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Author Topic: current and voltage along a non resonant dipole  (Read 18831 times)
W5DXP
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« Reply #15 on: November 12, 2013, 06:19:35 PM »

What would be the current distribution when the dipole is 3/4 wavelength? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?

That would be true for a straight wire but the feedpoint provides a phase shift that makes the current zero at both ends. Note that the current maximums are 1/4WL from each end. Here is the current pattern according to EZNEC:



This is a confusing and difficult concept for a lot of people, especially with all of the potential misunderstandings such as those that we have noted.

Many years ago on rec.radio.amateur.antenna, both W7EL and W8JI proved that they didn't understand the difference between standing wave current and traveling wave current. That prompted me to write the following article which explains the mathematical difference between the equation for standing wave current and the equation for traveling wave current:

http://w5dxp.com/current2.htm

The main point was that the phase of standing wave current on a standing wave antenna cannot be used to measure the velocity of propagation through a wire or through a 75m mobile loading coil as both W7EL and W8JI described and defended for their loading coil measurements.
« Last Edit: November 12, 2013, 06:37:32 PM by W5DXP » Logged
AE5EK
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« Reply #16 on: November 12, 2013, 09:47:16 PM »

This is great stuff and I appreciate your willingness to explain
I found this great graphic of standing waves.  I had never considered the difference between standing waves and the reflected waves on an antenna.
http://www.youtube.com/watch?v=ic73oZoqr70

Dennis
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AE5EK
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« Reply #17 on: November 12, 2013, 10:08:10 PM »

So it would seem that on any wire length with open ends that the voltage and current are 90 degrees out of phase. Is this correct? 

tks
Dennis
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W5DXP
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« Reply #18 on: November 13, 2013, 10:28:11 AM »

So it would seem that on any wire length with open ends that the voltage and current are 90 degrees out of phase. Is this correct?

Note that my previous postings contained the words, "pure standing wave". A pure standing wave only exists when the amplitude of the forward wave is exactly equal to the amplitude of the reflected wave. That almost never happens in the real world so almost all real-world waves are a combination of traveling waves and standing waves. A standing wave is a human mathematical construct that doesn't meet the definition of an "EM wave" because it carries no energy, no momentum, and doesn't move at the speed of light in the medium. Standing waves cannot exist without the forward and reflected traveling waves which do meet the definition of "EM waves".

"90 degrees out of phase" is only true for pure standing waves which exist only at the open-circuit ends of the dipole wire because the reflected wave is the same magnitude as the forward wave and the SWR on the standing wave antenna is (nearly) infinite at the ends of the wire. At the feedpoint of a 1/2WL dipole, the SWR on the standing wave antenna is in the ballpark of 20:1 which means that the forward power is about 20% greater than the reflected power on the antenna at the feedpoint. So about 80% of the energy is contained in the standing waves and is not radiated (until key up). The 20% of the energy that is radiated during steady-state key-down is contained in the traveling waves. On a standing wave antenna, there is (almost) always a mixture of traveling waves and standing waves which means that the angle between the total voltage and total current cannot be 90 degrees except at the open-circuits at the tip ends of the dipole.

This subject is easier to understand if one first uses lossless transmission lines as the example which eliminates the radiated power. Then stepping up to lossy transmission lines allows the radiated power from an antenna to be considered as just another loss.
« Last Edit: November 13, 2013, 10:30:50 AM by W5DXP » Logged
WS3N
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« Reply #19 on: November 13, 2013, 08:50:58 PM »

A standing wave is a human mathematical construct that doesn't meet the definition of an "EM wave" because it carries no energy, no momentum, and doesn't move at the speed of light in the medium. Standing waves cannot exist without the forward and reflected traveling waves which do meet the definition of "EM waves".

Nonsense. Why is a standing wave a mathematical construct while a traveling wave is not? Standing waves are perfectly good solutions of Maxwell's equations and are required to match certain boundary conditions. Their instantaneous energy density, energy flux and momentum flux are non-zero, while the time-averaged energy density is non-zero and the time-averaged fluxes are zero.

One can express standing waves as linear combinations of traveling waves or the other way round. The physical field is what it is and one is free to choose any decomposition that is convenient.
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RFRY
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« Reply #20 on: November 14, 2013, 04:55:49 AM »

At the feedpoint of a 1/2WL dipole, the SWR on the standing wave antenna is in the ballpark of 20:1 which means that the forward power is about 20% greater than the reflected power on the antenna at the feedpoint. So about 80% of the energy is contained in the standing waves and is not radiated (until key up).The 20% of the energy that is radiated during steady-state key-down is contained in the traveling waves.

Just to note that the maximum, free-space, far field radiated by a center-fed, 1/2-wave dipole driven with 1000 watts of Z-matched power is about 221.7 mV/m at a distance of 1 km - which is the value predicted by theory for virtually 100% radiation efficiency of that dipole.

This steady-state (key down) performance has been measured and proven for U.S. FM/TV broadcast antenna systems, whose radiated fields must meet a certain minimum value acceptable to the FCC.

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W5DXP
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« Reply #21 on: November 14, 2013, 05:47:16 AM »

Why is a standing wave a mathematical construct while a traveling wave is not?

Well for one thing, a standing wave doesn't meet the definition of a "wave". Here are a couple of quotes:

Quoting one of my college textbooks, "Electrical Communication", by Albert:

"Such a plot of voltage is usually referred to as a voltage standing wave or as a stationary wave. Neither of these terms is particularly descriptive of the phenomenon. A plot of effective values of voltage, appearing as in Fig. 6(e), is not a wave in the usual sense. However, the term "standing wave" is in widespread use."

From "College Physics", by Bueche and Hecht:

"These ... patterns are called standing waves, as compared to the propagating waves considered above. They might better not be called waves at all, since they do not transport energy and momentum."

In short, the physics definition of an EM wave requires that it transport energy and momentum and travel at the speed of light in the medium. A standing wave doesn't satisfy those terms of definition.

When we talk about standing waves, we have not broken things down to an elementary level. A pure standing wave consists of two equal amplitude coherent traveling waves propagating in opposite directions at the speed of light in the medium. They are commonly referred to as the forward wave and reflected wave and they indeed do satisfy the definition of EM waves. Since the standing wave cannot exist without the two elementary traveling waves and since it doesn't satisfy the definition of an EM wave at all and since it can be separated into its two elementary components, that's why I call it a mathematical construct.

Proof that a standing wave is not traveling at the speed of light in the medium is easy. Simply measure the phase shift from the feedpoint of a 64 ft 40m dipole to a point halfway to the tip end, i.e. 16 ft from the feedpoint. At the speed of light in the medium (about 0.94 ft per nanosecond) it would take about 15 ns to travel that 45 degrees of dipole. If we measure the total current group delay, we get about 2 degrees which is obviously unrelated to the speed of light. 16 ft in 0.7 ns is faster than the speed of light. Is the standing wave really traveling faster than the speed of light? The answer to that question is why I call it a mathematical construct because it fails to satisfy the laws of physics.

Indeed, between the nodes of a pure standing wave, there is zero phase shift. That's why a measurement of the phase shift through a loading coil in a standing wave antenna will never yield the propagation delay through the coil. A 3ns propagation delay through an 80m Texas Bugcatcher coil is impossible
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G8HQP
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« Reply #22 on: November 14, 2013, 06:33:52 AM »

Quote from: W5DXP
Proof that a standing wave is not traveling at the speed of light in the medium is easy.
. . . and quite unnecessary. The hint is in the name: 'standing wave'. It is not travelling at all.

Any linear superposition of Maxwell solutions is itself a Maxwell solution; just as physical as any other. Any argument must be about what it is called, not what it is.
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WS3N
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« Reply #23 on: November 14, 2013, 06:35:08 AM »

I see, it's the combination of "standing" and "wave" that you dislike. Call it a mode of oscillation or whatever you like, it's still the same thing.

Your example of the antenna, demonstrating the fallacy of faster-than-light propagation, is a straw man. You're the only one who suggested such a thing, immediately before calling it impossible.




sin(kx - wt) = sin(kx) cos(wt) - cos(kx) sin(wt)

Hmmm, looks like a pure traveling wave consists of two equal-amplitude standing waves, 90 degrees out of phase.  Wink
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W5DXP
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« Reply #24 on: November 14, 2013, 06:45:17 AM »

Just to note that the maximum, free-space, far field radiated by a center-fed, 1/2-wave dipole driven with 1000 watts of Z-matched power is about 221.7 mV/m at a distance of 1 km - which is the value predicted by theory for virtually 100% radiation efficiency of that dipole.

Nothing I said disagrees with that statement. After the standing wave energy has reached steady-state on a standing-wave antenna, no additional energy is required for the standing waves and the antenna radiates essentially all of the steady-state energy being delivered to the antenna. During the transient key-down state, it takes energy to charge up the antenna's standing waves. That energy is not radiated until key-up and the energy in the standing waves is approximately 4-6 times the energy being delivered to the antenna. Hence my assertion that only 20% of the energy contained in the antenna is radiated. The other 80% exists during steady-state in order to support the standing waves.

The dipole is a standing wave antenna. From key-down, it takes a certain amount of nanosecond transient time to charge up the antenna to the steady-state conditions. After the standing-wave antenna has been charged to steady-state, it is radiating all of the energy being supplied to the antenna but the energy contained in the forward waves and reflected waves on the antenna is 4-6 times that amount of energy.

Here's how I arrived at those numbers. The Z0 of a single horizontal #14 wire at a height of 30 ft above ground is 600 ohms according to the formula for single-wire transmission lines. The feedpoint impedance of the antenna is 50 ohms. The antenna element is 1/4WL long and open-circuited. It's just like a lossy transmission line open-stub problem. Given 100 watts steady-state to the stub, what is the forward power on the stub and what is the reflected power on the stub at the 50 ohm feedpoint. I get 352w forward power and 252w reflected power. It takes energy to support those forward and reflected waves and the energy in those waves is about 4-6 times the energy being applied to the stub during steady-state.


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W5DXP
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« Reply #25 on: November 14, 2013, 06:55:48 AM »

Your example of the antenna, demonstrating the fallacy of faster-than-light propagation, is a straw man. You're the only one who suggested such a thing, immediately before calling it impossible.

On the contrary, you apparently have not have seen W8JI's web page where he asserts a 3ns propagation delay through a 10" coil that has an axial propagation factor of 3 deg/inch on 80m. 30 deg in 3 ns on 80m is about seven times faster than the speed of light.

http://www.w8ji.com/inductor_current_time_delay.htm

Quote
Hmmm, looks like a pure traveling wave consists of two equal-amplitude standing waves, 90 degrees out of phase.  Wink

Thanks for supporting my assertion that it is only a mathematical construct.Smiley

The hint is in the name: 'standing wave'. It is not travelling at all.

Then, by definition, it is not a "wave" at all so the name is actually a misnomer which is what I said previously.
« Last Edit: November 14, 2013, 07:16:48 AM by W5DXP » Logged
WS3N
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« Reply #26 on: November 14, 2013, 07:48:02 AM »

Your example of the antenna, demonstrating the fallacy of faster-than-light propagation, is a straw man. You're the only one who suggested such a thing, immediately before calling it impossible.

On the contrary, you apparently have not have seen W8JI's web page where he asserts a 3ns propagation delay through a 10" coil that has an axial propagation factor of 3 deg/inch on 80m. That is certainly faster than the speed of light.

http://www.w8ji.com/inductor_current_time_delay.htm

No, I haven't. I didn't realize you were grinding an old axe.

Quote
Quote
Hmmm, looks like a pure traveling wave consists of two equal-amplitude standing waves, 90 degrees out of phase.  Wink

Thanks for supporting my assertion that it is only a mathematical construct.

Yes, both equally good, just two different ways of looking at things. I will use the tool that is suited to the task at hand.

Let's look at an example. Say the current in a wire is a "standing wave." At each point, the current will simply oscillate with a fixed amplitude and phase that depend on the position. Using what you say is the more fundamental description of its behavior,

i = sin(kx - wt) + sin(kx + wt).

If we use my "mathematical construct" we find

i = [sin(kx) cos(wt) - cos(kx) sin(wt)] + [sin(kx) cos(wt) + cos(kx) sin(wt)] = 2 sin(kx) cos(wt).

Which form more clearly shows what is the current actually doing?
« Last Edit: November 14, 2013, 08:04:42 AM by WS3N » Logged
W5DXP
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« Reply #27 on: November 14, 2013, 08:45:03 AM »

Which form more clearly shows what is the current actually doing?

Since the total current doesn't have an independent existence outside of the two component traveling-wave currents, I would say that

i = sin(kx - wt) + sin(kx + wt)

conveys more valid information closer related to a picture of the physical reality of two phasor quantities rotating in opposite directions than does the non-wave "mashed potatoes" short-cut version of the equation, an abstract mathematical construct somewhat divorced from reality. Since a standing wave has no moving parts, i.e. zero phase shift, zero transfer of energy, zero momentum, and zero velocity, how can it possibly cause additional losses on a transmission line?Smiley

Question: If the north-bound traffic on the Golden Gate Bridge equals the south-bound traffic, can we say there is zero net traffic? And since there is zero net traffic, no repairs are necessary? That's the kind of logic I have seen associated with standing waves.
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WS3N
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« Reply #28 on: November 14, 2013, 09:19:04 AM »

Since a standing wave has no moving parts, i.e. zero phase shift, zero transfer of energy, zero momentum, and zero velocity, how can it possibly cause additional losses on a transmission line?Smiley

The standing-wave current is non-zero, so there will clearly be losses if you choose to include them.

Quote
Question: If the north-bound traffic on the Golden Gate Bridge equals the south-bound traffic, can we say there is zero net traffic? And since there is zero net traffic, no repairs are necessary? That's the kind of logic I have seen associated with standing waves.

I'll agree, your logic is flawed. That analogy makes no sense.
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W5DXP
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« Reply #29 on: November 14, 2013, 09:43:45 AM »

The standing-wave current is non-zero, so there will clearly be losses if you choose to include them.

But pure standing wave current has no direction and is standing still so how can something that is not moving possibly cause losses? Could it be that the losses are actually caused by the existence and superposition of the forward and reflected waves?
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