Amplifier Efficiency >100% What do you think?

[W5DXP]

What has happened to the reflected power?

There are three things that can happen to a reflected wave when it is incident upon the source. Any and all of the following can occur in any percentage combination adding up to 100%:

1. Between 0% and 100% of the reflected wave energy is re-reflected at the source according to the rules of reflection.

2. The superposition of the forward wave with the reflected wave causes interference. Between 0% and 100% of the reflected energy is re-distributed and joins the forward wave back toward the load.

3. Between 0% and 100% of the reflected energy is dissipated in the source.

Of these 3 possibilities, #2 is the most ignored and least understood. But optical physicists understand it perfectly. I will hit the basics here:

Energy in a single transmission line is in a one-dimensional environment. Any unit of energy in a wave is traveling in one direction or the other and the principle of conservation of energy rules. Except for a few special cases, interference occurs when waves superpose at any impedance discontinuity point. Destructive interference results on one side of the impedance discontinuity while an equal magnitude of constructive interference occurs on the other side. Whatever energy is left over after the destructive interference event reverses direction and joins the constructive interference event on the other side of the discontinuity. Total energy into a discontinuity must equal the total energy out of the discontinuity. Here's an example:

100w---50 ohm coax---+====1/2WL 300 ohm feedline====50 ohm load

On the 50 ohm coax: Pfor = 100w, Pref = 0w

On the 300 ohm feedline: Pfor = 204w, Pref = 104w

The real-world reflection coefficient at the impedance discontinuity is 0.714

Destructive interference on the coax side of the discontinuity eliminates reflections on the coax. That exact magnitude of wave energy joins the forward wave on the 300 ohm feedline as constructive interference.

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If we superpose two (coherent) 100 watt out of phase RF waves, we know they cancel during the destructive interference event. We also know that energy cannot be destroyed so what happened to the 200 joules/sec? In a transmission line or inside a source, those 200 joules/sec reverse direction and become constructive interference in the opposite direction. It looks something like a re-reflection but technically, it isn't. It is a re-distribution of energy caused by superposition of coherent waves. In a one-dimensional environment, destructive interference in one direction must be matched by an equal magnitude of constructive interference in the opposite direction. For more information, here's an old Worldradio magazine article of mine:

http://www.w5dxp.com/energy.htm

http://micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html

[W5DXP]

I've been asked where one can learn about RF wave interference. The best reference I know of is Optics, by Eugene Hecht. There is an entire chapter devoted to EM wave interference. There is a good earlier chapter on superposition of EM waves.

RF waves and light waves are both EM waves, just at different frequencies. They both obey the laws of physics for EM waves including the fact that they both must move at the speed of light in the medium. Here's what Eugene Hecht had to say about standing waves:

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

[ZS6BIM]

The object of this exercise was to show that although an inline power meters display power in Watts it doesn’t directly measure power.

My understanding is - both the meters I used in my experiment measure voltage and current in the line and then use these to calculate and display power even when no power is passing through the line!
The meters are displaying the power that would be flowing in the forward or reflected direction if voltage and current happened to be in phase – not true with an open circuited load.

Makes one think, if there is a source (transmitter),  transmission line and a mismatched load do you actually get power flowing back (reflected power) towards the source?

If you are in doubt I believe the excellent & interesting article by Dr Gary Bold http://www.qsl.net/zl1an/Downloads/Bruene_explanation_V13.pdf explains all - check it out.

73
Mike
 

[W5DXP]

My understanding is - both the meters I used in my experiment measure voltage and current in the line and then use these to calculate and display power even when no power is passing through the line!

Power is a measure of the joules of energy per second passing a fixed measurement point on the line. The meters are not measuring power directly at the measurement point but are assuming a particular characteristic impedance (Z0) and then doing phasor addition and subtraction to obtain an indirect measurement of forward or reflected power. If the Z0 of the transmission line is equal to the calibration Z0, then the meters are giving an accurate reading of either forward power and/or reflected power. It is not possible for EM energy to stand still. It is not possible for an EM wave to posses zero energy. A standing "wave" does not satisfy the definition of an EM "wave". If you do a transient analysis, you will find there is exactly enough energy (joules) stored in the transmission line to support the energy in the forward and reflected waves.

As a conceptual exercise, make the transmission line one second long and lossless and track the joules in the transmission line to steady state. You will find that the source supplied two seconds of energy before the first reflections arrive. What happened to that energy that must necessarily be moving at the speed of light in the medium? Remember that energy must obey the laws of physics including the conservation of energy principle.

Pfor = Vfor*Ifor with Vfor and Ifor being in phase if Z0 is purely resistive. This is the magnitude of the Poynting vector associated with the forward power.

Pref = Vref*Iref with Vref and Iref being in phase if Z0 is purely resistive. This is the magnitude of the Poynting vector associated with the reflected power. (Ref: Fields and Waves in Communications Electronics; Ramo, Whinnery, Van Duzer; 3rd edition; page 277.)

According to the laws of physics, the forward wave and the reflected wave must necessarily be moving at the speed of light in the medium which is c times the velocity factor of the transmission line. The power in the forward wave is (Ef)x(Hf). The power in the reflected wave is (Er)x(Hr).

The net power transfer is (Efor x Hfor) - (Eref x Href)

In the case of no load and neglecting losses, those terms are equal so there is indeed no net power transfer. That does not give the forward EM wave and the reflected EM wave permission to violate the laws of physics.

If your statement, "no power is passing through the line", also means that no energy is passing through the line at the speed of light multiplied by the velocity factor, then it violates the laws of physics.

... do you actually get power flowing back (reflected power) towards the source?

Again, power doesn't flow - energy flows. Do you actually get energy flowing back (reflected waves) towards the source? Absolutely! I have already discussed the possible events that can happen to that reflected energy.

Consider these similar arguments:

If an equal number of vehicles are going north on the Golden Gate Bridge as are going south, then there are no vehicles on the bridge. Humorously false.

If an equal number of photons are going north on a transmission line as are going south, then there is no energy in the line. Also humorously false. What we can say is that there is no NET energy transfer.

[G8HQP]

The object of this exercise was to show that although an inline power meters display power in Watts it doesn’t directly measure power.

My understanding is - both the meters I used in my experiment measure voltage and current in the line and then use these to calculate and display power even when no power is passing through the line!
The meters are displaying the power that would be flowing in the forward or reflected direction if voltage and current happened to be in phase – not true with an open circuited load.

Makes one think, if there is a source (transmitter),  transmission line and a mismatched load do you actually get power flowing back (reflected power) towards the source?

On the contrary, this exercise demonstrates that the inline power meters correctly distinguish between forward and reflected power. 30W out and 30W back means net power transfer of zero, and net system efficiency of zero.

Forward current and voltage are combined to give forward power. Similarly for reverse power. As the line has 50 ohm impedance (i.e. pure resistance) the forward current and voltage will be perfectly in phase with each other, irrespective of termination. Similarly for reverse power; but forward and reverse will differ. The net voltage (sum) and net current (difference) will not necessarily be in phase everywhere except when the line is properly terminated.

[W5DXP]

The net voltage (sum) and net current (difference) will not necessarily be in phase everywhere except when the line is properly terminated.

30w of forward power and 30w of reflected power means that we are dealing with a pure standing wave. The total (net) voltage and the total (net) current are always 90 degrees out of phase at any particular point and V*I*cos(theta) is always zero because total net power equals zero, i.e. there is no NET transfer of energy. The equation for a pure standing wave is of the form:

I_(x,t) = [I_max(sin kx)](sin wt)

where 'w' is omega = 2*pi*freq and kx is the physical position measured from a reference point 'x' multiplied by 'k' to convert to degrees. Note that the physical position is part of the magnitude term and not part of the phase term. Over each 1/2WL of line (in between two nulls) the phase of the standing wave is constant and only the magnitude changes with time. Therefore, standing wave phase cannot be used to measure the delay in the line. Here are some animations of standing wave patterns. Click on "Classic Standing Wave" to observe a pure standing wave.

http://www.csupomona.edu/~ajm/materials/animations/stwaves.html

Note that the familiar textbook standing wave current plots are envelope patterns, not instantaneous current plots. As can be seen from the "Classic Standing Wave" animation, for every half-cycle of an RF standing wave, the instantaneous current magnitude at all points along the line is zero, i.e. the magnetic field is zero, and all of the energy in the entire standing wave has migrated to the electric field.

Please note that the forward traveling wave and reverse traveling wave are unaffected by the standing wave because they are the cause of the standing wave (not an effect of the standing wave). A standing wave has no existence apart from the forward and reverse traveling waves each of which must obey the laws of physics. One would be surprised at the amateur radio experts who do not comprehend that fact of physics. Incidentally, the equation for a current traveling wave is:

I_(x,t) = I_max(sin kx +/- wt) where '+/-' is "plus or minus" depending upon the direction of travel.

Note that the physical position 'x' has an effect on phase but not magnitude, just the opposite of a pure standing wave. Here is a graph that may make that clear:

http://www.w5dxp.com/travstnd.GIF

 

[G8HQP]

30w of forward power and 30w of reflected power means that we are dealing with a pure standing wave.

Yes, I should have noted that. My main point is that the forward power has voltage and current exactly in phase because the line impedance is pure resistance 50 ohms. Same for reverse power, because that also sees a 50ohm line. At any point on the line we see the superposition of forward and reverse waves. If they are equal, then as you say we get just a standing wave which is the net result of two travelling waves.

[W5DXP]

If they are equal, then as you say we get just a standing wave which is the net result of two travelling waves.

I wasn't disagreeing with you - just expanding on what you posted. I think we are both trying to point out that when one ignores the two traveling waves and tries to understand the energy system based only on standing waves, one is likely to become confused about where the component energy goes. If we know the exact source impedance encountered by the reflected waves, we can easily calculate where the energy goes. Trouble is, I'm not convinced that anyone has ever accurately measured the actual source impedance encountered by the reflected waves.