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Author Topic: • Favorite Old Wives' Tales •  (Read 63194 times)
AA4PB
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« Reply #15 on: November 12, 2011, 11:27:18 AM »

On JT65 and other narrow band digital modes you may not have to "hear" them with your ears but certainly the software decoder needs to "hear" the other station. Even with PSK31 you can sometimes "print" stations when you can't "hear" the station with your ear. That's because you are listening to the signal through a very wide filter thus the signal to noise ratio is very poor (lots of noise). The decoder however is essentially listening through a very narrow filter which strips away most of the noise, giving the decoder a much better signal to noise ratio.

CW ops have experienced this affect for years. Listening to a weak signal through a 3KHz filter the signal is buried in the noise and interference. Switch in a 200Hz filter and the CW signal appears to pop right out of the noise.
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Bob  AA4PB
Garrisonville, VA
W8JI
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« Reply #16 on: November 12, 2011, 03:06:11 PM »

On JT65 and other narrow band digital modes you may not have to "hear" them with your ears but certainly the software decoder needs to "hear" the other station. Even with PSK31 you can sometimes "print" stations when you can't "hear" the station with your ear. That's because you are listening to the signal through a very wide filter thus the signal to noise ratio is very poor (lots of noise). The decoder however is essentially listening through a very narrow filter which strips away most of the noise, giving the decoder a much better signal to noise ratio.

CW ops have experienced this affect for years. Listening to a weak signal through a 3KHz filter the signal is buried in the noise and interference. Switch in a 200Hz filter and the CW signal appears to pop right out of the noise.


A good CW op can filter with his brain. I know many people, including myself, who can decode CW when the bandwidth is wide and noise power greatly exceeds signal power. Like anything, it just takes practice. The human brain is a very good filter-decoder.

For those less trained, a narrow filter helps make the CW more obvious.








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G3TXQ
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« Reply #17 on: November 13, 2011, 01:25:45 AM »

Many years ago I was involved in assessing the performance of military CW operators under varying S/N conditions - the purpose was to set a "benchmark" for a new modulation scheme we were developing. The operators could choose whichever receiver bandwidth they felt most comfortable with; they varied from 3kHz down to 200Hz if I recall correctly. As a young engineer, I was quite surprised that few of them ever went narrower than 1kHz.

The other surprise was to watch one of them copying random 5-character groups at 25wpm and writing them down with one hand, whilst at the same time filling in "The Times" cryptic crossword with the other hand!

Steve G3TXQ
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W8JI
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« Reply #18 on: November 13, 2011, 04:33:04 AM »

Many years ago I was involved in assessing the performance of military CW operators under varying S/N conditions - the purpose was to set a "benchmark" for a new modulation scheme we were developing. The operators could choose whichever receiver bandwidth they felt most comfortable with; they varied from 3kHz down to 200Hz if I recall correctly. As a young engineer, I was quite surprised that few of them ever went narrower than 1kHz.

The other surprise was to watch one of them copying random 5-character groups at 25wpm and writing them down with one hand, whilst at the same time filling in "The Times" cryptic crossword with the other hand!

Steve G3TXQ

Absent QRM, BW I use depends on noise. For rough noise I use wider BW, maybe 1 kHz. For smooth noise, I use less BW. Maybe 200Hz or less for very smooth white noise.

There is a little distortion from level problems, but this is a stereo sample of a 9V1 QSO on 160 meters. He was running 10 watts :

http://www.w8ji.com/Sound/9v1goNov0303.wav 

more files:
http://www.w8ji.com/dx_sound_files.htm

I started experimenting with diversity in the 1970's.  I finally gave up on combining methods because of the slow rolling phase shift on 160, and started using phase locked (common oscillators) receivers in the late 70's and early 80's.

My antennas now are about 2500-3000 foot spacing for the two channels, with nearly identical patterns in both systems. At 3000 feet there is as much as 10 dB level difference at times, and phase rotates constantly but slowly. With training a brain can add the signals, making another 6 dB or so S/N improvement when both antennas are hearing the same signal at about the same level, even when it is below noise floor.

73 Tom
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W8JI
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« Reply #19 on: November 13, 2011, 05:02:23 AM »

By the way, as long as someone finally responded to the "bait thread" we might as well discuss the bait:


1.  A half-wave dipole radiates mostly from where the RF current is max.  

That is true. Radiation is cause by charge acceleration, and that is tied to ampere-feet of spatial distance.  An antenna, in the area where current is maximum and charges are moving over maximum spatial distance without opposing movement, radiates the most.

Radiated power is by the square of current for an equal spatial distance.

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2.  12" wide Cu strap has virtually zero L.


Silly bait. A 12" wide strap has zero L at zero length. It behaves as a transmission line at higher frequencies as length is longer, so it might even look capacitive.

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3.  Directly grounding grid pins with wide Cu strap completely eliminates VHF oscillations.


It will in many cases, but it depends on the system.

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4.  In order to suppress VHF parasitics a VHF suppressor has to "look right".  

It has to electrically "look right" to the system. Looking at the suppressor by itself and not the entire system is silly.

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5.  A 40% increase in P out is definitely a big advantage on HF.


It can be indeed. When a signal is near or in noise, even one dB can be major.


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6.  A HV fuse and a grid fuse will protect 8877s, 3cx800A7s,8939s. and similar tubes from gold evaporating from the grid during an "oscillation condition".  


The only place gold evaporates exclusively from parasitics is at AG6K's "laboratory". At all other places, it is from kinetic energy of electrons striking the grid.


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7.  You still have to hear 'em to work 'em.
 

Yes you do, unless a computing system is decoding.

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8.  Some transmitting tubes simply  do not have enough feedback-C to sustain oscillation.  

Absolutely.


 
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9.  One can obtain non-inductive resistors on this planet.


Relative to a given frequency, yes we can.

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10.  Receiving 160m DX signals requires a full-wave water well casing ground.


Poor bait that makes no sense.

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11.  When not enough space is available, 1/4 wave 160m meter radials can be coiled and laid on the ground under a bush.


Yes they can. They just won't be effective as a ground.


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12.  VHF parasitic oscillations will never take place with a recognized expert's amplifier layout.


Bad bait. That makes no sense because the layout is not defined.

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13.  The FWB rectifier is better than a FWD rectifier.


Depends on the application. Each has pro's and con's.


73 Tom
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AG6K
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« Reply #20 on: November 13, 2011, 01:02:39 PM »

By the way, as long as someone finally responded to the "bait thread" we might as well discuss the bait:


1.  A half-wave dipole radiates mostly from where the RF current is max.  

That is true.

  This is not what is taught at the University of California's schools of engineering.

Quote
Radiation is cause by charge acceleration, and that is tied to ampere-feet of spatial distance.  An antenna, in the area where current is maximum and charges are moving over maximum spatial distance without opposing movement, radiates the most.

Radiated power is by the square of current for an equal spatial distance.

  What about P=E^2/R ?

Quote
Quote
2.  12" wide Cu strap has virtually zero L.


Silly bait. A 12" wide strap has zero L at zero length.

  chortle.  So does #44awg. 

Quote
It behaves as a transmission line  at higher frequencies as length is longer, so it might even look capacitive.

  the hydra grows another head.

Quote
3.  Directly grounding grid pins with wide Cu strap completely eliminates VHF oscillations.


Quote
It will in many cases, but it depends on the system.

  when in doubt, add another "system".  .

Quote
Quote
4.  In order to suppress VHF parasitics a VHF suppressor has to "look right".  

It has to electrically "look right" to the system. Looking at the suppressor by itself and not the entire system is silly

  How does the VHF suppressor itself affect VHF amplification Tom ?

.
Quote
Quote
5.  A 40% increase in P out is definitely a big advantage on HF.


It can be indeed.

  So why can't anyone tell on the air?

Quote
When a signal is near or in noise, even one dB can be major.


Quote
6.  A HV fuse  and a grid fuse will protect 8877s, 3cx800A7s,8939s. and similar tubes from gold evaporating from the grid during an "oscillation condition".  


The only place gold evaporates exclusively from parasitics is at AG6K's "laboratory".

  http://www.somis.org/FooteL.GIF

Quote
At all other places, it is from kinetic energy of electrons striking the grid.

  now there's a new one.

Quote
7.  You still have to hear 'em to work 'em.
 

Yes you do, unless a computing system is decoding.[/quote]

  such sites have computer controlled HF \/MF receivers.  They listen where and how you tell them to,

Quote
Quote
8.  Some transmitting tubes simply  do not have enough feedback-C to sustain oscillation.  

Absolutely.

 The XC of the feedback path in an 8877 is 4200-ohms at its F-max rating.  Is it good engineering practice to ignore –j4200-ohms between the output and the input of a tube with an amplification factor of 200?


Quote
Quote
9.  One can obtain non-inductive resistors on this planet.


Relative to a given frequency, yes we can.

  not according to Terman. 

Quote
Quote
10.  Receiving 160m DX signals requires a full-wave water well casing ground.


Poor bait that makes no sense.

  It's what the dude claimed.  There were a # of witnesses Tom.


Quote
11.  When not enough space is available, 1/4 wave 160m meter radials can be coiled and laid on the ground under a bush.


Yes they can. They just won't be effective as a ground.[/quote]

  How effective?


Quote
12.  VHF parasitic oscillations will never take place with a recognized expert's amplifier layout.


.
Quote
Bad bait. That makes no sense because the layout is not defined

  I have never heard any of our recognized amplifier experts define what a good layout is.  Jim, VE7RF recently came up with the idea of moving the VHF parasitic resonance below the grid-resonant freq by increasing L between the DC blocker C and the Tune-C so that the parasitic-resonance would be in the tube's neutralized region.  I do not see any reason why it would not improve VHF stability. 

Quote
13.  The FWB rectifier is better than a FWD rectifier.


Quote
Depends on the application. Each has pro's and con's.

  agreed
Rich, ag6k


Quote
73 Tom
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G3RZP
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« Reply #21 on: November 13, 2011, 02:33:43 PM »

>This is not what is taught at the University of California's schools of engineering. <

What about Maxwell's displacement current? Take away the dipole, put two electrodes a half wave apart, feed them in antiphase with a voltage. It radiates in the same way as half wave dipole. But where's the current?

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W8JI
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« Reply #22 on: November 13, 2011, 03:11:04 PM »

Since it took about 5-10 pages with the photon silliness, and I'm not sure you ever agreed with everyone else on that, we better do this one misconception at a time.  :-)

1.  A half-wave dipole radiates mostly from where the RF current is max.  

That is true.


  This is not what is taught at the University of California's schools of engineering.

While California seems to have more than a fair share of odd ideas, I find that hard to believe. This is something defined way back in the 1800's that has never been proven wrong. All antenna modeling programs even use current in each segment to calculate EM fields.

Everything from physics textbooks to EM theory right down to the ARRL Handbooks tell everyone current defines EM radiation. That is because the force at a distance know as EM radiation is caused by charge acceleration.

This is why radiation resistance is directly tied to the ampere-feet.

So if you are right, and all the texbooks and engineers are wrong, that must mean efforts to make current high and uniform along a distance of space occupied by a conductor a waste of time. End loading or top loading is illogical or unnecessary, according to you.  

Radiation is cause by charge acceleration, and that is tied to ampere-feet of spatial distance.  An antenna, in the area where current is maximum and charges are moving over maximum spatial distance without opposing movement, radiates the most.

Radiated power is by the square of current for an equal spatial distance.

  What about P=E^2/R ?

That's fine for dissipation in a resistance, because it translates to current, but it is not the mechanism that causes EM radiation.

Chapter 2 of Jasik's  Antenna Engineering Handbook might do you some good, or any good physics book.

You probably can even find info on-line. You might search for Lamor's equation or Lamor formula.

Lamor's equation states a charged particle radiates when accelerated. It further quantifies total radiated power as being proportional to the square of the acceleration.

It's all about the ampere-feet. That's why, if we bend a dipole, we should bend it in the low current (high voltage) area to minimize impact on radiation resistance. EM radiation is proportional to the square of the current, or linear with spatial distance for a constant current.

73 Tom
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W8JI
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« Reply #23 on: November 13, 2011, 03:14:40 PM »

>This is not what is taught at the University of California's schools of engineering. <

What about Maxwell's displacement current? Take away the dipole, put two electrodes a half wave apart, feed them in antiphase with a voltage. It radiates in the same way as half wave dipole. But where's the current?



They don't have displacement current in California, at least according to Rich. :-)

They do have an abundance of grid dip meters that measure impedance, and photons that cause oscillations. :-)

This is gonna be a good one, Peter. If you think the Rp took forever, and the photons were a mess, wait 'til the conversation about displacement currents start. :-)

« Last Edit: November 13, 2011, 03:28:35 PM by W8JI » Logged
AF6LJ
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Posts: 582




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« Reply #24 on: November 13, 2011, 05:15:21 PM »

Tom;
And some people wonder why I want the Hell out of this state. There is something in the water, even the ground water, or maybe it's the air. You do know the state forced all the municipalities to put fluoride in the water, I wonder what else was added to the water to make this whole state nuts.

Some of the things I hear on 7.255 in the afternoon remind me of my Ex chatting to his good buddies on 27.395.

As for Amplifier Layout; it's been discussed maybe not here, I can recall one thread over on my main haunt where the subject was discussed in detail.

We don't have a lot of things in The People's Republic of Kalifornia, common sense and sound electronics theory seem to be in short supply. As for the rest of the bait; I keep coming here to see what is going to happen next.
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Take Care
Sue,
AF6LJ

Don't Kalifornicate My Life
W8JI
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« Reply #25 on: November 13, 2011, 06:00:13 PM »

And some people wonder why I want the Hell out of this state. There is something in the water, even the ground water, or maybe it's the air. You do know the state forced all the municipalities to put fluoride in the water, I wonder what else was added to the water to make this whole state nuts.

I know some really great people out there, but the people with totally whacked-out ideas sure are set in their ways. There is a fellow out there who tested an electronic bias system by running audio into the RF detector, and no amopunt of logic or reasoning can get him to understand that a 30 Hz audio signal into a 3.8 MHz SSB transmitter makes a 3.80003 signal, not a 30 Hz signal.

This antenna thing is going to be a good one, and then there is the magical GDO that can predict impedances in the future. Never mind measuring S21 or S12 with power off to look at grid shielding effectiveness. :-)

Quote
Some of the things I hear on 7.255 in the afternoon remind me of my Ex chatting to his good buddies on 27.395.

I don't know what 7.255 is supposed to be. I probably don't want to know. 

Quote
We don't have a lot of things in The People's Republic of Kalifornia, common sense and sound electronics theory seem to be in short supply. As for the rest of the bait; I keep coming here to see what is going to happen next.


I reckon after Maxwell and Faraday spin in their graves a few hundred times, one lone soul will remain convinced the rest of the world is wrong.

73 Tom
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KC9TNH
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« Reply #26 on: November 14, 2011, 05:27:42 AM »

5.  A 40% increase in P out is definitely a big advantage on HF.
Not when it was coffee-generated & you're in the middle of a net.
 Cheesy
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73
Wes -KC9TNH
"Don't get treed by a chihuahua." - Pete
AG6K
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« Reply #27 on: November 14, 2011, 06:57:42 AM »

Since it took about 5-10 pages with the photon silliness, and I'm not sure you ever agreed with everyone else on that, we better do this one misconception at a time.  :-)

1.  A half-wave dipole radiates mostly from where the RF current is max.  

That is true.


  This is not what is taught at the University of California's schools of engineering.

While California seems to have more than a fair share of odd ideas, I find that hard to believe. This is something defined way back in the 1800's that has never been proven wrong. All antenna modeling programs even use current in each segment to calculate EM fields.

Everything from physics textbooks to EM theory right down to the ARRL Handbooks tell everyone current defines EM radiation.

   RE:  A bundle of Electromagnetic / EM radiation - a.k.a. a photon : has two parts.  The Electric field - E - part and the Magnetic field - H - part .  Take away either the E (voltage) field or the H (amperage) field and there is no EM radiation / photons.
   Myths die a painfully slow death.  Example: "witch" burning.  The malpractice began c. 1231 and was not discontinued until 1830. 

Quote
That is because the force at a distance know as EM radiation is caused by charge acceleration.

This is why radiation resistance is directly tied to the ampere-feet.

So if you are right, and all the texbooks and engineers are wrong,

  The EE textbooks used at UCLA, UCSB, and UC Berserkley must be wrong too Tom R.

Quote
that must mean efforts to make current high and uniform along a distance of space occupied by a conductor a waste of time. End loading or top loading is illogical or unnecessary, according to you.

  the hydra grows yet another head.  

Radiation is cause by charge acceleration, and that is tied to ampere-feet of spatial distance.  An antenna, in the area where current is maximum and charges are moving over maximum spatial distance without opposing movement, radiates the most.

Radiated power is by the square of current for an equal spatial distance.

  say what?

  What about P=E^2/R ?

That's fine for dissipation in a resistance, because it translates to current, but it is not the mechanism that causes EM radiation[/quote]

  So I^2 x R is okay but E^2 / R is nonsense ?

Quote
Chapter 2 of Jasik's  Antenna Engineering Handbook might do you some good, or any good physics book.

  Modern physics books say that photons have an E field from electrical potential - a.k.a. voltage - and they have a magnetic/H field from electric current.
•••  Rich, ag6k
•  You can't teach old dogmatists new tricks. •


Quote
You probably can even find info on-line. You might search for Lamor's equation or Lamor formula.

Lamor's equation states a charged particle radiates when accelerated. It further quantifies total radiated power as being proportional to the square of the acceleration.

It's all about the ampere-feet. That's why, if we bend a dipole, we should bend it in the low current (high voltage) area to minimize impact on radiation resistance. EM radiation is proportional to the square of the current, or linear with spatial distance for a constant current.

73 Tom
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W8JI
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« Reply #28 on: November 14, 2011, 07:11:53 AM »

Rich,

This everywhere, including ARRL Handbooks if you don't have engineering books.

The cause of EM radiation is always an accelerating charge. Always.

Charges are current, and to satisfy Maxwell's equations with the fictitious electric dipole we have displacement currents. In real antennas, radiation is related directly to current and distance. This is why, to maintain resistance, antennas are top loaded. 

So apparently you also seem to think voltage causes heat in a resistor, and not current?

You seem to be confusing the fact an EM wave has a force on other charges that can be defined as an electric field or magnetic field, with how that effect is created. This puts you in line with the inventor of the hoax CFA antenna, and the hoax EH antennas. :-)

So the ARRL handbooks and all the people writing textbooks are wrong, and voltage and not charge acceleration causes EM radiation?? The high current areas of the antenna are not primarily responsible for radiation, so all the radiation resistance formulas based on current distribution are also incorrect. Also, none of the computer modeling programs for antennas work, because they calculate EM radiation exclusively by current?

When are you going to straighten out the rest of the world on this?   We can replace Maxwell's equations with Measure's equations soon?  :-)

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G3TXQ
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« Reply #29 on: November 14, 2011, 07:52:26 AM »

Quote from: W8JI
We can replace Maxwell's equations with Measure's equations soon?  :-)
Tom,

I don't think you need to worry yet! I see that UCLA's course EE162A "Wireless Communication Links and Antennas" still lists the very first 'course outcome' as the ability to "Recite Maxwell's equations, boundary conditions and their physical meaning."

Then lower down the list: "Understand how a method of moment code is written and use it to calculate the characteristics of a half-wave dipole antenna." and lower down again: "Know the current distribution and polarization states of a large loop antenna."

I wonder why they would think it important for students to understand the current distribution along an antenna?

Steve G3TXQ
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