[music-dsp] remove

Chris Paredes ChrisParedes at home.com
Mon Apr 30 08:53:31 EDT 2001 

remove
----- Original Message -----
From: "music-dsp-digest" <owner-music-dsp-digest at shoko.calarts.edu>
To: <music-dsp-digest at shoko.calarts.edu>
Sent: Sunday, April 29, 2001 9:00 AM
Subject: music-dsp-digest V1 #1052


>
> music-dsp-digest        Sunday, April 29 2001        Volume 01 : Number
1052
>
>
>
> In this issue:
>
>  [music-dsp] Pole Zero Placement on the Unit Circle
>  Re: [music-dsp] Pole Zero Placement on the Unit Circle
>  Re: [music-dsp] Pole Zero Placement on the Unit Circle
>  Re: [music-dsp] generating Pink Noise (again!)
>  [music-dsp] Re: SHARC
>  Re: [music-dsp] Pole Zero Placement on the Unit Circle
>  [none]
>  Re: [music-dsp] Pole Zero Placement on the Unit Circle
>  [music-dsp] The Making of the Moog Vocoder
>  Re: [music-dsp] generating Pink Noise (again!)
>  Re: [music-dsp] generating Pink Noise (again!)
>
> ----------------------------------------------------------------------
>
> Date: Sat, 28 Apr 2001 03:22:34 -0700 (PDT)
> From: Bagpuss The Cloth Cat <bagpussandco at yahoo.com>
> Subject: [music-dsp] Pole Zero Placement on the Unit Circle
>
> Does anybody know where I can find a good explaination
> (ie one with all the maths in a simple format) for
> pole-zero placement on the unit circle?
>
> Chris
>
> __________________________________________________
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>
>
>
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> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sat, 28 Apr 2001 10:21:30 -0700
> From: Nigel Redmon <earlevel at earlevel.com>
> Subject: Re: [music-dsp] Pole Zero Placement on the Unit Circle
>
> Hi Chris,
>
> I don't know if you're looking for ways to place poles and zeros to yield
a
> result you want, or to be able to look at a pole-zero diagram and find its
> magnitude and phase response. If it's the latter, the math is pretty
basic. I
> don't know of a link (this would be nicer with diagrams), but I'll give a
hack
> at the basic idea:
>
> For magnitude, traverse the unit circle starting at x=1 (frequency 0) to
x=-1
> (Nyquist freq). The magnitude response at any point along the circle is
the
> product of the distances from that point to the zeros, divided by the
product of
> the distances to the poles:
>
> magnitude = (dZ1 * dZ2 * dZ3...) / (dP1 * dP2 * dP3...)
>
> (If you have a diagram and see no poles or no zeros, don't panic--they are
at
> the origin, meaning the distance is always 1.)
>
> Some cool things become apparent--you can see why a pole placed on the
unit
> circle is a stability problem, because at that point in frequency the
> denominator in the above magnitude equation goes to zero--yikes!
>
> A simple example--a diagram with a single zero at x=1: Starting at an
angle of 0
> on the unit circle (x=1), the magnitude is 0 / 1 = 0 ("1" in the denom
because
> there is a corresponding pole at the original). At the other end, Nyquist
> (x=-1), it's 2 / 1 = 2. At half Nyquist (top of the unit circle), it's
1.414.
> Fill in the points in between, but its easy to see that DC is totally
blocked,
> and the magnitude response rises gently until it gets to a gain of 2 at
high
> frequencies. If you implemented this filter, it would just be a unit delay
of
> the input subtracted from the input (y[n] = x[n] - x[n-1]; if you want an
> explanation of why that's so, feel free to ask). Run that filter and
you'll see
> that it's a highpass filter that blocks DC and has a gain of 6 dB (2x) at
the
> top end, matching the curve from the pole-zero diagram.
>
> For phase, it's the sum of the angles formed relative to the zeros, minus
the
> sum of the sangles relative to the poles as you traverse the circle:
>
> phase = (aZ1 + aZ2 + aZ3...) - (aP1 + aP2 + aP3...)
>
> Hope that helps a little. If that's not what you're looking for, be a
little
> more specific and I'm sure someone can point you to some helpful info.
>
> Nigel
>
>
> > Does anybody know where I can find a good explaination
> > (ie one with all the maths in a simple format) for
> > pole-zero placement on the unit circle?
> >
> > Chris
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sat, 28 Apr 2001 10:59:23 -0700 (PDT)
> From: Bagpuss The Cloth Cat <bagpussandco at yahoo.com>
> Subject: Re: [music-dsp] Pole Zero Placement on the Unit Circle
>
> OK, what I mean was is that I'm trying to implement a
> 2nd order resonant IIR filter for my final year
> computer project.  What I wanted to know was - how do
> I calculate the co-efficients based on the positions
> of the poles and zeroes at the unit circle.
>
> I tried reading the Steiglitz book (A DSP Primer with
> Applications to Digital Audio and Computer Music) but
> I couldn't understand a lot of the maths.
>
> - --- Nigel Redmon <earlevel at earlevel.com> wrote:
> > Hi Chris,
> >
> > I don't know if you're looking for ways to place
> > poles and zeros to yield a
> > result you want, or to be able to look at a
> > pole-zero diagram and find its
> > magnitude and phase response. If it's the latter,
> > the math is pretty basic. I
> > don't know of a link (this would be nicer with
> > diagrams), but I'll give a hack
> > at the basic idea:
> >
> > For magnitude, traverse the unit circle starting at
> > x=1 (frequency 0) to x=-1
> > (Nyquist freq). The magnitude response at any point
> > along the circle is the
> > product of the distances from that point to the
> > zeros, divided by the product of
> > the distances to the poles:
> >
> > magnitude = (dZ1 * dZ2 * dZ3...) / (dP1 * dP2 *
> > dP3...)
> >
> > (If you have a diagram and see no poles or no zeros,
> > don't panic--they are at
> > the origin, meaning the distance is always 1.)
> >
> > Some cool things become apparent--you can see why a
> > pole placed on the unit
> > circle is a stability problem, because at that point
> > in frequency the
> > denominator in the above magnitude equation goes to
> > zero--yikes!
> >
> > A simple example--a diagram with a single zero at
> > x=1: Starting at an angle of 0
> > on the unit circle (x=1), the magnitude is 0 / 1 = 0
> > ("1" in the denom because
> > there is a corresponding pole at the original). At
> > the other end, Nyquist
> > (x=-1), it's 2 / 1 = 2. At half Nyquist (top of the
> > unit circle), it's 1.414.
> > Fill in the points in between, but its easy to see
> > that DC is totally blocked,
> > and the magnitude response rises gently until it
> > gets to a gain of 2 at high
> > frequencies. If you implemented this filter, it
> > would just be a unit delay of
> > the input subtracted from the input (y[n] = x[n] -
> > x[n-1]; if you want an
> > explanation of why that's so, feel free to ask). Run
> > that filter and you'll see
> > that it's a highpass filter that blocks DC and has a
> > gain of 6 dB (2x) at the
> > top end, matching the curve from the pole-zero
> > diagram.
> >
> > For phase, it's the sum of the angles formed
> > relative to the zeros, minus the
> > sum of the sangles relative to the poles as you
> > traverse the circle:
> >
> > phase = (aZ1 + aZ2 + aZ3...) - (aP1 + aP2 + aP3...)
> >
> > Hope that helps a little. If that's not what you're
> > looking for, be a little
> > more specific and I'm sure someone can point you to
> > some helpful info.
> >
> > Nigel
> >
> >
> > > Does anybody know where I can find a good
> > explaination
> > > (ie one with all the maths in a simple format) for
> > > pole-zero placement on the unit circle?
> > >
> > > Chris
> >
> > dupswapdrop -- the music-dsp mailing list and
> > website: subscription info,
> > FAQ, source code archive, list archive, book
> > reviews, dsp links
> > http://shoko.calarts.edu/musicdsp/
> >
>
>
> __________________________________________________
> Do You Yahoo!?
> Yahoo! Auctions - buy the things you want at great prices
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>
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> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sat, 28 Apr 2001 14:35:30 -0400
> From: "James Chandler Jr" <jchandjr at bellsouth.net>
> Subject: Re: [music-dsp] generating Pink Noise (again!)
>
> > Damn decibels. We oughta ditch'em right now. ;)
>
> dBs is dBs. 'Amplitude dB' and 'Power dB' are the same thing.
>
> Electrical power in the analog realm = Voltage^2 / Impedance.
>
> If you drive a one ohm load with 1 volt, and then boost the signal to 2
> volts--
>
> Power at 1 volt = 1^2 / 1 = 1 watt
> Power at 2 volt = 2^2 / 1 = 4 watts
>
> "amplitude dB" = 20 * log10(2/1) = 6 dB
> "power dB" = 10 * log10(4/1) = 6 dB
>
> The amplitude dB's completely agree with the power dB's.
>
> Since there are no loads in computer digital audio, I suppose the easiest
> way to calculate "virtual digital audio power" is to square the amplitude
> and assume the "imaginary load" is 1.0?
>
> For measurement purposes in digital audio, since the two forms of dB
agree,
> one could use whatever is convenient. If you were writing RMS meter code
by
> summing the squares of all the audio samples, it wouldn't be necessary to
> square root the sum to get amplitude dB. Just use the power version of the
> dB formula for the meter display. The same dBs properly represent both RMS
> amplitude and RMS "virtual digital audio power".
>
> > But seriously, isn't white
> > noise supposed to have a flat amplitude spectrum - after all, it's named
> > after white light which has precisely that
>
> AFAIK, white noise is equal power per Hz. The power in the range of 10 Hz
to
> 11 Hz would be equal to the power in the range of 10,000 Hz to 10,001 Hz.
> White noise is flat along a linear frequency scale.
>
> On an analog Real Time Analyzer, which uses an array of bandpass filters
> spaced along exponential frequencies, white noise displays a positive
slope.
> Ferinstance with an Octave analyzer, the top octave might span 10 KHz to
20
> KHz, and the bottom octave might span 20 Hz to 40 Hz. There is a twenty Hz
> span in the low band, and a 10,000 Hz span in the top band. With equal
power
> per Hz white noise, the top band has 500 X more power than the low band.
>
> For equal power per octave-- Perhaps the first octave spans 20 Hz to 40
Hz,
> and the second octave spans 40 Hz to 80 Hz. The second band of white noise
> has twice as much power, since it has twice as many Hz in its span. A
> doubling of power is +3 dB, so we LP filter the white noise at -3 dB per
> octave to get flat measurements. Pink noise.
>
> White noise looks flat when you analyze with a raw FFT, and Pink noise
looks
> flat when you analyze with a typical RTA.
>
> James Chandler Jr.
>
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sat, 28 Apr 2001 19:36:44 +0100
> From: "Paul Kellett" <paul.kellett at maxim.abel.co.uk>
> Subject: [music-dsp] Re: SHARC
>
> emerson tan <emtan at up.edu.ph> wrote:
> >
> > Hello! I would like to write my own programs that would communicate with
> > the SHARC EZ-KIT LITE's monitor program via the RS232. Much like what
the
> > Visual DSP++ Debugger does; except that I don't want to use it.
>
>
> I wrote a Visual Basic program to do this and it was quite easy - there
> is a web page out the somewhere telling you what comm port settings to
> use, and it makes using the EZ-KIT much easier. - BUT - unless you write
> your own RS232 handler for the EZ-KIT (and maybe even then) communicating
> with the board interrupts execution quite badly, so it's no use for
> streaming audio.
>
>
>
> Paul.
>
> _____________________________
>
>   m a x i m | digital audio
>
>      http://mda-vst.com
> _____________________________
>
>
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sat, 28 Apr 2001 13:50:17 -0700
> From: Nigel Redmon <earlevel at earlevel.com>
> Subject: Re: [music-dsp] Pole Zero Placement on the Unit Circle
>
> OIC. Well, I said "If you implemented this filter, it would just be...
(y[n] =
> x[n] - x[n-1]; if you want an
> explanation of why that's so, feel free to ask)", and that's basically
what
> you're asking, so I'll take a crack at it.
>
> The transfer function of the filter defines everything you need to know.
>From
> that, you can plot the poles and zeros or implement a difference equation
to
> realize it in DSP code or hardware.
>
> H(z) = a/b, where "a" is a polynomial in z that determines the zero
locations
> and "b" likewise for the poles. If we had the simple filter I described
before
> (zero at x=1), the transfer function would be:
>
> H(z) = (z-1)/1
>
> That is, the "z" on the right side is the zero value that makes the
numerator
> equal to 0--that is, 1. (I'll warn you right now, I don't have time to
think of
> a great way to explain this in few words with no drawings, so I hope you
either
> get the idea--I have no clue what level of understanding you have on all
> this--or someone else might be able to help you or give references where
you can
> get a better lesson.)
>
> OK, don't make me try to draw a circle in ascii. Here's as drawing of the
> pole-zero diagram with a pole at -0.4, and zeros at -0.7 and 1:
>
>          |
>          |
>          |
>  --0--x--+-------0
>          |
>          |
>          |
>
> - -1       0       1
>
> H(z) = (z - 1) * (z + 0.7) / ((z + 0.4) * (z - 0))
>
> [there's a pole at zero to match the number of zeros and keep the order
the same
> top and bottom--I could omit it, showing how it's equivalent to shifting
time by
> one later, but it's easier to describe this way]
>
> Again, the top corresponds to the zero, the bottom the poles. Similarly,
the top
> corresponds to the delayed input terms (feedforward paths), and the bottom
to
> the delayed output (feedback paths). You'll want everything in terms of
> polynomials if you want to implement this as a single filter of multiple
order
> (instead of cascading first order filters), so multiply out the top:
>
> H(z) = (z^2 + -0.3z + -0.7) / (z^2 + 0.4z)
>
> z^2 implies two samples into the future, so multiply everything by
z^-2/z^-2 to
> back up--easier to work with past samples than ones we haven't received
yet:
>
> H(z) = (1z^0 + -0.3z^-1 + -0.7z^-2) / (z^0 + 0.4z^-1)
>
> Now, transform that into a difference equation (not going to go through a
big
> explanation--basically, top is x's, bottom y's, z^-n is a delay of n):
>
> y[n] + 0.4y[n-1] = x[n] + -0.3x[n-1] + -0.7x[n-2]
>
> or, solving for the output y[n],
>
> y[n] = x[n] + -0.3x[n-1] + -0.7x[n-2] - 0.4y[n-1]
>
> The fun part begins when the poles aren't purely real. All the same rules
apply,
> but you get to do complex math--no biggie; once you multiply the terms
out, as
> above, the result will be real (because they always come in complex
conjugate
> pairs--you won't have a pole or zero with an imaginary part that isn't
mirrored
> in a "real" filter). For instance, say you have a pole at 0+.5i and 0-.5i.
You
> get (z - 0+.5i) * (z - 0-.5i) in the denominator. Multiplying gives you
z^2 -
> 0.25, and you go on.
>
> Tell me if this helps or I'm wasting my "breath" ;-)
>
> Nigel
>
> > OK, what I mean was is that I'm trying to implement a
> > 2nd order resonant IIR filter for my final year
> > computer project.  What I wanted to know was - how do
> > I calculate the co-efficients based on the positions
> > of the poles and zeroes at the unit circle.
> >
> > I tried reading the Steiglitz book (A DSP Primer with
> > Applications to Digital Audio and Computer Music) but
> > I couldn't understand a lot of the maths.
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sun, 29 Apr 2001 05:39:37 +0200 (CEST)
> From: integer at www.god-emil.dk
> Subject: [none]
>
> From: Scott Sunn <neurotransmitter at subdimension.com>
> To: 55 at bek.no
> MIME-Version: 1.0
> User-Agent: IMP/PHP3 Imap webMail Program 2.0.11
> X-WebMail-Company: subDIMENSION.com
> Subject: /55\live nato mix 4.30.01
> Sender: 55-admin at bek.no
> Errors-To: 55-admin at bek.no
> X-BeenThere: 55 at bek.no
> X-Mailman-Version: 2.0beta2
> Precedence: bulk
> Reply-To: 55 at bek.no
> List-Id:  <55.bek.no>
> X-UIDL: HM2!!*^N"!6;\!!XI6!!
> Status: O
>
> jello!
>
> if you find yourself in front of your komputer this coming monday at
> 9:00 pst check out http://www.oseao.com/   -  i will be streaming live
> nato collagework to some blippy detroit (and beyond) techno.  i will
> remind everyone tomorrow night...              frog saays, gonna ribbit
> up.
>
>
>
> - -ss
> http://www.tracervisuals.com/
>
>
>
>
>
>
>
>
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sun, 29 Apr 2001 00:56:05 -0700
> From: Nigel Redmon <earlevel at earlevel.com>
> Subject: Re: [music-dsp] Pole Zero Placement on the Unit Circle
>
> > Multiplying gives you z^2 - 0.25
>
> oops, make that z^2 + 0.25
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sun, 29 Apr 2001 13:26:29 +0200
> From: "Svante DSP" <svante_dsp at hotmail.com>
> Subject: [music-dsp] The Making of the Moog Vocoder
>
> <html><DIV>I've been working on a (new?) synthesis technique, and I was
wondering if any of you know if it's been done before, or your beliefs on
wheather it's doable or not.</DIV>
> <DIV>&nbsp;</DIV>
> <DIV>The whole idea is based on the fact that programming subtractive
synthesizers to sound like a 'real' sound is very hard. So I've been working
on a program that takes a wave file and tries to come as close to that sound
as possible with parameters being the classic Moog ones: pitch, filter
cutoff and resonance, envelopes, standard oscillators. The result, I
imagine, would be a "Moog Vocoder". I've come as far as detecting the pitch,
harmonic spectrum and envelope of the original sound. <BR>And now I guess I
need to dig up my old optimization book.</DIV>
> <DIV>&nbsp;</DIV>
> <DIV>Hit me with thoughts, comments, ideas, whatever.</DIV>
> <DIV>&nbsp;</DIV>
> <DIV>/Svante "anything is possible" Stadler</DIV><br clear=all><hr>Get
Your Private, Free E-mail from MSN Hotmail at <a
href="http://www.hotmail.com">http://www.hotmail.com</a>.<br></p></html>
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sun, 29 Apr 2001 16:03:00 +0300 (EEST)
> From: Sampo Syreeni <decoy at iki.fi>
> Subject: Re: [music-dsp] generating Pink Noise (again!)
>
> On Sat, 28 Apr 2001, James Chandler Jr wrote:
>
> >> Damn decibels. We oughta ditch'em right now. ;)
> >
> >dBs is dBs. 'Amplitude dB' and 'Power dB' are the same thing.
> >
> >Electrical power in the analog realm = Voltage^2 / Impedance.
>
> Never read my circuit design well enough. But isn't it that there will be
> some trouble with reactive loads, especially if we drive them with
something
> other than sinusoids?
>
> >Since there are no loads in computer digital audio, I suppose the easiest
> >way to calculate "virtual digital audio power" is to square the amplitude
> >and assume the "imaginary load" is 1.0?
>
> Probably. My remark was mainly motivated by the fact that decibels tend to
> make things a bit complicated compared to a linear scale, and they lead
> people's intuition astray. For instance, decibels have little to do with
the
> perceptual response of the ear, other than the fact that the logarithmic
> scale is needed to keep the numbers small. Most of the common DSP stuff is
> also LTI, so logarithmic plots are often a bit less than natural when
> analyzing the ops.
>
> >Ferinstance with an Octave analyzer, the top octave might span 10 KHz to
20
> >KHz, and the bottom octave might span 20 Hz to 40 Hz. There is a twenty
Hz
> >span in the low band, and a 10,000 Hz span in the top band. With equal
power
> >per Hz white noise, the top band has 500 X more power than the low band.
>
> Indeed. Somebody had pointed this out already, but I'd already managed to
> post my comment.
>
> Sampo Syreeni, aka decoy, mailto:decoy at iki.fi, gsm: +358-50-5756111
> student/math+cs/helsinki university, http://www.iki.fi/~decoy/front
>
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> Date: Sun, 29 Apr 2001 15:00:14 +0100
> From: Richard Dobson <RWD at cableinet.co.uk>
> Subject: Re: [music-dsp] generating Pink Noise (again!)
>
> Decibels have everything to do with the response of the ear! Compare an
> exponential decay to a linear amplitude decay, for example. Volume
> controls, whether on a domestic amplifier or on a pro mixing desk, have
> a logarithmic law. This means their performance is linear in dB: an
> equal movement of a fader gives an equal perceived change in intensity.
> This closely parallels the ear's logarithmic response to frequency - we
> hear ratios of intervals (ie.g the octave is a ~doubling~ of frequency).
> So just as you would find a graphic equalizer arranged in lnear
> frequency very intractable, so a linear volume control would be very
> un-intuitive, as most of the audble effect would be confined to the
> bottom quarter or so of movement. Panning controls similarly use an
> 'equal power' law, e.g. using sine/cosine quadrant envelopes, to avoid
> the 'hole in the middle' effect when a sound is panned to the centre.
>
> And the LTI aspect of DSP is a quite separate issue from the way in
> which data is displayed. LTI, broadly speaking, means that additive and
> subtractive synthesis works - when you add one sinusoid to another (and
> whever you add it), it doesn't alter what you've added it to, so you can
> take it away again!
>
> I use a display program that can draw waveforms with a dB vertical
> scale. Sinewaves look very odd, but you can see the parts of the signal
> that are otherwise below pixel resolution. A (good) reverb decay appears
> as a very nice straight line slope.
>
>
> Richard Dobson
>
>
> Sampo Syreeni wrote:
>
> >
> > Probably. My remark was mainly motivated by the fact that decibels tend
to
> > make things a bit complicated compared to a linear scale, and they lead
> > people's intuition astray. For instance, decibels have little to do with
the
> > perceptual response of the ear, other than the fact that the logarithmic
> > scale is needed to keep the numbers small. Most of the common DSP stuff
is
> > also LTI, so logarithmic plots are often a bit less than natural when
> > analyzing the ops.
> >
>
>
> - --
> Test your DAW with my Soundcard Attrition Page!
> http://www.bath.ac.uk/~masrwd (LU: 3rd July 2000)
> CDP: http://www.bath.ac.uk/~masjpf/CDP/CDP.htm (LU: 23rd February 2000)
>
> dupswapdrop -- the music-dsp mailing list and website: subscription info,
> FAQ, source code archive, list archive, book reviews, dsp links
> http://shoko.calarts.edu/musicdsp/
>
> ------------------------------
>
> End of music-dsp-digest V1 #1052
> ********************************
>
>
> dupswapdrop: the music-dsp mailing list and website
> http://shoko.calarts.edu/music-dsp


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