1
$\begingroup$

There is a technique called clipping in sound synthesis. It is explained on Wikipedia .

I make music and like this technique: You make an extreme "fat" noise, maybe with a lot of resonance, but then it sounds somehow too loud. Then you can use clipping to make the sound less loud.

Now, I ask myself whether clipping to a value $m$ really guarantees that there is no "particle" oscilating with more peak-to-peak amplitude than $m$. Let's say the speaker has reached the part where clipping shall start. Then it suddenly stands still. However, due to inertia, won't the particles in air keep the movement, and thus reach a higher amplitude?

Practical application of the question is whether clipping kind of protects your ears. For me, clipped sounds often sound louder than non-clipped. E.g. a square wave (which is close to a clipped very loud sine wave) sounds louder than a sine wave, both with the same amplitude. Why?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Clipping to m means the maximum (zero-topeak) value does not exceed m. Clipping is usually the reult of a nonlinear component which saturates beyond some point. For example, the dynamic gain of an amplifier 'flattens out' to zero when the input signal exceeds its maximum. In practice, this does not occur instantaneously, but very quickly. Similarly, the cone of a speaker is very light (has low inertia) so whilst it doesn't stop instantaneously, it stops very quickly. In any case, if the value does not exceed m, it is 'clipped to m'.

Now the 'loudness' of a sound signal is not determined by the peak voltage, but by the average power in the signal, which is proportional to the rms of the voltage. A square wave of a given peak magnitude will have a higher average power (rms voltage) than a sine wave with the same peak-to-peak magnitude.

In a sense, the 'loudness' of the sound is governed by the volume of air that is moved during every cycle, which is proportional to the area under the curve of the half-cycle, that is, the rms (square root of the mean of the square) of the voltage signal.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! I think I got it. Just for interest: Do you know whether ear damage only depends on the maximum volume amplitude, or also on the power? $\endgroup$
    – Johannes
    Dec 22, 2012 at 20:43
  • $\begingroup$ Ear damage is the result of accumulated exposure, that is, it depends on the total time the ear is exposed to high sound power levels. For example, damage to the ear may begin with 2 hours of exposure to 100dB at a loud concert, and builds up with say 8 hours of exposure to 85dB at work from using power tools. The damage to the ear is cumulative over a lifetime. $\endgroup$
    – theo
    Dec 22, 2012 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.