Author Topic: Amateur radio operators here?  (Read 10392 times)
Medved
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Re: Amateur radio operators here? « Reply #60 on: December 07, 2020, 05:29:34 AM » Author: Medved
One could say "good old spark transmitter". They are radiating all over the band, disturbing pretty everything...

And banned since about 100 years ago...
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Binarix128
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Re: Amateur radio operators here? « Reply #61 on: December 07, 2020, 06:48:59 AM » Author: Binarix128
One could say "good old spark transmitter". They are radiating all over the band, disturbing pretty everything...

And banned since about 100 years ago...
So I better keep that thing out of the air.  :o But it's very unlikely that an aircraft or military radar could listen to it, the thing barely broadcasted 3 meters and I had to turn the volume all the way up lo listen something, the signal was weak because the battery was dead.

The reason I think why it disturbs all the frequencies is because it starts with a huge spike from the ballast collapse and then the signal looks like the one in the picture, creating harmonics all over the band from a certain point.
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Medved
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Re: Amateur radio operators here? « Reply #62 on: December 07, 2020, 04:04:18 PM » Author: Medved
It is wide band because laws of mathematics:
If you make something limited in frequency band, it will become infinite in time.
If you make something limited in time, it will become infinite in frequency band.

A single pulse is limited in time and has very limited in time edges, it therefore must occupy the whole band.
If you restrict the frequency by e.g. resonant circuit filter, you get prolongation in time (aka the ringing you see on the oscilloscope)...

The good thing about disturbing others is, practically all modulation use some form of bandwidth limitation, alias averaging over longer time. There the single pulse, with all its energy spread all over the band, does not create much. But if the pulses would be repeated (at few 100's Hz as was the case with the real spark gap transmitters), the energy would start to concentrate into narrow bands/peaks in the spectrum
Another math thing: Infinitely periodic in time means spectrum are spikes 1/period apart across the frequency band, so all energy concentrated i to those spikes. That is the reason for the ban: The spark gap transmitters operated at 100's frequencies at the same time, so occupied 100's frequencies, but the receiver could pick the energy just from a single one. So the spectrum was very wasted that way. And once electronics came into radio technology, and with that all the filtering and a clean signal generation capability, it was really pointless to occupy the whole spectrum by just few transmitters, when 100's could make connections with the then new technology. So such waste of the valuable band space was banned by the agreements...
 
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Binarix128
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Re: Amateur radio operators here? « Reply #63 on: December 07, 2020, 04:49:39 PM » Author: Binarix128
Quote
It is wide band because laws of mathematics:
If you make something limited in frequency band, it will become infinite in time.
If you make something limited in time, it will become infinite in frequency band.
So that's why square signals are horrendously noisy, because it passes from 1 to 0 in practically no time, and probably why current spikes creates infinite harmonics, because of the short time the event happens.

Quote
A single pulse is limited in time and has very limited in time edges, it therefore must occupy the whole band.
If you restrict the frequency by e.g. resonant circuit filter, you get prolongation in ti e (aka the ringing you see on the oscilloscope)...
Note that the signal you in the oscilloscope is not actually the one happening on my circuit, I took it from internet for illustrate what I thought was happening. What I think it's happening is that the signal starts with a huge spike which actually causes the infinite harmonic, caused by the collapse of the magnetic field into the coil when I disconnect the battery, then due to the huge spike, the inductance of the ballast and the capacitor I added they start to oscillate and the iron core adds extra spikes increasing the harmonic effect. So the signal would start with a huge negative spike and a spiky R-C oscillation.
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Re: Amateur radio operators here? « Reply #64 on: December 07, 2020, 05:13:32 PM » Author: Medved
So that's why square signals are horrendously noisy, because it passes from 1 to 0 in practically no time, and probably why current spikes creates infinite harmonics, because of the short time the event happens.
Note that the signal you in the oscilloscope is not actually the one happening on my circuit, I took it from internet for illustrate what I thought was happening. What I think it's happening is that the signal starts with a huge spike which actually causes the infinite harmonic, caused by the collapse of the magnetic field into the coil when I disconnect the battery, then due to the huge spike, the inductance of the ballast and the capacitor I added they start to oscillate and the iron core adds extra spikes increasing the harmonic effect. So the signal would start with a huge negative spike and a spiky R-C oscillation.

The main signal source is disconnection of the battery, so the contacts. That is a sharp turn off of a current. So it could be seen as a turn on of the same current (same in value, opposite in polarity to the original, so since the switching time these two cancel out) into the impedance formed by the rest of the circuit in the switch-off state. Andyou may do some math to realize the single edge of a function (the current vs time) means the frequency spectrum (amplitude vs frequency) would be proportional to 1/f (so a continuum in frequency, so why it excites every resonant filter). Again a math.
All the rest is just a filtering of this single event. High inductance with low capacitance means very high impedance, hence the high voltage.
Because nearly everything tends to resonate somewhere, the filter will have many resonance spikes, each forming its own damped sinewave fromthe initial pulse.
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Re: Amateur radio operators here? « Reply #65 on: December 07, 2020, 05:35:38 PM » Author: Binarix128
So you mean that the current of the battery and the current reaction of the ballast cancel and then many sinewaves start to oscillate by their own causing a mixture? But I'm a bit confused. How can the current generated by the ballast cancel the current of the battery that already passed? Would be good if you post a drawing of how you think the signal would look like, for understand it better.
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Re: Amateur radio operators here? « Reply #66 on: December 07, 2020, 06:03:30 PM » Author: Medved
So you mean that the current of the battery and the current reaction of the ballast cancel and then many sinewaves start to oscillate by their own causing a mixture? But I'm a bit confused. How can the current generated by the ballast cancel the current of the battery that already passed? Would be good if you post a drawing of how you think the signal would look like, for understand it better.

It is an aid to think about the math behind, to help you to grab the math more intuitively.
Because it is the impedance after the switch off what matter, but then there is no current. Quite unintuitive to think about what causes the current once it does not flow anymore.

So the aid works like this:

If you have two current sources in parallel, their currents sum up.
If the first source is a constant, lets say +1A,
and the second source is a step from 0 to -1A,
their sum would be a step from 1A to zero, so matching what really your battery and switch are doing.
Now because we have a linear system (voltages are proportional to currents exciting them), we may analyze the system separately for each virtual source and sum up the resulting voltage waveforms and we get the same response:
The constant 1A makes nothing on an inductor except the DC shift onthe wire resistance, so its response is a DC voltage. But that wont radiate, so we may quite easily neglect it at all.
The second source then forms 0 -> -1A step, flowing into the bunch of wire with R, L, C all along it. So for evaluating the radiation, we need to evaluate how the circuit (with the switch open) responds to this step.
So you may form a bunch of differential equations and solve them,
Or you may transform everyrhing into a frequency domain and just multiply the impedance/transfer/radiation impedances (depends what you want to calculate) at each frequency by the given frequency component of the source signal.
Now we know the source is a (unity) step, so has a 1/freq spectrum.
The system itself is just a bunch of resonances, so a bunch of spikes with their respective Q and peak value.
By multiplying these two functions you get the spectrum of either the coil voltage or of what is being radiated.
Because there is always one frequency with the highest impedance, sticking above the others (the fundamental resonant frequency), its response will be what plays the dominant role on the output, so dictates the decaying oscillations, similar as on the scope picture you presented.
And if caring about radiation, the picture of both E, as well as H field component will look very alike.
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Re: Amateur radio operators here? « Reply #67 on: March 08, 2021, 11:40:10 AM » Author: Binarix128
I'm bout to set up my own experimental tv channel, I'm waiting for the camera and signal amplifier to arrive.
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