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Author Topic: For W8JI: key clicks and amplifier non linearity  (Read 64679 times)
AC7ZN
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« Reply #375 on: September 10, 2015, 12:56:01 PM »

Glenn,

Fourier analysis of an aperiodic 3 minute CW QSO may test my abilities  Wink

Steve G3TXQ

Yeah, me too.  I'm reminded of a time early in my career when I would attend meetings where spectrum analyzer measurements were presented to show runout and vibration.  I couldn't make heads or tails of them but the other engineers around me would nod sagely.  I wondered if I would ever get this stuff.  I later realized it was all gibberish and it was a case of the emperor's new clothes.

Anyway, no three minute QSOs (I get little enough info from those anyway...my signal always seems to be 599). We will be going the other direction...here is my evil plan:

I am attempting to answer the unwritten (but I think implied) question of this thread:  "I get it that an abrupt keying transition causes clicking sounds in my receiver, and know that adding a low pass filter to the keying waveform  cures it,  but what makes the actual click?  Looking at the keying waveform there is sure enough a sharp edge, but why does it sound like a click?  Can a spectrum of the click itself help me understand this?"

Now, you certainly know what makes the click, but to my knowledge, no one in the ump-t-ump posts here has presented the actual spectrum of the click.  This, then, is my intent:  to zero in on the click itself with the Fourier integral and get a spectrum that gives insight as to why we hear a clicking sound, and why we can hear it many 100's of Hz or even kHz away.  This spectrum is difficult to catch with a spectrum analyzer, even the new gated ones, because it comes and goes so fast.

If nothing else, the exercise will give us (me especially) a review of the Fourier integral and what we can do with it.

73,
Glenn AC7ZN


 
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SM0AOM
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« Reply #376 on: September 10, 2015, 01:40:00 PM »

Very early in this thread there was a brief discussion of what constitutes a "key click".
To my knowledge, there is no formal definition, but i tried to predict what I believe that the ITU would answer if an Interim Working Party of the Study Group 1 should be set up:

"Higher order keying sidebands, created by deviations from the optimum shaping of the keying envelope,  causing the occupied bandwidth of an A1A or A1B radiotelegraph emission to become much greater than its necessary bandwidth as established by the relevant relations between signalling rate and bandwidth set out in Appendix 1 of the Radio Regulations and in ITU-R Recommendation SM.328"

This may serve as a starting point.

73/
Karl-Arne
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G3TXQ
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« Reply #377 on: September 10, 2015, 01:47:51 PM »

Now, you certainly know what makes the click, but to my knowledge, no one in the ump-t-ump posts here has presented the actual spectrum of the click.  This, then, is my intent:  to zero in on the click itself with the Fourier integral and get a spectrum that gives insight as to why we hear a clicking sound, and why we can hear it many 100's of Hz or even kHz away.  This spectrum is difficult to catch with a spectrum analyzer, even the new gated ones, because it comes and goes so fast.

If nothing else, the exercise will give us (me especially) a review of the Fourier integral and what we can do with it.
Glenn,

I will certainly be interested in the results - particularly in how you would then seek to apply them more generally to a complete Morse character.

The more I ponder this the more it becomes clear what separates the two "camps". I'll go back to periodic functions again (because I find the DFT easier than the more general Fourier Integral) and consider repeated Dots:

I think we all agree that if we look at the DFT of a short time slice taken around the leading edge of the Dot we'll see a wideband spectrum; let's call it Spectrum A. Then if we look at a time slice of the Dot period we'll see a different spectrum - it'll be a single spectral line at the carrier frequency; call it Spectrum B. Then we look at a time slice around the trailing edge of the Dot and get a wideband Spectrum C. And finally look at the spectrum of the Off period - nothing - call it Spectrum D.

We can look at that analysis and say that the spectrum of the periodic signal is clearly a sequence of changing spectra:
Spectrum A > Spectrum B > Spectrum C > Spectrum D > Spectrum A > Spectrum B > Spectrum C > Spectrum D ....... etc

However we can also look at that signal, recognise that it is periodic, and take a time slice equal to a complete period (Dot plus gap). In that case we get a completely different Spectrum - let's call it Spectrum X.

The spectrum of the periodic signal is then: Spectrum X > Spectrum X > Spectrum X ...... etc
In other words a steady, continuous, spectrum with constant amplitude spectral line for as long as the signal exists.

Some may then ask: "Which is the real one?" To which the answer - as so often - is "it all depends"!

W6RZ usefully reminded us earlier of the duality between time resolution and bandwidth. If we observe that signal with a wide-bandwidth spectrum analyser we will get good time resolution and will see the sequencing, changing, spectra. But if we look at it with a very narrow bandwidth spectrum analyser we would see just one continuous spectrum.

So the answer to the question  "Which is the real one?" is:  "It depends on how you observe it"!

Most of our differences have been because we've been viewing the same signal from those two different perspectives.

I have a feeling another contributor said something very similar a zillion posts ago and we ignored him. My apologies!

Steve G3TXQ

« Last Edit: September 10, 2015, 01:53:27 PM by G3TXQ » Logged
K9AXN
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« Reply #378 on: September 10, 2015, 01:56:08 PM »

You'll have to be patient here Steve,

Reference #302

As I understand it, you have a carrier generator A, connected to a diode switch B, and the output from the switch is represented by the top trace.

The bottom trace is rendered by connecting a crystal filter to the output of the switch which is pulsed carrier and probing at the output of the filter.

Is this correct and if so, what is the time duration of the on off bursts?  Also is the filter the 1.4Mc crystal filter that you referred earlier?


Kindest regards Jim
« Last Edit: September 10, 2015, 01:58:30 PM by K9AXN » Logged
G3TXQ
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Posts: 1845




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« Reply #379 on: September 10, 2015, 02:12:14 PM »

As I understand it, you have a carrier generator A, connected to a diode switch B, and the output from the switch is represented by the top trace.

The bottom trace is rendered by connecting a crystal filter to the output of the switch which is pulsed carrier and probing at the output of the filter.

Is this correct and if so, what is the time duration of the on off bursts?  Also is the filter the 1.4Mc crystal filter that you referred earlier?
All correct!

The filter is centred on 1.4MHz and has a bandwidth 2.6kHz [Post #338 refers]
The keying frequency was 5kHz [Post #353 refers]
The impulse response of the filter is shown in #354

Steve G3TXQ
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K9AXN
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« Reply #380 on: September 10, 2015, 02:52:51 PM »

Steve,

Don't believe I can key that fast. 

I believe you just explained the carrier being present between keying periods.  I asked you earlier if it could have been ringing and you said no.

You drove that filter with 1.4Mc 5Kc bursts of full amplitude square wave modulation.  That's 10us on and 10us off with a gap of only 10us between exposure to full carrier amplitude.  The experiment and measurement is patently invalid.

Isn't It a bit optimistic to think that a filter with a Q exceeding 10,000 would not ring through a 10us period.  In fact it did, and that is what your basing your supposition.

Need to call the troops back and regroup.

Please have a good day.  Kindest regards Jim
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G3TXQ
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« Reply #381 on: September 10, 2015, 02:59:21 PM »

It's getting late here, so if I may, I'll try to guess where Jim's questions are leading and pre-empt them.

Edit: posted this just before I saw Jim's question - looks like I guessed correctly Wink

We know we can extract energy from the continuous signal at the output of that 1.4MHz filter, even when the switch is Off; so to that extent it is "real" - but is it an artefact of the filter "ringing" through the Off period, excited by the trailing edge of the input signal?

I say "no" and offer the following evidence:

1) The time-domain equation says we should see a time-invariant signal without needing to invoke explanations of filter ringing.

2) There is no evidence in the filter output waveform of any "decay" during the Off period that would be consistent with the measured impulse response of the filter.

3) The amplitude of the filter output signal is exactly one half of the carrier amplitude at all times. How can that be if the filter is simply "allowing the carrier through" during the On period, and then ringing during the Off period? Why would the filter output be half the carrier amplitude during the On period of the switch?

4) The amplitude of the filter output signal is exactly one half of the carrier amplitude, confirming the 0.5xAcxsin(ωct) predicted by the maths.

5) More generally, if the existence of individual spectral components is attributed to filter "ringing", how is it that we can filter out the carrier from a sine-wave modulated AM signal and find that it is constant amplitude even through the periods when the input signal is at a trough? It's very difficult to believe that the filter is ringing as a result of being excited by a sine-wave modulated input signal.

Steve G3TXQ

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G3TXQ
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« Reply #382 on: September 10, 2015, 03:45:39 PM »

Isn't It a bit optimistic to think that a filter with a Q exceeding 10,000 would not ring through a 10us period.

Maybe, but since the Q is 538 that's irrelevant Wink

Q= fc/Δf = 1.4MHz/2.6kHz = 538

Steve G3TXQ
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K9AXN
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« Reply #383 on: September 10, 2015, 05:15:29 PM »

Isn't It a bit optimistic to think that a filter with a Q exceeding 10,000 would not ring through a 10us period.

Maybe, but since the Q is 538 that's irrelevant Wink

Q= fc/Δf = 1.4MHz/2.6kHz = 538

Steve G3TXQ

Steve,

Your calculation regarding the Q of a crystal filter is patently incorrect --- no absurd.

The apparent Q of a crystal filter is based on the dimensions of the skirts not the 2.4Kc width of the 3db pass band.
If your calculation method was correct, the shape factor of your filter would be astronomical and resemble a pair LC tuned circuits which would not even be light years from distinguishing the difference between a carrier and 5Kc  sideband @1.4Mc.  

Now that you are aware of the behavior of a crystal filter, you can retrace your steps and rerun your test using the appropriate inputs and measurement techniques.  

I would suggest a 100 cycle on off, depending on the skirt selectivity of your filter, so you don't get tangled up in your calculations.  Using 100 cycle on off times will provide a good idea as to the rise and quench time of that filter.  Note there is a spool up time in the filter as well.   Remember we were addressing key clicks not audio.  if you need further help to understand crystal filter behavior G3RZP or W8JI are good tutors.

Now lets look at #336 the, the photo you copied from a book.  You seem to understand the concept.  Now replace the red modulation line with a pair of two positive Square waves separated by an off condition.  What do you think it will morph to?  Do it in your head it's simple.  You don't have to draw it just explain it.

BTW, how do you add a photo to the text?   Someone please.

G-day to you --- Kindest regards Jim      
« Last Edit: September 10, 2015, 06:04:42 PM by K9AXN » Logged
K9AXN
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« Reply #384 on: September 10, 2015, 06:01:21 PM »

It's getting late here, so if I may, I'll try to guess where Jim's questions are leading and pre-empt them.

Edit: posted this just before I saw Jim's question - looks like I guessed correctly Wink

We know we can extract energy from the continuous signal at the output of that 1.4MHz filter, even when the switch is Off; so to that extent it is "real" - but is it an artefact of the filter "ringing" through the Off period, excited by the trailing edge of the input signal?

I say "no" and offer the following evidence:

1) The time-domain equation says we should see a time-invariant signal without needing to invoke explanations of filter ringing.

2) There is no evidence in the filter output waveform of any "decay" during the Off period that would be consistent with the measured impulse response of the filter.

3) The amplitude of the filter output signal is exactly one half of the carrier amplitude at all times. How can that be if the filter is simply "allowing the carrier through" during the On period, and then ringing during the Off period? Why would the filter output be half the carrier amplitude during the On period of the switch?

4) The amplitude of the filter output signal is exactly one half of the carrier amplitude, confirming the 0.5xAcxsin(ωct) predicted by the maths.

5) More generally, if the existence of individual spectral components is attributed to filter "ringing", how is it that we can filter out the carrier from a sine-wave modulated AM signal and find that it is constant amplitude even through the periods when the input signal is at a trough? It's very difficult to believe that the filter is ringing as a result of being excited by a sine-wave modulated input signal.

Steve G3TXQ



#1 You are apparently misinterpreting the Time domain equation.  I see nothing that excludes crystal filter behavior when they are used to confirm a theory.  On the contrary, simple logic would include any components that would subvert the outcome of a test.

#2  Once you get your arms around the behavior of a crystal filter you'll change your mind.

#3  A 3db loss in a crystal filter is not so bad --- most are more.

#4  Answered with #3.  You have again forgotten to acknowledge all of the relevant considerations to a simple equation.  The math works but you have to provide correct input.

#5  You need a tutor regarding crystal filters --- Again try G3RZP he may be local.

Please have a good evening and Kindest regards Jim
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AC7CW
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« Reply #385 on: September 10, 2015, 07:55:00 PM »

Glenn,

Fourier analysis of an aperiodic 3 minute CW QSO may test my abilities  Wink

Steve G3TXQ

Yeah, me too.  I'm reminded of a time early in my career when I would attend meetings where spectrum analyzer measurements were presented to show runout and vibration.  I couldn't make heads or tails of them but the other engineers around me would nod sagely.  I wondered if I would ever get this stuff.  I later realized it was all gibberish and it was a case of the emperor's new clothes.

Anyway, no three minute QSOs (I get little enough info from those anyway...my signal always seems to be 599). We will be going the other direction...here is my evil plan:

I am attempting to answer the unwritten (but I think implied) question of this thread:  "I get it that an abrupt keying transition causes clicking sounds in my receiver, and know that adding a low pass filter to the keying waveform  cures it,  but what makes the actual click?  Looking at the keying waveform there is sure enough a sharp edge, but why does it sound like a click?  Can a spectrum of the click itself help me understand this?"

Now, you certainly know what makes the click, but to my knowledge, no one in the ump-t-ump posts here has presented the actual spectrum of the click.  This, then, is my intent:  to zero in on the click itself with the Fourier integral and get a spectrum that gives insight as to why we hear a clicking sound, and why we can hear it many 100's of Hz or even kHz away.  This spectrum is difficult to catch with a spectrum analyzer, even the new gated ones, because it comes and goes so fast.

If nothing else, the exercise will give us (me especially) a review of the Fourier integral and what we can do with it.

73,
Glenn AC7ZN


 


If the click comes from the ramped waveform of the keying envelope then it's pretty much a straight line ramp and it will be a fundamental and all the harmonics. How you get a fundamental frequency of a ramping signal is beyond me though. What if it just ramps forever, where is the fun(damental) in that?
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G3TXQ
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« Reply #386 on: September 11, 2015, 02:22:39 AM »

Your calculation regarding the Q of a crystal filter is patently incorrect --- no absurd.

From one of my engineering text books: "A band-pass filter can be characterised by its Q factor. The Q-factor is the inverse of the fractional bandwidth." Fractional bandwidth = Δf/fc, so Q= fc/Δf

I would suggest a 100 cycle on off, depending on the skirt selectivity of your filter, so you don't get tangled up in your calculations.  Using 100 cycle on off times will provide a good idea as to the rise and quench time of that filter.  Note there is a spool up time in the filter as well.   Remember we were addressing key clicks not audio.

I already showed you the result of that measurement!

Now lets look at #336 the, the photo you copied from a book.  You seem to understand the concept.  Now replace the red modulation line with a pair of two positive Square waves separated by an off condition.  What do you think it will morph to?  Do it in your head it's simple.  You don't have to draw it just explain it.

That's easy - it will morph to the same diagram but with many more spectral components, each spaced by the modulation frequency. Each spectral component will be continuous even through the period when the modulated carrier is Off.

Steve G3TXQ

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G3TXQ
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« Reply #387 on: September 11, 2015, 02:41:26 AM »

#1 You are apparently misinterpreting the Time domain equation.  I see nothing that excludes crystal filter behavior when they are used to confirm a theory.  On the contrary, simple logic would include any components that would subvert the outcome of a test.

You miss my point - the time domain equation predicts the existence of a constant amplitude carrier without invoking any crystal filter behaviour.

Here's the time domain formula again:
fo(t) = Ac.sin(ωct) x [0.5 + 2/π [sin(ωst) + 1/3 sin(3ωst) + 1/5 sin(5ωst) + ......  etc  ........]]

The Ac.sin(ωct) term is the carrier; the terms in brackets are a signal toggling between 0 and +1 at the switching rate. We are multiplying the two together so that we get a carrier which toggles on and off at the switching rate. Do you notice that there is nothing here representing "filter ringing" and yet the equation predicts the existence of a 0.5 x Ac x sin(ωct) term?

I really don't know how to make it any plainer!

#3  A 3db loss in a crystal filter is not so bad --- most are more.

I'm comparing the filter output amplitude with a keyed carrier, to the filter output amplitude when the carrier is not keyed. The filter loss is common to both and therefore irrelevant to the discussion.

[PS: although it's irrelevant, since when has a halving of amplitude equated to a 3dB loss HuhHuh]

#5  You need a tutor regarding crystal filters --- Again try G3RZP he may be local.

It would be helpful if you try to answer the questions rather than make personal comments. I'll try it again:

Consider the very simple case of a carrier AM modulated with a sine-wave. If we filter that signal to exclude the two sidebands we get a constant amplitude spectral line at the carrier frequency. It's real, we can extract energy from it, it's not "just a mathematical construct"! Try it with a receiver tuned across an AM signal. Now:

Q1: How can that carrier component be constant amplitude when the input signal is a carrier that is varying in amplitude?
Q2: How can the instantaneous power in the filtered carrier component be constant, even when the instantaneous power in the input signal is zero?
Q3: Over the first modulation half-cycle, how can the energy in the filtered carrier component be so much less than the energy in the input signal.
Q4: Over the second modulation half-cycle, how can the energy in the filtered carrier component be so much higher than the energy in the input signal.


If you can think through satisfactory answers to those basic questions you'll have a much better understanding of the underlying principles.

Steve G3TXQ



« Last Edit: September 11, 2015, 05:29:07 AM by G3TXQ » Logged
K9AXN
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« Reply #388 on: September 11, 2015, 10:06:09 AM »

Your calculation regarding the Q of a crystal filter is patently incorrect --- no absurd.

From one of my engineering text books: "A band-pass filter can be characterised by its Q factor. The Q-factor is the inverse of the fractional bandwidth." Fractional bandwidth = Δf/fc, so Q= fc/Δf

I would suggest a 100 cycle on off, depending on the skirt selectivity of your filter, so you don't get tangled up in your calculations.  Using 100 cycle on off times will provide a good idea as to the rise and quench time of that filter.  Note there is a spool up time in the filter as well.   Remember we were addressing key clicks not audio.

I already showed you the result of that measurement!

Now lets look at #336 the, the photo you copied from a book.  You seem to understand the concept.  Now replace the red modulation line with a pair of two positive Square waves separated by an off condition.  What do you think it will morph to?  Do it in your head it's simple.  You don't have to draw it just explain it.

That's easy - it will morph to the same diagram but with many more spectral components, each spaced by the modulation frequency. Each spectral component will be continuous even through the period when the modulated carrier is Off.

Steve G3TXQ



Point #1 The calculation of Q.  Your engineering textbook is correct for simple resonant circuits.  It in no way considers complex clustered filters i.e crystal filters.  I was not being facetious when I recommended a tutor regarding crystal filters.  If you would kindly cite the engineering text book that you found the Q formula, I'll be happy to point you to the chapter regarding Dampened oscillations --- ringing and the formulas used to calculate the durations. 

Your assertion that your 1.4Mc crystal filter has a Q of 538 is totally absurd.  A simple way to calculate the Q is to use the shape factor of your 1.4Mc filter which I would expect to be no more than 2/1.  Separate the width of the rise and fall  from 3db to 60db.  For a shape factor of 2/1 the combined skirt width would be 2.4Kc.  Now design a single 1.4Mc series resonant LC circuit that has a skirt width of 2.4Kc at -60db.  Nope, you won't be able to do it.  but if you could you would find the 3db point to be microscopically narrow and the Q in the 10,000 range.

The net of what I'm saying is that you are using a crystal filter to validate your assertions, --- the wrong component.  The formulas are correct --- your experiment is badly flawed.  Use a heterodyne scheme, it will render correct results.

You need to think before doing proof of concept.  Like it or not that filter is ringing through with a 10us gap between 10us square bursts.  Look at your example where you used a .5 second interval.  Note the carrier did in fact go to zero.  That was because the gap exceeded the ringing effect of the filter --- or does that mean there must be some flaw in the formula or maybe the test setup was incorrect.  Formulas look good to me.     

Try an analogy.  You attach a generator to a motor and wire the generator to the motor.  Then spin it up --- voila perpetual motion.  sounds good feels good its exciting but just doesn't keep going --- resistance will stop it.  What kept it going?  Stored inertial energy ---- sorta like RINGING.  Nothing is free not even carrier Steve.

If your engineering or physics book doesn't have the chapter regarding dampening oscillations and spin up time calculations, let me know and I'll send a copy of the chapter.  You actually can do the rough calculations using a theoretical series resonant LC circuit.   


 

Have a good day Steve.  Kindest regards Jim       
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G3TXQ
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« Reply #389 on: September 11, 2015, 11:28:09 AM »

Jim,

Are you going to answer my questions? If we can agree the answers we don't need to debate Q definitions.

Look at my 0.5 second interval results - there is a delay (not ringing) common to both the leading and the trailing edges; but the "ringing" drops significantly in the 100uSec Off period of the keyed signal. Why do you keep talking about 10uSec gaps - a 5kHz square wave is 100uSec On followed by 100uSec Off?

Steve G3TXQ
« Last Edit: September 11, 2015, 11:40:16 AM by G3TXQ » Logged
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