossless transmission line. http://www.w5dxp.com/1secsgat.gif -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
... So what does this really mean? on the surface it would tend to support the assertion that there is no power flow at the points where the current is zero, and the impedance measures is infinite. BUT there is a catch. ... Only "on the surface?" Really? Huh, I like to think the "cold" and "hot spots" in a microwave oven, where these very standing waves are happening, is an absolute proof ... oh yeah, now I remember--that is only my imagination too! :-D Regards, JS |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
"Tom Donaly" wrote: Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. ARGH! i was too nice saying that the ancient guys that started the name 'standing' waves didn't understand superposition, neither does everyone in this group! YES, superposition works in this case, why would it not work??? It was a strawman, Dave. The non-linear failures/events described by Yuri are, by definition, not covered by superposition which is a linear process. If Yuri had described the superposed hot spots preceding the non- linear failures, he would have been entirely correct. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
clip .... In the setup above used for "standing waves" it can be seen that there is zero power in the line every 90 degrees back from the open end. At a zero power point, no energy is being transferred. Therefore, the forward and reverse waves can not be transferring energy across these points. Conclusion: forward and reverse waves do not always transport energy. ....Keith Hi Keith, You are basing this conclusion on the observation that Power = V*I, and because we can not detect V or I at some points in the standing wave, then V*I is zero at these points. Correct math, but wrong conclusion. What you are forgetting is that power is also found from Power = V^2/Zo and Power = I^2*Zo. More accurately, on the standing wave line, Power = (V^2 + I^2)/Zo. This is why a SWR power meter detects both current and voltage from the standing wave. This will also be true on the quarter wave stub, which is really 1/2 wave length long electrically, when you consider the time required for the wave to go from initiation to end and back to beginning point. Power is stored on the stub during the 1/2 cycle energized, and then that stored power acts to present either a high or low impedance to the next 1/2 cycle, depending upon whether the stub is shorted or open. I think you did a very good job in building your theory. It was only at the end (where I think we need to consider additional ways of measuring power) that we disagree. 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
clip...... Now we know from basic electricity that Power is Volts times Amps, so we have: P(x,t) = V(x,t) * I(x,t) P(x,t) is the instantaneous power at any point and time on the line. Power being the rate of energy flow, P(x,t) is the instantaneous energy flow at that point and time on the line. If you disagree with any of the above please read no further and post any objections now. Good! Agreement. Hi again Keith, I would suggest that you add a caveat here. The power equation is true if the measurements are across a resistance. If we are also measuring reactive power (or reflected power), then we need to account for that. 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
"Tom Donaly" wrote in message t... Dave wrote: "Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH ARGH! i was too nice saying that the ancient guys that started the name 'standing' waves didn't understand superposition, neither does everyone in this group! YES, superposition works in this case, why would it not work??? Evidently, you haven't done enough reading. Yuri is right this time. 73, Tom Donaly, KA6RUH |
Standing-Wave Current vs Traveling-Wave Current
Roger wrote:
Keith Dysart wrote: clip .... In the setup above used for "standing waves" it can be seen that there is zero power in the line every 90 degrees back from the open end. At a zero power point, no energy is being transferred. Therefore, the forward and reverse waves can not be transferring energy across these points. Conclusion: forward and reverse waves do not always transport energy. ....Keith Hi Keith, You are basing this conclusion on the observation that Power = V*I, and because we can not detect V or I at some points in the standing wave, then V*I is zero at these points. Correct math, but wrong conclusion. What you are forgetting is that power is also found from Power = V^2/Zo and Power = I^2*Zo. More accurately, on the standing wave line, Power = (V^2 + I^2)/Zo. This is why a SWR power meter detects both current and voltage from the standing wave. This will also be true on the quarter wave stub, which is really 1/2 wave length long electrically, when you consider the time required for the wave to go from initiation to end and back to beginning point. Power is stored on the stub during the 1/2 cycle energized, and then that stored power acts to present either a high or low impedance to the next 1/2 cycle, depending upon whether the stub is shorted or open. I think you did a very good job in building your theory. It was only at the end (where I think we need to consider additional ways of measuring power) that we disagree. 73, Roger, W7WKB Haste makes waste, and errors as well. The standing wave power equation is incorrect. It should read "Power = V^2 / Zo + I^2 * Zo" Sorry for any inconvenience, and for the several postings it will probably stimulate. 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
"Tom Donaly" wrote in message . net... Dave wrote: "Tom Donaly" wrote in message t... Dave wrote: "Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH ARGH! i was too nice saying that the ancient guys that started the name 'standing' waves didn't understand superposition, neither does everyone in this group! YES, superposition works in this case, why would it not work??? Evidently, you haven't done enough reading. Yuri is right this time. 73, Tom Donaly, KA6RUH Yuri is trying to say that standing waves have real power, they do not. I have shown that in my last big post on here. The one part he properly states is that the effects are due to standing wave voltage. The voltage is indeed real, as i have said. you can measure the 'standing' wave voltage, that has been known for a long time... but the effects are NOT due to power in standing waves. |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
[tons of utter crap deleted] So what do you take away from this? 1. Standing waves have no physical significance, they do not represent power or energy, they do not obey ohms law, they are ONLY a result of superposition of the voltage and/or current waves in the line. 2. can you measure standing waves? Yes, of course. that is how they got their name, you could measure them and they didn't seem to move on the line. but this is only because simple measurement tools can't distinguish the forward and reflected components that make them up. 3. if you want to talk about power and energy you MUST use the individual traveling waves. "Dave" It appears to be useful that you choose to be anonymous. Otherwise it might be embarrassing. Rather than all of that handwaving nonsense about Ohm's Law and such, why don't you show us how standing waves fail to satisfy the Maxwell equations? So what do you take away from this? Standing waves and traveling waves have equal legitimacy. There are many cases where multiple model descriptions completely capture the physical reality. There is no reason to say that one description is more fundamental than the other. 73, Gene W4SZ |
Standing-Wave Current vs Traveling-Wave Current
"Gene Fuller" wrote in message ... Dave wrote: [tons of utter crap deleted] So what do you take away from this? 1. Standing waves have no physical significance, they do not represent power or energy, they do not obey ohms law, they are ONLY a result of superposition of the voltage and/or current waves in the line. 2. can you measure standing waves? Yes, of course. that is how they got their name, you could measure them and they didn't seem to move on the line. but this is only because simple measurement tools can't distinguish the forward and reflected components that make them up. 3. if you want to talk about power and energy you MUST use the individual traveling waves. "Dave" It appears to be useful that you choose to be anonymous. Otherwise it might be embarrassing. Rather than all of that handwaving nonsense about Ohm's Law and such, why don't you show us how standing waves fail to satisfy the Maxwell equations? So what do you take away from this? Standing waves and traveling waves have equal legitimacy. There are many cases where multiple model descriptions completely capture the physical reality. There is no reason to say that one description is more fundamental than the other. 73, Gene W4SZ 'standing' voltage and current waves DO satisfy maxwell's equations, ohm's law, and superposition. you have failed completely to see my point, its power waves that are the illegitimate children and must be bansished forever. maxwell's equations are overkill for this exercise, ohm's law and the simple power equations are all that is necesary to show the inconsistency, why muddle it with details. after all if you can't understand ohms law and calculate power properly you have no hope of understanding maxwell. |
Standing-Wave Current vs Traveling-Wave Current
"Gene Fuller" wrote in message ... It appears to be useful that you choose to be anonymous. Otherwise it might be embarrassing. i prefer to limit the exposure of my email address to limit spam and direct email outside of the group. if you really want to know who i am use your great deductive skills and figure it out. |
Standing-Wave Current vs Traveling-Wave Current
Cecil Moore wrote:
[big snip] The *only* energy in a transmission line with standing waves is EM energy. Standing wave energy does NOT meet the definition of EM energy. Therefore, standing waves may be a useful math model but that model has a built in technical contradiction when forced upon reality. Cecil, Where do you get so many goofy ideas? Do you have any references at all that support your contention that standing wave energy does not meet the definition of EM energy? I have been in the wave business professionally for about 40 years, and I have read many technical papers, reference books, and text books. I have yet to encounter anything that indicated the inferior nature of standing waves in the energy community. 73, Gene W4SZ |
Standing-Wave Current vs Traveling-Wave Current
Dave wrote:
"Gene Fuller" wrote in message ... Dave wrote: [tons of utter crap deleted] So what do you take away from this? 1. Standing waves have no physical significance, they do not represent power or energy, they do not obey ohms law, they are ONLY a result of superposition of the voltage and/or current waves in the line. 2. can you measure standing waves? Yes, of course. that is how they got their name, you could measure them and they didn't seem to move on the line. but this is only because simple measurement tools can't distinguish the forward and reflected components that make them up. 3. if you want to talk about power and energy you MUST use the individual traveling waves. "Dave" It appears to be useful that you choose to be anonymous. Otherwise it might be embarrassing. Rather than all of that handwaving nonsense about Ohm's Law and such, why don't you show us how standing waves fail to satisfy the Maxwell equations? So what do you take away from this? Standing waves and traveling waves have equal legitimacy. There are many cases where multiple model descriptions completely capture the physical reality. There is no reason to say that one description is more fundamental than the other. 73, Gene W4SZ 'standing' voltage and current waves DO satisfy maxwell's equations, ohm's law, and superposition. you have failed completely to see my point, its power waves that are the illegitimate children and must be bansished forever. maxwell's equations are overkill for this exercise, ohm's law and the simple power equations are all that is necesary to show the inconsistency, why muddle it with details. after all if you can't understand ohms law and calculate power properly you have no hope of understanding maxwell. "Power waves" is a standing joke around here. The last person to seriously consider such things is Cecil, and he now denies ever saying such. 73, Gene W4SZ |
Standing-Wave Current vs Traveling-Wave Current
"Dave" wrote in message news:3Uwbj.9477$_o6.6702@trndny06... "Tom Donaly" wrote in message . net... Dave wrote: "Tom Donaly" wrote in message t... Dave wrote: "Yuri Blanarovich" wrote in message ... the REAL answer is that the 'standing' wave is a creation of experimenters 100 years ago who didn't have the impedance, current, and voltage measurement tools we have today, and didn't know of or understand superposition. 'standing' waves are nothing but a result of superposition of the forward and reflected waves, they have no physical significance beyond that. it is worthless to talk about power or energy in them since they can always be broken down into the component waves which make more sense to work with. Dave Whoa! No physical significance? Like there is no frying the Hustler loading coil from the bottom up (due to standing wave current) or corona flames from the tip (due to high SW voltage) when applying a bit of "worthless" power? Yuri not due to 'standing' waves... that is due to the superposition of the forward and reflected waves. They are the real waves, the 'standing' ones are just figments of your imagination. Superposition doesn't work in the environment Yuri described. You've been hanging around Cecil too long. 73, Tom Donaly, KA6RUH ARGH! i was too nice saying that the ancient guys that started the name 'standing' waves didn't understand superposition, neither does everyone in this group! YES, superposition works in this case, why would it not work??? Evidently, you haven't done enough reading. Yuri is right this time. 73, Tom Donaly, KA6RUH Yuri is trying to say that standing waves have real power, they do not. I have shown that in my last big post on here. The one part he properly states is that the effects are due to standing wave voltage. The voltage is indeed real, as i have said. you can measure the 'standing' wave voltage, that has been known for a long time... but the effects are NOT due to power in standing waves. So you are trying to say that there is standing wave voltage but no standing wave current and therefore no power associated with current??? OK, explain to me where I went wrong. Back to our standing wave quarter wave coil loaded antenna, aka Hustler 80m mobile whip. I understand that it is standing wave resonant antenna, with maximum current at the base and maximum voltage at the tip, in between sinusoidal distribution of them. Inserted loading coil exhibits decrease of the current along the coil, diminishing at the top, even if W8JI et other gurus do not believe so. W9UCW measured the current (standing wave) at top of the coil to be about 40 - 60% less than on the bottom. K3BU found out that when he put 800W into the antenna, the bottom of the coil started to fry the heatshrink tubing, demonstrating more power to be dissipated at the bottom of the coil, proportional to the higher current there, creating more heat and "frying power" (RxI2). This is in perfect agreement with W9UCW measurements. So the way I understand it, forward wave is reflected off the tip, reflected wave on the way back superimposes with forward wave, creates standing wave, which at any point can be measured and has current and voltage magnitudes proportional to their position on the radiator. They seem to be real current and voltage, current heats up resistance, voltage lights up the neons and power is consumed, portion is radiated. The larger the current containing portion, the better antenna efficiency. Where am I wrong? I have a hard time to swallow statement that there is no power in standing wave, when I SAW standing wave's current fry my precious coil and tip burned off with spectacular corona Elmo's fire due to standing wave voltage at the tip. Antenna (quarter wave) radiator is a standing wave circuit exhibiting the above properties, if the formulas say it ain't so...... Merry Christmas to al believers and unbelievers! Yuri, K3BU |
Standing-Wave Current vs Traveling-Wave Current
Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy across these points." A wave is defined as a progressive vibrational disturbance propagated through a medium, such as air, without progress or advance of the parts or particles themselves, as in the transmission of sound, light, and an electromagnetic field. Light, for example, is also calld luminous or radiant energy. Sound and radio waves are also examples of energy in motion. Waves in motion are transporting energy no matter how their constituents seem to add at a particular point. Best regards, Richard Harrison, KB5WZI |
Standing-Wave Current vs Traveling-Wave Current
"Yuri Blanarovich" wrote in message ... So you are trying to say that there is standing wave voltage but no standing wave current and therefore no power associated with current??? i am saying that there is standing wave voltage, and standing wave current, but there is no physical sense in multiplying them to calculate a power. hence the concept of standing power waves is meaningless. K3BU found out that when he put 800W into the antenna, the bottom of the coil started to fry the heatshrink tubing, demonstrating more power to be dissipated at the bottom of the coil, proportional to the higher current there, creating more heat and "frying power" (RxI2). This is in perfect right R*I^2 makes perfect sense. you are talking about ONLY the current standing wave which makes perfectly good sense. and the R*I^2 losses associated with it make perfect sense. BUT, resistive losses ARE NOT a result of power in the standing wave, they are resistive lossed resulting from the current. remember the initial assumptions of my analysis, a LOSSLESS line, hence there are not resistive losses. If this requirement is changed then you can start to talk about resistive heating at current maximums and all the havoc that can cause. They seem to be real current and voltage, current heats up resistance, voltage lights up the neons and power is consumed, portion is radiated. The larger the current containing portion, the better antenna efficiency. Where am I wrong? again on the voltage, it is exactly analogous to the current above. voltage waves capacitively coupled to a neon bulb are losses and not covered in my statements. But again note, that type of measurement is measuring the peak voltage of the voltage standing wave, and any power dissipated is sampled from that wave and is not a measure of power in the standing wave. I have a hard time to swallow statement that there is no power in standing wave, when I SAW standing wave's current fry my precious coil and tip burned off with spectacular corona Elmo's fire due to standing wave voltage at the tip. YOU HAVE IT RIGHT! the standing wave current causes heating. the standing wave voltage causes corona. but neither one represents POWER in the standing waves. remember the relationship that must apply within the coax. and can be extended to antennas, though it gets real messy taking into account the radiated part and the change in capacitance and inductance along the length of the radiating elements. P=V^2/Z0=I^2*Z0 now remember that you only have to look at one or the other and at all times the relationship in each wave must obey V=I*Z0. Now consider the infamous shorted coax. at the shorted end the voltage standing wave is always ZERO, by your logic the power would always be zero at that point, but how can that be?? conversly, at a point where the current standing wave is always zero there can be no power in the standing wave, but at that point the voltage is a maximum so would say the power was a maximum... an obvious contradiction. |
Standing-Wave Current vs Traveling-Wave Current
"Dave" wrote in message news:mozbj.844$si6.691@trndny08... "Yuri Blanarovich" wrote in message ... So you are trying to say that there is standing wave voltage but no standing wave current and therefore no power associated with current??? i am saying that there is standing wave voltage, and standing wave current, but there is no physical sense in multiplying them to calculate a power. hence the concept of standing power waves is meaningless. K3BU found out that when he put 800W into the antenna, the bottom of the coil started to fry the heatshrink tubing, demonstrating more power to be dissipated at the bottom of the coil, proportional to the higher current there, creating more heat and "frying power" (RxI2). This is in perfect right R*I^2 makes perfect sense. you are talking about ONLY the current standing wave which makes perfectly good sense. and the R*I^2 losses associated with it make perfect sense. BUT, resistive losses ARE NOT a result of power in the standing wave, they are resistive lossed resulting from the current. remember the initial assumptions of my analysis, a LOSSLESS line, hence there are not resistive losses. If this requirement is changed then you can start to talk about resistive heating at current maximums and all the havoc that can cause. They seem to be real current and voltage, current heats up resistance, voltage lights up the neons and power is consumed, portion is radiated. The larger the current containing portion, the better antenna efficiency. Where am I wrong? again on the voltage, it is exactly analogous to the current above. voltage waves capacitively coupled to a neon bulb are losses and not covered in my statements. But again note, that type of measurement is measuring the peak voltage of the voltage standing wave, and any power dissipated is sampled from that wave and is not a measure of power in the standing wave. I have a hard time to swallow statement that there is no power in standing wave, when I SAW standing wave's current fry my precious coil and tip burned off with spectacular corona Elmo's fire due to standing wave voltage at the tip. YOU HAVE IT RIGHT! the standing wave current causes heating. the standing wave voltage causes corona. but neither one represents POWER in the standing waves. So we have better than perpetual motion case - we can cause heating without consuming power. I better get patent for this before Artsie gets it! There must be some equilibrium somewhere. ....like, there is standing wave current, there standing wave voltage, but no standing wave and no power in it? Richard, wake up your english majorettes and snort it out :-) remember the relationship that must apply within the coax. and can be extended to antennas, though it gets real messy taking into account the radiated part and the change in capacitance and inductance along the length of the radiating elements. P=V^2/Z0=I^2*Z0 now remember that you only have to look at one or the other and at all times the relationship in each wave must obey V=I*Z0. Now consider the infamous shorted coax. at the shorted end the voltage standing wave is always ZERO, by your logic the power would always be zero at that point, but how can that be?? conversly, at a point where the current standing wave is always zero there can be no power in the standing wave, but at that point the voltage is a maximum so would say the power was a maximum... an obvious contradiction. I know one thing, when standing wave current is high, I get loses in the resistance, wires heating up - power being used. When standing wave voltage is high, I get dielectric loses, insulation heats up and melts - power being used, ergo there is a power in standing wave and is demonstrated by certain magnitude of current and voltage at particular distance and P = U x I. I don't know what you are feeding your standing wave antennas, but I am pumping power into them and some IS radiated, some lost in the resistive or dielectric loses. 73 Yuri |
Standing-Wave Current vs Traveling-Wave Current
Evidently, you haven't done enough reading. Yuri is right this time. 73, Tom Donaly, KA6RUH Whoaaaa! When wasn't Yuri right? :-))))) Yuri, da BUm |
Standing-Wave Current vs Traveling-Wave Current
"Yuri Blanarovich" wrote in message ... "Dave" wrote in message news:mozbj.844$si6.691@trndny08... "Yuri Blanarovich" wrote in message ... So you are trying to say that there is standing wave voltage but no standing wave current and therefore no power associated with current??? i am saying that there is standing wave voltage, and standing wave current, but there is no physical sense in multiplying them to calculate a power. hence the concept of standing power waves is meaningless. K3BU found out that when he put 800W into the antenna, the bottom of the coil started to fry the heatshrink tubing, demonstrating more power to be dissipated at the bottom of the coil, proportional to the higher current there, creating more heat and "frying power" (RxI2). This is in perfect right R*I^2 makes perfect sense. you are talking about ONLY the current standing wave which makes perfectly good sense. and the R*I^2 losses associated with it make perfect sense. BUT, resistive losses ARE NOT a result of power in the standing wave, they are resistive lossed resulting from the current. remember the initial assumptions of my analysis, a LOSSLESS line, hence there are not resistive losses. If this requirement is changed then you can start to talk about resistive heating at current maximums and all the havoc that can cause. They seem to be real current and voltage, current heats up resistance, voltage lights up the neons and power is consumed, portion is radiated. The larger the current containing portion, the better antenna efficiency. Where am I wrong? again on the voltage, it is exactly analogous to the current above. voltage waves capacitively coupled to a neon bulb are losses and not covered in my statements. But again note, that type of measurement is measuring the peak voltage of the voltage standing wave, and any power dissipated is sampled from that wave and is not a measure of power in the standing wave. I have a hard time to swallow statement that there is no power in standing wave, when I SAW standing wave's current fry my precious coil and tip burned off with spectacular corona Elmo's fire due to standing wave voltage at the tip. YOU HAVE IT RIGHT! the standing wave current causes heating. the standing wave voltage causes corona. but neither one represents POWER in the standing waves. So we have better than perpetual motion case - we can cause heating without consuming power. I better get patent for this before Artsie gets it! There must be some equilibrium somewhere. ...like, there is standing wave current, there standing wave voltage, but no standing wave and no power in it? Richard, wake up your english majorettes and snort it out :-) remember the relationship that must apply within the coax. and can be extended to antennas, though it gets real messy taking into account the radiated part and the change in capacitance and inductance along the length of the radiating elements. P=V^2/Z0=I^2*Z0 now remember that you only have to look at one or the other and at all times the relationship in each wave must obey V=I*Z0. Now consider the infamous shorted coax. at the shorted end the voltage standing wave is always ZERO, by your logic the power would always be zero at that point, but how can that be?? conversly, at a point where the current standing wave is always zero there can be no power in the standing wave, but at that point the voltage is a maximum so would say the power was a maximum... an obvious contradiction. I know one thing, when standing wave current is high, I get loses in the resistance, wires heating up - power being used. When standing wave voltage is high, I get dielectric loses, insulation heats up and melts - power being used, ergo there is a power in standing wave and is demonstrated by certain magnitude of current and voltage at particular distance and P = U x I. I don't know what you are feeding your standing wave antennas, but I am pumping power into them and some IS radiated, some lost in the resistive or dielectric loses. 73 Yuri you are SO CLOSE... open your eyes, turn off your preconceived notions and read what i wrote again slowly and carefully. FIRST remember the assumption was a LOSSLESS line so there are no dielectric or resistive losses. But this is only useful because it makes it easier to see that the power given by V*I in the standing waves doesn't make sense. YOU ARE CORRECT when you say that the R*I^2 loss is REAL... and YOU ARE CORRECT when you say the V^2/R loss in dielectric is REAL. Where you lose it is that the V*I for the standing waves is not correct... this is because V and I are related to each other and you can't apply superposition to a non-linear relationship. If you think you have a way to do it then please take the given conditions of a lossless transmission line, shorted at the end, in sinusoidal steady state, and write the equation for power in the standing wave at 1/4 and 1/2 a wavelength from the shorted end of the line. |
Standing-Wave Current vs Traveling-Wave Current
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Standing-Wave Current vs Traveling-Wave Current
There's been a lot of confusion between average and instantaneous power.
Let me try to clarify a little. Keith Dysart wrote: . . . Now power is really interesting. Recall that P(x,t) = V(x,t) * I(x,t) . . . This is correct. Cecil and others have often muddled things by considering only average power, and by doing this, important information is lost. (As was the case of the statistician who drowned crossing a creek whose average depth was only two feet.) Roger wrote: I would suggest that you add a caveat here. The power equation is true if the measurements are across a resistance. If we are also measuring reactive power (or reflected power), then we need to account for that. And here's where one of the common errors occurs. The fundamental equation given by Keith does "account for" reactive power. If I(t) and V(t) are sinusoidal in quadrature, for example, then V(t) * I(t) is a sinusoidal function (at twice the frequency of V or I), with zero offset. This tells us that for half the time, energy is moving in one direction, and for the other half the time, energy is moving in the opposite direction. The average power is zero, so there is no net movement of energy over an integral number of cycles. This is what is called "reactive power". On the other hand, if V(t) and I(t) are in phase, the product is again a sinusoid with twice the frequency of V or I, but this time with an offset equal to half the peak value of V times the peak value of I. What this tells us is that energy is always moving in the same direction, although its rate (the power) increases and decreases -- clear to zero, in fact, for an instant -- over a cycle. The average power equals this offset. So here the "reactive power" is zero. I've described this as energy "sloshing back and forth", which Cecil has taken great delight in disparaging. But that's exactly what it does, as the fundamental equations clearly show. When V(t) and I(t) are at other relative angles, the result will still be the sinusoid at twice the frequency of V and I, but with an amount of "DC" offset corresponding to the net or average power and therefore the average rate of energy flow. The periodic up and down cycles represent energy moving back and forth around that net value. No additional equation or correction is needed to fully describe the power or the energy flow, or to calculate "real" (average) power or "reactive power". A careful look at what the power and energy are doing on an instantaneous basis is essential to understanding what the energy flow actually is in a transmission line. Attempting to ignore this cyclic movement and looking only at average power can lead to some incorrect conclusions and the necessity to invent non-existent phenomena (such as waves bouncing off each other) in order to hold the flawed theory together. Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
Roy Lewallen wrote:
... This is correct. Cecil and others have often muddled things by considering only average power, and by doing this, important information is lost. (As was the case of the statistician who drowned crossing a creek whose average depth was only two feet.) ... This: 2. Microwave ovens use standing waves to cook food. This means that nodes, where the amplitude is zero (where the wave crosses the x-axis), remain at nearly fixed locations in the oven, and cooking won't occur at those locations. From he http://faculty.fortlewis.edu/tyler_c..._microwave.htm Now, you can argue that any damn way you wish, but "standing waves of no power" is a myth for idiots! Regards, JS |
Standing-Wave Current vs Traveling-Wave Current
"Dave" wrote in message news:UTAbj.1170$OH6.803@trndny03... "Yuri Blanarovich" wrote in message ... "Dave" wrote in message news:mozbj.844$si6.691@trndny08... "Yuri Blanarovich" wrote in message ... So you are trying to say that there is standing wave voltage but no standing wave current and therefore no power associated with current??? i am saying that there is standing wave voltage, and standing wave current, but there is no physical sense in multiplying them to calculate a power. hence the concept of standing power waves is meaningless. K3BU found out that when he put 800W into the antenna, the bottom of the coil started to fry the heatshrink tubing, demonstrating more power to be dissipated at the bottom of the coil, proportional to the higher current there, creating more heat and "frying power" (RxI2). This is in perfect right R*I^2 makes perfect sense. you are talking about ONLY the current standing wave which makes perfectly good sense. and the R*I^2 losses associated with it make perfect sense. BUT, resistive losses ARE NOT a result of power in the standing wave, they are resistive lossed resulting from the current. remember the initial assumptions of my analysis, a LOSSLESS line, hence there are not resistive losses. If this requirement is changed then you can start to talk about resistive heating at current maximums and all the havoc that can cause. They seem to be real current and voltage, current heats up resistance, voltage lights up the neons and power is consumed, portion is radiated. The larger the current containing portion, the better antenna efficiency. Where am I wrong? again on the voltage, it is exactly analogous to the current above. voltage waves capacitively coupled to a neon bulb are losses and not covered in my statements. But again note, that type of measurement is measuring the peak voltage of the voltage standing wave, and any power dissipated is sampled from that wave and is not a measure of power in the standing wave. I have a hard time to swallow statement that there is no power in standing wave, when I SAW standing wave's current fry my precious coil and tip burned off with spectacular corona Elmo's fire due to standing wave voltage at the tip. YOU HAVE IT RIGHT! the standing wave current causes heating. the standing wave voltage causes corona. but neither one represents POWER in the standing waves. So we have better than perpetual motion case - we can cause heating without consuming power. I better get patent for this before Artsie gets it! There must be some equilibrium somewhere. ...like, there is standing wave current, there standing wave voltage, but no standing wave and no power in it? Richard, wake up your english majorettes and snort it out :-) remember the relationship that must apply within the coax. and can be extended to antennas, though it gets real messy taking into account the radiated part and the change in capacitance and inductance along the length of the radiating elements. P=V^2/Z0=I^2*Z0 now remember that you only have to look at one or the other and at all times the relationship in each wave must obey V=I*Z0. Now consider the infamous shorted coax. at the shorted end the voltage standing wave is always ZERO, by your logic the power would always be zero at that point, but how can that be?? conversly, at a point where the current standing wave is always zero there can be no power in the standing wave, but at that point the voltage is a maximum so would say the power was a maximum... an obvious contradiction. I know one thing, when standing wave current is high, I get loses in the resistance, wires heating up - power being used. When standing wave voltage is high, I get dielectric loses, insulation heats up and melts - power being used, ergo there is a power in standing wave and is demonstrated by certain magnitude of current and voltage at particular distance and P = U x I. I don't know what you are feeding your standing wave antennas, but I am pumping power into them and some IS radiated, some lost in the resistive or dielectric loses. 73 Yuri you are SO CLOSE... open your eyes, turn off your preconceived notions and read what i wrote again slowly and carefully. FIRST remember the assumption was a LOSSLESS line so there are no dielectric or resistive losses. But this is only useful because it makes it easier to see that the power given by V*I in the standing waves doesn't make sense. YOU ARE CORRECT when you say that the R*I^2 loss is REAL... and YOU ARE CORRECT when you say the V^2/R loss in dielectric is REAL. Where you lose it is that the V*I for the standing waves is not correct... this is because V and I are related to each other and you can't apply superposition to a non-linear relationship. If you think you have a way to do it then please take the given conditions of a lossless transmission line, shorted at the end, in sinusoidal steady state, and write the equation for power in the standing wave at 1/4 and 1/2 a wavelength from the shorted end of the line. So let me get this straight: I describe real antenna situation, which is a standing wave circuit, with real heating of the coil, which is consuming power demonstrably proportional to the amount of standing wave current. You are not answering rest of my argument. You bring in lossless transmission line to argue that there can not be power and no standing waves. I still don't get it. 73 Yuri |
Standing-Wave Current vs Traveling-Wave Current
Roger wrote:
What you are forgetting is that power is also found from Power = V^2/Zo and Power = I^2*Zo. More accurately, on the standing wave line, Power = (V^2 + I^2)/Zo. This is why a SWR power meter detects both current and voltage from the standing wave. This will also be true on the quarter wave stub, which is really 1/2 wave length long electrically, when you consider the time required for the wave to go from initiation to end and back to beginning point. Power is stored on the stub during the 1/2 cycle energized, and then that stored power acts to present either a high or low impedance to the next 1/2 cycle, depending upon whether the stub is shorted or open. I think you did a very good job in building your theory. It was only at the end (where I think we need to consider additional ways of measuring power) that we disagree. 73, Roger, W7WKB Haste makes waste, and errors as well. The standing wave power equation is incorrect. It should read "Power = V^2 / Zo + I^2 * Zo" Sorry for any inconvenience, and for the several postings it will probably stimulate. 73, Roger, W7WKB Sorry, neither is correct. Z0 is the ratio of V to I of a traveling wave. It's not the ratio of V to I of the total of a forward and reverse wave (which has been carelessly called "the standing wave" in this thread). If you want to calculate power as V^2 / R or I^2 * R, you have to use Re(V/I) as the value of R -- then you will, as you must, get the same result using V * I, I^2 * R, or V^2 / R. As any text can tell you, the value of Z (ratio of V to I) varies along a line which has reflected waves (i.e., has a standing wave). If you use Z0 for the calculation in those cases, you get results which have no meaning or physical significance. V^2 / R + I^2 * R doesn't give power, using either Z0 or V/I as R. I'm curious as to where that equation came from or how it was derived. Roy Lewallen, W7EL |
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