[sdiy] A question about Chorus
Richie Burnett
rburnett at richieburnett.co.uk
Sun Sep 1 02:12:49 CEST 2013
Sampling is equivalent to amplitude modulation (multiplication) with an impulse train for the carrier.
An impulse train has a spectrum that is a series of impulses in the frequency domain. So when you modulate audio by the impulse train carrier (sampling) you get a spectrum that consists of infinitely repeating images of the audio centred around multiples of the sampling rate. That's why you usually follow DACs with anti-imaging filters to remove the excess images from the output spectrum.
http://fourier.eng.hmc.edu/e161/lectures/sampling/node3.html
With this picture in mind it's easy to see why aliasing takes place if the audio input's bandwidth exceeds half the sampling rate. The multiple images all start to overlap. And there is now nothing that any anti-imaging filter can do to separate them afterwards and reconstruct the original input signal.
Digital sampling performs quantization and discretization. I guess BBD sampling just does the discretization.
-Richie
Sent from my Sony Ericsson Xperia ray
rsdio at sounds.wa.com wrote:
>You're probably right, at least on some points. I certainly didn't
>mean to imply that FM, RM, and sampling are all exactly the same in
>every respect.
>
>Digital sampling does introduce quantization error, which is not part
>of FM or RM.
>
>I disagree with your characterization of a BBD as having analog
>quantization. A BBD works by storing a voltage in a capacitor, and
>there is nothing about that circuit which forces quantization. A
>capacitor can hold any possible voltage (within its maximum limit),
>only quantized at the Coulomb level when you consider discrete
>electrons. However, any analog circuit is quantized at the discrete
>electron level, so a BBD is no more quantized than a single-pole low-
>pass filter. There is certainly error when reading the voltage out of
>a capacitor stage in a BBD, and this error is compounded 256 or 1024
>or 4096 times as the signal passes through the long brigade of
>buckets. However, this signal error is not a quantization error.
>There is no mapping of values into a smaller set, nor is there rounding.
>
>There is utility is looking at the ideal sampling process, as if
>quantization did not occur. In that case, sampling is basically the
>same as ring modulation, at least in terms of what happens in the
>frequency domain and not the amplitude (quantization). Analog
>sampling is closer to RM than digital sampling.
>
>Also, within small ranges, phase modulation and frequency modulation
>overlap. Of course, they do not overlap over the full range of
>possible depths of modulation.
>
>Brian
>
>
>On Aug 31, 2013, at 15:13, Richard Wentk wrote:
>> Nope.
>>
>> Sampling is the same as passing a signal through a sample and hold.
>> This is not the same as ring-mod with a square wave or FM, no
>> matter what frequencies you use, because the sampling introduces
>> quantisation errors and resulting noise.
>>
>> Ring-modulation is amplitude modulation - i.e. instantaneous four-
>> quadrant signal multiplication.
>>
>> FM is (more or less) phase modulation. If you do wide FM on a
>> general audio signal, you have to do it +/- an average delay.
>>
>> Just because they all add extra frequencies *does not* mean they're
>> the same process.
>>
>> You can of course use a delay line for phase mod. That's why it's
>> so easy to use delay lines for phasing and flanging. With a faster
>> modulation rate and wider depth this does indeed turn into FM.
>>
>> The others aren't FM.
>>
>> BBDs are complex things. The sound of a BBD is a mix of analog
>> quantisation slop and noise, sample aliasing, the analog filtering
>> around the BBD, other non-linear distortions - I suspect these play
>> a big part in the sound, as does the fact that the BBD cells are
>> slightly leaky - and the 'pure' theoretical output of a delay effect.
>>
>> I suspect there's also some frequency drift in most modulation
>> oscillators, which adds some further detail. As does the fact that
>> modulation sine waves are probably not all that perfect, so dm/dt
>> is going to be more complex than a plain cos curve.
>>
>> Richard
>
>
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