I find that the best way to avoid paying for access to scholarly knowledge -
both in legal terms and in effectiveness - is to email the author, express
interest, and ask for any material they can send you. Most times, you'll get
something free, legal, newer and with helpful insight straight
IIRC this is one of those situations where you get to define the value of the
integral at that point however you like (i.e., either the limit from one side
or the other, or a convex combination of both) and all the math then works as
expected under that convention.
Or just point out that it's
The analog bench gear in that video made me nostalgic... I can almost smell
the solder fumes :]
E
On Thu, Feb 13, 2014 at 6:01 PM, Andrew Simper a...@cytomic.com wrote:
I was really enjoying the proportional Q thread, so in an attempt to
keep in on topic here is a thread where people can
I recall from the first measurement shown on the analog spectrum analyzer
the source sine wave had 2d or 3d harmonic distortion component of -70 dB.
My power amps are noticeably better than that.
But then they don't have the heterodyne circuits you need for an analogue
frequency analyser
, Theo Verelst theo...@theover.org wrote:
Ethan Duni wrote:
I recall from the first measurement shown on the analog spectrum analyzer
the source sine wave had 2d or 3d harmonic distortion component of -70
dB.
My power amps are noticeably better than that.
But then they don't have
Seems to me that you could just reformulate an EM algorithm to work
directly on the actual (magnitude) responses of the biquad stages, and so
bypass the step of converting from a Gaussian response to the actual biquad
response (along with its attendant error). The only obvious wrinkle that
occurs
In the mathematical sense, we could take a Fourier Integral of these
signals neatly added together, and the outcome would be 3 spikes at the
respective frequencies, with the proper amplitude.
No, you get *6* spikes, since these are real-valued signals. So there are
components at both the
And an AD converter with 2.5 MS/S and 24 bits accuracy sigma delta which
should be great for measurements and undisputably great sampling behavior
for perfect delay lines, etc.:
http://www.analog.com/en/analog-to-digital-converters/ad-converters/ad7760/products/product.html
(eval board:
Hi Doug-
To address some of your general questions about Fourier analysis and
relationship to sampling theory:
Broadly speaking any reasonably well-behaved signal can be decomposed into
a sum of sinusoids (actually complex exponentials but don't worry about
that detail for now). There are
Hi Doug-
Regarding this:
Terms like well behaived when applied to the functon make me wonder
what
stipulations might be implied by the language that you'd have to be a formal
mathmatician to interpret. As an example, I don't even know what the
instrinsic properties of a function may be in this
Yeah this is sometimes called bandpass sampling or under sampling (
http://en.wikipedia.org/wiki/Undersampling) and is commonplace in the
context of like RF communications and such.
But it can also come up in audio applications, for example critically
sampled filter banks. I.e. say you split a
it is, at least, if you accept the EE notion of the Dirac delta function
and not worry so much about it not really being a function, which is
literally what the math folks tell us.
I may be misremembering, but can't non-standard analysis be used to make
that whole Dirac delta function approach
...@audioimagination.com wrote:
On 3/27/14 5:27 PM, Ethan Duni wrote:
it is, at least, if you accept the EE notion of the Dirac delta function
and not worry so much about it not really being a function, which is
literally what the math folks tell us.
I may be misremembering, but can't non
Not to be overly antagonistic, but:
I can easily hear the difference between a 192 or 96kHz 24 (or 22 bits +
exponent) bit and downgrading to 48 or 44.1 / 24 bit OR to 192 or 96 kHz 16
bits. Let alone both, easily audible.
If you are hearing obvious differences between those settings, it's a
So this talk of compressors in the playback chain brings up an important
point. The usual results about CD rate/depth being sufficient are referring
to the signal delivered to the final analog audio output. We all know that
higher rates and depths are appropriate/required for intermediate
It would appear to me that the human hearing system is an LTI system.
It doesn't react in a linear fashion to frequency or loudness, but it
perceives
the same signal the same way at all times, disregarding aging, hearing
loss, etc.
One of the easiest ways to see that hearing must be nonlinear is
the human hearing system is kind of an LTI... only at very low level
processing. The consistency of measured signal (= perceiving the same
signal the same way at all time as somebody wrote here) is present in
the ear canal up to brainstem - inferior colliculus.
My understanding is that there are
So in the digital sense, or in combination with the analog domain,
is it reasonable to think about correctable operations, which as
it were can be inverted, so that applying a digital signal transformation
*and* it's converse, we end up with the same signal or something similar.
I think you mean
rbj
another semantic to be careful about is transfer function.
we mean something different when it's applied to LTI systems
(the H(z) or H(s)) than when applied to a diode. the latter
semantic i don't use. i would say volt-amp characteristic
of the diode or vacuum tube. or if it was a
rbj
Urs
Regarding the iterative method, unrolling like you did
y0 = y[n-1]
y1 = g * ( x[n] - tanh( y0 ) ) + s
y2 = g * ( x[n] - tanh( y1 ) ) + s
y3 = g * ( x[n] - tanh( y2 ) ) + s
y[n] = y3
is *not* what I described in general.
it *is* precisely equivalent to the example you were
This means that in principle any piecewise polynomial signal
with bandlimited discontinuities of the signal and its derivatives
is also band limited.
Sorry if I'm missing something obvious, but what is a bandlimited
discontinuity?
E
On Tue, Jul 1, 2014 at 1:17 AM, Vadim Zavalishin
without the minimum phase approach. Even at 44.1 kHz, the lowest
rate you’re likely to make a synth, it would take quite a few samples to
result in enough delay to matter.
On Jul 1, 2014, at 12:29 PM, Ethan Duni ethan.d...@gmail.com wrote:
If you reset the output of a sine wave generator so
What are the alternatives to the FFT? Have wavelets been
used for real world solutions?
Sure, wavelets get used. Maybe more in image/video than audio, but I'm
certain someone can come up with some examples of wavelet audio
applications.
If an app needs much higher time resolution
and there are
Sinc interpolation would be theoretically correct, but, remember,
that this thread is not about strictily theoretically correct frequency
recognition, but rather about some more intuitive version with the
concept of instant frequency.
What is instant frequency? I have to say that I find this
Maybe I'm missing something obvious, but shouldn't the filter bank itself
be constant? I.e., no change in the overlap or windowing. The time
stretch/compression is obtained by extrapolating/interpolating the analysis
parameters, not by shifting around the synthesis filter bank relative to
the
to the one exposed here
http://www.ece.uvic.ca/~peterd/48409/Bernardini.pdf, but modified to
work
in real time processing frames of fixed size.
2014-08-19 21:30 GMT+01:00 Ethan Duni ethan.d...@gmail.com:
Maybe I'm missing something obvious, but shouldn't the filter bank itself
be constant
Well, your standard options for computing 2 to a fractional power are
either polynomial approximation or table look-up. If I'm reading it
correctly, the approach you quoted there is a second-order polynomial
approximation. You can gain accuracy at the cost of complexity by dialing
up the
values in the time
domain is pretty much irrelevant to the question of what its entropy is
(particularly when those values themselves are taken as variable, as you
are doing here).
E
On Thu, Oct 9, 2014 at 9:35 AM, Peter S peter.schoffhau...@gmail.com
wrote:
On 09/10/2014, Ethan Duni ethan.d
to strong conclusions like this. You'll get a lot farther if
you invest a bit into understanding the theoretical framework for this
stuff.
E
On Thu, Oct 9, 2014 at 9:48 AM, Peter S peter.schoffhau...@gmail.com
wrote:
On 09/10/2014, Peter S peter.schoffhau...@gmail.com wrote:
On 09/10/2014, Ethan
- since there is no dependence on previous
samples, there is a constant amount of surprise in each sample,
corresponding to the entropy of the distribution the noise is drawn from.
E
On Thu, Oct 9, 2014 at 10:00 AM, Peter S peter.schoffhau...@gmail.com
wrote:
On 09/10/2014, Ethan Duni ethan.d
capacity of human
audio perception.
E
On Thu, Oct 9, 2014 at 11:11 AM, Peter S peter.schoffhau...@gmail.com
wrote:
On 09/10/2014, Ethan Duni ethan.d...@gmail.com wrote:
Let's assume I have a sinusoidal signal.
Let's assume I amplify it to 10x.
Where does new entropy come from?
It comes from
... since the original question of r b-j was: how can the human ear
convey the high amount of digital PCM information contained on a CD?
Right, my point is that the digital PCM info on a CD typically contains a
*lot* of data that is redundant to human audio perception, and which gets
I did not claim anything about entropy of
continuous signals,
Aren't we talking about impulses in auditory nerves (among other things)?
Those things live in the analog domain.
I was only talking about the entropy
content of digital PCM signals that could be estimated using standard
digital,
own is fine,
but your response to attempts to enlighten you is counterproductive. If and
when you get serious about understanding this stuff, this list is here to
help.
E
On Fri, Oct 10, 2014 at 4:21 AM, Peter S peter.schoffhau...@gmail.com
wrote:
n 09/10/2014, Ethan Duni ethan.d...@gmail.com
Sounds like a fun project Scott.
One question though:
Sample rate is approximately 44.6 kHz.
What's with the non-standard sampling rate?
E
On Sun, Oct 12, 2014 at 5:25 PM, Scott Gravenhorst music.ma...@gte.net
wrote:
I've been working on a MIDI Karplus-Strong synthesizer using a Microchip
Although, it's interesting to me that you might be able to get some
surprising value out of information theory while avoiding any use of
probability ...
Hartley entropy doesn't avoid any use of probability, it simply
introduces the assumption that all probabilities are uniform which greatly
The Hartley entropy
is invariant to the actual distribution (provided all the
probabilities are non-zero, and the sample space remains unchanged).
No, the sample space does not require that any probabilities are nonzero.
It's defined up-front, independently of any probability distribution.
The relevant limit here is:
lim x*log(x) = 0
x-0
It's pretty standard to introduce a convention of 0*log(0) = 0 early on
in information theory texts, since it avoids a lot of messy delta/epsilon
stuff in the later exposition (and since the results cease to make sense
without it, with empty
I might find myself in the situation where I am given a nonuniform
distribution. Then Shannon and Hartley formulas would give a different
answer.
And the Hartley formula would be inapplicable, since it assumes a uniform
distribution. So you'd have to use the Shannon framework, both to calculate
Of course it is only a theoretical matter, but we have a
sensible spectrum for our repeating envelope generator,
say at 1 Hertz, with a low pass filter set to 2Hz, possibly
with an actually limited spectrum, that we can multiply
with another wave, which can also be frequency limited
in such a way
Say we're only taking one length of the FFT transform, and
are only interested in the volume of the various output bins.
Now, how probable is it that we get all equal frequency amounts as
the output of the this FFT transform (without regarding phase), taking
for instance 256 or 4096 bins, and 16
I am not sure if the PDFs are preserved across
transforms from one orthonormal basis to
another, and the answer to your question would
depend on that (Of course it would also depend
on several other parts of the phrasing of your question
that aren't clear to me). My intuition is that PDFs
are
There is a theorem that goes something like this:
If you have white noise expressed in one orthonormal basis, and you
transform it to another orthonormal basis, the result will still be white
noise.
There is certainly no such theorem. For any noise signal you can define
a basis that contains
. You're talking about
something quite different.
E
On Fri, Oct 31, 2014 at 3:42 PM, Andreas Tell li...@brainstream-audio.de
wrote:
On 31 Oct 2014, at 23:31, Ethan Duni ethan.d...@gmail.com wrote:
If you have Gaussian i.i.d. noise, you can apply any unitary transform
you
want and you
On Fri, Oct 31, 2014 at 4:16 PM, Andreas Tell li...@brainstream-audio.de
wrote:
On 01 Nov 2014, at 00:06, Ethan Duni ethan.d...@gmail.com wrote:
The correct statement would be
that an arbitrary unitary transform of a Gaussian white noise signal is
*expected* to give a gaussian white noise
...@iki.fi wrote:
On 2014-10-31, Ethan Duni wrote:
Transforms between orthogonal bases are basically rotations. I.e., they
are linear operators that produce each component of the output as a linear
combination of input components. Generally, then, the Central Limit Theorem
tells us
, Andreas Tell li...@brainstream-audio.de
wrote:
On 01 Nov 2014, at 02:39, Ethan Duni ethan.d...@gmail.com wrote:
The expectation value of the signal over the ensemble of all unitary
transforms
with a suitable measure (like Haar).
The expected value you describe is equal to the zero signal
what other presumption is there? i, personally, have never seen a
sequence of
samples of audio or music that was not equidistant and linearly
sampled. it's
what we call uniform sampling.
Some of this new stuff in compressive sensing/sparse reconstruction
involves non-uniform sampling. Not that
zeros happen to line up with the bin frequencies.
E
On Mon, Dec 8, 2014 at 3:28 PM, Jerry lancebo...@qwest.net wrote:
On Dec 8, 2014, at 2:33 PM, Ethan Duni ethan.d...@gmail.com wrote:
what other presumption is there? i, personally, have never seen a
sequence of
samples of audio
it takes a little less than a millisecond to receive a 3-byte MIDI
message.
should we put in some kinda delay (like a sample-and-hold) on that which
prevents updating the net control value until 1 or 2 milliseconds after
either
the MSB or LSB is received? or slewing the net control value (slewing
My filter has 2 poles and 1 zero. Unlike the Cookbook filter, which
has 2 poles and 2 zeros.
I think that automatically assumes, the transfer function cannot be
equivalent.
No, that does not follow. A filter with two zeros can produce all of the
transfer functions that a filter with one zero can,
I completely agree! I find it mentally easier to think of energy
stored in each component rather than state variables even though
they are the same. So for musical applications it is important that a
change in the cutoff and resonance doesn't change (until you process
the next sample) the energy
, as a special case of the general 2 pole biquad
filter.
On 05/02/2015, Ethan Duni ethan.d...@gmail.com wrote:
You just stick the extra zero(s) off at z=infinity.
Does 'z=infinity' mean it's at the origin? I'm not 100% sure of the
terminology used here.
- Peter
--
dupswapdrop -- the music-dsp
P.S. Anyone who knows how to effectively turn ideas into money while
everyone can benefit, let me know. Patenting stuff doesn't sound like
a viable means to me.
Well, that's exactly what patents are for. I'm not sure why you don't
consider that viable. Is it to do with the costs and time required
There is just no way A/B testing on a sample of listeners,
at loud, but still realistic listening levels, would show that
dithering to 16bit makes a difference.
Well, can you refer us to an A/B test that confirms your assertions?
Personally I take a dim view of people telling me that a test would
Thanks for the reference Vicki
What they are hearing is not noise or peaks sitting at the 24th
bit but rather the distortion that goes with truncation at 24b, and
it is said to have a characteristic coloration effect on sound. I'm
aware of an effort to show this with AB/X tests, hopefully it
when filters blow up because of varying coefficients, don't they settle
down some finite time after the coefs stop varying?
In a digital implementation? Probably not. Once you have some Inf or
NaN values in your state, you're going to keep getting garbage out
forever unless you explicitly reset
wrote:
On 2/3/15 2:01 PM, Ethan Duni wrote:
I completely agree! I find it mentally easier to think of energy
stored in each component rather than state variables even though
they are the same. So for musical applications it is important that a
change in the cutoff and resonance doesn't change
questions.
E
On Tue, Feb 3, 2015 at 5:13 PM, robert bristow-johnson
r...@audioimagination.com wrote:
On 2/3/15 7:07 PM, Ethan Duni wrote:
well, the output states, y[n-1] and y[n-2], will change if coefs change.
No, those have already been computed and (presumably) output at that
point.
It's
So to you, that Pono player isn't snake oil?
It's more the 192kHz sampling rate that renders the Pono player into snake
oil territory. The extra bits probably aren't getting you much, but the
ridiculous sampling rate can only *hurt* audio quality, while consuming
that much more battery and
.
On 2/10/2015 1:13 PM, Ethan Duni wrote:
I'm all for releasing stuff from improved masters. There's a trend in my
favorite genre (heavy metal) to rerelease a lot of classics in full
dynamic range editions lately. While I'm not sure that all of these
releases really sound much better (how much
bristow-johnson
r...@audioimagination.com wrote:
On 2/10/15 1:51 PM, Ethan Duni wrote:
So to you, that Pono player isn't snake oil?
It's more the 192kHz sampling rate that renders the Pono player into snake
oil territory. The extra bits probably aren't getting you much, but the
ridiculous
How do the crest factors of these different sawtooth waveforms compare?
I'd expect one with randomized phase to have a much lower crest factor.
Which is to say that I'd expect the in-phase sawtooth to activate a lot
more nonlinearity in the playback chain, which explains why that one is
easy to
Yeah if you simply pick out a few peaks you can get the general shape of
the reverb decay, but you miss all of the dense reflections. Multi-tap
delay on its own is fine for the early reflections, but the rest of the
reverb response is more dense as Steffan says.
Keun Sup Lee did some work along
If you just want higher frequency resolution, you don't need to do any
oversampling in the time domain. Just zero-pad the time domain signal out
to whatever long length you want, and then use an FFT of that size.
E
On Tue, Mar 31, 2015 at 3:12 PM, MF ukel...@gmail.com wrote:
given a N-point
Your intuition is correct that SIMD can't use the output of one
sub-operation as an input to another in the same vector. This poses a
problem for stuff like recursive filters particularly. For FIR filters it
is not as much of an obstacle, since the output at any particular time is
only a function
For the theoretically inclined: approximating a full Fourier Transform
requires
time interpolation of the samples to a (possibly much higher) sampling
frequency, and on top of that a very long FFT, and proper analysis of the
results of the FFT.
It sounds like you are talking about trying to
Also a good starting place for beginners are the xiph show-and-tell videos
(probably been posted here before, but whatever):
https://xiph.org/video/vid2.shtml
E
On Wed, Jun 3, 2015 at 3:05 PM, Ethan Duni ethan.d...@gmail.com wrote:
Perfect sinusoids/square waves/etc. only exist
Perfect sinusoids/square waves/etc. only exist as mathematical
abstractions. A good starting point would be to get a feel for what, say,
the square wave coming out of an analog synthesizer actually looks like -
the noise floor, the distribution of harmonics, frequency jitter,
under/overshoot, etc.
Now the assignment is as follows: can we, given the output signal
coming from our filter which was fed the input signal, and the filter
coefficients, compute the input signal ?
Invertible digital filters are invertible, up to numerical precision. Are
you wanting to talk about finite word length
If you try to take the Fourier transform integral of a exp(j*omega_0*t),
it will not
converge in the sense, how an improper integral's convergence is usually
understood.
You will need to employ something like Cauchy principal value or Cesaro
convergence
to make it converge to zero at
Could you give a little bit more of a clarification here? So the
finite-order polynomials are not bandlimited, except the DC? Any hints
to what their spectra look like? How a bandlimited polynomial would look
like?
Any hints how the spectrum of an exponential function looks like? How
does a
Wow, good answer!
E
On Sat, Jun 6, 2015 at 4:34 PM, Sampo Syreeni de...@iki.fi wrote:
On 2015-06-06, Alan Wolfe wrote:
I am so sorry... meant to send this to myself to investigate later, my
name starts with A and my address book has this as A for some reason.
Please ignore... or feel
at 12:49 PM, Sampo Syreeni de...@iki.fi wrote:
On 2015-06-12, Ethan Duni wrote:
Thanks for expanding on that, this is quite interesting stuff. However,
if I'm following this correctly, it seems to me that the problem of
multiplication of distributions means that the whole basic set-up
, and then downsampling? Is there mileage to be had by
combining oversampling with BLEP?
E
On Thu, Jun 25, 2015 at 1:34 AM, Vadim Zavalishin
vadim.zavalis...@native-instruments.de wrote:
On 24-Jun-15 21:30, Ethan Duni wrote:
Could you expand a bit on exactly what it means to apply the BLEP method
vadim.zavalis...@native-instruments.de wrote:
On 09-Jun-15 19:23, Ethan Duni wrote:
Could you give a little bit more of a clarification here? So the
finite-order polynomials are not bandlimited, except the DC? Any hints
to what their spectra look like? How a bandlimited polynomial would look
to
sampling/reconstruction of well-tempered distributions in the first place.
No?
E
On Thu, Jun 11, 2015 at 2:00 AM, Sampo Syreeni de...@iki.fi wrote:
On 2015-06-09, Ethan Duni wrote:
The Fourier transform does not exist for functions that blow up to +-
infinity like that. To do frequency domain
https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
E
___
music-dsp mailing list
music-dsp@music.columbia.edu
https://lists.columbia.edu/mailman/listinfo/music-dsp
a first-order interpolator.
quite familiar with it.
Yeah that was more for the list in general, to keep this discussion
(semi-)grounded.
E
On Wed, Aug 19, 2015 at 9:15 AM, robert bristow-johnson
r...@audioimagination.com wrote:
On 8/18/15 11:46 PM, Ethan Duni wrote:
for linear
Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1,
-1...
The sampling theorem requires that all frequencies be *below* the Nyquist
frequency. Sampling signals at exactly the Nyquist frequency is an edge
case that sort-of works in some limited special cases, but there is no
a nyquist frequency
sinusoid when you run it through a DAC.
E
On Tue, Aug 18, 2015 at 1:28 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 18/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1,
-1...
The sampling
no bearing on the frequency
response of fractional interpolators. I'd suggest dropping this whole
derail, if you are no longer hung up on this point.
E
On Tue, Aug 18, 2015 at 2:08 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 18/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
That class
, the aliasing issue
works like this: I add two numbers together, and find that the answer is X.
I tell you X, and then ask you to determine what the two numbers were. Can
you do it?
E
On Tue, Aug 18, 2015 at 2:13 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 18/08/2015, Ethan Duni ethan.d
be used there.
But the example of the weird things that can happen when you try to sample
a sine wave right at the nyquist rate and then process it is orthogonal to
that point.
E
On Tue, Aug 18, 2015 at 1:16 PM, robert bristow-johnson
r...@audioimagination.com wrote:
On 8/18/15 3:44 PM, Ethan Duni
In order to reconstruct that sinusoid, you'll need a filter with
an infinitely steep transition band.
No, even an ideal reconstruction filter won't do it. You've got your
+Nyquist component sitting right on top of your -Nyquist component. Hence
the aliasing. The information has been lost in the
and it doesn't require a table of coefficients, like doing higher-order
Lagrange or Hermite would.
Well, you can compute those at runtime if you want - and you don't need a
terribly high order Lagrange interpolator if you're already oversampled, so
it's not necessarily a problematic overhead.
for linear interpolation, if you are a delayed by 3.5 samples and you
keep that delay constant, the transfer function is
H(z) = (1/2)*(1 + z^-1)*z^-3
that filter goes to -inf dB as omega gets closer to pi.
Note that this holds for symmetric fractional delay filter of any odd order
(i.e.,
.
E
On Wed, Aug 19, 2015 at 3:55 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 20/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
I don't dispute that linear fractional interpolation is the right choice
if
you're going to oversample by a large ratio. The question is what
, not a
discrete time signal of whatever sampling rate.
E
On Fri, Aug 21, 2015 at 2:09 AM, Peter S peter.schoffhau...@gmail.com
wrote:
On 21/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
In this graph, the signal frequency seems to be 250 Hz, so this graph
shows the equivalent of about 22000/250 = 88x
, 2015 at 1:24 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 21/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
It shows *exactly* the aliasing
It shows the aliasing left by linear interpolation into the continuous
time
domain. It doesn't show the additional aliasing produced
...@gmail.com
wrote:
On 21/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
Creating a 22000 Hz signal from a 250 Hz signal by interpolation, is
*exactly* upsampling
That is not what is shown in that graph. The graph simply shows the
continuous-time frequency response of the interpolation
The details of how the graphs were generated don't really matter. The point
is that the only effect shown is the spectrum of the continuous-time
polynomial interpolator. The additional spectral effects of delaying and
resampling that continuous-time signal (to get fractional delay, for
example)
to
what I'm saying in the first place. It is indeed a waste of your time to
invent equivalent ways to generate graphs, since that is not the point.
E
On Fri, Aug 21, 2015 at 2:56 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 21/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
The details
1) Olli Niemiatalo's graph *is* equivalent of the spectrum of
upsampled white noise.
We've been over this repeatedly, including in the very post you are
responding to. The fact that there are many ways to produce a graph of the
interpolation spectrum is not in dispute, nor is it germaine to my
of the noisiness no matter how
much data you throw at it).
E
On Fri, Aug 21, 2015 at 5:47 PM, Peter S peter.schoffhau...@gmail.com
wrote:
On 22/08/2015, Ethan Duni ethan.d...@gmail.com wrote:
We've been over this repeatedly, including in the very post you are
responding to. The fact
In this graph, the signal frequency seems to be 250 Hz, so this graph
shows the equivalent of about 22000/250 = 88x oversampling.
That graph just shows the frequency responses of various interpolation
polynomials. It's not related to oversampling.
E
On Thu, Aug 20, 2015 at 5:40 PM, Peter S
Yeah I am also curious. It's not obvious to me where it would make sense to
spend resources compensating for interpolation rather than just juicing up
the interpolation scheme in the first place.
E
On Mon, Aug 17, 2015 at 11:39 AM, Nigel Redmon earle...@earlevel.com
wrote:
Since compensation
If all you're trying to do is mitigate the rolloff of linear interp
That's one concern, and by itself it implies that you need to oversample by
at least some margin to avoid having a zero at the top of your audio band
(along with a transition band below that).
But the larger concern is the
rbj
and it doesn't require a table of coefficients, like doing higher-order
Lagrange or Hermite would.
Robert I think this is where you lost me. Wasn't the premise that memory
was cheap, so we can store a big prototype FIR for high quality 512x
oversampling? So why are we then worried about the
Well, I was thinking about this as well. How about a 1bit square wave then?
Such a signal is deterministic and so has entropy rate of zero.
Your bitflip counter would not be sensitive to duty cycle.
The simpler bit counter would be.
I don't see why entropy should change with duty cycle since I
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