[sdiy] samp. freq for ADSR ?

[Robert Holmström]

Thanks a lot for you answer, although now I want to know more than I thought 
I did... I will be back.

Tack igen!

Robert

>
>Hej Robert,
>
> > > > > Creating a 500Hz sampling frequency. Nyquist's theory is that the
> > > > > maximum sampled frequency is half the sample rate, leaving the 
>Buchla
> > > > > MARF with a maximum of 250Hz (real low for audio).
> > > >
> >
> > I am not an expert in this so maybe somebody can explain it (simple) but
> > this is how I understand it.
>
>I may fail on the simple part for parts of my discussion, but maybe you can
>survive by skiping some of the text (theory corner) on the first reading.
>
> > As written above the maximum sampled frequency is half the sample rate. 
>This
> > means that if I want a 1Hz (low, just as an example) signal sampled, the
> > minimum samplingfrequency should be 2Hz.
>
>That's correct.
>
> > That sampled signal would be a 1Hz "square waveform" because only two
> > samples are taken in one full wave cycle. If I need any precision in the
> > sampled signal (like if the source was a saw waveform), I need higher
> > samplerate.
>
>Actually, it wouln't be a squarewave, since we already established that you
>need an anti-aliasing filter, so the overtones would be wiped out and you 
>would
>end up with a single sinewave anyway.
>
><theory corner>
>
>However, this assumes a key thing, namely that the sample clock (this is 
>the
>signal establishing the signal rate) is not synchronous with the incomming
>signal and if it is _exactly_ half the sampling rate, it is infact 
>synchronous,
>being of same clock (which is what synchronous breaks up into when looking 
>at
>it's original greek words, as explained in ITU-T Rec. G.701). When they are
>synchronous, the phase relationship between half the sampling clock (i.e. 
>the
>Nyquist frequency) and the incomming sine will determine the amplitude
>responce as being the cosine of the phase differance. Near frequency hit 
>will
>create a wobble in amplitude as the phase differance changes with the
>difference in frequency unless the anti-aliasing filter is able to 
>completely
>remove the mirror frequency. Sufficient damping removes the wobbeling.
>
>Now, most of that is theoretically correct, but usually not much of a 
>problem.
>There is another flaw with the above discussion which is also apparent in 
>the
>original Nyquist frequency discussion, since it assumes constant sines.
>Actually, the assumptions show themselfs when looking at the proof provided
>by B.M.Oliver, J.R.Pierce and C.E.Shannon in their article "The Philosophy 
>of
>PCM" from 1948, the proof is in the Appendix I. Besides being a very good
>article which makes it points very clearly and IMHO is still valid reading 
>for
>many today dealing with PCM and sampling, it touches on the subject we are
>discussing here. The thing is, this proof only discusses a function f(t) 
>being
>bandwidth limited by the bandwidth W0 (i.e. the Nyquist frequency) and uses
>the Fourier transform. However, the Fourier transform isn't very good in
>handling impulse responce type of signals (the Fourier transform is greatly
>missused for doing it, even if it is part of the full solution). So, my 
>point
>here is that not all transients survive as well in a too tigh bandwidth 
>limited
>system. It takes time to build up the full picture.
>
>If you translate the function into thinking in the LaPlacian domain, we see
>that not only the frequency is relevant, but also the allowed range of
>amplitude acceleration, which is the real part of the LaPlacian variable s.
>Converting the signal into exponential functions of time only (similar to 
>only
>study the sine waveforms) we now must ask us how quick and slow rises we 
>can
>handle. And, for a bandwidth limited system there is indeed an upper limit 
>to
>the maximum speed that we can replicate over such a system! This means that
>any signal having such acceleration would also be incorrectly transfered 
>over
>the system, i.e. loss of information.
>
></theory corner>
>
>So, the conclusion is that transient functions also suffer information loss 
>in
>a bandwidth limited system, such as a sampled system. Additional "headroom" 
>is
>needed both for frequency as such and for transients as such or distorsion 
>of
>signal (i.e. loss of information) will occur.
>
>For transient responses time will not repair impairments as they will with
>constant amplitude signals such as sine (or a cluster of sines). This means
>that the impairments needs to be reduced when the signal actually apears in
>time.
>
> > If the above is a correct explanation (?) that would explain why bass 
>sounds
> > sound "good" on lower sample rates while high frequencys sounds 
>"metallic".
> > The bass frequency waveforms are so long that a lower sample rate still
> > gives us an "accurate" sampled wave.
>
>More or less, yes. For bassier sounds, you use up less of the given 
>bandwidth
>and transient responce of the system. This means that there is alot of
>redundancy in there from a theoretical point of view. You may lower the
>sampling rate to remove that redundancy for the benefit of better usage of
>bits, without greatly reducing the actual information of the sound. When 
>you
>are trying to cut the corners too tight, you start to suffer more and more.
>
> > Maybe I went a bit outside the ADSR subject, but I have wanted to ask 
>this
> > for soo long that I thought it was good.
>
>It's only until you try to design something that you learn how little you 
>know
>and what could possibly be an issue. This motivates your question very well 
>and
>you are correct in asking it. Hell, that's really a good road to increasing
>ones knowledge!
>
>Cheers,
>Magnus - who is gonna toss himself in bed for a finite amount of time

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