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Author Topic: A Modulator for Rice Boxes  (Read 17262 times)
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wavebourn
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« Reply #25 on: April 19, 2006, 08:24:35 PM »

Wavebourne - plot cos(f0)+((cos(2f0))/2)

Bacon - I can plot the entire Fourie transformation for a rectangular signal of the big porosity. Then, I can multiply it on a function of the single R-C differentiating network to get rid of asymmetry.

Asymmetrical vaveform means DC presence. It is gone after the 1'st coupling capacitor or transformer. Due to non - linearity of output tubes and of a tube you modulate it is back again because of distortions. Also, dynamic resistance of a power supply gives us some. It is simple like 2x2.

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Bacon, WA3WDR
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« Reply #26 on: April 19, 2006, 09:19:21 PM »

No, this is just a time plot of two cosine waves added together, one twice the frequency of the other, and the higher frequency one about 1/2 the amplitude of the lower frequency one.  When you do this, there is no DC offset.  The waveform has a relatively wide, flat negative cycle at a medium amplitude, and a relatively narrow spike of higher amplitude on the positive side.  Here there is no distortion, just two sine waves added as they would be in an audio mixer, but you get an asymmetrical output waveform.  But if you do this with sine waves instead of cosine waves, you get a strange but symmetrical output waveform.  With harmonic combination, the relative phase is important.

Or think of a square wave.  Its peak value is equal to its RMS value.  But a sine wave's peak value is 1.414 times its RMS value.  Take the positive peak of a sine wave of peak amplitude of +1.414V, and then switch to the negative peak of a square wave of -1V peak amplitude.  You have the same energy in the positive and negative peak, which means no DC offset, but you have a 1.414:1 waveform asymmetry.  But in this case you have nasty distortion.

The key is equal watt-seconds per half-cycle.  If they are the same, you have no DC, but the peaks do not have to be the same.  When you have even-harmonic addition in the correct amplitudes and relative phase, you get significant asymmetry, and the voice has a lot of that.
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Truth can be stranger than fiction.  But fiction can be pretty strange, too!
wavebourn
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« Reply #27 on: April 19, 2006, 09:32:47 PM »

No, this is just two cosine waves added together, one twice the frequency of the other.  There is no DC offset.  The waveform has a relatively wide, flat negative cycle at a medium amplitude, and a relatively narrow spike of higher amplitude on the positive side.

Or think of a square wave.  Its peak value is equal to its RMS value.  but a sine wave's peak value is 1.414 times its RMS value.  Take the positive peak of a sine wave of peak amplitude of +1.414V, and then switch to the negative peak of a square wave of -1V peak amplitude.  You have the same energy in the positive and negative peak, which means no DC offset, but you have a 1.414:1 waveform asymmetry.

What? Positive half-sine of 1.414 and negative half-sine of 1 and no dc? Do not make laugh my slippers!, please. Cheesy

Quote
The key is equal watt-seconds per half-cycle.  If they are the same, you have no DC, but the peaks do not have to be the same.  When you have even-harmonic addition in the correct amplitudes and relative phase, you get significant asymmetry, and the voice has a lot of that.

You do not transmit just a lalf of a cycle, you transmit the whole Hhhyaaallloooooowwwwwfffff! Smiley

Try to invert your voice signal and ask some DX-er to compare S-readings.

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Bacon, WA3WDR
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« Reply #28 on: April 20, 2006, 02:33:50 AM »

No - positive half of sine, negative half of square, same energy in each, but different peak amplitudes.

Anyway, Bob's circuit soft-clips peaks on one side, which is set up to be the negative side, and some distortion does result.  We all find that we need to do something like that to prevent overmodulation by occasional peaks.  This is the only way a symmetrical wave like a sine wave can be made asymmetrical, but most voice waveforms are asymmetrical.  When the asymmetry of the voice is aligned with the negative peak soft-clipping of a circuit like Bob shows us, distortion is light, and modulation is louder.

When a waveform is soft-clipped on one side like this, there is some rectification, and a DC component is introduced.  Bob's circuit is DC coupled after the asymmetrical soft-clipper, so this will result in some positive carrier shift.  This DC offset is sometimes removed by AC coupling after the soft-clipping.  Peak control is reduced when this is done, but extra aymmetry is still produced, because one peak is dented.

Natural voice aymmetry is reduced if low frequency response rolls off and attenuates the lower voice frequencies.  Also, some voices are not very asymmetrical.  But with low frequency response down to 50 to 80 Hz or so, most voices are significantly asymmetrical.
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Truth can be stranger than fiction.  But fiction can be pretty strange, too!
wavebourn
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« Reply #29 on: April 20, 2006, 01:46:29 PM »

Bacon;
to align asymmetry of voice you do not need pre-distortions, you need just one coupling capacitor or one transformer. You do not need even a capacitor; the entire mic is acting as a high-pass filter. Sometimes when it's low end frequency is too low so called pop-screen helps.

For better sound quality predistortion should have different characteristics than just rectification to increase positive peaks. If you want to fool FCC and transmit more energy it is fine. But you may transmit no carrier neither one of sidebands, and it will be absolutely legal. But please do not call it AM then.

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