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

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

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  1. G3TXQ

    G3TXQ Ham Member QRZ Page

    Tim,

    I applaud your attempts to analyse this in the frequency domain :)

    I hope you can accept that it is what I see on a 'scope. Here's a very crude qualitative way of looking at what's happening:

    Assume the capacitor is discharged.
    Apply a 5v "step" to the input of the filter.
    The capacitor cannot charge instantaneously, so that 5v step is communicated immediately to the output. i.e. the voltage drop across the capacitor is zero.
    The voltage across the capacitor now begins to rise exponentially as it charges with time constant RC.
    So the output voltage drops exponentially from 5v to zero


    Now step the input voltage back to zero. Again the capacitor cannot charge instantaneously, so the -5v step is communicated to the output. The voltage across the capacitor now begins to rise exponentially, so the output voltage rises exponentially to zero.

    Now repeat 10 times a second and you'll see exactly the same pulses. Do it at 100 Hz and you'll still see the same pulses.

    Only when the keying rate gets high enough that the C can't charge or discharge within the time period does anything begin to change.

    Now here's an even sillier qualitative way of looking at it. As we reduce the keying rate the power in these high-order components gets less and less; but the pesky little sidebands, even though they are all weaker, conspire to produce the same pulse amplitude and shape - it's just that, being weaker, they can't manage it as often. That should have Fourier turning in his grave :)

    I'll leave you to think about a rigorous frequency-domain explanation, but I hope you can see from my description that is what must happen in the time domain.

    73,
    Steve
     
  2. AB0WR

    AB0WR Ham Member QRZ Page

    Steve,

    Here is a web site that should show how this works.

    http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/diff/diff.html

    Change the resistor value to 3300 ohms and the capacitor to 100nf. That's pretty close to what you have.

    This is what is called a differentiating circuit.

    What you will see as you lower the frequency below 480hz is the impulse spike at the leading and falling edge followed by the exponential.

    The rise and fall time of that exponential is determined by the high freq components your filter lets through.

    Those components appear to be dominated by the harmonics of the square wave.

    If they were not being dominated by the harmonics of the square wave but, instead, by the Fourier components of the step change then the rise and fall time after each impulse would never change since exactly the same high freq components would be let through on each step-up and step-down edge of the square wave.

    Instead, as you lower the frequency, the harmonic-number of the components getting through become higher and higher thus resulting in a steeper slope and a quicker fall-off toward Zero.

    My guess is that, if you could expand the range on the graph, that you wouldn't see the step-change freq components dominate until you are well below 1hz.

    I still think it is those impulses that should be concentrated on as the culprits for causing "key clicks". That impulse can cause all kinds of problems in amplifiers with non-linearities.

    tim ab0wr
     
  3. G3TXQ

    G3TXQ Ham Member QRZ Page

    Tim, just to be clear - the shape of that exponential recovery does not change provided the frequency is below, say 200Hz. It is determined solely by the value of the R and C.

    Confusingly, the on-line demo doesn't alter the way it displays the input square wave, so the pulse looks like it gets shorter as the frequency drops. But it doesn't! What's really happening is that the pulse is staying the same and the square wave is getting longer.

    Which is then entirely consistent with the pulse being determined by the "Fourier components of the step change".

    73,
    Steve
     
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