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JonnieCache

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JonnieCache
·10 năm trước·discuss
Indeed. I think that's what I'm trying to describe in my sibling comment to yours.

However, what it will successfully avoid is the loudspeaker driver attempting to instantaneously snap from some nonzero x-position back to it's origin, which is what causes the really nasty clicks.
JonnieCache
·10 năm trước·discuss
Indeed, I imagine that is what is going on. The straight line case is what the loudspeaker driver will do in reality if you send it a discontinuity, and that's what produces the pop.

Smooth interpolation will avoid a really nasty pop, but in the real world, musical waveforms are highly complex, so any interpolation algorithm, however smooth, will produce some kind of artefact if you chop the wave in the middle of a cycle and smooth it to zero.

This can be observed when setting loop points in a sampler - you are usually provided with tools to help you match the loop points to the zero crossings. This is not enough however to remove all artefacts. Only some zero crossings will do: one has to match the higher-order cycles in the waveform as well. I don't really have the mathematical vocabulary to really describe what I mean here, but hopefully it's clear.

(BTW when I say driver in these posts I mean the magnet-and-cardboard-cone assembly in the speaker, not any kind of software.)
JonnieCache
·10 năm trước·discuss
That's what I mean by interpolating towards a zero crossing.
JonnieCache
·10 năm trước·discuss
Quoting from the Waldorf microwave XT synthesizer FAQ:

The brightness of the click depends on the speed of the level change. The faster the level changes, the brighter is the click. So, the level change speed can be compared with the cutoff of a lowpass filter. There is an easy formula for it:

Let's consider a level change from full to zero (or from zero to full) output from one sample to another on a machine that uses 44.1kHz sample rate. So, we first transfer the sample to milli seconds:

1 sample equals 1/44100 second, which is = 0.02267573696ms.

To calculate the cutoff frequency of the click, just use this formula:

Cutoff (Hz) = 1000 / Level Change Time (ms)

which in the example results in:

44100Hz = 1000 / 0.02267573696ms

Whoops? This the sampling frequency and, err, very bright.


http://faq.waldorfian.info/faq-browse.php?product=xt#116
JonnieCache
·10 năm trước·discuss
Its certainly true that different hardware handles buffer underruns differently, my high-quality RME interface emits fairly muted pops when the buffer runs dry, whereas I've had cheaper units make absolutely horrendous noises as the speakers flail around trying to cope with the signal discontinuities. The RME is interpolating the signal towards a zero crossing.
JonnieCache
·12 năm trước·discuss
This is an amazing comment. The only way it could have been funnier is if you had made up the word "leptokurtic."

You should get Malcolm Gladwell's people on the phone and set up an interview.