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#1
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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 |
#2
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#3
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Hi Walt,
I'm a little confused here. I hope you can straighten me out. Walter Maxwell wrote: It appears to me that even with all the successive posts on the subject of power in the standing wave, you all seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this condition results in zero power. I've always regarded a "standing wave" as being a description of the envelope caused by the interference between forward and reverse traveling waves. But you're saying there are currents and voltages "in" the standing wave. Are you referring to the total current and voltage at any point along the line? If so, why are they always in quadrature? 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. Or do you regard a line as having a "standing wave" with its own voltage and current which are different from the total V and I? If so, how do you define a "standing wave"? Are there separate equations for "standing wave" V and I that are different than for total V and I? . . . Roy Lewallen, W7EL |
#4
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Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the envelope caused by the interference between forward and reverse traveling waves. If you have "Fields and Waves ..." by Ramo and Whinnery, take a look at the equation for standing wave voltage and current on page 285 of the 2nd edition. Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz) Roy, that is NOT the equation for an envelope. -- 73, Cecil http://www.w5dxp.com |
#5
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Cecil Moore wrote:
Roy Lewallen wrote: I've always regarded a "standing wave" as being a description of the envelope caused by the interference between forward and reverse traveling waves. If you have "Fields and Waves ..." by Ramo and Whinnery, take a look at the equation for standing wave voltage and current on page 285 of the 2nd edition. Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz) Roy, that is NOT the equation for an envelope. It's too bad you don't have the foggiest notion as to just what it is the equation of. Much of the thousands of posts could have been avoided if that were the case. 73, Tom Donaly, KA6RUH |
#6
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Tom Donaly wrote:
Cecil Moore wrote: Roy Lewallen wrote: I've always regarded a "standing wave" as being a description of the envelope caused by the interference between forward and reverse traveling waves. If you have "Fields and Waves ..." by Ramo and Whinnery, take a look at the equation for standing wave voltage and current on page 285 of the 2nd edition. Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz) Roy, that is NOT the equation for an envelope. It's too bad you don't have the foggiest notion as to just what it is the equation of. Much of the thousands of posts could have been avoided if that were the case. The technical content of your posting is noted. Here is what it is the equation of: http://www.chemmybear.com/standing.html -- 73, Cecil http://www.w5dxp.com |
#7
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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 |
#8
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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 |
#9
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Roy Lewallen wrote:
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? The standing wave voltage is *ALWAYS* in quadrature with the standing wave current at all points and at all times. If the total voltage is not in quadrature with the total current, there is a traveling wave in the mixture. -- 73, Cecil http://www.w5dxp.com |
#10
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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. 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. Seems you two are arguing about two different things. If Z0 is purely resistive: Pure standing waves are *ALWAYS* in quadrature, i.e. the sine of the angle between V and I is always 1.0. Pure traveling waves are are *ALWAYS* in phase or 180 degrees out of phase, i.e. the cosine of the angle between V and I is always 1.0. In a mixed environment of standing waves and traveling waves, the angle between V and I can assume any value. -- 73, Cecil http://www.w5dxp.com |
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