# HF Digital Error correcting? Also, what's up with PSK31?

Discussion in 'General Technical Questions and Answers' started by N0NS, Oct 9, 2008.

Not open for further replies.
1. ### AB0WRHam MemberQRZ Page

go to:

http://www.kepcopower.com/nomoslew.htm

and take a look at the nomograph.

tim ab0wr

2. ### AB0WRHam MemberQRZ Page

What you are saying makes sense. But see if the following analysis makes sense as well.

The maximum slope of a sin wave occurs when t=0 or t=pi. (i.e. a zero crossing)

So let's start with a sin wave of the form (A)sin(wt) where A is the amplitude.

d(Asin wt)/dt = (Aw)cos(wt)

and at t = 0 or t=pi, cos(wt) = 1 or -1 (depending on whether the voltage is rising or falling).

That means the slope of the sin wave is (Aw).

For a fixed w, i.e a constant frequency, the slope of the sin wave varies as A, the amplitude.

If you raise the amplitude you also raise the slope of the sin wave.

This keeps the rise time constant for any amplitude of a fixed frequency sin wave.

However ---
in answer to LYK's question. You *can* lower the bandwidth if you keep the (Aw) product constant. If you double 'A' then you can halve 'w' in order to keep the same slope. Lowering 'w' results in a lower bandwidth but will, of course, result in doubling the rise time.

This is where I have a problem with the assertion that the slope of rise time drives the sidebands. It's *not* just a rise time slope issue. It's a lot more complex than that.

tim ab0wr

Last edited: Oct 28, 2008
3. ### AB0WRHam MemberQRZ Page

Yep, if you double the rise time then you halve the bandwidth required.

I think we've been talking at cross-purposes here.

I've probably mis-stated some conclusions in my hurry to get things answered.

Can we agree on this:

Certainly the slope of the rise time *is* dependent on the amplitude *and* frequency, the 'Aw" product. Therefore saying that the sidebands generated from a keying pulse is dependent on the slope of the rise time is a questionable assertion at best. The exact same frequency component can provide a significantly different slope depending on the amplitude of the signal.

The *rise time* is a different issue. As a first approximation it is 1/3 the period of the highest effective frequency component in the signal.

tim ab0wr

4. ### AB0WRHam MemberQRZ Page

As RAN continues to point out, there is no "time to put up or shut up" with you, Cecil.

You've sidetracked the discussion just about as far as you can. You've marginalized yourself in your passion to distract everyone from the assertions you originally made.

I'm tired of feeding the troll.

tim ab0wr

5. ### G3TXQHam MemberQRZ Page

Tim, glad to see you are now agreeing with me after so many posts. I have never used the term "slope of the risetime" - I don't even know what it means. How can a "time" have a "slope"? I have only ever talked about the "risetime" which has an unambiguous engineering definition, and which has a one-to-one relationship with "cut-off frequency".

Remember, I began my contribution to this thread to try to answer your challenge:
"I would love to see a coherent answer to WA0LYK's question about why, if rise time causes the sidebands, don't we just raise the keying voltage in order to extend the rise time and lessen the harmonics?"

No mention there of "slope of the risetime". Of course it may be you were using your own definition of "risetime" rather than the normal engineering definition.

I think we are now agreeing, but just to be sure let me conduct a "thought experiment" and see if you agree my observations.

Let's say I apply a perfect square wave to a simple low-pass RC filter and examine the output on a 'scope and a spectrum analyser. Lest there be any misunderstandings I will use the usual engineering definition of risetime, and I will take bandwidth to mean the point by which the high frequency components have fallen in amplitude to some prescribed fraction of the fundamental.

First I examine the leading edge of the square wave on the 'scope and note the risetime, and I also note the bandwidth on the SA. Now I increase the amplitude of the squarewave by a factor of 10 and repeat the measurements; for convenience I will also reduce the Y sensitivity on the 'scope by a factor of 10. This is what I observe:

1) The waveform shape hasn't changed
2) The risetime hasn't changed
3) The slope of the leading edge (measured in volts/sec) has increased by a factor of 10.
4) The bandwidth hasn't changed

Now I double the value of the resistor R, and I observe:

5) The waveform shape has changed
6) The risetime has doubled
7) The bandwidth has reduced

Conclusions:

8) Increasing the amplitude had no effect on the risetime, indicating that the phrase "..... we just raise the keying voltage in order to extend the rise time ......" in the original question was flawed.
9) Increasing the amplitude had no effect on bandwidth, so raising the keying voltage will not "lessen the harmonics".
10) Increasing the risetime had a direct impact on bandwidth.

Are there any of the numbered observations with which you disagree? If not, I believe I have met your original challenge.

73,
Steve

6. ### W5DXPHam MemberQRZ Page

I can't believe you are passing up that easy \$1000 bet won by finding a tree whose circumference to diameter ratio is pi accurate to ten decimal places.

When a human being looks at a cloud and says, "There's a rabbit! See the ears!", there is not really a rabbit there - the rabbit is completely in that human mind's eye. When you look at a tree stump and say, "There's pi!", that pi is completely in your mind's eye existing only as an abstraction. In a way, the "rabbit" in the clouds is more real than the pi that you see. One can at least take a picture of the "rabbit" in the clouds.

I have been trying to get you (and others) to recognize the difference in something existing as real matter/energy in the real world and something existing as an abstraction in a collection of human minds. Abstractions and concepts did not exist before humans evolved as, to the best of my knowledge, we are the only species capable of abstractions and concepts.

Last edited: Oct 28, 2008
7. ### AB0WRHam MemberQRZ Page

Oh, we agree. I was certainly confused for a time in what I was saying.

Did I make a challenge? I forget.

What is leading this discussion are statements like these:

As I think you've shown, the shape of the rise and the fall time is not a determining factor of the bandwidth of the signal since the shape (i.e. the slope) is dependent on both amplitude and frequency. The duration is the main determining factor.

The only unresolved issue now is whether the sidebands exist only for the rise time or if they exist for the duration of the signal.

Does the sequency domain give any insight into this or is it like the frequency domain where there is no time factor?

tim ab0wr

8. ### KI4NGNPremium SubscriberQRZ Page

I'll jump in and say only for the rise time.

No math for me: if I start a carrier with a 1khz tone attached, and then stop the tone, the sidebands disappear when the tone stops. It would be impossible to communicate otherwise.

It doesn't matter if I transmit several wavelengths of that tone, or just a fraction of a wavelength: as soon as the tone signal is gone, so is the sideband.

Mike

9. ### G3TXQHam MemberQRZ Page

Phewwww!!
Well, I was always with you on this one. I'm well versed in a similar argument: when I first began work I had a senior engineer who contentiously claimed that he'd never seen a DC signal - it must have started at some time and was likely to stop at some time. It therefore generated a component in the frequency domain, albeit a very low frequency

All I've been trying to do is assert that the risetime (usual engineering definition) does determine the bandwidth in cases such as a CW signal which is filtered sub-optimally i.e. it is not band limited to the minimum necessary for the information rate.

Well, the sequency components have to exist for the duration of the signal, as do the frequency components. It has to be so because that's the premise on which a Fourier Transform works - the components it identifies are sinusoids that are present and constant for the duration of the analysis period. Take away the high-order sidebands for any part of the analysis period and you'll destroy the mid part of the square wave in addition to changing the leading and trailing edges.

For anyone who doesn't like the math, take a look at Jim's (WA0LYK) posting earlier. I found it an overwhelmingly convincing, pratical, argument.

Now for a practical experiment. I fed a square wave into a first-order RC high-pass filter this morning and listened to the output on a loudspeaker. The filter had a cut-off of 500Hz, so the low-order "sidebands" were lost and I was listening to just the high-order components - akin to being 500Hz away from an unfiltered keying signal.

Here's a recording you can listen to:
http://www.karinya.net/g3txq/sidebands.mp3

0-9 seconds is a square wave at about 2Hz - it's clearly a "click" folowed by silence. It sounds like 4Hz because you get a click off both the leading and trailing edge.

11-22 seconds. Same thing at about 5Hz - still discernable as a "click" followed by silence.

25-48 seconds. Gradually increasing the frequency - eventually it is discerned as a "rasping noise"

50 seconds onwards. Hand keying - for homework see if you can read the CW

This confirms my earlier statement that the high-order components clearly contain the keying cadence.

73,
Steve

10. ### G3TXQHam MemberQRZ Page

Mike - I presume you are talking about AM modulation of the carrier? If so, of course what you say is correct. But your example doesn't represent the debate we're having.

Take the example where your carrier is being AM modulated with a square wave. On a spectrum analyser you would see high-order sidebands that are contributing to the "sharp" leading edge. They are also contributing to the overall composition of the whole square wave. They don't just exist for the duration of the leading edge.

Your example is akin to saying that when I stop modulating with the squarewave the sidebands disappear - which is of course correct - but that is a different issue.

73,
Steve