LINES THEY ARE ALWAYS MATCHED WITH LOADS!

[PY1LL]

This happens to real lines and with not resistive loads too, but the analysis is only a bit more complicated.
Suppose we have these hypotheses:
1 - the line is ideal
2 - the load is resistive
3 - the generator is perfectly matched with the line
With these assumptions, we see that all the generated power is delivered to the load, no matter which the load mismatch is. This because, if all the power from the generator is delivered with no return to the line, as the latter is ideal, that power can be dissipated only in the load, with no considerations about the reflected wave, that is, independently of VSWR
We can see this from another point of view. Let the load value R be different from the cable surge impedance Zo. The forward wave has its maximum voltage or current values as Vf or If, with the valid ratio Vf÷If = Zo. But on the resistor R, V÷I = R, due the Ohm law that always holds. Suppose Zo > R (a similar consideration can be done for the opposite case). As the cable is connected to the resistor, it has to decrease its voltage and increase its current to satisfy the resistor condition. Then Vf becomes Vf - Vr and If becomes If + Ir such that (Vf - Vr) ÷ (If + Ir) = R. So, the voltage Vr has the opposite sign of Vf and Ir has the same sign as If.
When we change the signal only of the voltage (or only of the current) in a cable, we really have a wave traveling in the opposite direction of the original one. This means that Vr and Ir represent the voltage and the current in the load end of the cable of a returning wave. So, the mismatch between Zo and R, results in a reflected wave from the load to the generator. As the traveling impedance of any wave in the cable is Zo, we have the relationships:
(a) (Vf - Vr) ÷ (If + Ir) = R, to be possible the maximum power transfer.
(b) Vf ÷ If = Vr ÷ Ir = Zo, both waves are independent and travel under a surge impedance Zo.
Now consider the load end of the cable as a generator connected to the load R.
Which are the impedances of the generator and of the load? Both are R, because of the relationship (a) and, thus, we have a perfect match between the cable and the load.
We continue having a mismatch between Zo and R (or between the forward wave and the load), but the line, now, is perfectly matched to the load. The cable is an ideal impedance transformer (between the generator and the load) and, to do so, the cable adjusts itself via a reflected wave to guarantee the correct match to the load (remember to keep the generator well matched to the line).
The forward power is given by Pf = Vf² ÷ Zo (or R x If²), the reflected one by Pr = Vr² ÷ Zo (or by R x Ir²) and the transmitted one by Pt = Vt² ÷ R (or R x It²), where Vt and It are the voltage and current on the load R. Those three powers are positive numbers. As Pf = Pr + Pt (the total power arriving to the cable end is equal to the total power leaving it, with Pf arriving and Pr and Pt leaving), we see that Pf > Pt. As we saw, Pt = Pg (the generated power) and the forward power is greater than the generated one (for example, the generator delivers 100 W to the ideal cable that creates a forward power of 120 W, with a returned power of 20 W, the load receiving just 120 - 20 = 100 W, that is, just the generated power)!.
Is it shocking? No, because the source of the forward power is not only the generated, but also the reflected one. As the total arriving power is equal to the leaving one also on the generator end of the cable, that is, Pg + Pr = Pf (Pg and Pr arrive and Pf leaves that point), we see that just the difference between Pf and Pg is guaranteed by Pr. Pr, conversely, comes from the same Pf at the load end of the cable. Thus, there is no power creation, with only a 'circulation' of the excess power.
This suggests that the cable is 'filled' with power waves (Pf and Pr), so that the generated power feeds the load via part of the forward power. The other part of it is used to create Pr only to be possible the total delivery of power to the load, as the cable is ideal.
Thus, in spite of the fact that Zo is mismatched to R, the cable itself is perfectly MATCHED to the load.
The load may be a resistor, a loudspeaker or an antenna.
Taking the real losses into account or an eventual load reactance, the calculi only become more complicated, but the involved concepts are the same.
The reflected power IS NOT the loss itself, as we saw in the ideal cable case. VSWR only enhances an ALREADY EXISTING cable loss (the loss increases with VSWR in a lossy cable, so VSWR must be always minimized in the real world).

Luiz - PY1LL

[kl7aj]

Good job!

Of course, this was the original assertion of W2DU in his classic "REFLECTIONS" series. But we can never be reminded too often about these truths!

Thanks!

eric

[n6ajr]

OR is a few words. all the power goes some where, either out the load, or as heat, even the reflections, it all goes somewhere..

[K5USS]

As a VERY wise man once told me after I bragged about my latest antenna venture being a perfect 1:1 as soon as I put it up, "a dummy load has a VSWR of 1:1, what is your point?"...Something I think about regularly when building an antenna and it turns out "flat" the first time!

Charlie
K5USS

[K5MDM]

If pie are square, am cornbread then round?javascript:emoticon('http://www.qrz.com/iB_html/non-cgi/emoticons/tounge.gif')
smilie

[N2EY]

Quote[/b] (PY1LL @ Feb. 17 2006,06:48)]
Quote[/b] ]This happens to real lines and with not resistive loads too, but the analysis is only a bit more complicated.


Not really. See below.

Quote[/b] ]Suppose we have these hypotheses:
1 - the line is ideal
2 - the load is resistive
3 - the generator is perfectly matched with the line
With these assumptions, we see that all the generated power is delivered to the load, no matter which the load mismatch is. This because, if all the power from the generator is delivered with no return to the line, as the latter is ideal, that power can be dissipated only in the load, with no considerations about the reflected wave, that is, independently of VSWR


That's true because the line is ideal and the generator is perfectly matched to the line through some sort of ideal network.

In the real world things aren't so ideal.

Quote[/b] ]We can see this from another point of view. Let the load value R be different from the cable surge impedance Zo. The forward wave has its maximum voltage or current values as Vf or If, with the valid ratio Vf÷If = Zo. But on the resistor R, V÷I = R, due the Ohm law that always holds. Suppose Zo > R (a similar consideration can be done for the opposite case). As the cable is connected to the resistor, it has to decrease its voltage and increase its current to satisfy the resistor condition. Then Vf becomes Vf - Vr and If becomes If + Ir such that (Vf - Vr) ÷ (If + Ir) = R. So, the voltage Vr has the opposite sign of Vf and Ir has the same sign as If.
When we change the signal only of the voltage (or only of the current) in a cable, we really have a wave traveling in the opposite direction of the original one. This means that Vr and Ir represent the voltage and the current in the load end of the cable of a returning wave. So, the mismatch between Zo and R, results in a reflected wave from the load to the generator. As the traveling impedance of any wave in the cable is Zo, we have the relationships:
(a) (Vf - Vr) ÷ (If + Ir) = R, to be possible the maximum power transfer.
(b) Vf ÷ If = Vr ÷ Ir = Zo, both waves are independent and travel under a surge impedance Zo.
Now consider the load end of the cable as a generator connected to the load R.
Which are the impedances of the generator and of the load? Both are R, because of the relationship (a) and, thus, we have a perfect match between the cable and the load.
We continue having a mismatch between Zo and R (or between the forward wave and the load), but the line, now, is perfectly matched to the load. The cable is an ideal impedance transformer (between the generator and the load) and, to do so, the cable adjusts itself via a reflected wave to guarantee the correct match to the load (remember to keep the generator well matched to the line).
The forward power is given by Pf = Vf² ÷ Zo (or R x If²), the reflected one by Pr = Vr² ÷ Zo (or by R x Ir²) and the transmitted one by Pt = Vt² ÷ R (or R x It²), where Vt and It are the voltage and current on the load R. Those three powers are positive numbers. As Pf = Pr + Pt (the total power arriving to the cable end is equal to the total power leaving it, with Pf arriving and Pr and Pt leaving), we see that Pf > Pt. As we saw, Pt = Pg (the generated power) and the forward power is greater than the generated one (for example, the generator delivers 100 W to the ideal cable that creates a forward power of 120 W, with a returned power of 20 W, the load receiving just 120 - 20 = 100 W, that is, just the generated power)!.
Is it shocking? No, because the source of the forward power is not only the generated, but also the reflected one. As the total arriving power is equal to the leaving one also on the generator end of the cable, that is, Pg + Pr = Pf (Pg and Pr arrive and Pf leaves that point), we see that just the difference between Pf and Pg is guaranteed by Pr. Pr, conversely, comes from the same Pf at the load end of the cable. Thus, there is no power creation, with only a 'circulation' of the excess power.
This suggests that the cable is 'filled' with power waves (Pf and Pr), so that the generated power feeds the load via part of the forward power. The other part of it is used to create Pr only to be possible the total delivery of power to the load, as the cable is ideal.
Thus, in spite of the fact that Zo is mismatched to R, the cable itself is perfectly MATCHED to the load.
The load may be a resistor, a loudspeaker or an antenna.


In other words, all the power winds up in the load because there's nowhere else for it to go. The generator is ideal and matched, so it can't dissipate any power. The line is ideal and lossless, so it can't dissipate any power. The load is the only thing left.

Quote[/b] ]Taking the real losses into account or an eventual load reactance, the calculi only become more complicated, but the involved concepts are the same.
The reflected power IS NOT the loss itself, as we saw in the ideal cable case. VSWR only enhances an ALREADY EXISTING cable loss (the loss increases with VSWR in a lossy cable, so VSWR must be always minimized in the real world).


No.

In the real world, the loss can come from the line itself or from the network(s) used for matching.

Depending on the situation, the loss added by SWR and matching networks maybe considerable, or it may be negligible. A low-loss line operated with high SWR and a matching network can have less loss than a higher-loss line operated with unity SWR and no matching networks.

It all depends on the situation.

73 de Jim, N2EY

[pp8kwa]

Olá Luiz.

Se você tem uma versão em Português??? Gostaria de lê-la.

Frederico PP8KWA
http://www.pp8kwa.hpg.com.br
pp8kwa@hotmail.com

[kf4vgx]

Quote[/b] (pp8kwa @ Feb. 17 2006,06:56)]Olá Luiz.

Se você tem uma versão em Português??? Gostaria de lê-la.

Frederico PP8KWA
http://www.pp8kwa.hpg.com.br
pp8kwa@hotmail.com
I agree http://www.qrz.com/iB_html/non-cgi/emoticons/biggrin.gif .

[PY1LL]

The important things I tried to call attention to are just the facts that, in an ideal line well matched to the generator,
1 - no matter the VSWR is, the forward power is greater than the delivered by the generator.
2 - the line is always matched to the load, using the reflected wave to perform that match. The surge impedance of the line can be mismached to the load, but NOT the line, because the 'output' impedance of the line is always equal to the load R.

I said this also occurs to lossy lines, because, the ratio V/I is also the same on the resistor or on the load end of the cable, as the ideal case. Clearly, not all generated power is delivered to the load, but the match is still perfect.
These concepts, unfortunately, are not generally accepted by our ham community, generating much controversy.

Para você Frederico, a versão em português está em luizamaral.tecnociencia.com.br
http://www.qrz.com/iB_html/non-cgi/emoticons/biggrin.gif

[kl7aj]

Quote[/b] (PY1LL @ Feb. 18 2006,10:02)]The important things I tried to call attention to are just the facts that, in an ideal line well matched to the generator,
1 - no matter the VSWR is, the forward power is greater than the delivered by the generator.
2 - the line is always matched to the load, using the reflected wave to perform that match. The surge impedance of the line can be mismached to the load, but NOT the line, because the 'output' impedance of the line is always equal to the load R.

I said this also occurs to lossy lines, because, the ratio V/I is also the same on the resistor or on the load end of the cable, as the ideal case. Clearly, not all generated power is delivered to the load, but the match is still perfect.
These concepts, unfortunately, are not generally accepted by our ham community, generating much controversy.

Para você Frederico, a versão em português está em luizamaral.tecnociencia.com.br
http://www.qrz.com/iB_html/non-cgi/emoticons/biggrin.gif
I always use TWO directional wattmeters to demonstrate this at ham club meetings. It's amazing how many hams don't believe the forward power can be greater than the transmitter power, even when they see it for themselves. Of course, if you subtract the reverse power from the forward power, the numbers all work out okay. http://www.qrz.com/iB_html/non-cgi/emoticons/smile.gif

And yet the myths persist. *sigh


eric

[KD7WHQ]

Then there is the bane of the power companies; the massively reactive load. These do exist in the realm of power generation and distribution.

The change in phase requires intervention through several means, which, is dependant entirely on circumstance, to prevent the load from affecting the entire system.

And, the equipment owner pays for it...usually.

Large industries are generally the ones creating this scenario, but.

Tesla had his day ;)

[K7JEM]

This whole thing is a confusement to me. I think something is being lost in the translation.

Eric- how are you measuring "forward" power and "reflected" power, and how is it that the reflected power acts as "negative" power?

I get more baffled with each post.

Joe

[W5DXP]

Quote[/b] (PY1LL @ Feb. 18 2006,11:02)]The important things I tried to call attention to are just the facts that, in an ideal line well matched to the generator,
1 - no matter the VSWR is, the forward power is greater than the delivered by the generator.
2 - the line is always matched to the load, using the reflected wave to perform that match. The surge impedance of the line can be mismached to the load, but NOT the line, because the 'output' impedance of the line is always equal to the load R.
I recently wrote an energy analysis article about this topic that was published in "WorldRadio". It is available on my web page at: http://www.qsl.net/w5dxp/energy.htm

The key to what you are saying is to achieve a 50 ohm Z0-match where no reflections reach the generator, i.e. (Vf+Vr)/(If+Ir) = 50 ohms; Vr is 180 degrees out of phase with Vf and Ir is in phase with If. There is an easy-to-understand example at the end of my article.

[W8JI]

Quote[/b] (K7JEM @ Feb. 18 2006,21:48)]This whole thing is a confusement to me. I think something is being lost in the translation.

Eric- how are you measuring "forward" power and "reflected" power, and how is it that the reflected power acts as "negative" power?

I get more baffled with each post.

Joe
I agree with you, something is being lost in the translation. This is common with abstract technical ideas. Even two experts with the correct idea of how things work, both with the same primary technical language, often disagree about the words used.

As with many things in life, there are multiple ways of looking at this problem. All can be correct. All can be misused.

Try looking at it this way....

When the SWR on a transmission line is not unity, reactance causes voltage and current to not appear in step with current peaking in step with voltage. This effect allows the product of voltage times current to exceed the real "work power". In other words a 100 watt transmitter might be able to produce 1000 volts and also 1 ampere into a load, if the voltage was read on a meter and current read on a meter. The VAR (volt-amperes reactive) power can be 1000 watts, or nearly any other amount more than 100 watts, if reactance is present at that point in the system. This can occur with a real power of only 100 watts, becuase the voltage and current are not in step.

The real power that can actually do work is only 100 watts, but the voltage that causes a capacitor to fail or the current that causes a conductor to overheat might both be much higher than we think.

Real power that does work or makes heat is still 100 watts, if that's all the source can deliver, but the reactive power can make it look like power is very high.

The base impedance of an electrically short mobile whip is one example. The whip radiation resistance, conductor, and ground losses might total 50 ohms. We would have a real current of 1.414 amperes at 70.7 volts RMS into that resistance (and real power would be 1.414 times 70.7= 100 watts), but with the reactance voltage can be thousands of volts. Just beyond a feedpoint loading coil with negligible stray capacitance current would still be 1.414 amperes and voltage might be 2000 volts. Current and voltage are no longer in phase. The reactive power is 2828 watts, but the real power useful for things like heat or radiation is still 100 watts.

[VK1OD]

Some comments:

1. It seems that the writer describes a scenario where the source exhibits an Thevenin equivalent source impedance equal to Zo and Rload. Under that situation, half the power from the Thevinin equivalent generator is lost in the source's internal resistance, and the other half in the load resistance. This issue may be one of semantics about the meaning of the term "generator". But then the text seems to treat the generator as some kind of generator that delivers a constant power to any load (so it cannot be characterised by a Thevenin or Norton equivalent).

2. The power flow at a point in a line is the product of the time varying voltage at a point (x) and the time varying current at that point, ie v(x)*i(x). If v(x)=vf(x)+vr(x) and i(x)=if(x)+ir(x), then the product has four terms. The discussion refers to one of the four terms as Pf, another as Pr, and ignores the other two terms (the cross products) without explaining why it does both.

The "magic" being discussed (as to how a 100W generator can "create" 120W of forward power) raises in my mind a question of the legitimacy of considering the forward and reflected waves to be separate "power waves", ie the labelling of Vf*If as Pf and Vr*Ir as Pr (even with lossless lines).

Owen

[W8JI]

Quote[/b] (VK1OD @ Feb. 19 2006,19:49)]Some comments:

1. It seems that the writer describes a scenario where the source exhibits an Thevenin equivalent source impedance equal to Zo and Rload. Under that situation, half the power from the Thevinin equivalent generator is lost in the source's internal resistance, and the other half in the load resistance. This issue may be one of semantics about the meaning of the term "generator". But then the text seems to treat the generator as some kind of generator that delivers a constant power to any load (so it cannot be characterised by a Thevenin or Norton equivalent).

2. The power flow at a point in a line is the product of the time varying voltage at a point (x) and the time varying current at that point, ie v(x)*i(x). If v(x)=vf(x)+vr(x) and i(x)=if(x)+ir(x), then the product has four terms. The discussion refers to one of the four terms as Pf, another as Pr, and ignores the other two terms (the cross products) without explaining why it does both.

The "magic" being discussed (as to how a 100W generator can "create" 120W of forward power) raises in my mind a question of the legitimacy of considering the forward and reflected waves to be separate "power waves", ie the labelling of Vf*If as Pf and Vr*Ir as Pr (even with lossless lines).

Owen
Owen,

Thevinen and Norton equivalents were never meant to deal with anything going on in the source, except transfer of maximum available energy.

Consider a water wheel with water flowing driving a high efficiency generator. Let the water pressure limit power. When the generator is delivering maximum available power to the load, it is not dissipating an equal amount of power.

The "resistance" in the model is very often not a dissipative resistance, it simply represents a limited source. Most sources also only behave like a Norton or Thevenin equivalent over a very narrow range of output levels.

Those models are NOT able to let us look at anything going on inside the source. They are only to let us "black box" the source under certain operating conditions.

Let's not drag the behavior of the source into this, especially a non-linear source like a power amplifier!

It will go on forever and ever and confuse everyone!!

I do agree the power model is incomplete as written, because it assume the source always delivers 100 watts regardless of load impedance. That could happen though if the source had a matching network on the output, and so I assumed the author intended it to have one.

73 Tom

[W5DXP]

Quote[/b] (w8ji @ Feb. 19 2006,13:06)]Just beyond a feedpoint loading coil with negligible stray capacitance current would still be 1.414 amperes and voltage might be 2000 volts.
The magnitude of the net current is NOT the same on both sides of a loading coil. This is explained on my web page at: http://www.qsl.net/w5dxp/current.htm at the bottom of the page.

Such an antenna is a standing-wave antenna and has a forward current and reflected current flowing through the coil. The phasor sum of those currents is the net current, i.e. (Inet = Ifor + Iref). Here are the laws of physics being obeyed by the antenna currents.

The forward current is approximately the same magnitude on both sides of the coil but suffers a phase shift through the coil, i.e. a coil *delays* the current, causing it to lag. That delay can easily be measured on the bench.

The reflected current is approximately the same magnitude on both sides of the coil but suffers an equal phase shift through the coil in the opposite direction.

The forward and reflected current phasors are rotating in opposite directions so the sum of their phases results in approximately equal and constant phase for the net current. I suspect this fact is the origin of the constant current myth.

The magnitude of the net current at the bottom of a base-loading coil is the sum of the forward and reflected currents which are in phase at resonance so the net current is a maximum at that point, e.g. Ifor+Iref = 1.414 amps.

The magnitude of the net current at the top of a base-loading coil is the sum of the forward and reflected currents which are NOT in phase. The coil guarantees a delay for any component current. For an 8 foot 75m mobile antenna, the forward and reflected currents are considerably out of phase at the top of the coil, e.g. Ifor+Iref = ~0.3 amps.

We know the forward and reflected currents are 180 degrees out of phase at the tip of the antenna. Let's assume the forward current is at +90 degrees which will make the reflected current at -90 degrees. An eight foot whip is about 11 degrees on 3.8 MHz. That means the forward current at the top of the coil is at 79 degrees and the reflected current at the top of the coil is at -79 degrees. Since cos(79)=0.19, the magnitude of the net current at the top of the coil is about 20% of its value at the base of the coil.

In a real-world standing-wave antenna loading coil, it is impossible for the *net* current to exhibit both constant magnitude and phase at each end of the coil. Eight feet of whip is ~11 degrees on 3.8 MHz. There's just no getting around that fact. If we assume you are right and the phase of the forward and reflected current are the same at the top and bottom of the loading coil, then somehow you have managed to stuff 90 degrees of phase shift into 11 degrees of whip and I just don't believe in magic.

[W5DXP]

Quote[/b] (VK1OD @ Feb. 19 2006,20:49)]The "magic" being discussed (as to how a 100W generator can "create" 120W of forward power) raises in my mind a question of the legitimacy of considering the forward and reflected waves to be separate "power waves", ie the labelling of Vf*If as Pf and Vr*Ir as Pr (even with lossless lines).
One of my favorite subjects, Owen. Consider the following one-second long lossless transmission line.

100W--50 ohm line--+--one second long 291.4 ohm line--50 ohm load

The system is Z0-matched to 50 ohms at the '+' point. The forward power on the one second long line is 200 watts and the reflected power is 100 watts. It can be easily shown that after steady-state has been reached, there are 300 joules of energy stored in the one second long line that have not reached the load. The question arises: Where are those 300 joules assuming conservation of energy? If the 200w forward power is real, 200 joules would exist in the forward wave. If the 100w reflected power is real, 100 joules would exist in the reflected wave. 200+100=300 joules. If the 300 joules are not in the forward and reflected waves, where are they? Seems to me, Occam's razor would indicate that the 300 joules are indeed in the forward and reflected waves.

[W8JI]

Quote[/b] (w5dxp @ Feb. 19 2006,20:24)]Quote[/b] (w8ji @ Feb. 19 2006,13:06)]Just beyond a feedpoint loading coil with negligible stray capacitance current would still be 1.414 amperes and voltage might be 2000 volts.
The magnitude of the net current is NOT the same on both sides of a loading coil. This is explained on my web page at: http://www.qsl.net/w5dxp/current.htm at the bottom of the page.
Read a bit more careful before you leap Cecil.

I specifically said a loading inductor with negligable stray capacitance.

73 Tom

[w5ah]

Even though a directional wattmeter will show more forward power than delivered by the source when the Zo of the line and the load Z are different, it is interesting to note that if one measures the voltage, current and phase angle (theta) at any point on the line and then does the math... E x I x Cos (theta), the result is always equal to the power applied to the line.

Bob

[W5DXP]

Quote[/b] (w8ji @ Feb. 20 2006,18:53)]I specifically said a loading inductor with negligable stray capacitance.
Sorry, I missed that caveat but the truth is that the great majority of real-world loading coils do not have negligible stray capacitance. A Texas Bugcatcher coil, for instance, has a self-resonant frequency not extremely far removed from the operating frequency where the phase shift through the coil increases to a whopping 180 degrees. Such is the nature of distributed networks as opposed to lumped circuits.

[W5DXP]

Quote[/b] (w5ah @ Feb. 20 2006,21:39)]Even though a directional wattmeter will show more forward power than delivered by the source when the Zo of the line and the load Z are different, it is interesting to note that if one measures the voltage, current and phase angle (theta) at any point on the line and then does the math... E x I x Cos (theta), the result is always equal to the power applied to the line.
E*I*cos(theta) yields the *net* power at the measurement point. If we have 200 watts of forward power and 100 watts of reflected power, the *net* power is 200-100 = 100 watts which is the steady-state power supplied by the source and delivered to the load. But there's more energy in the transmission line than the net power reading would indicate by itself. If one makes the transmission line one second long and lossless, one will discover that it contains 300 joules of energy that have been sourced but have not yet reached the load. If those 300 joules are not being used to sustain the 200 watts of forward power and 100 watts of reflected power, then what are they being used for? That's a rhetorical question. http://www.qrz.com/iB_html/non-cgi/emoticons/smile.gif

[W8JI]

Quote[/b] (w5dxp @ Feb. 20 2006,21:37)]Quote[/b] (w8ji @ Feb. 20 2006,18:53)]I specifically said a loading inductor with negligable stray capacitance.
Sorry, I missed that caveat but the truth is that the great majority of real-world loading coils do not have negligible stray capacitance. A Texas Bugcatcher coil, for instance, has a self-resonant frequency not extremely far removed from the operating frequency where the phase shift through the coil increases to a whopping 180 degrees. Such is the nature of distributed networks as opposed to lumped circuits.
I've never seen an inductor a very small part of a wavelength long that doesn't have a grounded center tap shift phase 180 degrees.

At self-resonance phase shift is zero degrees, and Q is zero. It is a resistor.

At anything else, the maximum is about 90. If the capacitance from the inductor to surroundings is pretty high conmpared to load capacitance, it can shift more than 90. But that's a poor way to mount a loading coil.

The problem is one of flux linkage between turns and ratio of capacitance to the outside world compared to load impedance, not self-resonance.

73 Tom

[W5DXP]

Quote[/b] (w8ji @ Feb. 21 2006,05:44)]At self-resonance phase shift is zero degrees, and Q is zero. It is a resistor. At anything else, the maximum is about 90.

You are again talking about lumped circuits. A 75m bugcatcher coil is NOT a lumped circuit. It is a distributed network and should be treated as such as explained on the following web page: http://www.ncbi.nlm.nih.gov/entrez
"Cylindrical coils near self-resonance", Harpen MD., Department of Radiology, University of South Alabama, Mobile.

"We present a model of birdcage resonator operation when the size of the resonator approaches a quarter wave length resulting in SIGNIFICANT PHASE SHIFTS along the current conducting elements. In the model resonators are treated as generalized transmission lines." e.g. exactly as they should be for 75m mobile loading coils which do indeed "approach a quarter wave length".

Quoting "Antennas for all Applications", by Kraus and Marhefka, 3rd edition, page 824: "A coil (or trap) can also act as a 180 degree phase shifter as in the collinear array of 4 in-phase 1/2WL elements in Fig. 23-21b. Here the elements present a high impedance to the coil which MAY BE *RESONATED* WITHOUT AN EXTERNAL CAPACITANCE DUE TO ITS DISTRIBUTED CAPACITANCE. The coil may also be thought of as a coiled-up 1/2WL element." (Emphasis mine) Kraus says self-resonance can cause a 180 degree phase shift, certainly more than your maximum of 90 possible in a lumped circuit.

A shorted 1/4WL stub is self-resonant and provides a 180 degree phase shift for the current such that the reflected current cancels the forward current resulting in close to an open circuit. This is akin to Kraus's 180 degree current shift caused by the distributed capacitance in his "phase-reversing coils".

[ab9lz]

Brilliant thread guys! Thanks! Sure beats the typical code no code dross. Please continue!

73's
Mark.
AB9LZ

[N7YA]

Quote[/b] (ab9lz @ Feb. 21 2006,12:33)]Brilliant thread guys! Thanks! Sure beats the typical code no code dross. Please continue!

73's
Mark.
AB9LZ
Yes! agreed! I enjoy threads like this...a little island in the middle of a murky lake.

73...Adam, N7YA

[W8JI]

Quote[/b] (w5dxp @ Feb. 21 2006,09:16)]Quote[/b] (w8ji @ Feb. 21 2006,05:44)]At self-resonance phase shift is zero degrees, and Q is zero. It is a resistor. At anything else, the maximum is about 90.

You are again talking about lumped circuits. A 75m bugcatcher coil is NOT a lumped circuit. It is a distributed network and should be treated as such as explained on the following web page: http://www.ncbi.nlm.nih.gov/entrez
"Cylindrical coils near self-resonance", Harpen MD., Department of Radiology, University of South Alabama, Mobile.

"We present a model of birdcage resonator operation when the size of the resonator approaches a quarter wave length resulting in SIGNIFICANT PHASE SHIFTS along the current conducting elements. In the model resonators are treated as generalized transmission lines." e.g. exactly as they should be for 75m mobile loading coils which do indeed "approach a quarter wave length".
Sorry Cecil, but a bug catcher coil is NOT 1/4 wl long on 80 meters.

1/4 wl long is about 65 feet. The bug catcher coil is at best a foot long, or about 1.36 degrees long. That is a far cry from 90 degrees of electrical length.

No wonder we disagree so much!


Actual measurements and description (http://www.w8ji.com/mobile_and_loaded_antenna.htm)

You'll find in links from that page independent verification of my measurements.

73 Tom

[W8JI]

Quote[/b] (w5dxp @ Feb. 21 2006,09:16)][Quoting "Antennas for all Applications", by Kraus and Marhefka, 3rd edition, page 824: "A coil (or trap) can also act as a 180 degree phase shifter as in the collinear array of 4 in-phase 1/2WL elements in Fig. 23-21b. Here the elements present a high impedance to the coil which MAY BE *RESONATED* WITHOUT AN EXTERNAL CAPACITANCE DUE TO ITS DISTRIBUTED CAPACITANCE. The coil may also be thought of as a coiled-up 1/2WL element." (Emphasis mine) Kraus says self-resonance can cause a 180 degree phase shift, certainly more than your maximum of 90 possible in a lumped circuit.
A dimensionally small inductor (in terms of wavelength) with reasonably tight mutual coupling from end-to-end cannot have anywhere near 180-degree phase shift unless it has a ground referenced center tap.

I've unsuccessfully tried to build collinear arrays using just such a scheme, and they never worked. The reason is the inductor at resonance acts just like a resistor, and away from resoance acts like a simple lumped reactance and resistance combination.

I could only make it work as a phase shifting network when I stretched the inductor way out, reducing flux linkage from end-to-end to negligable values.

I've also built delay lines for various applications, and found exactly the same thing. The only way to make an electrically compact size (in relation to wavelength) delay line is to cancel mutual coupling from turn to turn or to add significant distributed capacitance to ground. A small coil of wire 30 degrees long when unwound does NOT look like a 30 degree wire when closely wound. Flux linkage from end to end travels at light speed, and when current appears at one end it is closely matched at the other end through flux linkage.

The 1/4 wave stub is an entirely different matter, because it has the physical size to have significant distributed capacitance, especially to the outside world.

Real world measurements and circuit behaviors disagree with what you are saying, and they are not just my measurements. The problem is you are omitting limits caused by size and shape of the inductor, and effects of mutual coupling.

See this link:

W7EL's measurements (http://www.w8ji.com/agreeing_measurements.htm)


73 Tom

[W5DXP]

Quote[/b] (w8ji @ Feb. 22 2006,07:07)]1/4 wl long is about 65 feet. The bug catcher coil is at best a foot long, or about 1.36 degrees long. That is a far cry from 90 degrees of electrical length.

No wonder we disagree so much!
That pretty funny, Tom. Nobody in their right mind would make a coil dimension approach 1/4 wavelength. When Harpen says, "We present a model of birdcage resonator operation when the size of the resonator approaches a quarter wave length ...", he is talking about the length of the wire used to make the coil. The length of wire used in my 75m Texas Bugcatcher coil is about 42 feet. That's 0.64 of 1/4 wavelength. That's what Harpen means when he says, "approaches a quarter wave length".

About a year ago, we went over what is wrong with your web page experiments. In case you don't remember, here it is again.

The current through a coil made from an appreciable percentage of a wavelength of wire is not constant and cannot be analyzed using your lumped circuit analysis methods. A lumped circuit analysis simply doesn't work for distributed network problems. That's why the distributed network analysis was developed.

Here's what EZNEC says is the current through a 16 turn 40m center loading coil containing 128 segments. Segment 2 is the bottom of the coil and segment 131 is the top of the coil. Here are the current values at a spacing of every ten segments.

seg 2 = 1.042 a --- seg 12 = 1.107a --- seg 22 = 1.144a --- seg 32 = 1.165a --- seg 42 = 1.172a --- seg 52 = 1.169a --- seg 62 = 1.155a --- seg 72 = 1.132a --- seg 82 = 1.099a --- seg 92 = 1.055a --- seg 102 = 0.9972a --- seg 112 = 0.9219a --- seg 122 = 0.819a --- seg 131 = 0.6765a

The current at the feedpoint is 1.0 amp and increases to a maximum of 1.172 amps inside the coil.

It is obvious that this is a standing wave current pattern within the coil complete with a current maximum point about 1/3 of the way up from the base of the coil. THE CURRENT MAXIMUM POINT IS INSIDE THE COIL!!!

The current at the bottom of the coil is 1.042 amps. The maximum current inside the coil is 1.172 amps. The current at the top of the coil is 0.6765 amps. The current standing wave pattern would be even more obvious if presented as a graph. Someone did post a similar graph during the last argument.

It's easy to be fooled by your experimental results. For your toroidal coil, the current maximum point is in the center of the coil which makes Ifor+Iref equal at each end. I can do exactly the same thing with a piece of transmission line with reflections. At point A, the current is one amp. At point B, the current is one amp. Does that mean the current is constant between point A and point B? Of course not! The truth goes something like this: At the bottom of the coil, Ifor=0.5a at -X degrees and Iref=0.45a at +X degrees so Inet=Ifor+Iref=0.8a at zero degrees. At the top of the coil, Ifor = 0.5a at +X degrees and Iref = 0.45a at -X degrees so Inet = Ifor+Iref=0.8a at zero degrees. Yes, the currents are equal because you are measuring them at equal distances from the current maximum point of 0.95 amps which is in the middle of the coil. That's a well-known characteristic of standing waves.

If you will make the middle loop of your toroidal coil big enough to get an RF current probe around it, you will discover that the current in the middle of the coil is considerably higher than the current at either end, i.e. the current through the coil is NOT constant.

[W5DXP]

Quote[/b] (w8ji @ Feb. 22 2006,08:19)]Real world measurements and circuit behaviors disagree with what you are saying, ...
Tom, it is easy to disprove your conclusions. Set up your toroidal loading coil experiment again but this time make the center loop of wire large enough to get a current probe around it. You will discover, as I did, that even though the current at each end of the coil is equal, the current at the center of the coil is considerably higher. W7EL's *conclusions* suffer from the same lack of data mid-point measurements as your *conclusions*.

Neither you nor Roy seem to realize that a standing wave current maximum point exists inside the loading coil. That's what makes the currents equal at the ends of the coil but it doesn't tell us anything about the current inside the coil which is considerably higher than the current at each end.

You guys seem to fail to understand that the net current phasor is the sum of the forward current phasor and reflected current phasor that are rotating in opposite directions. If a coil has a phase delay of 60 degrees, Ifor could be at +30 degrees while Iref is at -30 degrees at one end of the coil. At the other end of the coil Ifor would be at -30 degrees while Iref is at +30 degrees. Ifor+iref would be of identical magnitude and phase at each end of the coil. But in the center of the coil, Ifor+Iref would be 15% higher.

If I tell you a feedline has an SWR of 10:1, the current at point A is one amp, and the current at point B is one amp, could I convince you that the current is a constant one amp between point A and point B? That's exactly the conceptual mistake that you and Roy are making. You guys are simply ignoring the current readings inside the coil.

In logic it's called "petitio principii". You assume the current is constant through the coil and when you measure the end points and find equal currents, your assumption of proof is fulfilled. By only measuring the end point currents you are missing all the data points in between the end points.

[W5DXP]

For those who are interested in following the loading-coil current discussion, here's a brainteaser. Given the following center-loaded mobile antenna (where to the right is really vertical):

Source-----------a-////////////-b--------------

------ is the shaft portion of the antenna. /////// is the loading coil portion. 'a' and 'b' are current measuring points.

Given: The forward current at 'a' is 0.55 amps at -30 degrees. The reflected current at 'a' is 0.45 amps at +30 degrees. The forward current at 'b' is 0.55 amps at +30 degrees. The reflected current at 'b' is 0.45 amps at -30 degrees. The net current at 'a' and 'b' is equal to the phasor sum of the forward and reflected currents.

Question: What is the magnitude and phase of the net current at 'a'? What is the magnitude and phase of the net current at 'b'? What is the estimated magnitude and phase of the net current at a midpoint inside the coil?

[wq7j]

Nice to see a good thread about tech issues, instead of rants.:p

[W8JI]

Quote[/b] (w5dxp @ Feb. 22 2006,08:42)]Quote[/b] (w8ji @ Feb. 22 2006,08:19)]Real world measurements and circuit behaviors disagree with what you are saying, ...
Tom, it is easy to disprove your conclusions. Set up your toroidal loading coil experiment again but this time make the center loop of wire large enough to get a current probe around it. You will discover, as I did, that even though the current at each end of the coil is equal, the current at the center of the coil is considerably higher. W7EL's *conclusions* suffer from the same lack of data mid-point measurements as your *conclusions*.
I'm pretty busy with work, but think I understand this correctly......

1.) I say the current entering and leaving the coil at each end is about the same, I measure it. It is.

2.) Roy Lewallen finds the same thing in an entirely different test.

3.) You now claim current peaks in the middle of a coil because you moved a sampling link over it, which doesn't change anything Roy or I said.


I know you think the link directly measures current in the area of the inductor immediately below the link, but it doesn't.

Assuming your indicator is good and does not respond significantly to electric field gradient around the inductor, the link really only responds to the net magnetic flux cutting the link.

This flux level is by definition maximum at the center area of a single layer coil. That's true even in a Txx-2 material powdered iron core, since a core with such low permability has considerable flux leakage.

If I want build an old-style link coupled transmitter tank coil fed from a constant current source and look for maximum secondary (link) current, it is easy to find. I simply slide the link to the center area of the coil. If I move the link towards either end, secondary current decreases.

This isn't because current is highest in the middle area, it is because of the very simple commonly understood fact that flux fringing is maximum at the inductor ends, and so fewer time-varying flux lines cut the secondary at the inductor ends.

The most concentrated area of flux is at the inductor middle.

It's pretty easy just to remove the coil from the antenna and test it on the bench and see how at 1 MHz or 10 MHz, so long as you are below self-resonance, maximum flux density is always at the winding center area.

That isn't rocket science, and we don't need to invoke "reflected wave theory" to prove flux density is highest when the link is surrounded by as much coil as possible.

I'm really amazed you would even offer that test as any type of evidence, because almost anyone who understands inductors and transformers would also understand the link only measures flux density, and we can't measure turn-by-turn distributed current by moving a link over a solenoid that has tight mutual coupling from end-to-end.

The current taper in the inductor has little to do the "wire length" of the antenna it replaces. In other words if I have a six foot whip on 80 meters, and "replace" the 60 missing feet with an inductor at the whip base, the current reduction through the inductor is very small. The current reduction is really just a function of the distributed capacitance to the outside world and the series reactance of the coil. The capacitance provides a place for the current to go, via displacement currents.

It's a shunt reactance vs. series reactance issue. Nothing more, nothing less.

By the way, I swept S12 phase with my network analyzer on a 100uH inductor a few hours ago while working on a phasing system. The phase shift through that series inductor was about -60 or -70 degrees on 1 MHz, crossing ZERO phase at self resonance (where loss became very high) near 18 MHz, and gradually increasing leading phase above 18MHz reaching 90 degrees and staying there well above resonance.

If I parallel it with a capacitor and move resonance to 5 MHz, it does the same thing there. There is no 180 shift, unless I add enough additional shunt capacitance to dominate load impedance. In that case I have a series L (-90 shift in delivered current to the capacitance) and shunt C (another -90 shift in load voltage driving the measurement device). Or I can make it a pi, or a distributed network. In all cases it requires shunt capacitance to "ground" where that reactance is low compared to load impedance.

A floating parallel L/C at resonance simply appears as a pure resistance with zero phase shift, and inductive below resonance and capacitive above. The only time it fails to follow these rules is when the distributed capacitive reactance leaking displacement current to the outside world starts to be a major part of net system current.

73 Tom

[W5DXP]

Quote[/b] (w8ji @ Feb. 22 2006,21:15)]1.) I say the current entering and leaving the coil at each end is about the same, I measure it. It is.
I smell a straw man. http://www.qrz.com/iB_html/non-cgi/emoticons/smile.gif Of course they are about equal and I show exactly why that is true with my brainteaser. Your and my main argument is whether the forward current through the coil exhibits a phase shift and whether the reflected current flowing the other direction exhibits a phase shift. If they do then the coil occupies that part of a wavelength. You apparently haven't looked at my brainteaser where the forward current is 0.55 amps, the reflected current is 0.45 amps, and they both exhibit a 60 degree phase shift while flowing through the coil.

At the bottom of the coil:
0.55 amps at -30 degrees plus 0.45 amps at +30 degrees equals one amp at zero degrees.

At the top of the coil:
0.55 amps at +30 degrees plus 0.45 amps at -30 degrees equals one amp at zero degrees.

They *are* equal even though both currents exhibited a 60 degree phase shift in opposite directions. That 60 degree phase shift is 2/3 of 1/4 wavelength. Your concept that a 75m bugcatcher coil doesn't occupy any electrical wavelength is simply wrong. I show how it indeed does occupy a good portion of 1/4 wavelength while maintaining equal currents at both ends of the coil. I don't know how you can sincerely argue that there's not 90 degrees in an electrical 1/4 wavelength antenna.

The net current that you and Roy are measuring is a standing wave current made up of forward and reflected currents. The forward current flows into the bottom and out the top of the coil but the reflected current flows into the top and out the bottom of the coil. You and Roy talk about the net current as if it were flowing only in one direction even though it is a standing wave made up of two currents flowing in opposite directions. That's simply confusing and misleading.

Roy's confusion is apparent in his statement: "I found that the difference in current between input and output of the inductor was 3.1% in magnitude and with no measurable phase shift ..."

Of course there's no measurable phase shift because IT'S A STANDING WAVE. The phase shift is in the forward current and reflected current. You and Roy don't seem to realize that it is a standing wave which naturally exhibits something like a cosine curve inside and outside of the coil. In "Antennas for all Applications", by Kraus and Marhefka, 3rd edition, page 464 it shows the phase of a standing wave over 1/2 wavelength. It does indeed exhibit no measurable phase shift. That's a well known characteristic of standing waves. However, the component forward and reflected currents certainly exhibit the phase shifts predicted by the laws of physics.

[W5DXP]

Quote[/b] (w8ji @ Feb. 22 2006,21:15)]By the way, I swept S12 phase with my network analyzer on a 100uH inductor a few hours ago while working on a phasing system. The phase shift through that series inductor was about -60 or -70 degrees on 1 MHz, ...
Sorry, I forgot to reply to this part in my last posting. That measurement supports my position, Tom. A 60 degree phase shift through a coil is 2/3 of 1/4 wavelength. In fact, a 75m bugcatcher coil does indeed contribute part of the electrical 1/4 wavelength by causing that same phase shift in the forward and reflected currents. Seems you have just proved that a loading does occupy part of a wavelength in an antenna system.

[ae4tr]

Quote[/b] (K5MDM @ Feb. 17 2006,22:10)]If pie are square, #am cornbread then round?javascript:emoticon('http://www.qrz.com/iB_html/non-cgi/emoticons/tounge.gif')
smilie
http://www.qrz.com/iB_html/non-cgi/emoticons/biggrin.gif
#In this part of the country, pie are not square.. pie are round, cornbread are square....
http://www.qrz.com/iB_html/non-cgi/emoticons/cool.gif

[W8JI]

Quote[/b] (w5dxp @ Feb. 22 2006,21:42)]Of course there's no measurable phase shift because IT'S A STANDING WAVE. The phase shift is in the forward current and reflected current. You and Roy don't seem to realize that it is a standing wave which naturally exhibits something like a cosine curve inside and outside of the coil.
So we say there is no significant phase delay, we measure no significant phase delay, you agree there isn't significant phase delay, but when you break the system apart in your analysis it has phase shift.

It seems like to only thing you are arguing about is how youdecided you wanted to analyze the system.

Since we all agree on results, that current flowing through the coil that occupys a very small fraction of a wavelenth is essentially equal, then I guess we agree on results.

I'm going to go do something else more productive now.

73 Tom

[W5DXP]

Quote[/b] (w8ji @ Feb. 23 2006,06:42)]So we say there is no significant phase delay, ...
No, you completely missed the point, Tom. THERE IS A 60 DEGREE DELAY THROUGH THE COIL. The laws of physics do not allow anything else. What I have shown is that even with a 60 degree delay through the coil, the net current, which is the phasor sum of the forward and reflected current, can be equal in magnitude and phase at both ends of the coil because IT IS A STANDING WAVE. Like a transmission line, a coil doesn't have to have zero phase shift for the net current to be equal at both ends. In fact, it is just as *physically impossible* for a 75m bugcatcher coil to exhibit a zero phase shift to traveling wave currents as it is for a one wavelength transmission line to exhibit a zero phase shift. Hint: it shifts by 360 degrees, not by zero.

Do you understand vector math? Please pay attention this time.

A certain loading coil exhibits a 60 degree delay which causes the following conditions:

The forward current at the bottom of the coil is 0.55 amps at -30 degrees. The reflected current at the bottom of the coil is 0.45 amps at +30 degrees. The net current is the vector sum of those two phasors and is equal to one amp at zero degrees.

(0.55a /_ -30) + (0.45a /_ +30) = 1.0a /_0

The forward current at the top of the coil is 0.55 amps at +30 degrees. The reflected current at the top of the coil is 0.45 amps at -30 degrees. The net current is the vector sum of those two phasors and is equal to one amp at zero degrees.

(0.55a /_ +30) + (0.45a /_ -30) = 1.0a /_0

To summarize: The phase shift through the loading coil is 60 degrees, i.e. 2/3 of 1/4 wavelength, i.e. the loading coil occupies 2/3 of the electrical 1/4 wavelength of the antenna. If the entire rest of of an electrical 1/4WL antenna occupies 30 degrees, the other 60 degrees cannot magically disappear. It has to be associated with the coil since there's nothing else in the system.

The forward current at the bottom of the coil is 0.55 amps at -30 degrees and 0.55 amps at +30 degrees at the top.
The forward current undergoes a phase shift of 60 degrees. The forward current magnitude is constant through the coil.

The reflected current at the top of the coil is 0.45 amps at -30 degrees and 0.45 amps at +30 degrees at the bottom. The reflected current undergoes a phase shift of 60 degrees. The reflected current magnitude is constant through the coil.

The net current is equal at both ends of the coil, not because the coil exhibits a zero phase shift, but because IT IS A STANDING WAVE and the SUM of the phase shifted forward and reflected currents has a phase angle of zero. Hint: the forward and reflected current phasors are rotating in opposite directions as is normal for a standing wave.

The problem with your concepts seems to lie in your treatment of a standing wave current as a lumped circuit current. You cannot treat a standing wave transmission line current as a lumped circuit current. For exactly the same reason, you cannot treat a standing wave antenna current as a lumped circuit current. Doing so leads you to some outrageous conclusions like there is no phase shift through a 75m bugcatcher loading coil. Your own measurements on 100 uH at 1 MHz showed a 60 degree phase shift. How can you then argue that 70 uH at 4 MHz exhibits zero phase shift?

[W5DXP]

For those of you wondering what the argument is all about, here it is in a nutshell. Assume 1/2 #wavelength of a straight wire standing wave antenna. One end is -90 degrees and the other end is +90 degrees. The net current at each end is obviously zero. The net current is maximum in the middle at zero degrees. Let's assume it is 1a /_ 0 (one amp with a phase angle of zero degrees) at the middle of the 1/2 wavelength element.

-90 ------------ -30 ------ 0 ------ +30 ------------ +90

The net current at -30 degrees is 1a*cos(-30) = ~0.87a /_ 0.
The net current at +30 degrees is 1a*cos(30) = ~0.87a /_ 0.

The constant phase angle all up and down the antenna is a characteristic of the standing wave net current, NOT JUST A CHARACTERISTIC OF A COIL. W8JI and W7EL make a big deal out of that constant phase but the constant phase exists whether a coil in used or not, i.e. the constant phase has absolutely nothing to do with the coil and is certainly NOT caused by the coil.

The net current at points 1/12 of a wavelength from the center are equal in magnitude and phase.

Now we physically shorten the antenna by replacing that 1/6 of a wavelength of wire with a coil that brings the antenna to resonance, i.e. an electrical 1/2 wavelength. The antenna is now physically approximately 1/2 - 1/6 = 1/3 wavelength long. We measure the current on one side of the coil to be the same magnitude and phase as on the other side of the coil. We have replaced 1/6 of a wavelength of the antenna with a coil without appreciably changing anything.

-90 ------------ -30 ////// +30 ------------ +90

W8JI's web page says we didn't replace 1/6WL (60 degrees) of the antenna with a coil yet that's exactly what we did do. The antenna is still electrically 180 degrees long since it is resonant and it contains only 120 degrees of straight wire. Where does the other 60 degrees come from if not from the coil? That's the crux of the argument.

[k2dc]

There has been a consistent math error in most of these posts. The original post in section 3a states that
(Vf-Vr)/(If+Ir)=R. I have not read all the posts, but no one has seemed to pick up on the fact that the forward and reflected voltages and currents are vector quantities, with their signs following direction.
The correct equation is (Vf-Vr)/(If-Ir)=R. The power delivered to the load is (Vf-Vr)^2/R or (If-Ir)^2 * R. The reflected power (Vr/Ir) is either dissipated in the source, or is stored in the line as multiple reflections (and if the source is perfectly matched to an ideal line it has to be dissipated in the source).
One post suggests that the reflected power is 180 degrees out of phase with the forward power. That only happens when the load is a short circuit. If the load is purely resistive or an open circuit, the reflected power is in phase with the forward. And in all cases, voltage and current (forward or reflected) are 90 degrees apart. You certainly can see voltages or currents along the line that are much higher than those due strictly to the generator, but the total power at any given point on the line is a constant, as is the VSWR.
Conservation of energy dictates that you just can't get more than you make. Some of the posts suggest that you could configure a source perfectly matched with an ideal line along with a load mismatched to that line and realize a power gain somewhere along the line. If that's the case, I've been paid an awful lot of money for an awfully long time to deal with the WRONG laws of physics.
If you believe, as the original post suggests, that "LINES THEY ARE ALWAYS MATCHED TO THE LOAD"
there should have been a stampede to the patent office.

Don, K2DC

[W5DXP]

Quote[/b] (k2dc @ Feb. 24 2006,11:36)]There has been a consistent math error in most of these posts. The original post in section 3a states that
(Vf-Vr)/(If+Ir)=R. The correct equation is (Vf-Vr)/(If-Ir)=R.
Don, your posting shows the tricky part of phasor math. I avoid any confusion by assuming that superposition is the *addition* of two waves of often differing phases. The negative sign in phasor math implies a 180 degree phase shift. Assuming that each term in the following equation possesses a phase angle, the impedance in a standing-wave environment is simply:
(Vfor + Vref)/(Ifor + Iref) = impedance
As long as the phase angles are properly associated with the magnitudes, there is no need to use a minus sign. If Vref is 180 degrees out of phase with Vfor, the phasor superposition, i.e. addition, will take care of itself. One at zero degrees *plus* one at -180 degrees equals zero. Or 1*cos(0) + 1*cos(180) = 0

Public Challenge to W8JI: Tom, why are you afraid to discuss this topic to its logical conclusion?

[k2dc]

Cecil,

You're exactly right about phasor math, but I was referring to vector math such that (without regard to signal phase) using the forward power direction as a reference and treating everything as real numbers, the reflected voltage and current both need a minus sign - insted of one and not the other as the original post implied. The relative phase of the forward and reflected signals are entirely dependant on the nature of the load (180 for a short, 0 otherwise) and the voltages and currents are 90 degrees apart maintaining constant power everywhere on the line.
In any event, on a linear line (one dimension), if you fail to treat the forward and reflected signals as vectors and maintain the proper sign convention you could probably talk yourself into all kinds of things. Just like the original poster did.

73,

Don, K2DC

[k2dc]

Cecil - on relative phase of the forward and reflected, I should have said 180 degrees for a short, 0 for an open or purely resistive load, any other case depends on the
reactance of the load.

73,
Don K2DC

[W5DXP]

Quote[/b] (k2dc @ Feb. 24 2006,15:10)]You're exactly right about phasor math, but I was referring to vector math such that (without regard to signal phase) using the forward power direction as a reference and treating everything as real numbers, the reflected voltage and current both need a minus sign ...
I'm sorry for having to point this out to you Don, but the open-circuit end of a dipole causes the reflected current to experience a 180 degree phase shift while the reflected voltage experiences zero phase shift. This has to do with the conventions adopted by RF engineers. Thus, the reflected current is 180 degrees out of phase with the reflected voltage. So, for a 1/2 wavelength resonant dipole, the correct formula for the feedpoint impedance would be:
Zfp = (|Vfor| - |Vref|)/(|Ifor| + |Iref|)
The voltages are out of phase and subtract while the currents are in phase and add. This results in a low voltage and a high current, i.e. a low feedpoint resistance. This is really complicated stuff so I apologize if I have misunderstood your posting.

[nf0a]

Quote[/b] (n6ajr @ Feb. 17 2006,16:25)]OR #is a few words. all the power goes some where, either out the load, or as heat, even the reflections, it all goes somewhere..
All that verbiage distilled in your words here is priceless!

[W5DXP]

Quote[/b] (k2dc @ Feb. 24 2006,15:13)]Cecil - on relative phase of the forward and reflected, I should have said 180 degrees for a short, 0 for an open or purely resistive load, any other case depends on the reactance of the load.
The IEEE RF Engineering convention is that, for a purely resistive characteristic impedance, the reflected voltage is always 180 degrees out of phase with the reflected current, no matter what is the impedance of the load. Optics engineers have a different convention since visible EM light waves do not flow through a copper wire. http://www.qrz.com/iB_html/non-cgi/emoticons/smile.gif

[KM4XG]

Uhh...I don't think so...

Any impedance, if not the same value as the transmission line, will reflect power to the generator, where it will be re-absorbed by that generator if it is a perfect match for the line. #If not, a portion of the reflected power will be reflected again, back towards the load. #This process continues until all the power is consumed by the total dissipation of the load, the line, and the generator.

This is why mis-matched loads are bad. #They cause the generator (your nice new HF hot-rod) to get hot, smoke, and die on occasion.

The assertion that a line is matched to any load can be disproved easily by experiment. #It is such a fundamental principal that it almost goes without saying. #Taking your logic to the limit, it should #be possible to operate any transmitter at any phase angle into any impedance without consequence. #Absurd. #Try it.

The point made earlier in the thread regarding the SWR of a dummy load is very illustrative of the mis-understanding persistent among hams these days, and about which our predecessors were not encumbered.

A perfect load is one which is entirely resistive and matches the characteristic impedance, Zo, of the line to which it is connected. #In this condition, all of the power is dissipated in the load as heat. #If the load gains a reactive component, it then becomes an impedance. #If the reactive part of this impedance is a transducer, such as an antenna, that couples energy to space, then part of the total power is consumed in the real resistance of the load as before, and the remainder is delivered to space. #Simple.

The trick is to increase the power delivered to space, and decrease the power consumed in the resistive component of the total impedance, while at the same time maintaining a good match for the transmission line.

Case in point, opposite to the perfect dummy load, is the perfect antenna: #This is a radiator that delivers all of the input power to space. #This antenna has no resistive component of impedance...it is totally reactive. #What this means is that all forward power is radiated...there are no resistive losses in the antenna itself. #But it is still possible that not all the power incident on this DX champion is radiated...if even a perfect resonant antenna's input impedance is different from the line Zo, then some power will bounce off the antenna and head back to your expensive output transistors.


To further illustrate the point, consider the ubiquitous Trans-Match: #This device, when adjusted properly, presents a perfect match between your transmitter and your transmission line ( NOT the antenna). #What this means is that any reflected power from the antenna is RE-reflected BACK to the antenna! #This is called a state of conjugate match: #The impedance presented to the line by the transmatch, in the reverse direction, is of equal magnitude but opposite in sign ( reactance ) to the impedance presented to the line by the antenna.

So what this allows you to do is operate into a mis-match and still: #A) Keep your expensive hot-rod happy, and B) dissipate ALL forward power in the antenna, line, and Trans-match. #Power reflects back and forth between the antenna and the matchbox until it is all radiated (or until the coax melts). #The trick is to operate into an antenna that is efficient enough that, each time power returns to the antenna, a substantial portion is radiated away as RF, not heat. #This property of the antenna is independent of transmission line SWR. #Any given resonant antenna design will have an input impedance that is finite, and may be different from the Zo of your transmission line. #An efficient antenna is just as efficient regardless of transmission line Zo. #But the mis-match between the Zo of the antenna and line will mean that some incident power will be reflected. #Obviously, its important to keep this percentage as low as possible.

But that's only important insofar as protecting the generator and line from damge. #It has nothing to do with antenna efficiency. In fact, if you run balanced line behind your tuner such as the commonly available 150-ohm or 300-ohm plastic ladder line, then its possible to operate well in a situation where you have a resonant antenna with high radiation resistance and good efficiency, that nevertheless presents a #bad mismatch to the transmission line. #In this case, so what? #The SWR on the line can be sky-high and yet, all of the power generated by the transmitter is dissipated in the antenna as power coupled to air. #This is due to the fact that any reflected power from the antenna is re-reflected back to the antenna by the tuner until it is all radiated to space ( minus the resistive losses in the ladder line). #Result? #Transmitter happy, gud DX. #

How can this be? #Because SWR on the line and efficiency of the antenna as a radiator are not related. #They just aren't. #You can have a perfect SWR with a dummy load. #You can have a really bad SWR but still work the World. #See Radiation Resistance.

You can get away with running SWR on the line of 20:1 easily without damaging your ladder line or your tuner. #Try that with coax and it will puncture at a high-voltage node...it will just arc right over through the dielectric, becoming a long piece of trash.

I have a single, 120 -foot long wire antenna center fed with 150-ohm balanced, and work all HF bands no problem using an MFJ-989C tuner. #It works.

Watch out for scholarly and learned texts like the one I've responded to here. #The concepts are just wrong. #And the math is fuzzy, too.

Cheers and 73,
KM4XG

[W5DXP]

Quote[/b] (KM4XG @ Feb. 24 2006,20:06)]Uhh...I don't think so...
I'm assuming that you are replying to my posting. For a transmission line that has a purely resistive characteristic impedance, the forward voltage is in phase with the forward current and the ratio of forward voltage to forward current is equal to Z0. The reflected current is 180 degrees out of phase with the reflected voltage and the ratio of reflected voltage to reflected current is equal to Z0. In a nutshell, those are the boundary conditions set by the reflection model. Vfor/Ifor = Vref/Iref = Z0. A horizontal 1/2WL dipole antenna is just a lossy single-wire transmission line broken in the middle and open at each end. For #14 wire about 30 feet from the ground, the Z0 is approximately 600 ohms. The formula is: Z0 = 138*log(4D/d) where D is the height above ground and d is the wire diameter.

For a center-fed 1/2WL dipole, the thing that guarantees the low feedpoint impedance is that the reflected voltage arrives 180 degrees out of phase with the forward voltage and therefore interferes destructively. The reflected current, having undergone a 180 degree phase shift at the open end of the dipole, arrives in phase with the forward current and therefore interferes constructively. So the feedpoint impedance is ~(|Vfor|-|Vref|)/(|Ifor|+|Iref|), essentially the same formula used for a lossy transmission line with reflections.

[KM4XG]

Hello W5DXP de KM4XG,

No, I wasn't replying to your post, but to the original posting.

My point was to dispell the meritless assertions of the original author of the thread and to possibly shed some light on the issue for the benefit of the average ham. Somehow, what used to be common knowledge is not so common anymore.

There are practical benefits to be realized from a proper appreciation of the established theoretical foundation supporting the art.

By that I mean I would hate for some kid to waste his transmitter that he worked all summer to afford because he tried to apply the principles outlined in the original post.

So please, if you are not sure (this is QST), don't add to the confusion by re-inventing the wheel in a forum like this. Thousands of folks who don't know better might be reading it. Let's be sure we give them all the help they deserve by keeping metaphysics out of ham radio and in the Ouiga parlor, where it belongs.

Cheers and 73,
KM4XG

[KM4XG]

As I re-read the original posting by PY1LL, I noticed that his sub-title is, "Concepts, generally not well understood".

As you may gather from my previous input here, I agree.

KM4XG
http://www.qrz.com/iB_html/non-cgi/emoticons/cool.gif

[KE4IKY]

Thanks KM4XG....

The original post sounded quite strange and there didn't seem to be a lot of data challanging the actual original assertions.

[W5DXP]

Quote[/b] (KM4XG @ Feb. 25 2006,01:04)]No, I wasn't replying to your post, but to the original posting. Any impedance, if not the same value as the transmission line, will reflect power to the generator, where it will be re-absorbed by that generator if it is a perfect match for the line. If not, a portion of the reflected power will be reflected again, back towards the load. This process continues until all the power is consumed by the total dissipation of the load, the line, and the generator.
I apologize. With no quote from you, I assumed you were disagreeing with the previous posting. I stand corrected but next time it would help me (and maybe others) if you quoted at least a small portion of what you are disagreeing with. Just a friendly suggestion.

I think the original poster was trying to describe a system with a mismatched load and a Z0-match at the source provided by a tuner or other means. If the system is lossless and no reflected energy is allowed to reach the source, then there is no wasted power. Walter Maxwell said essentially that when he described lossless conjugately matched systems. That's just my take on what the original poster was trying to say.

[PY1LL]

VK1OD ,

There are two real power flows, each one carried by one wave, f and r, Vf.IF and Vr.Ir. The product Ir.Vf is not a power flow because each variable pertains to two independent waves and that type of product leads to an zero average power.

Luiz - PY1LL

[PY1LL]

I have considered the load a perfect resistance, with no reactance just to simplify things.
As the voltage wave speed is equal to the current wave one and, on the load, V is in phase with I, they are in phase in all points of the line for that wave. Then, no cos(phase) in the discussion.

Luiz = PY1LL

[PY1LL]

The current at one end of a coil is equal to the current at the other end as long as it can be considered as a lumped component, that is, its length is small enough to be possible to forget propagation times. If not, the component is not a simple inductance because it is a distributed component.

Luiz - PY1LL

[PY1LL]

For a 200 W of forward power and 100 W of reflected one, there is no 300 J of extra energy in the line.
After the steady state being established (for the lossless case), the source of the forward excess power is just the reflected one and the source of the latter is the former one. The forward power is formed by the generated and the reflected one. The latter is used only to create a perfect match between the line and the load.
Its true that, before the steady state, complicated things occur to 'fill' the line with both forward and reflected powers. And if you go off the air, the line takes some time (greater than in the 1:1 case) to empty the line.

Luiz - PY1LL

[PY1LL]

Suppose the case where Vfor/Ifor = Zo is > R.
Than, Vfor + Vref)/(Ifor + Iref) is still greater and NEVER we can get it equal to R, that is imposed by Ohm law. In this case, the phase of the reflected voltage wave is opposite of that of the forward wave, so that the maximum value of the voltage in the load end of the cable is Vfor - Vref. The current reflected wave is not inverted (we can only invert one of them) and the maximum current value at that point is Ifor + Iref, with the conditions:
Vfor/Ifor = Vref/Iref = Zo (independent waves)
And (Vfor - Vref)/(Ifor + Iref) = R
In the case of Zo < R it occurs just the opposite:
(Vfor + Vref)/(Ifor - Iref) = R

[W5DXP]

Quote[/b] (PY1LL @ Feb. 25 2006,10:39)]There are two real power flows, each one carried by one wave, f and r, Vf.IF and Vr.Ir. The product Ir.Vf is not a power flow because each variable pertains to two independent waves and that type of product leads to an zero average power.
There is the special case of very lossy lines where the Vf*Ir and Vr*If interference terms are not negligible. It occurs when the load reactance and the transmission line reactance are equal and opposite in sign thus resulting in a series resonance between them. Reference: "Transmission Lines", by Robert A. Chipman, page 138: "Consideration of equation (7.34) reveals that the coefficient on the right, multiplied by the first term (unity) in the braces, represents the real power that would be calculated for the incident wave alone" (the forward wave) ..."while the coefficient multiplied by the second term represents the real power that would be calculated for the reflected wave alone, at the same point. The third term on the right represents interaction between the two waves." i.e. there can be a real power component associated with the interference interaction of forward and reflected waves under special conditions. (I'm going to try to attach equation (7.34) as a GIF file attachment)

[PY1LL]

"For purely resistive characteristic impedance lines...."
We have to mention that, in spite of the fact the lossless cable has a real impedance, it doesn't mean dissipation on it and we have to take care with this words 'resistive characteristic impedance'.
Why a real line impedance doesn't mean dissipation on it? Because, as we saw, the lossless cable only 'brings the load to the generator', that means, it is only a lossless impedance transformer. When the cable is infinitely long, it doesn't give back the power that was delivered to it, and, so, it is dissipative. In this case, there is no reflected wave and the impedance seen by the generator must be Zo AND REAL (to be a pure dissipative element).
Dissipation is defined by the generator. If the component doesn't give the power back, it is dissipative (and may be represented by a pure resistor). If a component/circuit is pure reactive, instead, it gives back ALL the energy delivered to it.
In the case of the infinitely long lossless cable, the generator 'sees' an element that doesn't give any part of the delivered energy and is pure dissipative. Which is the ratio V/I on this element? It must be real to be dissipative and, because of no refleted wave, its value is Zo.

Luiz - PY1LL

[PY1LL]

"Any impedance, if not the same value as the transmission line, will reflect power to the generator, where it will be re-absorbed by that generator if it is a perfect match for the line. If not, a portion of the reflected power will be reflected again, back towards the load. This process continues until all the power is consumed by the total dissipation of the load, the line, and the generator."

The generator is considered to be perfectly matched to the line, but the reflected power is not dissipated by the generator just because the forward power is greater than the generated one. Let's analyse the case.
Which is the power reflection coefficient? By definition, it is the ratio back power/arriving power. At that cable end the back power is the forward one and the arriving power is the reflected wave one.
As Pfor = Pref + Pgen, Pfor/Pref = 1 + Pgen/Pref or the reflection coefficient is greater than 1. Is this possible? Yes, because the transmission coefficient Pgen/Pref is negative as Pgen is not really transmitted, but also arrives at that point, so that the sum 'reflection coefficient + transmission coefficient = 1' still holds.
This occurs because, indeed, the arriving power is not only the reflected one and we can't say that the reflected power is delivered to the generator and dissipated on it.
I have an article called "Generalized Reflection and Transmission Coefficients" where I try to explain all of this.
I said 'I try', please...

Luiz - PY1LL

[PY1LL]

w5dxp, you are perfect, but remember that I have chosen lossless lines just to avoid new things that would mask the main concepts involved, like Pfor > Pgen and the 'always line to load perfect match'.

Luiz - PY1LL

[KM4XG]

W5DXP Writes:

"I apologize. With no quote from you, I assumed you were disagreeing with the previous posting. I stand corrected but next time it would help me (and maybe others) if you quoted at least a small portion of what you are disagreeing with. Just a friendly suggestion."


Hey, no problem.

#Just for your edification, one may also discern the meaning or intent of a particular post by actually reading it. #Sometimes, there are enough contextual clues contained within a given work to allow even the unconversant to acquire the gist.

Cheers & 73,
KM4XG
http://www.qrz.com/iB_html/non-cgi/emoticons/cool.gif

[KM4XG]

PY1LL Writes:

"The generator is considered to be perfectly matched to the line, but the reflected power is not dissipated by the generator just because the forward power is greater than the generated one. Let's analyse the case."

That's gobbledygook.

Reflected power hurts generators. #That's why it is a bad thing. #Reflected power is re-absorbed by the generator unless there is a conjugate match prior to the generator output port to reflect the incident power back towards the load. #Have you ever seen an isolator or circulator? #What are you trying to prove...besides the fact that anything you say is highly suspect? #Please...stop the madness....! (geeesh)



KM4XG

http://www.qrz.com/iB_html/non-cgi/emoticons/rock.gif

[W5DXP]

Quote[/b] (PY1LL @ Feb. 25 2006,11:10)]For a 200 W of forward power and 100 W of reflected one, there is no 300 J of extra energy in the line.
Yes, during steady-state, there are 300 joules stored in the line that haven't reached the load. There are a number of ways to illustrate that fact. The easiest to understand is to use a fast autotuner that doesn't allow any reflected energy to reach the source. In the following mental exercise, the FATuner is a "fast autotuner". No matter what impedance it is presented with at its output, it immediately establishes a 50 ohm Z0-match so that reflected energy is prevented from reaching the source.

100W Source---50 ohm coax---FATuner---one second long lossless 291.42 ohm transmission line---50 ohm load

The source will always see 50 ohms and therefore will deliver a constant 100 joules/second into the system. This is simply to make the math easy to understand. Now, we can perform a second by second accounting of the energy. Note that the power reflection coefficient at the load is 0.5, i.e. half the incident power is reflected from the load.

After the first second, the line contains 100 joules and zero joules have been delivered to the load. The load rejects half the power so after the second second, 50 joules have been delivered to the load and the line contains 150 joules of the 200 joules output so far by the source. The 50 joules of reflected energy is re-reflected by the FATuner and joins the forward wave during the third second so at the end of the third second, there's 200 joules stored in the line and 100 joules have been delivered to the load. Already there more joules in the line that is being sourced per second. Simply continue this pattern until steady-state has been reached and you will find that 200 joules are contained in the forward wave, 100 joules are contained in the reflected wave, and of the total joules delivered by the source, 300 joules have not reached the load. Someone may be able to sweep forward and reflected power under the rug, but the conservation of energy principle will not allow that to be done with joules. That's the purpose of the lossless one second long line. The joules in that line cannot simply be ignored.

I'm going to extend the above sequence for you but this proportional font makes it difficult to present in a table. The first line is the number of seconds and subsequent lines contain terms for which there is a one to one correspondence to those seconds.

seconds: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... steady-state

total energy sourced in joules: 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, ... X

energy in the forward wave in joules: 100, 100, 150, 150, 175, 175, 187.5, 187.5. 193.75, 193.75, ... 200

energy in the reflected wave in joules: 0, 50, 50, 75, 75, 87.5, 87.5, 93.75, 93.75, 96.875 ... 100

total energy delivered to the load in joules: 0, 50, 100, 175, 250, 337.5, 425, 518.75, 612.5, 709.375... X-300

You can see that the total energy delivered to the load equals total energy sourced minus the joules in the forward wave minus the joules in the reflected wave. Conservation of energy strikes again.

[W5DXP]

Quote[/b] (PY1LL @ Feb. 25 2006,12:10)]w5dxp, you are perfect, but remember that I have chosen lossless lines just to avoid new things that would mask the main concepts involved, like Pfor > Pgen and the 'always line to load perfect match'.
Yes, and I have chosen the one second long line for the same reason. In that mental example of mine, Pfor > Pgen and the source is always perfectly matched. Yet there are still 300 joules of energy that have been sourced and not dissipated in the load.

[W5DXP]

Quote[/b] (KM4XG @ Feb. 25 2006,12:32)]> PY1LL Writes: "The generator is considered to be perfectly matched to the line, but the reflected power is not dissipated by the generator just because the forward power is greater than the generated one. Let's analyse the case." <

That's gobbledygook.
I think maybe you are not understanding what he is assuming which is that the generator is matched to the virtual load that it is seeing. I've presented a mental example where a fast autotuner accomplishes that feat. No reflections reach the generator.

[W5DXP]

Quote[/b] (KM4XG @ Feb. 25 2006,12:15)]Just for your edification, one may also discern the meaning or intent of a particular post by actually reading it. Sometimes, there are enough contextual clues contained within a given work to allow even the unconversant to acquire the gist.
I did read it and decided that if you were disagreeing with me, you didn't understand what I was saying. Sorry about that.

[KM4XG]

W5DXP writes:

"I think maybe you are not understanding what he is assuming which is that the generator is matched to the virtual load that it is seeing. I've presented a mental example where a fast autotuner accomplishes that feat. No reflections reach the generator. "

No, I think I understand just fine. #I think you two are pretending to be something you are not. #I am embarassed for you both.

http://www.qrz.com/iB_html/non-cgi/emoticons/rock.gif

[KM4XG]

W5DXP Writes:

"Yes, during steady-state, there are 300 joules stored in the line that haven't reached the load. There are a number of ways to illustrate that fact. The easiest to understand is to use a fast autotuner that doesn't allow any reflected energy to reach the source. In the following mental exercise, the FATuner is a "fast autotuner". No matter what impedance it is presented with at its output, it immediately establishes a 50 ohm Z0-match so that reflected energy is prevented from reaching the source."


Here is a perfect example of what is wrong with this type of hand-waving argument: #How does the FATuner know what the impedance at the far end is?

Do you know how those devices work? #They MUST receive a reflected power from the load in order to detect the mis-match at the far end. #Only then can the matchbox determine how to present the complex conjugate match to the line. #There is no instantaneous match. #During the time the tuner is trying to tune itself, it is absorbing some part of the reflected power. #If the tuner fails to find a good match, it could smoke if not hefty enough to dissipate that power safely. #Have either of you two actually ever operated a radio station of any kind before?

'Nuff said.
KM4XG
http://www.qrz.com/iB_html/non-cgi/emoticons/rock.gif

[W5DXP]

Quote[/b] (KM4XG @ Feb. 25 2006,15:36)]Here is a perfect example of what is wrong with this type of hand-waving argument: How does the FATuner know what the impedance at the far end is?

Do you know how those devices work? They MUST receive a reflected power from the load in order to detect the mis-match at the far end.
Yes, yes, yes, you finally have "got it". The fast autotuner doesn't know (and doesn't care) and doesn't need to know what the mismatch at the "far end" is, i.e. the SWR is what it is and cannot be changed by a tuner at the source. It reacts ONLY to the SWR it sees on its *INPUT* and tunes itself to a 50 ohm Z0-match which allows zero reflected energy to reach the source. If you manage to re-reflect the reflected energy at the local tuner point, you don't need to know or do anything else. My SGC-230 doesn't "know" when I switch antennas. It simply re-establishes a Z0-match to 50 ohms every time it detects an SWR above about ~1.5:1 on the source side of the line.

So please tell us what it is that you are getting all emotional about. Are you afraid that some sacred cow is about to be slaughtered? This ain't rocket science. It's simply conservation of energy in action.

[KM4XG]

W5DXP writes:

"You can see that the total energy delivered to the load equals total energy sourced minus the joules in the forward wave minus the joules in the reflected wave. Conservation of energy strikes again. "


Not only is this ridiculous, it's not even mathematically consistent with your own thesis.

Here's what Conservation of Energy means: #regardless of the generator, the line, or the load, the total output power from all sources of radiation...RF or heat...can never be greater than the input power. #If that weren't so, we'd all be getting rich generating power we sucked out of the 23rd dimension for free (where did I put that Star Trek lunch box...?)

I think I've demonstrated the following:

Lines, they are NOT always matched to loads.
Lines, they can smoke if not matched to loads.
Lines, understanding them, saves $, Good DX
PY1LL, generally, not well understanding

(crash and burn, lid....) http://www.qrz.com/iB_html/non-cgi/emoticons/mad.gif
http://www.qrz.com/iB_html/non-cgi/emoticons/laugh.gif
http://www.qrz.com/iB_html/non-cgi/emoticons/cool.gif

de KM4XG sk #. ... .

[W5DXP]

Quote[/b] (KM4XG @ Feb. 25 2006,16:23)]Not only is this ridiculous, it's not even mathematically consistent with your own thesis.
If you think that, you are mistaken. Given that you were messing your diapers when I graduated from Texas A&M with a EE degree, it seems that you need to be carefull about misunderstanding what I am saying. I believe in the conservation of energy principle. Do you reject that principle?

What I find amazing is that you seem to be willing to accept a one-second long lossless transmission line, but not willing to accept an autotuner that requires zero time to achieve a Z0-match. You accept one ideal and reject another ideal? Seems to me you need to reject any mental exercise that doesn't exactly match reality. What's up with that?

[KM4XG]

W5DXP writes, laboriously:

"If you think that, you are mistaken. Given that you were messing your diapers when I graduated from Texas A&M with a EE degree, it seems that you need to be carefull about misunderstanding what I am saying. I believe in the conservation of energy principle. Do you reject that principle?

What I find amazing is that you seem to be willing to accept a one-second long lossless transmission line, but not willing to accept an autotuner that requires zero time to achieve a Z0-match. You accept one ideal and reject another ideal? Seems to me you need to reject any mental exercise that doesn't exactly match reality. What's up with that? "


What part didn't you understand?

I like Aggies, even Old Aggies, so I'll cut you some slack. #But I don't recall accepting anything posted here by anyone yet.

Allow me to digress about your belief, or my belief, in any scientific principle: It doesn't matter what you or I believe. #The only thing that matters is repeatable, reproducible results. #The proof is in the pudding.

And no, I don't accept lossles transmission lines, nor instantaneous auto-tuners, nor wires that exhibit gain, nor cold fusion, nor sham-job mathematical sleight of hand however skillfully conjured.

I think if you really are an engineer, you want to be very careful about what you say. #Stick to what you can prove. #Be conservative. #Don't perpetuate pseudo-technical dogma. #Your reputation as an accurate and careful practitioner of the art depends upon it. #If you don't know something, resist the temptation to save #face by effluviating jabberwocky. #There is always going to be someone in your audience that knows you are full of...well, you know. #

If you are a theoretical researcher, with some fantastic and un-provable ideas ala Michio Kaku or other such dabblers, then you should say so, and not present your theories as fact.

If you really want to be useful, why don't you use your EE right here by explaining basic transmission line theory to the average ham out there who just wants to know the straight poop, by the book. #I'm sure many would be interested, and would benefit from having the information. #Whadd'ya say, Professor? #

KM4XG
http://www.qrz.com/iB_html/non-cgi/emoticons/wink.gif

[W5DXP]

Quote[/b] (KM4XG @ Feb. 25 2006,22:33)]And no, I don't accept lossles transmission lines, nor instantaneous auto-tuners, ...
Idealized mental experiments are a conceptual method of separating the relevant concepts from the background noise. Half of the transmission line theory presented in any textbook deals with lossless lines. It is an extremely common and powerful technique. Since you don't accept lossless transmission lines, even for purpose of discussion, then don't read those books and don't read my postings. And please don't try to interfere with the learning process of others who are willing to read such books and postings and discuss them in objective abstract terms. In addition, to be consistent, you probably should also avoid any use of abstractions like words and numbers.

I could make the transmission line one tenth of a microsecond long and lossy. I could use a signal generator with a circulator load instead of a fast autotuner. The math would be magnitudes more difficult but would yield exactly the same conclusions. You are advising me to make this stuff much more difficult to understand than it already is. Sorry, that's going the wrong way. People already complain that it is too complicated. Do you also object to Einstein's mental pictures of himself riding a light beam to see what relativistic effects he would experience?

[KM4XG]

W5DXP Writes,

"I think the original poster was trying to describe a system with a mismatched load and a Z0-match at the source provided by a tuner or other means. If the system is lossless and no reflected energy is allowed to reach the source, then there is no wasted power. Walter Maxwell said essentially that when he described lossless conjugately matched systems. That's just my take on what the original poster was trying to say."

OK...given that is what he meant to say...the concept is just wrong. #Even the best commercial high-power gas purged systems exhibit substantial losses, especially under conditions of load mis-match. #In a commercial TV installation for example, where the HPA output power might be 50 kilowatts, SWR on the line must be kept very low, averaging 1.2:1, to prevent expensive losses due to heating of the transmission line components.

The reason is simple. #As power flows back and forth between a mismatched load and an isolator, it experiences attenuation. #Every time the wave travels up or down the line, it drops by the full value of the inherent loss in the line measured in decibels per foot. #If you loose 3 dB each time the wave bounces, it doesn't take too many reflections before that power is gone...making your coax warm. So if you have a bad match when transmitting 50 KW you could end up paying a very high electric bill for the honor of heating the neighborhood with the transmission line. # That is definitely wasted power. #

There is always wasted power, in any system. #See Carnot Engine.

By postulating a lossless line, any further work based on the premise is reduced to forever being unverfiable experimentally. #Unless of course you intend to cool your coax to a superconducting state...mmm, now wouldn't that be interesting?


Face it....lines, they are not always matched to loads.

KM4XG

[KM4XG]

W5DXP writes,

"Do you also object to Einstein's mental pictures of himself riding a light beam to see what relativistic effects he would experience? "


Of course not. But Einstein never said he actually had ridden a light beam.

What I object to, and what prompted me to respond to the thread, was the initial assertions at the opening of the topic by PY1LL, "Lines, they are always matched to loads". I was alarmed by the postings of congratulations for the effort, and other signs of seeming acceptance. The work is just wrong, and is written in a style to induce the reader into believing that the work is not theoretical, nor even allegorical, but factual. It isn't. Can we agree on that?

http://www.qrz.com/iB_html/non-cgi/emoticons/rock.gif

[W5DXP]

Quote[/b] (KM4XG @ Feb. 26 2006,00:12)]What I object to, and what prompted me to respond to the thread, was the initial assertions at the opening of the topic by PY1LL, "Lines, they are always matched to loads". I was alarmed by the postings of congratulations for the effort, and other signs of seeming acceptance. The work is just wrong, and is written in a style to induce the reader into believing that the work is not theoretical, nor even allegorical, but factual. It isn't. Can we agree on that?
I think most of us recognized that the original poster was not posting in his native language and interpreted his posting within that framework. Seems to me, Luiz was simply trying to state some variation of, "My feedline tunes my antenna system", or as Walter Maxwell said, "My transmatch really does tune my antenna". But I also understand your objection. Discussing these technical concepts in my second language would be so difficult that I wouldn't even try for fear of failure.

[KM4XG]

roger.

[W5DXP]

Quote[/b] (w8ji @ Feb. 23 2006,06:42)]Since we all agree on results, that current flowing through the coil that occupys a very small fraction of a wavelenth is essentially equal, then I guess we agree on results.
One more try on this topic. Anyone can make measurements and report results. It's the CAUSE of those results upon which we disagree.

The point is: Essentially the same conditions exist in a wire carrying standing wave currents - a wire with no coil in sight. If replacing part of the wire with a coil doesn't much change those conditions, then the cause of those conditions cannot be the coil, wouldn't you agree? In fact, since the conditions don't change by much, your argument that the coil is not replacing the wire is false.

The things that are fooling you and W7EL about your measurements is that you are measuring standing wave current at the current maximum point. Standing wave current exhibits very little phase shift WHETHER IT IS IN A WIRE OR A COIL so measuring zero phase shift through the coil is meaningless. The phase shift is the same value through 1/3WL of wire. The slope of the standing wave curve is zero at the current maximum point WHETHER IT IS IN A WIRE OR A COIL so measuring equal currents at the ends of the coil is meaningless. Those things have absolutely nothing to do with the presence of a coil. ESSENTIALLY THE SAME THING HAPPENS IN A WIRE WITHOUT A COIL so the coil is NOT the cause of those conditions. In fact, one could argue that standing wave current doesn't flow at all - it just stands there, being the net sum of two traveling wave currents flowing in opposite directions. Standing wave current DOES NOT rotate like a normal phasor. It's phase is always near zero degrees for 1/2WL or 1/4WL standing-wave antennas.

Your own measurements indicate a 60 degree phase shift in the traveling wave current in a 100 uH coil at 1 MHz, contradicting your statement that there is no phase shift through a coil. The 75m bugcatcher coil causes that same large phase shift in the traveling wave currents, the forward current and reflected current.

Your own statements here say the flux density is greatest at the middle of the coil. Since current is proportional to flux density, how can you say the current in the middle of the coil is equal to the current at each end? Another contradiction.

If you locate the loading coil somewhere else besides where the slope of the standing wave current is zero, the current won't be the same at both ends of the coil. If the coil is located at the top of the antenna, the current at the top end will obviously be zero while the current at the bottom depends upon how much of 1/4WL the coil has replaced. The coil does distort the net current somewhat from the normal cosine curve but not by nearly as much as you say.

The fact that the magnitude and phase of a standing wave current is essentially the same in a WIRE WITHOUT A COIL makes your assertion that those conditions are caused by the coil a little ridiculous, wouldn't you say?

[PY1LL]

I'm sorry, but it seems clearly that the problem is very poorly understood by some guys here!
I said only simple things:
1 - The energy conservation imposes that, for lossless lines, all power delivered by the generator will reach the load and, thus, dissipated on it, and so, independently of the SWR value. The reflected power doesn't matter.
2 - for those lines, when a impedance mismatch occurs between the impedances of the line and load, that is, Zo not equal to R, the forward power is greater that the generated one.
3 - for those lines, considering the upper end of the line as a generator output, it is perfectly matched to the load, as the line output impedance is equal to R; although Zo, the intrinsic line impedance is Zo, is different from R, the line output impedance is equal to it, to guarantee the power transfer (for item 1 to hold). That why I said that "lossless lines are always matched to the load, although intrinsic impedances are different".
4 - The lossless line, so, is solely a lossless impedance transformer, but only distributed rather than lumped.

If somebody disagree with some of these items, let's discuss them, one by one, but limiting the discussion to them, or we will lose the control of the thread.

Luiz - PY1LL

[PY1LL]

KM4XG,
When you lead with common circuits, C, L, transformers, etc are elements many times considered as ideal, although they are never in real world. Lines may be analysed as ideal too, like any other circuit element. It is not a question of reality, but a conceptual issue.
But if you take a very low attenuation line, that is, very close to the ideal case, it is perfectly possible to measure all I am saying.

Luiz - PY1LL

[W5DXP]

Quote[/b] (PY1LL @ Feb. 27 2006,07:38)]1 - The energy conservation imposes that, for lossless lines, all power delivered by the generator will reach the load and, thus, dissipated on it, and so, independently of the SWR value. The reflected power doesn't matter.
Discussing only one topic: "Power delivered by the generator" is defined as the forward power minus the reflected power at the generator output. The Zg of the generator may or may not be matched at that point and the generator may be suffering massive losses at that point if considerably mismatched.

For a transmitter (generator) designed to drive a 50 ohm load, if there's not a 50 ohm Z0-match in the antenna system, the transmitter can suffer heavy losses. For instance, if a 50 ohm transmitter is driving a 300 ohm feedline terminated in a 300 ohm load, the load is matched to the line, the SWR is 1:1, but a terrible mismatch exists at the transmitter output. It is forced by the laws of physics and by the final amp designers to run at reduced power output because of that mismatch.

Defining a system as "lossless", does not include the generator.

[KM4XG]

PY1LL Writes:

"I'm sorry, but it seems clearly that the problem is very poorly understood by some guys here!"

(Yes, but I'm doing my best to try and fix that. KM4XG)

PY1LL continues:

"I said only simple things:
1 - The energy conservation imposes that, for lossless lines, all power delivered by the generator will reach the load and, thus, dissipated on it, and so, independently of the SWR value. The reflected power doesn't matter."


KM4XG replies: For analysis of power consumption and efficiency issues, ( which by the way is the entire point in radio....its all about POWER MANAGEMENT) loss in the line and SWR does matter. #You are just wrong here.


PY1LL again:
"2 - for those lines, when a impedance mismatch occurs between the impedances of the line and load, that is, Zo not equal to R, the forward power is greater that the generated one."


That's impossible. #The voltages and currents in a mis-matched line may be greater or less than the source vector quantities of voltage or current, but the total power, forward and reverse, can never exceed that provided by the generator. #Remember, a typical wattmeter measures current, not power. #It only indicates power if certain conditions exist. The meter movement itself responds only to current flow through the movement coil. #Its scale indicates power only because it has been calibrated and marked to perform the conversion from amperes to Watts. For that calibration to be correct, the line must be properly terminated in its charactersistic impedance, Zo, already. #Not understanding that can lead to all kinds of ridiculous results...such as the foregoing.


"3 - for those lines, considering the upper end of the line as a generator output, it is perfectly matched to the load, as the line output impedance is equal to R; although Zo, the intrinsic line impedance is Zo, is different from R, the line output impedance is equal to it, to guarantee the power transfer (for item 1 to hold). That why I said that "lossless lines are always matched to the load, although intrinsic impedances are different"."



This is gobbledygook ; #The result of a poorly designed experiment, incorrect assumptions, and a dash of wishfulness.



"4 - The lossless line, so, is solely a lossless impedance transformer, but only distributed rather than lumped."



The conclusion is without merit, either experimentally or theoretically, and at a stroke eliminates important other factors from consideration of the reader. #It is intellectually dishonest.

But just for the sake of completeness: #If everything you say were true, then what? #How do we use this new knowledge? #What does it enable us to do that we could not do before?

Even if what you are doing is just blue-sky conceptualizing to prove another point...what is that point? #That losses can be disregarded, that reflected power is not important, that power can be created out of nothing?

Eh?

# http://www.qrz.com/iB_html/non-cgi/emoticons/rock.gif

[PY1LL]

<"Power delivered by the generator" is defined as the forward power minus the reflected power at the generator output.>

You are right, but I considered, by hypothesis, the generator (no matter how) perfectly matched to its load, the line input, there is no reflected power to the generator.

<The Zg of the generator may or may not be matched at that point and the generator may be suffering massive losses at that point if considerably mismatched.>

How? The system here is linear and time-independent, so, if the Zg of the generator is well matched to the load, this will receive the maximum power transfer from the former. It doesn't matter what is happening 'within' the load. I guessed there were no doubts about that!

<For a transmitter (generator) designed to drive a 50 ohm load, if there's not a 50 ohm Z0-match in the antenna system, the transmitter can suffer heavy losses. For instance, if a 50 ohm transmitter is driving a 300 ohm feedline terminated in a 300 ohm load, the load is matched to the line, the SWR is 1:1, but a terrible mismatch exists at the transmitter output. It is forced by the laws of physics and by the final amp designers to run at reduced power output because of that mismatch.>

No! You are forgetting my starting hypotheses!
The generator is correctly matched to the line and this is lossless.
The mismatch that exists, is between the surge impedance of the line and the load R. But this IS NOT a mismatch between the 'generator connected to the line' and the latter. This 'generator' is what the load 'sees', or just the upper end of the line. Reflected waves correct the line output impedance to get just the value R and, so, transfer the maximum power to the load. This is forced by Ohm law: on the resistance, V/I is well defined and equal to R. As the upper end of the line is connected to R, V/I of this line end MUST be equal to R.
This is confirmed by physics: if some power is delivered from the generator to the line and this is lossless, all that the power must be dissipated by the load R, There is no other answer for this.
If you have a common lossless impedance transformer with a secondary connected to a load R, it is ALWAYS matched to the load R and generates, in the primary, a resulting impedance to the generator. It works using the turns ratio between primary and secondary to generate this resulting impedance. The latter, by hypothesis, is adjusted (no matter how) to match the generator output. In this sense, we may say that the transformer is always matched to the load.
The same occurs with a lossless line, but instead of turns ratio, it uses the internal forward and reflected waves to correctly couple energy to R and generate the resulting impedance on the other end of the line. The resulting impedance presented to the generator is matched to the generator by hypothesis. So, the line is always matched to the load as in the transformer case.
There is no difference between both cases but only the internal method the components use to transfer the resulting impedance to the generator. Both are lossless impedance transformers.

Luiz - PY1LL

[W5DXP]

Quote[/b] (PY1LL @ Feb. 27 2006,11:42)]You are forgetting my starting hypotheses!
The generator is correctly matched to the line and this is lossless.
Your starting hypotheses were not clear. If they had been clear, you would not have anyone arguing with you. Let me restate your starting hypotheses:

1. The generator sees zero reflected energy because it is Zg (or Z0) matched to the load it sees.

2. The transmission line system, connected to the generator at one end and to the load at the other end, is lossless, so this is a hypothetical example that cannot exist in the real world.