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Standing-Wave Current vs Traveling-Wave Current
Cecil, W5DXP wrote:
However the VARS in the standing wave require energy which can be converted to watts---." VARS is an acronym for Volt Amps Reactive. Apparent power can include real power and VARS. I would think that VARS all have volts and amps in quadrature (at 90 degrees). If so, power is VI cos theta. WI cos 90 degrees = VI (0)= 0, thus the power in VARS is 0. |
Standing-Wave Current vs Traveling-Wave Current
Richard Harrison wrote:
Roy Lewallen, W7EL wrote: "As any text can tell you, the value of Z (ratio of V to I) varies along a line which has a reflected waves (i.e., has a standing wave)." Not exactly, maybe the apparent Z. Uniform line is assumed and it has a Zo determined only by line structure. Zo is identical for a signal traveling in either direction. However, a directional coupler must be used to measure voltage and current traveling in one direction while ignoring voltage and current traveling in the opposite direction. A Bird wattmeter uses a directional coupler. I stand by my statement that Z (the ratio of V to I) varies along a line which has reflected waves. Voltage to current ratio on a line with reflections and standing waves is of little practical value except for determinimg whether the capacity of the line is exceeded or nearly so. It's of essential use in the design of stub matching, impedance transformation, and a host of other transmission line applications. A Smith chart is a tool which shows this impedance, so the impedance (voltage to current ratio) is of practical value in any application for which the Smith chart is used. . . . Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is terminated by an open, short, or purely reactive load. But not in any other case." Something else is at work. The reflection reverses direction of the wave producing a 180-degree phase shift in either voltage or current, but not both, if there is a reflection. Because the waves are traveling at the sane speed in approaching each other, they produce a phase reversal in a distance of only 90-degrees instead of 180-degrees. This places the waves in quadrature to stay. Terman shows the vector diagrams of incident and reflected waves combined to produce a voltage distribution on an almost lossless transmission line (Zo=R) for an open circuit case and for a resistive load case where the load is Zo in Fig. 4-3 on page 91 of his 1955 opus. Indeed, the angle between the incident and reflected voltages is 90-degrees in either case. In Fig. 4-4 on page 92, Terman shows voltage and current distributions produced on low-loss transmission lines by different load impedances and in every case volts and amps are in quadrature. Best regards, Richard Harrison, KB5WZI |
Standing-Wave Current vs Traveling-Wave Current
Owen Duffy wrote:
"If that is the meaning, then the situation you describe of 90-degree phase difference between E and I is rather specific, it can only occur with a distortionless line AND a load that is (s/c OR o/c OR purely reactive). No. Walter is exactly right, and don`t drag any distortionless line into the discussion. That is a device for audio circuits. For RF, you only need a low-loss (Zo = R) transmission line. Best regards, Richard Harrison, KB5WZI |
Standing-Wave Current vs Traveling-Wave Current
"Yuri Blanarovich" wrote in message ... "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 ok, last try.. YOU ARE RIGHT!!! "consuming power demonstrably proportional to the amount of standing wave current". LISTEN, YOU ARE RIGHT!!! the loss in your coil is proportional to the square of the standing wave CURRENT. as long as you keep saying that you are RIGHT. now repeat after me... the loss in your coil is proportional to the square of the standing wave CURRENT. emphasize CURRENT every time. DO NOT start talking about POWER in the standing wave, then you will be wrong. Use the CURRENT young Yuri, Use the CURRENT... forget the POWER of the standing wave. |
Standing-Wave Current vs Traveling-Wave Current
Cecil, W5DXP wrote:
"Therefore, A standing wave is not an EM wave - it is something else, by definition." Cecil hit the nail on its head. A standing wave is an interference display of two waves of the same frequency traveling in opposite directions. Best regards, Richard Harrison, KB5WZI |
Standing-Wave Current vs Traveling-Wave Current
Richard Harrison wrote:
Roy Lewallen, W7EL wrote: "Certainly, the total V and I are in quadrature if the line is terminated by an open, short, or purely reactive load. But not in any other case." Something else is at work. The reflection reverses direction of the wave producing a 180-degree phase shift in either voltage or current, but not both, if there is a reflection. Yes. Because the waves are traveling at the sane speed in approaching each other, they produce a phase reversal in a distance of only 90-degrees instead of 180-degrees. This places the waves in quadrature to stay. ?? Which waves? Forward voltage and reverse current? Forward and reverse voltage? Terman shows the vector diagrams of incident and reflected waves combined to produce a voltage distribution on an almost lossless transmission line (Zo=R) for an open circuit case and for a resistive load case where the load is Zo in Fig. 4-3 on page 91 of his 1955 opus. Indeed, the angle between the incident and reflected voltages is 90-degrees in either case. Please look carefully at those diagrams. The horizontal axis is the distance along the line. The diagrams are showing the relationship between voltage and current envelopes as a function of position. The graphs aren't showing the time phase of V and I, which is the matter under discussion. In Fig. 4-4 on page 92, Terman shows voltage and current distributions produced on low-loss transmission lines by different load impedances and in every case volts and amps are in quadrature. I don't have that diagram in my 1947 Third Edition, but I'm sure that if you'll study the diagram and accompanying text you'll find that it's also a graph of peak amplitude vs position, not V or I as a function of time. Let me pose a very simple problem. Suppose you have a quarter wavelength of 50 ohm transmission line terminated in 25 ohms and driven with a 100 volt RMS sine wave source. Consider the phase of the voltage source to be the reference of zero phase angle. 1. What is the current at the line input? [Answer: 1 ampere, at 0 degree phase] 2. What is the ratio of V to I at the line input? [Answer: 100 at an angle of zero divided by 1 at an angle of zero = 100 + j0 ohms] 3. How do you resolve this with the graphs in Terman and your explanation of the voltage and current being in quadrature everywhere along the line? Feel free to repeat this with any other line length. I'll wait very patiently for the length which produces V and I in quadrature at the input. A very interesting result of that would be that no power would be consumed from the source, so if any reaches the load then we've created power. I'd be glad to post the equation relating Z (the ratio of V to I) at the line input or any point along the line to the load and characteristic impedances. But I'm afraid it would be wasted effort, since there's a great reluctance here to actually work an equation or understand its meaning. But good and accurate graphs of what Richard has claimed the Terman graphs show (but don't) can be found in the _ARRL Antenna Book_. In the 20th and 21st Editions, the graphs are Fig. 12 on p. 24-9. In other editions, they're probably also Fig. 12 in the Transmission Lines chapter. In the graphs, the angle between the I and E vectors is the relative phase angle between the two, and also the angle of the impedance of the point where the vectors are shown. Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
On *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.- OK, that was the least word infested post I can find... 1. A standing wave is not 'standing' in time... It phase rotates at the same rate as the excitation frequency... If the phase is rotating then V and I are changing - else Feynman is rotating in his grave... To do so requires the surface electrons at that point on the line to oscillate back and forth or whatever the heck surface electrons do at rf frequency... To excite these electrons from one energy state to another requires power/energy/joules/whatever-you-want-to-call-it... 2. It is real because I can measure it with a volt meter and I can extract power from it with a lamp, simultaneously.. Also: it will perforate the insulating jacket on the line if the power level is high enough... I used to maintain a herd of 100KW RF generators, and they would blow a hole through the side of a quarter inch thick copper bar in an instant when the load failed in my youthfull ignorance, I thought it was the standing wave RATIO that blew the line, silly me ... Try this thought... A Tesla coil (automobile spark coil) with no load on the output is all standing wave voltage and no current - so according to some has no power... Touch your finger to it... Also try: If an open ended line has 100.000 volts of DC on it, the standing wave DC contains no power because the current is zero? And, for whoever it was accused me of bashing someone - reread my post, carefully.... I absolutely do not bash or flame anyone... cheers ... denny |
Standing-Wave Current vs Traveling-Wave Current
Roy Lewallen wrote:
I'm surprised that standing waves seem so hard for people to understand. They're simply a spatial pattern formed by the interference between forward and reflected waves or, if you prefer, from the solution to a general transmission line problem with boundary values applied. I'd hope that even a brief look at a text would make it clear what standing waves are, and what they are not. Roy, if you would take a "brief look at a text", you would know NOT to try to use standing wave current phase to measure the delay through a 75m loading coil. At any point in time, the standing wave current phase is essentially the same value all up and down a 1/2WL dipole including through any loading coils. Phase does NOT equate to delay in a standing wave environment. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
On Dec 24, 6:49*am, "Dave" wrote:
"Yuri Blanarovich" wrote in message ... "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 ok, last try.. *YOU ARE RIGHT!!! *"consuming power demonstrably proportional to the amount of standing wave current". *LISTEN, YOU ARE RIGHT!!! *the loss in your coil is proportional to the square of the standing wave CURRENT. *as long as you keep saying that you are RIGHT. *now repeat after me... the loss in your coil is proportional to the square of the standing wave CURRENT. emphasize CURRENT every time. *DO NOT start talking about POWER in the standing wave, then you will be wrong. *Use the CURRENT young Yuri, Use the CURRENT... forget the POWER of the standing wave.- Hide quoted text - It seems to me that a bit more precision in the use of language might help. So, strictly: It is consuming power proportional to the square of the RMS current at that point on the line, and the constant of proportionality is the resistance of the line over the length of interest. When an author writes "standing wave current", do they always mean the RMS current at a point on the line? Or do they mean the envelope of the RMS current at each point along the line? Or the envelope of the peak current at each point? Or the RMS value of the spatially distributed peak currents along the line? Or? What is THE standing wave current? No wonder there is so much dispute. And could someone who likes to write "standing wave power" (Yuri perhaps?) please provide an unambiguous definition? It does not have to be the "right" definition, or agreed by all, just any definition which is unambiguous. ...Keith |
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