Is this claim actually true?
My understanding is that if you take a waveform and clip it, the resulting waveform actually carries less energy (think of the corresponding integral), but more of that energy is pushed into the higher frequencies. It's this – the unexpectedly large amount of high-frequency energy – that kills speakers because their crossover networks push it into the tiny, tiny tweeters, and they are utterly unprepared for it.
> Is this claim actually true?
Yes, it is -- but it depends on how we define "clipped".
> My understanding is that if you take a waveform and clip it, the resulting waveform actually carries less energy (think of the corresponding integral)
Only if the clipping reduces the peak value. If you compare a sinewave with a peak value of 1, and a square wave with a peak value of 1, the square wave has a substantially higher average level (with a ratio of pi / 2).
> It's this – the unexpectedly large amount of high-frequency energy – that kills speakers because their crossover networks push it into the tiny, tiny tweeters, and they are utterly unprepared for it.
Yes -- the rate at which the speaker cones are required to move is an additional factor. But for a "clipping" definition that clips by means of trying to exceed the available voltage, these two effects add.
http://i.imgur.com/oE5NFZ9.png
In the above linked image, the red trace is sin(x), the integral for the interval 0 < x < pi is 2. The green trace produces an integral of pi. The ratio of the two is pi/2, and the speaker power difference is (pi/2)^2 = 2.46 (because the speaker's power is the square of the applied voltage).
The green trace is what you would get if you simply turned up the volume beyond any reasonable setting -- the amplifier produces a clipped version of the sine wave and the peak value is equal to the supply voltage.
I don't know of any widely used definition of "clipping", nor any definition that fits the context of this discussion, that allows for a signal's peak value not to be reduced. It's called clipping because the extreme values look to have been clipped away, as if by scissors. The parts that have been clipped away contain energy, don't they? So won't the clipped signal will carry less energy than the original?
When you take a sine wave of peak value 1 and clip it, what you get is not a square wave with a peak value of 1. Rather, you get a peak-truncated sine wave in which values in excess of the clipping threshold V are replaced with the clipping threshold. There's less power in this clipped sine wave than in the original because min(V, |sin t|) <= |sin t| for all t.
But there is one. In electrical engineering, it's any process that arbitrarily limits a signal's amplitude. By far the most common meaning is a signal that exceeds the voltage range of an amplifier or signal pathway.
> nor any definition that fits the context of this discussion, that allows for a signal's peak value not to be reduced.
See above.
> It's called clipping because the extreme values look to have been clipped away, as if by scissors.
Yes, but this can result from trying to pass a signal too large for the circuit, or it can mean an intentional scheme in which a fixed signal amplitude is truncated, the meaning you're discussing.
In the present discussion, in which a volume setting is increased until the speakers are jeopardized, the meaning is clear -- it's an increase in signal amplitude that the amplifier cannot support, resulting in the waveform being clipped at the maximum available amplifier voltage.
> The parts that have been clipped away contain energy, don't they? So won't the clipped signal will carry less energy than the original?
Not if the signal amplitude is increased. In the present discussion, the problem is being caused by raising the volume level too high, which causes the signal to exceed the available amplifier voltage. My diagram shows this case:
http://i.imgur.com/oE5NFZ9.png
The red trace is the maximum volume setting that the amplifier can support without distorting the signal. The green trace is a much higher volume setting that essentially reduced the output to a square wave. In both cases, the peak voltage is the same.
> When you take a sine wave of peak value 1 and clip it, what you get is not a square wave with a peak value of 1.
That depends on how you define "clip". If you increase the size of the sinewave, clipping takes place at the maximum voltage. If you clip by reducing the possible range of voltages, the sinewave remains the same size but maximum amplitude goes down. The present discussion revolves around the first of these choices.
