Re: [music-dsp] Compensate for interpolation high frequency signal loss
Interesting story about the interpolation noise from very high oversampled signal approximations. I tend to think ïf it doesn't concern an actual sinc function of significant width and accuracy then the up-sampling is wrong unless the signal is prepared for it. I can imagine in sample processing machines and software that you could do effect processing in an oversampled domain which is arrived at by having knowledge of the sample signals and possibly their detune filters, and that it is possible to compute sample sets which allow oversampling by relatively simple or cheap DSP operations to anyway fulfill certain accuracy criteria, such as low noise, frequency accuracy, or effect accuracy (like the number of fractional sampling accuracy in a phaser effect for instance). The linear interpolation I've used a decade ago for chorussing on a moderately strong DSP of the time was ok bvecause of signal properties and the tunings of the chorus I programmed, of course there's a lot to be said for using more than zero order interpolations in general. I've looked at Taylor expansions for the sinc function and it's possible accuracies, for instance. In mechanical design, it was one of the early computer math issues to use all kinds of interpolation schemes for a variety of purposes, with some terminology I suppose from the early days of the industrial revolution. However, a good understanding of these should be based on an understanding of what they are for. Some interpolations are for minimal stress, some for minimal distance given a certain curvature, others are statistically neutral in some sense, etc. More interesting is to look at more dimensional curves and surfaces or try out these in functional analysis or computations, which is far outside the scope here, and not very useful in normal audio subjects. Unfortunately the averaging and continuity considerations of the various interpolation curves and their mathematical properties aren't very well correlated with audi signals, and certainly not necessarily with sampling issues. So think about what some suggested here: what is the filte rkernel that you're putting over your signal when using them, and what does the sampled nature add in terms of misery by the side ? T.V. ___ dupswapdrop: music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/26/15 9:47 PM, Ethan Duni wrote: 15.6 dB + (12.04 dB) * log2( Fs/(2B) ) Oh I see, you're actually taking the details of the sinc^2 into account. really, just the fact that the sinc^2 has nice deep zeros at every integer multiple of Fs (except 0). What I had in mind was more of a worst-case analysis where we just call the sin() component 1 and then look at the 1/n^2 decay (which is 12dB per octave). Which we see in the second term, but of course the sine's contribution also whacks away a certain portion of energy, hence the 15.6dB offset. well, it's more just how the strengths of the images add up. On the other hand if you're interested in something like the spurious-free dynamic range, then the simple 12dB/octave estimate is appropriate. The worst-case components aren't going to get attenuated at all by the sin(), just the 1/k^2. I tend to favor that in cases where we can't be confident that the noise floor in question is (at least approximately) flat. but whether you're assuming a flat spectrum up to B or just a single sinusoidal component at the maximum frequency of B, it's the sin() (or the sin^4 in the power spectrum of the images resulting from linear interpolation) that is the mathematical force in reducing the image strength in the oversampled case where 2B Fs. so it's *both* the sin^4 *and* the 1/k^4 is used. the 1/k^4 thing is needed for the power of all those images to add up to a decent finite number. but it's the sin() that puts a stake in the heart of the image and that is quantitatively quite useful. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ dupswapdrop: music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/25/15 7:08 PM, Ethan Duni wrote:
if you can, with optimal coefficients designed with the tool of your
choice, so i am ignoring any images between B and Nyquist-B, upsample
by 512x and then do linear interpolation between adjacent samples for
continuous-time interpolation, you can show that it's something like
12 dB S/N per octave of oversampling plus another 12 dB. that's 120
dB. that's how i got to 512x.
Wait, where does the extra 12dB come from? Seems like it should just
be 12dB per octave of oversampling. What am I missing?
okay, this is painful. in our 2-decade old paper, Duane and i did this
theoretical approximation analysis for drop-sample interpolation, and i
did it myself for linear, but we did not put in the math for linear
interpolation in the paper.
so, to satisfy Nyquist (or Shannon or Whittaker or the Russian guy) the
sample rate Fs must exceed 2B which is twice the bandwidth. the
oversampling ratio is defined to be Fs/(2B). and in octaves it is
log2(Fs/(2B)). all frequencies in your baseband satisfy |f|B and if
it's highly oversampled, 2B Fs.
now, i'm gonna assume that Fs is so much (like 512x) greater than 2B
that i will assume the attenuation due to the sinc^2 for |f|B is
negligible. i will assume that the spectrum between -B and +B is
uniformly flat (that's not quite worst case, but it's worser case than
what music, in the bottom 5 or 6 octaves, is). so given a unit height
on that uniform power spectrum, the energy will be 2B.
so, the k-th image (where k is not 0) will have a zero of the sinc^2
function going right through the heart of it. that's what's gonna kill
the son-of-a-bitch. the energy of that image is:
k*Fs+B
integral{ (sinc(f/Fs))^4 df }
k*Fs-B
since it's power spectrum it's sinc^4 for linear and sinc^2 for
drop-sample interpolation.
changing the variable of integration
+B
integral{ (sinc((k*Fs+f)/Fs))^4 df }
-B
+B
integral{ (sinc(k+f/Fs))^4 df }
-B
sinc(k+f/Fs) = sin(pi*(k+f/Fs))/(pi*(k+f/Fs))
= (-1)^k * sin(pi*f/Fs)/(pi*(k+f/Fs))
=approx (-1)^k * (pi*f/Fs)/(pi*k)
since |f| B Fs
raising to the 4th power gets rid of the toggling polarity. so now it's
+B
1/(k*Fs)^4 * integral{ f^4 df } = (2/5)/(k*Fs)^4 * B^5
-B
now you have to sum up the energies of all of the bad images (we are
assuming that *all* of those images, *after* they are beaten down, will
somehow fall into the baseband during resampling and their energies will
team up). there are both negative and positive frequency images to add
up. (but we don't add up the energy from the image at the baseband,
that's the good image.)
+inf +inf
2 * SUM{ (2/5)/(k*Fs)^4 * B^5 } = B*(4/5)*(B/Fs)^4 * SUM{1/k^4}
k=1k=1
the summation on the right is (pi^4)/90
so the energy of all of the nasty images (after being beaten down due to
the application of the sinc^2 that comes from linear interpolation
between the subsamples) is
B*(4/5)*(B/Fs)^4 * (pi^4)/90
and the S/N ratio is 2B divided by that.
( (2/450) * (2B/Fs)^4 * (pi/2)^4 )^-1
in dB we use 3.01*log2() because this is an *energy* ratio, not a
voltage ratio.
-3.01*log2( (2/450) * (2B/Fs)^4 * (pi/2)^4 )
= 3.01*log2(225) + 12.04*log2(2/pi) + 12.04*log2( Fs/(2B) )
= 15.6 dB + (12.04 dB) * log2( Fs/(2B) )
so, it seems to come out a little more than 12 dB. i think Duane did a
better empirical analysis and he got it slightly less.
but, using linear interpolation between subsamples, you should get about
12 dB of S/N for every octave of oversampling plus 15 dB more.
but the difference in price in memory only, *not* in computational burden.
Well, you don't get the full cost in computational burden since you
can skip computing most of the upsamples.
exactly and it's the same whether you upsample by 32x or 512x. but
upsampling by 512x will cost 8 times the memory to store coefficients.
But the complexity still goes up with increasing oversampling factor
since the interpolation filter needs to get longer and longer, no?
no. that deals with a different issue, in my opinion. the oversampling
ratio determines the number of discrete (and uniformly spaced)
fractional delays. there is one FIR filter for each fractional delay.
the number of coefs in the FIR filter is a performance issue regarding
how well you're gonna beat down them images in between baseband and the
next *oversampled* image. in the analysis above, i am assuming all of
those in-between images are beaten down to zero. it's a crude analysis
and i just wanted to see what the linear interpolation (on the upsampled
signal) does for us.
So there is some balancing of computational
Re: [music-dsp] Compensate for interpolation high frequency signal loss
15.6 dB + (12.04 dB) * log2( Fs/(2B) )
Oh I see, you're actually taking the details of the sinc^2 into account.
What I had in mind was more of a worst-case analysis where we just call the
sin() component 1 and then look at the 1/n^2 decay (which is 12dB per
octave). Which we see in the second term, but of course the sine's
contribution also whacks away a certain portion of energy, hence the 15.6dB
offset.
On the other hand if you're interested in something like the spurious-free
dynamic range, then the simple 12dB/octave estimate is appropriate. The
worst-case components aren't going to get attenuated at all by the sin(),
just the 1/n^2. I tend to favor that in cases where we can't be confident
that the noise floor in question is (at least approximately) flat.
so, it seems to come out a little more than 12 dB.
I long ago adopted an informal rule that when an engineer says 6dB he
means 20*log10(2), and not exactly 6dB. And likewise for 3dB, 12dB, etc.
Doubly so when talking about the rolloff of linear systems, nobody ever
splits that hair... IIRC the prof in my freshman linear circuits class
instructed us to fudge it this way immediately after introducing the
concept of dB :]
the number of coefs in the FIR filter is a performance issue regarding how
well you're gonna
beat down them images in between baseband and the next *oversampled* image
Right I see what you mean. I had mixed up my arithmetic on the lengths of
the filters as a function of oversampling ratio.
i think we're on the same page. ain't we?
Yeah, I was unclear on which scenario(s) the aliasing analysis was supposed
to apply to.
E
On Wed, Aug 26, 2015 at 12:53 PM, robert bristow-johnson
r...@audioimagination.com wrote:
On 8/25/15 7:08 PM, Ethan Duni wrote:
if you can, with optimal coefficients designed with the tool of your
choice, so i am ignoring any images between B and Nyquist-B, upsample by
512x and then do linear interpolation between adjacent samples for
continuous-time interpolation, you can show that it's something like 12 dB
S/N per octave of oversampling plus another 12 dB. that's 120 dB. that's
how i got to 512x.
Wait, where does the extra 12dB come from? Seems like it should just be
12dB per octave of oversampling. What am I missing?
okay, this is painful. in our 2-decade old paper, Duane and i did this
theoretical approximation analysis for drop-sample interpolation, and i did
it myself for linear, but we did not put in the math for linear
interpolation in the paper.
so, to satisfy Nyquist (or Shannon or Whittaker or the Russian guy) the
sample rate Fs must exceed 2B which is twice the bandwidth. the
oversampling ratio is defined to be Fs/(2B). and in octaves it is
log2(Fs/(2B)). all frequencies in your baseband satisfy |f|B and if it's
highly oversampled, 2B Fs.
now, i'm gonna assume that Fs is so much (like 512x) greater than 2B that
i will assume the attenuation due to the sinc^2 for |f|B is negligible. i
will assume that the spectrum between -B and +B is uniformly flat (that's
not quite worst case, but it's worser case than what music, in the bottom 5
or 6 octaves, is). so given a unit height on that uniform power spectrum,
the energy will be 2B.
so, the k-th image (where k is not 0) will have a zero of the sinc^2
function going right through the heart of it. that's what's gonna kill the
son-of-a-bitch. the energy of that image is:
k*Fs+B
integral{ (sinc(f/Fs))^4 df }
k*Fs-B
since it's power spectrum it's sinc^4 for linear and sinc^2 for
drop-sample interpolation.
changing the variable of integration
+B
integral{ (sinc((k*Fs+f)/Fs))^4 df }
-B
+B
integral{ (sinc(k+f/Fs))^4 df }
-B
sinc(k+f/Fs) = sin(pi*(k+f/Fs))/(pi*(k+f/Fs))
= (-1)^k * sin(pi*f/Fs)/(pi*(k+f/Fs))
=approx (-1)^k * (pi*f/Fs)/(pi*k)
since |f| B Fs
raising to the 4th power gets rid of the toggling polarity. so now it's
+B
1/(k*Fs)^4 * integral{ f^4 df } = (2/5)/(k*Fs)^4 * B^5
-B
now you have to sum up the energies of all of the bad images (we are
assuming that *all* of those images, *after* they are beaten down, will
somehow fall into the baseband during resampling and their energies will
team up). there are both negative and positive frequency images to add
up. (but we don't add up the energy from the image at the baseband, that's
the good image.)
+inf +inf
2 * SUM{ (2/5)/(k*Fs)^4 * B^5 } = B*(4/5)*(B/Fs)^4 * SUM{1/k^4}
k=1k=1
the summation on the right is (pi^4)/90
so the energy of all of the nasty images (after being beaten down due to
the application of the sinc^2 that comes from linear interpolation between
the subsamples) is
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 2015-08-19, Ethan Duni wrote: 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? In my (admittedly limited) experience these sorts of tradeoffs come when you need to resample generally, so not just downwards from the original sample rate but upwards as well, and you're doing it all on a dedicated DSP chip. In that case, when your interpolator approaches and goes beyond the Nyquist frequency of the original sample, you need longer and longer approximations of the sinc(x) response, with wonkier and wonkier recursion formulas for online calculation of the coefficients of the interpolating polynomial. Simply because of aliasing suppression, and because you'd like to calculate the coefficients on the fly to save on memory bandwidth. However, if you suitably resample both in the output sampling frequency and in the incoming one, you're left with some margin as far as the interpolator goes, and it's always working downwards, so that it doesn't actually have to do aliasing suppression. An arbitrary low order polynomial is easier to calculate on the fly, then. The crucial part on dedicated DSP chips is that they can generate radix-2 FFT coefficients basically for free, with no table lookup and severely accelerated inline computation as well. That means that you can implement both the input and the output side anti-aliasing/anti-imaging filters via polyphase Fourier methods, for much less effort than the intervening arbitrary interpolation step would ever require. When you do that right, the code is still rather complex since it needs to dynamically mipmap the input sample for larger shifts in frequency, but when done right, you can also get essentially perfect and perfectly flexible results from a signal chain with perhaps 3x the computational load of a baseband third order interpolation polynomial, absent hardware acceleration. -- Sampo Syreeni, aka decoy - de...@iki.fi, http://decoy.iki.fi/front +358-40-3255353, 025E D175 ABE5 027C 9494 EEB0 E090 8BA9 0509 85C2 ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/24/15 11:18 AM, Sampo Syreeni wrote: On 2015-08-19, Ethan Duni wrote: 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? that was my premise for using linear interpolation *between* adjacent oversampled (by 512x) samples. if you can, with optimal coefficients designed with the tool of your choice, so i am ignoring any images between B and Nyquist-B, upsample by 512x and then do linear interpolation between adjacent samples for continuous-time interpolation, you can show that it's something like 12 dB S/N per octave of oversampling plus another 12 dB. that's 120 dB. that's how i got to 512x. some apps where you might care less about inharmonic energy from images folding over (a.k.a. aliasing), you might not need to go that high of whatever-x. but the difference in price in memory only, *not* in computational burden. whether it's 64x or 512x, the computational cost is separating the index into integer and fractional parts, using the integer part to select the N samples to combine and the fractional part to tell you how to combine them. if it's 512x, the fractional part is broken up into the top 9 bits to select your N coefficients (and the neighboring set of N coefficients) and the rest of the bits are for the linear interpolation. with only the cost of a few K of words (i remember the days when 4K was a lotta memory :-), you can get to arbitrarily good with the cost of 2N+1 MAC instructions. with drop-sample interpolation between fractional delays (6 dB per octave of oversampling), then you need another 10 octaves of oversampling, 512K*N words of memory, but only N MAC instructions per output sample. when it's using Hermite or Lagrange then the S/N is 24 dB per octave of oversampling, i don't think it's worth it that you need only 16x or 32x oversampling (that saves only memory and the cost of computation becomes 4 times worse or worser). maybe in an ASIC or an FPGA, but in DSP code or regular-old software, i don't see the advantage of cubic or higher-order interpolation unless memory is *really* tight and you gotta lotta MIPs to burn. In my (admittedly limited) experience these sorts of tradeoffs come when you need to resample generally, so not just downwards from the original sample rate but upwards as well, and you're doing it all on a dedicated DSP chip. In that case, when your interpolator approaches and goes beyond the Nyquist frequency of the original sample, you need longer and longer approximations of the sinc(x) response, you need that to get sharper and sharper brick-wall LPFs to whack those 511 images in between the baseband and 512x. then the sinc^2 function in the linear interpolation blasts the hell outa all them images that are at multiples of 512x (except the 0th multiple of course). drop-sample interpolation would have only a sinc function doing it whereas and Mth-order B-spline would have a sinc^(M+1) function really blasting the hell outa them images. with wonkier and wonkier recursion formulas for online calculation of the coefficients of the interpolating polynomial. Simply because of aliasing suppression, and because you'd like to calculate the coefficients on the fly to save on memory bandwidth. However, if you suitably resample both in the output sampling frequency and in the incoming one, you're left with some margin as far as the interpolator goes, and it's always working downwards, so that it doesn't actually have to do aliasing suppression. An arbitrary low order polynomial is easier to calculate on the fly, then. The crucial part on dedicated DSP chips is that they can generate radix-2 FFT coefficients basically for free, with no table lookup yeah, but you get accumulated errors as you compute the twiddle factors on-the-fly. either in linear or bit-reversed order. and severely accelerated inline computation as well. That means that you can implement both the input and the output side anti-aliasing/anti-imaging filters via polyphase Fourier methods, the way that Bob Adams did it, with a single side (the input side) was to step through the sinc-like LPF impulse response at smaller steps (and interpolate) to LPF the input further for down-sampling. admittedly, i haven't worried about the natural foldover from downsampling and just assume there wasn't stuff below the original Nyquist that you had to worry about. or maybe my optimal filter kernel started to whack stuff at a frequency below 20 kHz. but i didn't change it for downsampling. i guess i was lazy. for much less effort than the intervening arbitrary interpolation step would ever require. When you do that right, the code is still rather complex since it needs to dynamically mipmap the input sample for larger
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 22/08/2015, Ethan Duni ethan.d...@gmail.com wrote: So your whole point is that it's not *exactly* sinc^2, but a slightly noisy version thereof? My point was that there are no effects of resampling visible in the graphs. And you're wrong - all those 88 alias images are effects of resampling... That has nothing to do with exactly how the graphs were generated, nor does insisting that the graphs are slightly noisy address the point. Well, it was *you* who insisted that it displays a graphed sinc^2 curve, and not a resampled signal... And you were wrong. Indeed, you've already conceded that the resampling effects are not visible in the graphs several posts back. Aren't all those 88 alias images effects of resampling? What are those, if not effects of resampling? You claimed no upsampling is involved, yet when I upsample noise, I get exactly that graph. So it seems you were wrong. It seems like you're just casting about for some other issue that you can tell yourself you won, and then call me names, to feed your fragile ego. Well, if you do not see that the curve is NOT a graphed sinc^2, but rather, a noisy curve seemingly from resampled noise, then you have some underlying problem. Honestly, it's a pretty sad spectacle and I'm embarrassed for you. I'm embarrassed for you. It really would be better for everyone - including you - if you could interact in a good-faith, mature manner. Please make an effort to start doing so, or you're pretty soon going to find that nobody here will interact with you any more. Yet - for some reason - you keep interacting with me for the past 22 mails you wrote. Maybe to feed your fragile ego and prove that you won... (?) By the way, there's no reason for any jaggedness to appear in the plots, given the lengths of data you were talking about. There *is* reason for jaggedness to appear in the plots. If you don't believe, try it yourself - take some white noise sampled at 500 Hz, and resample it to 44.1 kHz. The shorter the length, the more jagged the spectrum will look. Besides, we do not know how much data Olli processed, so you cannot say there's no reason for jaggedness in his graph - as you do not know how he derived his graph. So your argument is invalid again. Producing a very smooth graph from a long enough segment of data is straightforward, if you use appropriate techniques (not just one big FFT of the whole thing, that won't ever get rid of the noisiness no matter how much data you throw at it). Exactly. And that's what I used (spectral averaging over a long segment), yet it is STILL noisy, if the white noise segment is not very long. So your argument is wrong again... -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
So you claim that the graph depicts a sinc^2 graph, and it shows the frequency response of a continuous time linearly interpolated signal, and involves no resampling. That is false. That is not how Olli created his graph. First, the continuous time signal (which, by the way, already contains an infinite amount of aliases of the original spectrum) exists only in your imagination - I'm almost 100% certain Olli made his graph by resampling noise. The telltale signs of this are: - the curves on the graph are jagged/noisy, typical of averaged white noise spectrum - if you watch closely, the same jaggedness repeats at a 2*PI frequency interval, showing that they are aliases of the original spectrum, which was noisy. Therefore, Ollis graph does *not* depict a continuous time signal, but rather, a noisy signal that was resampled to 44.1 kHz. Therefore, what you see on the graph, is the artifacts from the resampling. Therefore, all your arguments are invalid. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
So let me get this straight - you have an *imaginary* graph in your head, depicting the frequency response of a continuous time linearly interpolated signal, and you keep arguing about this *imaginary* graph (maybe to feed your fragile ego and to prove that you won). That is *not* what you see on Olli's graph, as been discussed in depth. So what you're arguing about, is not Olli's graph that was presented, but rather, an *imaginary* graph, that exists only in your head. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
And besides, no one ever said that Olli's graph depicts analyitical frequency responses of continuous time interpolators. The graphs come from a musicdsp.org code entry: http://musicdsp.org/archive.php?classid=5#49 There's no comment whatsover, just the code and the graphs. If you read his 65 page long paper on interpolators, he doesn't discuss analytical continuous time interpolator frequency responses whatsoever. He just shows their graphs, and tells where they have zeros in the response. No formulas for analytical frequency responses, at all - seemingly he is not interested in that. I just skimmed through his paper again, and the closest thing that he has in it, are polynomial approximations for frequency responses in the passband. About that's all, other than that, no frequency response formulas. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 2015-08-18, Tom Duffy wrote: In order to reconstruct that sinusoid, you'll need a filter with an infinitely steep transition band. You've demonstrated that SR/2 aliases to 0Hz, i.e. DC. That digital stream of samples is not reconstructable. The conjugate sine to +1, -1, +1, -1, ... is 0, 0, 0, 0... Just phase shift the original sine at the Nyquist frequence. That'll show you that that precise signal cannot be reconstructed without resorting to complex continuation of the signal, on the Fourier plane. -- Sampo Syreeni, aka decoy - de...@iki.fi, http://decoy.iki.fi/front +358-40-3255353, 025E D175 ABE5 027C 9494 EEB0 E090 8BA9 0509 85C2 ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 22/08/2015, Sampo Syreeni de...@iki.fi wrote: The conjugate sine to +1, -1, +1, -1, ... is 0, 0, 0, 0... Just phase shift the original sine at the Nyquist frequence. Let me ask what do you mean by conjugate sine ? If you mean complex conjugate, and assume the sine to be the real part of complex phasor rotating around the complex unit circle, then isn't the conjugate of that phasor also +1, -1, +1, -1,... ? The only difference is that the phasor is mirrored around the X axis (so the imaginary part +i becomes -i), so it rotates in the opposite direction (negative frequency). Since the frequency of that phasor is pi, the complex conjugate phasor rotating at the other direction is also +1, -1, +1, -1... Either direction, the phasor toggles between positions z=1 and z=-1. Maybe you meant quadrature sine ? That'll show you that that precise signal cannot be reconstructed without resorting to complex continuation of the signal, on the Fourier plane. Let me ask, what do you mean by Fourier plane? I never heard that term, and Google only gives me optics-related pages. Maybe you mean complex plane? ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Okay, I'll risk exceeding my daily message limit. If the administrators think it is inappropriate, dealing with that is at their discretion. Here is another proof that the alias images in the spectrum are caused by the sampling/upsampling, not the interpolation: Let's replace linear interpolation with simply stuffing zeros between samples. So that means, we upsample the signal without applying interpolation or filtering. Let's try this on an ~50 Hz sine wave sampled at 44100/88 ~= 501 Hz, upsampled to 44.1 kHz by stuffing 87 zeros between each sample. The resulting waveform looks like individual impulses, spaced 88 samples apart: http://morpheus.spectralhead.com/img/sine_upsampled_waveform.png Here is the spectrum: http://morpheus.spectralhead.com/img/sine_upsampled_spectrum.png We can see the usual alias frequencies at 450 Hz, 550 Hz, 950 Hz, 1050 Hz, 1450 Hz, 1550 Hz, 1950 Hz, 2050 Hz, ... This is because the upsampling causes the original spectrum to repeated infinite times, causing these alias frequencies to appear in the resulting spectrum. Therefore, it is NOT the interpolation that is causing these alias images, but rather, the upsampling... More precisely, they're already present in the original signal sampled at 500 Hz, the upsampling just makes them visible. I used no interpolation at all, yet all this aliasing appeared on the spectrum. All the interpolation does, is it filters out some of this aliasing... Since the impulse response of linear interpolation is a triangle, applying linear interpolation is equivalent to convolving the resulting upsampled signal with a triangular kernel filter. Since the Fourier transform of a rectangle is a sinc function, and a triangular kernel is equivalent to convolving two rectangular kernels, the Fourier spectrum of a triangular kernel will look like a sinc^2 function. But that's not what causes the aliasing... it's there already after the upsampling, before you apply the interpolation/convolution. You can take a discretized version of a continuous triangular kernel sampled at the upsampled rate, and convolving the upsampled signal with that kernel will be equivalent to linear interpolation. You do not actually need a continuous time signal to be present, and the aliasing/imaging is there already before doing the triangular convolution at the upsampled rate. Several authors discuss the equivalence of linear interpolation and convolution with a triangular filter, examples: 1) linear interpolation can be expressed as convolving the sampled function with a triangle function[1] http://morpheus.spectralhead.com/img/linear_interpolation1.png 2) The first-order hold [= linear interpolation] corresponds to an impulse response for the reconstruction filter that is a triangle of duration equal to twice the sampling period.[2] http://morpheus.spectralhead.com/img/linear_interpolation2.png 3) http://morpheus.spectralhead.com/img/linear_interpolation3.png [1] Oliver Kreylos, Sampling Theory 101 http://idav.ucdavis.edu/~okreylos/PhDStudies/Winter2000/SamplingTheory.html [2] Alan V. Oppenheim, Signals and Systems, ch. 17. Interpolation http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/lecture-notes/MITRES_6_007S11_lec17.pdf [3] Ruye Wang, Sampling Theorem, Reconstruction of Signal by Interpolation http://fourier.eng.hmc.edu/e101/lectures/Sampling_theorem/node3.html -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 polynomials, graphed up to 22kHz. No resampling is depicted, or the frequency responses would show the aliasing associated with that. It's just showing the sinc^2 response of the linear interpolator, and similar for the other polynomials. This is what you'd get if you used those interpolation polynomials to convert a 250Hz sampled signal into a continuous time signal, 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 oversampling. That graph just shows the frequency responses of various interpolation polynomials. It's not related to oversampling. Creating a 22000 Hz signal from a 250 Hz signal by interpolation, is *exactly* upsampling - the sampling rate changes by a factor of 88x. It's not bandlimited interpolation (using a windowed sinc interpolator), hence there is a lot of aliasing above Nyquist. Irregardless, it's still oversampling - the resulting signal is sampled with a 88x higher frequency than the original. It's equivalent to creating a 3,880,800 Hz signal from a 44100 Hz signal. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Also, you even contradict yourself. You claim that: 1) Olli's graph was created by graphing sinc(x), sinc^2(x), and not via FFT. 2) The artifacts from the resampling would be barely visible, because the oversampling rate is quite high. So, if - according to 2) - the artifacts are not visible because the oversampling is high and the graph doesn't focus on that, then how do you know that 1) is true? You claim that the resampling artifacts wouldn't be visible anyways. If that's true, then how would you prove that FFT was not used for creating Olli's graph? Also, even you yourself acknowledge that It shows the aliasing left by linear interpolation into the continuous time domain. So, we agree that the graph shows aliasing, right? I do not know where you get your idea of additional aliasing - it's the very same aliasing, except the resampling folds it back... ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 21/08/2015, Ethan Duni ethan.d...@gmail.com wrote: So you agree that the effects of resampling are not shown, and all we see is the spectrum of the continuous time polynomial interpolators. I claim that they are aliases of the original spectrum. Just as you also call them: It shows the aliasing left by linear interpolation into the continuous time domain. I never claimed anything about the folding back of alias frequencies from the resampling at 44.1k rate. If I were you, I'd quit haranguing people over irrelevancies and straw men, It's rather you who argue about irrelevant things and straw man arguments - for that matter, I never claimed that the folded back aliases from the resampling at 44.1k are visible on Olli's graph. It's you who is forcing this irrelevant argument. So maybe you should listen to your own advice, first. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
A sampled signal contains an infinte number of aliases: http://morpheus.spectralhead.com/img/sampling_aliases.png the spectrum is replicated infinitely often in both directions These are called aliases of the spectrum. You do not need to fold back the aliasing via resampling for them to become aliases... They're aliases already - when sampled at the original rate, they would all alias back to the original signal. This is because exp(i*x) is periodic, and after 2*PI radians, you get back to the same frequency... hence, frequencies that are 2*PI apart from each other, are all aliases... If you fail to understand that, I think you fail to understand even the basics of sampling theory. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Since that image is not meant to illustrate the effects of resampling, but rather, to illustrate the effects of interpolation, *obviously* it doesn't focus on the aliasing from the resampling. So you agree that the effects of resampling are not shown, and all we see is the spectrum of the continuous time polynomial interpolators. I'm going to accept that concession of my point and move on. If I were you, I'd quit haranguing people over irrelevancies and straw men, and generally trying to pretend to superiority. Nobody is buying it, and it just highlights your insecurity. E On Fri, Aug 21, 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 by then delaying and sampling that signal. I.e., the images that would get folded back onto the new baseband, disturbing the sinc^2 curve. This image doesn't involve any fractional delay. Those differences would be quite small for resampling to 44.1kHz with no delay, since the oversampling ratio is considerable, so you'd have to look carefully to see them. I think they're actually on the image: http://morpheus.spectralhead.com/img/resampling_aliasing.png They're hard to notice, because the other aliasing masks it. This is a big hint that they are not portrayed: Ollie knows what he is doing, so if he wanted to illustrate the effects of the resampling, he would have constructed a scenario where they are easily visible. Since that image is not meant to illustrate the effects of resampling, but rather, to illustrate the effects of interpolation, *obviously* it doesn't focus on the aliasing from the resampling. Therefore, it is not a hint at all, and your argument is invalid. And probably mentioned a second sample rate, explicitly shown both the sinc^2 and its aliased counterpart, etc. The effect would be shown in a visible, explicit manner, if that was what the graph was supposed to show. The fact that this graph is not supposed to demonstrate the aliasing from the resampling, does not mean that 1) it's not there on the graph (it's just barely visible) 2) the images of the continuous time interpolated signal are not aliasing. That's also called aliasing!!! But all of those things depend on parameters like oversampling ratio and delay, so it would be a much more complicated picture. Yes, and that's all entirely irrelevant here... Because the images in the continuous time signal before the resampling are also called aliasing!!! They're all aliases of the original spectrum, and they all alias back to the original spectrum when sampled at the original sampling rate! They're called aliasing even before you resample them! What we're shown here is just the effects of polynomial interpolation to get to the continuous time domain. False. I've shown the FFT frequency spectra of actual upsampled signals. The additional effects of delaying and then sampling that signal back into the discrete time domain are not visible. There was no delaying involved at all. The effects of sampling that signal back are not visible, because there's 88x oversampling, just as I pointed out. If you want, you can repeat the same with less oversampling, and present us your results. It seems that you have assumed that some resampling must be happening because the graph only goes up to 22kHz. But that's just the range of the graph, you don't need to do any resampling of anything to graph sinc^2 over any particular range of frequencies. I never said you need do to resampling of the continuous time signal to graph sinc^2. I said: the images in the frequency spectrum of the continuous time signal are aliases of the original spectrum, and they alias back to the original spectrum when the continuous time signal is sampled at the original rate! But that's not quite the exact same graph. It's essentially the exact same graph. And why are you putting a sound card in the loop? That was the most convenient way to record the signal. This is all just digital processing in question here. You don't even need to process any signals, there are analytic expressions for all of the quantities involved. That's just one way of drawing fancy graphs. FFT is another way of drawing fancy graphs. Why would I restrict myself to one method? That's how Ollie generated graphs of them without reference to any particular signals. How do you know? Prove it! I'm convinced he generated it via numerical means and FFT. Again, the differences in question are small due to the high oversampling ratio, so it's going to be quite difficult to see them in macroscopic graphs like this. Let me point out again, that all those spectral images in the continunous time signal
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Let's repeat the same with a 50 Hz sine wave, sampled at 500 Hz, then linearly interpolated and resampled at 44.1 kHz: http://morpheus.spectralhead.com/img/sine_aliasing.png The resulting alias frequencies are at: 450 Hz, 550 Hz, 950 Hz, 1050 Hz, 1450 Hz, 1550 Hz, 1950 Hz, 2050 Hz, 2450 Hz, 2550 Hz, ... I think it should be obvious that these are all alias frequencies of 50 Hz, since if you sample any of these sinusoids at 500 Hz rate, they will all alias to 50 Hz. Hence, they are - by definition - aliases of the 50 Hz sinusoid. Welcome to sampling theory 101. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 by then delaying and sampling that signal. I.e., the images that would get folded back onto the new baseband, disturbing the sinc^2 curve. This is how we end up with a zero at Nyquist when we do half-sample delay, for example. And also how we end up with a perfectly flat response if we do the trivial resampling (original rate, no delay). Those differences would be quite small for resampling to 44.1kHz with no delay, since the oversampling ratio is considerable, so you'd have to look carefully to see them. This is a big hint that they are not portrayed: Ollie knows what he is doing, so if he wanted to illustrate the effects of the resampling, he would have constructed a scenario where they are easily visible. And probably mentioned a second sample rate, explicitly shown both the sinc^2 and its aliased counterpart, etc. The effect would be shown in a visible, explicit manner, if that was what the graph was supposed to show. But all of those things depend on parameters like oversampling ratio and delay, so it would be a much more complicated picture. What we're shown here is just the effects of polynomial interpolation to get to the continuous time domain. The additional effects of delaying and then sampling that signal back into the discrete time domain are not visible. It seems that you have assumed that some resampling must be happening because the graph only goes up to 22kHz. But that's just the range of the graph, you don't need to do any resampling of anything to graph sinc^2 over any particular range of frequencies. Oh, it's the *exact* same graph! (Minus some difference above 20 kHz, due to my soundcard's anti-alias filter.) You get the same graph if you sample that continuous time signal at a 44.1 kHz sampling rate (with some further aliasing from the sampling). But that's not quite the exact same graph. And why are you putting a sound card in the loop? This is all just digital processing in question here. You don't even need to process any signals, there are analytic expressions for all of the quantities involved. That's how Ollie generated graphs of them without reference to any particular signals. Again, the differences in question are small due to the high oversampling ratio, so it's going to be quite difficult to see them in macroscopic graphs like this. If you want to see the differences, just make a plot of both sinc^2 and its aliased versions (for whatever oversampling ratios and/or delays), and look at the differences. It won't be interesting for high oversampling ratios and zero delay - which is exactly why that scenario is a poor choice for illustrating the effects in question. The fact that sampling a continuous time signal at a very high rate results in a spectrum that closely resembles the continuous time spectrum (over the sampled bandwidth) is beside the point. It just means that you're operating in a regime where the effects are very hard to spot. It doesn't follow from that resemblance that resampling must be occurring to get a plot of the spectrum of the continuous time signal. E On Fri, Aug 21, 2015 at 10:51 AM, Peter S peter.schoffhau...@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 polynomials, graphed up to 22kHz. No resampling is depicted, or the frequency responses would show the aliasing associated with that. It shows *exactly* the aliasing http://morpheus.spectralhead.com/img/interpolation_aliasing.png There are about 88 alias images visible on the graph. The linear interpolation curve is not smooth, so it contains aliasing. It's just showing the sinc^2 response of the linear interpolator, and similar for the other polynomials. If the signal you interpolate is white noise, and the spectrum of the signal is a flat spectrum rectangle like the one displayed, then after resampling, you get *exactly* the spectrum you see on the graph, showing 88 alias images. Proof: I created 60 seconds of white noise sampled at 500 Hz, then resampled it to 44.1 kHz using linear interpolation. After the upsampling, it sounds like this: http://morpheus.spectralhead.com/wav/noise_resampled.wav Its spectrum looks like this: http://morpheus.spectralhead.com/img/noise_resampled.png Looks familiar? Oh, it's the *exact* same graph! (Minus some difference above 20 kHz, due to my soundcard's anti-alias filter.) It is an FFT graph of the upsampled white noise, and it shows *exactly* the aliasing. Good morning! This is what you'd get if you used those interpolation polynomials to convert a 250Hz sampled signal into a continuous time signal,
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 by then delaying and sampling that signal. I.e., the images that would get folded back onto the new baseband, disturbing the sinc^2 curve. This image doesn't involve any fractional delay. Those differences would be quite small for resampling to 44.1kHz with no delay, since the oversampling ratio is considerable, so you'd have to look carefully to see them. I think they're actually on the image: http://morpheus.spectralhead.com/img/resampling_aliasing.png They're hard to notice, because the other aliasing masks it. This is a big hint that they are not portrayed: Ollie knows what he is doing, so if he wanted to illustrate the effects of the resampling, he would have constructed a scenario where they are easily visible. Since that image is not meant to illustrate the effects of resampling, but rather, to illustrate the effects of interpolation, *obviously* it doesn't focus on the aliasing from the resampling. Therefore, it is not a hint at all, and your argument is invalid. And probably mentioned a second sample rate, explicitly shown both the sinc^2 and its aliased counterpart, etc. The effect would be shown in a visible, explicit manner, if that was what the graph was supposed to show. The fact that this graph is not supposed to demonstrate the aliasing from the resampling, does not mean that 1) it's not there on the graph (it's just barely visible) 2) the images of the continuous time interpolated signal are not aliasing. That's also called aliasing!!! But all of those things depend on parameters like oversampling ratio and delay, so it would be a much more complicated picture. Yes, and that's all entirely irrelevant here... Because the images in the continuous time signal before the resampling are also called aliasing!!! They're all aliases of the original spectrum, and they all alias back to the original spectrum when sampled at the original sampling rate! They're called aliasing even before you resample them! What we're shown here is just the effects of polynomial interpolation to get to the continuous time domain. False. I've shown the FFT frequency spectra of actual upsampled signals. The additional effects of delaying and then sampling that signal back into the discrete time domain are not visible. There was no delaying involved at all. The effects of sampling that signal back are not visible, because there's 88x oversampling, just as I pointed out. If you want, you can repeat the same with less oversampling, and present us your results. It seems that you have assumed that some resampling must be happening because the graph only goes up to 22kHz. But that's just the range of the graph, you don't need to do any resampling of anything to graph sinc^2 over any particular range of frequencies. I never said you need do to resampling of the continuous time signal to graph sinc^2. I said: the images in the frequency spectrum of the continuous time signal are aliases of the original spectrum, and they alias back to the original spectrum when the continuous time signal is sampled at the original rate! But that's not quite the exact same graph. It's essentially the exact same graph. And why are you putting a sound card in the loop? That was the most convenient way to record the signal. This is all just digital processing in question here. You don't even need to process any signals, there are analytic expressions for all of the quantities involved. That's just one way of drawing fancy graphs. FFT is another way of drawing fancy graphs. Why would I restrict myself to one method? That's how Ollie generated graphs of them without reference to any particular signals. How do you know? Prove it! I'm convinced he generated it via numerical means and FFT. Again, the differences in question are small due to the high oversampling ratio, so it's going to be quite difficult to see them in macroscopic graphs like this. Let me point out again, that all those spectral images in the continunous time signal before the resampling, *are* aliasing, as they're aliases of the original spectrum, and are *very* visible on the graph! If you want to see the differences, just make a plot of both sinc^2 and its aliased versions (for whatever oversampling ratios and/or delays), and look at the differences. It won't be interesting for high oversampling ratios and zero delay - which is exactly why that scenario is a poor choice for illustrating the effects in question. And you're entirely missing the point what it is supposed to illustrate. The fact that sampling a continuous time signal at a very high rate results in a spectrum that closely resembles the continuous time spectrum (over the sampled bandwidth) is beside the point. Exactly. It's
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 polynomials, graphed up to 22kHz. No resampling is depicted, or the frequency responses would show the aliasing associated with that. It shows *exactly* the aliasing http://morpheus.spectralhead.com/img/interpolation_aliasing.png There are about 88 alias images visible on the graph. The linear interpolation curve is not smooth, so it contains aliasing. It's just showing the sinc^2 response of the linear interpolator, and similar for the other polynomials. If the signal you interpolate is white noise, and the spectrum of the signal is a flat spectrum rectangle like the one displayed, then after resampling, you get *exactly* the spectrum you see on the graph, showing 88 alias images. Proof: I created 60 seconds of white noise sampled at 500 Hz, then resampled it to 44.1 kHz using linear interpolation. After the upsampling, it sounds like this: http://morpheus.spectralhead.com/wav/noise_resampled.wav Its spectrum looks like this: http://morpheus.spectralhead.com/img/noise_resampled.png Looks familiar? Oh, it's the *exact* same graph! (Minus some difference above 20 kHz, due to my soundcard's anti-alias filter.) It is an FFT graph of the upsampled white noise, and it shows *exactly* the aliasing. Good morning! This is what you'd get if you used those interpolation polynomials to convert a 250Hz sampled signal into a continuous time signal, not a discrete time signal of whatever sampling rate. Nope. You get the same graph if you sample that continuous time signal at a 44.1 kHz sampling rate (with some further aliasing from the sampling). Just as I've shown. Besides, I think the graph was created via numerical means using FFT, because it has noise at the low ampliutes (marked on the image). Therefore, it doesn't show a continuous time sinc^2 graph, because that wouldn't be noisy. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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) are not shown. There is no resampling to be seen in the graphs. I claim that they are aliases of the original spectrum. What we see in the graph is simply the spectra of the continuous-time interpolators. Since the spectra extend beyond the original nyquist rate, there will indeed be images of the original signal weighted by the interpolator spectrum present in the continuous-time interpolated signal. Whether those are ultimately expressed as aliases depends on what you then do with that continuous time signal. If you resample to the original rate (in order to implement a fractional delay, say), then those weighted images will be folded back to the same place they came from. In that case, there is no aliasing, you just end up with a modified frequency response of your fractional interpolator. This is where the zero at Nyquist comes from when we do a half-sample delay - the linear phase term corresponding to a half-sample delay causes the signal images to become out of phase with each other as you approach Nyquist, so they cancel out and you get a zero. It is only if the interpolated continuous-time signal is resampled at a different rate, or just used directly, that those signal images end up expressed as aliases. The rest of your accusations are your usual misreadings and straw men. I won't be legitimating them by responding, and I hope you will accept that and give up on these childish tactics. It would be better for everyone if you could make a point of engaging in good faith and trying to stick to the subject rather than attacking the intellects of others. E On Fri, Aug 21, 2015 at 2:05 PM, Peter S peter.schoffhau...@gmail.com wrote: Also, you even contradict yourself. You claim that: 1) Olli's graph was created by graphing sinc(x), sinc^2(x), and not via FFT. 2) The artifacts from the resampling would be barely visible, because the oversampling rate is quite high. So, if - according to 2) - the artifacts are not visible because the oversampling is high and the graph doesn't focus on that, then how do you know that 1) is true? You claim that the resampling artifacts wouldn't be visible anyways. If that's true, then how would you prove that FFT was not used for creating Olli's graph? Also, even you yourself acknowledge that It shows the aliasing left by linear interpolation into the continuous time domain. So, we agree that the graph shows aliasing, right? I do not know where you get your idea of additional aliasing - it's the very same aliasing, except the resampling folds it back... ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Which contains alias images of the original spectrum, which was my point. There is no original spectrum pictured in that graph. Only the responses of the interpolators. There is no reference to any input signal at all. No one claimed there was fractional delay involved. Fractional delay is a primary topic of this thread, and a major motivation for interest in polynomial interpolation in dsp in general. Then how do you explain that taking noise sampled at 500 Hz, and resampling it to 44.1 kHz gives an identical FFT graph? We've been over this already. It's because you're resampling the signal at such a large rate that the effects of the sampling are not visible. And you've chosen a signal with a flat spectrum, so there are no features of the signal spectrum visible - only the interpolator response. This goes exactly to the point that no resampling effects are present in the graphs. All we see are the interpolator spectra. The fact that there are various ways to generate a graph of an interpolator spectrum is entirely beside the point. If you resample to the original rate (in order to implement a fractional delay, say), then those weighted images will be folded back to the same place they came from. That's exactly why they're called aliases. No, if you fold the images back to the same spots they originated, they are not aliases. All of the frequencies are mapped back to their original locations, none end up at other frequencies. Aliases are when signal images end up in new locations corresponding to different frequency bands. This distinction is crucial to understanding the operation of fractional delay interpolators: it's why they don't produce aliasing at their output. We just get a fractional delay filter with an imperfect spectrum. It's only the frequency response of the interpolator that gets aliased (introducing the zero at Nyquist for half-sample delay, for example), not the underlying signal content. That's why it's important to graph the frequency response of the interpolators directly, without worrying about signal spectra - to figure out what happens in the final digital interpolator, you take that continuous time interpolator spectrum, add a linear phase term for whatever delay you want, and then alias it according to your new sampling rate to get the final response of the digital interpolation filter. Signal aliasing only results if that involves a change in sampling rate. Which is not the case on Olli's graph. Right, Ollie's graph shows only the intermediate stage, the spectrum of the polynomial interpolator in continuous time. This is an analytical convenience, we never actually produce any such signal. It's used as an input to figure out what the final response of a digital interpolator based on one of these polynomials will be. You can of course sample that at a very high rate and so neglect the aliasing of the interpolator response, but what is the point of that? You wouldn't use any of these interpolators if what you're trying to do is upsample a 500Hz sampled signal to 44.1kHz, the graphs show that they're crap for that. I spent (wasted?) a considerate amount of time creating various demonstrations and FFT graphs showing my point. Your time would be better spent figuring out a point that is relevant 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 of how the graphs were generated don't really matter. Then why do you keep insisting that they're generated by plotting sinc^2(x) ? The point is that the only effect shown is the spectrum of the continuous-time polynomial interpolator. Which contains alias images of the original spectrum, which was my point. The additional spectral effects of delaying and resampling that continuous-time signal (to get fractional delay, for example) are not shown. No one claimed there was fractional delay involved. There is no resampling to be seen in the graphs. I recreated the exact same graph via resampling a signal, proving that is one method of generating that graph. I claim that they are aliases of the original spectrum. What we see in the graph is simply the spectra of the continuous-time interpolators. Then how do you explain that taking noise sampled at 500 Hz, and resampling it to 44.1 kHz gives an identical FFT graph? How do you explain that an 50 Hz sine wave, resampled to 44.1 kHz, contains alias frequencies at 450 Hz, 550 Hz, 950 Hz, 1050 Hz, 1450 Hz, 1550 Hz, etc. ? What are those, if not aliases ? Whether those are ultimately expressed as aliases depends on what you then do with that continuous time signal. They're already aliases... You may filter them out, or do whatever you want with them - that doesn't change the fact that
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 point. I'm not sure what you're trying to accomplish by harping on this point, while ignoring everything I say. Certainly, it is not convincing me that you have some worthwhile response to my points, or even that you are understanding them in the first place. It's seems like you are trying to avoid my point entirely, in favor of some imaginary dispute of your own invention, which you think you can win. Have you actually looked at Olli Niemitalo's graph closely? Here is proof that it is NOT a graph of sinc(x)/sinc^2(x): http://morpheus.spectralhead.com/img/other001-analysis.gif It is NOT sinc(x)/sinc^2(x), and you're blind as a bat if you do not see that. I have no idea what you think you are proving by scrutinizing graph artifacts like that, but it's a preposterous approach to signal analysis on its face. It's also in extremely poor taste to use retard as a term of abuse. People with mental disabilities have it hard enough already, without others treating their status as an insult to be thrown around. I'd appreciate it if you would compose yourself and refrain from these kinds of ugly outbursts. Meanwhile, it seems that you are suggesting that the spectrum of white noise linearly interpolated up to a high oversampling rate is not sinc^2. Is your whole point here that generating such a plot by FFTing the interpolation of a finite segment of white noise will produce finite-data artifacts in the resulting graph? Because that's not relevant to the subject, and only goes to show that it's better to just graph the sinc^2 curve directly and so avoid all of the excess computation and finite-data effects. Are you claiming that those wiggles in the graph represent aliasing of the spectrum from resampling at 44.1kHz? If so, that is unlikely. You do agree that the spectrum of a continuous-time linear interpolator is given by sinc^2, right? E On Fri, Aug 21, 2015 at 4:59 PM, Peter S peter.schoffhau...@gmail.com wrote: Since you constantly derail this topic with irrelevant talk, let me instead prove that 1) Olli Niemiatalo's graph *is* equivalent of the spectrum of upsampled white noise. 2) Olli Niemitalo's graph does *not* depict sinc(x)/sinc^2(x). First I'll prove 1). Using palette modification, I extracted the linear interpolation curve from Olli's figure: http://morpheus.spectralhead.com/img/other001b.gif Then I sampled white noise at 500 Hz, and resampled it to 44.1 kHz using linear interpolation. I got this spectrum: http://morpheus.spectralhead.com/img/resampled_noise_spectrum.gif To do a proper A/B comparison between the two spectra, I tried to align and match them as much as possible, and created an animated GIF file that blinks between the two graphs at a 500 ms rate: http://morpheus.spectralhead.com/img/olli_vs_resampled_noise.gif Although the alignment is not 100% exact, to my eyes, they look like totally equivalent graphs. This proves that upsampled white noise has the same spectrum as the graph shown on Olli's graph for linear interpolation. Second, I'll prove 2). Have you actually looked at Olli Niemitalo's graph closely? Here is proof that it is NOT a graph of sinc(x)/sinc^2(x): http://morpheus.spectralhead.com/img/other001-analysis.gif It is NOT sinc(x)/sinc^2(x), and you're blind as a bat if you do not see that. Since I proved both 1) and 2), it is totally irrelevant what you say, because none of what you could ever say would disprove this. Sinc(x) does not have a jagged/noisy look, therefore it is 100% certain it is not what you see on Olli's graph. Point proven, end of discussion. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Naturally, there's going to be some jaggedness in the spectrum because of the noise. So, obviously, that is not sinc^2 then. So your whole point is that it's not *exactly* sinc^2, but a slightly noisy version thereof? My point was that there are no effects of resampling visible in the graphs. That has nothing to do with exactly how the graphs were generated, nor does insisting that the graphs are slightly noisy address the point. Indeed, you've already conceded that the resampling effects are not visible in the graphs several posts back. It seems like you're just casting about for some other issue that you can tell yourself you won, and then call me names, to feed your fragile ego. Honestly, it's a pretty sad spectacle and I'm embarrassed for you. It really would be better for everyone - including you - if you could interact in a good-faith, mature manner. Please make an effort to start doing so, or you're pretty soon going to find that nobody here will interact with you any more. By the way, there's no reason for any jaggedness to appear in the plots, given the lengths of data you were talking about. You might want to look into spectral density estimation methods to trade off frequency resolution and bin accuracy. It's pretty standard statistical signal processing 101 stuff. Producing a very smooth graph from a long enough segment of data is straightforward, if you use appropriate techniques (not just one big FFT of the whole thing, that won't ever get rid 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 that there are many ways to produce a graph of the interpolation spectrum is not in dispute, nor is it germaine to my point. Earlier you disputed that there's no upsampling involved. Apparently you change your mind quite often... It's seems like you are trying to avoid my point entirely, in favor of some imaginary dispute of your own invention, which you think you can win. I claimed something, and you disputed it. I proved that what I claimed, is true. Therefore, all your further arguments are invalid... (and are boring) I have no idea what you think you are proving by scrutinizing graph artifacts like that I am proving that what you see on the graph is not sinc(x) / sinc^2(x), but rather some noisy curve, like the spectrum of upsampled noise. Therefore, my original argument is correct. It's also in extremely poor taste to use retard as a term of abuse. Well, if you do not see that the graph pictured on Olli's figure is not sinc(x), then you're retarded. Meanwhile, it seems that you are suggesting that the spectrum of white noise linearly interpolated up to a high oversampling rate is not sinc^2. Naturally, there's going to be some jaggedness in the spectrum because of the noise. So, obviously, that is not sinc^2 then. Are you claiming that those wiggles in the graph represent aliasing of the spectrum from resampling at 44.1kHz? If so, that is unlikely. Nope, the wiggles in the graph are from the noise. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Since you constantly derail this topic with irrelevant talk, let me instead prove that 1) Olli Niemiatalo's graph *is* equivalent of the spectrum of upsampled white noise. 2) Olli Niemitalo's graph does *not* depict sinc(x)/sinc^2(x). First I'll prove 1). Using palette modification, I extracted the linear interpolation curve from Olli's figure: http://morpheus.spectralhead.com/img/other001b.gif Then I sampled white noise at 500 Hz, and resampled it to 44.1 kHz using linear interpolation. I got this spectrum: http://morpheus.spectralhead.com/img/resampled_noise_spectrum.gif To do a proper A/B comparison between the two spectra, I tried to align and match them as much as possible, and created an animated GIF file that blinks between the two graphs at a 500 ms rate: http://morpheus.spectralhead.com/img/olli_vs_resampled_noise.gif Although the alignment is not 100% exact, to my eyes, they look like totally equivalent graphs. This proves that upsampled white noise has the same spectrum as the graph shown on Olli's graph for linear interpolation. Second, I'll prove 2). Have you actually looked at Olli Niemitalo's graph closely? Here is proof that it is NOT a graph of sinc(x)/sinc^2(x): http://morpheus.spectralhead.com/img/other001-analysis.gif It is NOT sinc(x)/sinc^2(x), and you're blind as a bat if you do not see that. Since I proved both 1) and 2), it is totally irrelevant what you say, because none of what you could ever say would disprove this. Sinc(x) does not have a jagged/noisy look, therefore it is 100% certain it is not what you see on Olli's graph. Point proven, end of discussion. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 that there are many ways to produce a graph of the interpolation spectrum is not in dispute, nor is it germaine to my point. Earlier you disputed that there's no upsampling involved. Apparently you change your mind quite often... It's seems like you are trying to avoid my point entirely, in favor of some imaginary dispute of your own invention, which you think you can win. I claimed something, and you disputed it. I proved that what I claimed, is true. Therefore, all your further arguments are invalid... (and are boring) I have no idea what you think you are proving by scrutinizing graph artifacts like that I am proving that what you see on the graph is not sinc(x) / sinc^2(x), but rather some noisy curve, like the spectrum of upsampled noise. Therefore, my original argument is correct. It's also in extremely poor taste to use retard as a term of abuse. Well, if you do not see that the graph pictured on Olli's figure is not sinc(x), then you're retarded. Meanwhile, it seems that you are suggesting that the spectrum of white noise linearly interpolated up to a high oversampling rate is not sinc^2. Naturally, there's going to be some jaggedness in the spectrum because of the noise. So, obviously, that is not sinc^2 then. Are you claiming that those wiggles in the graph represent aliasing of the spectrum from resampling at 44.1kHz? If so, that is unlikely. Nope, the wiggles in the graph are from the noise. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Upsampling means, that the sampling rate increases. So if you have a 250 Hz signal, and create a 22000 Hz signal from it, that is - by definition - upsampling. That's *exactly* what upsampling means... You insert new samples between the original ones, and interpolate between them (using whatever interpolation filter of your preference). And that is often used synonymously with 'oversampling', and that's what happens in an oversampled D/A converter. (Though 'oversampling' has a different meaning in A/D context.) -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Hi, A suggestion for those working on practical implementations, and lighten up the tone of the discussion with some people I know from worked on all kinds of (semi-) pro implementations when I wasn't even into more than basic DSP yet. The tradeoffs about engineering and implementing on a platform with given limitations (or for advanced people making filters: possibly even trading off the computation properties required for a self-designed DSP unit) including memory use, required clock speed, and heat build-up (not so important nowadays for simple filters) can be more accurately met by being specific about the requirements in terms of the quality and the quantification of the error bounds, as in this case how much high frequency loss can I prevent, at which engineering (or research) cost, and how many extra clock cycles of my DSP/CPU. In some cases, it can pay to do the extra effort of separating your audio frequency range in a couple of bands, so say you make an interpolator for low frequencies (e.g. simple zero-order) for mid-frequencies (with some attention for artifacts in the oh so sensitive m3 kHz range, and for instance for frequencies above 10kHz, where you can then pay most attention to the way the damping of the higher frequencies come across more than the exact accuracy of the short time convolution filter you use. Such a in this case limited multi-band approach costs a few filters and a little thinking about how those bands will later add back up properly to a decent signal, but it can make audio quality higher without requiring extreme resources. T.V. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 20/08/2015, Ethan Duni ethan.d...@gmail.com wrote: 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 table space for the fractional interpolator? For the record, wasn't it you who said memory is often a constraint? Quote from you: There are lots of dsp applications where memory is very much the main constraint. So apparently your premise is that memory can be expensive and a constraint, and now you ask why are we worried about using extra memory. At least make up your mind, whether you consider memory cheap or expensive... -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Let's analyze your suggestion of using a FIR filter at f = 0.5/512 = 0.0009765625 for an interpolation filter for 512x oversampling. Here's the frequency response of a FIR filter of length 1000: http://morpheus.spectralhead.com/img/fir512_1000.png Closeup of the frequency range between 0-0.01 (cutoff marked with black line): http://morpheus.spectralhead.com/img/fir512_1000_closeup.png Apparently that's a pretty crappy anti-alias filter, the transition band is very wide. So let's try a FIR filter of length 5000: http://morpheus.spectralhead.com/img/fir512_5000_closeup.png Better, but still quite a lot of aliasing above the cutoff freq. FIR filter of length 20,000: http://morpheus.spectralhead.com/img/fir512_2_closeup.png Now this starts to look like a proper-ish anti-alias filter. The problem is - its length is 20,000 samples, so assuming 32-bit float representation, the coefficients alone need about 80 KB of memory... meaning that there's a high chance that it won't even fit into the L1 cache, causing a lot of cache misses, so this filter will be extra slow, since your CPU will be constantly waiting for the coefficients to arrive from the L2 cache and/or RAM. Also consider how much CPU power you need to do convolution with a 20,000 sample long kernel at 512x oversampled rate... I bet you're not trying to do this in realtime, are you? So, that's not exactly the brightest way to do 512x oversampling, unless you prefer to waste a lot of resources and spend a week on upsampling. In that case, it is ideal. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 peter.schoffhau...@gmail.com wrote: In the case of variable pitch playback with interpolation, here are the frequency responses: http://musicdsp.org/files/other001.gif (graphs by Olli Niemitalo) In this case, there's no zero at the original Nyquist freq, rather there are zeros at the original sampling rate and its multiplies. So it's useful to specify what you mean by high frequency signal loss due to interpolation, beacause that term is ambiguous and can mean various things. In this graph, the signal frequency seems to be 250 Hz, so this graph shows the equivalent of about 22000/250 = 88x oversampling. At that oversampling rate, gain of alias images of linear interpolation is -84 dB. High amounts of oversampling for high SNR ratios may be necessitated by the slow rolloff of aliasing. (This was not mentioned in the question in this thread, but is relevant.) -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
In the starting post, it was not specified that resampling was also used - the question was: Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Without specifying that variable rate playback is involved, that could be understood in various ways - for example, at first I thought the interpolation was for the purpose of a (static or modulated) fractional delay line. A third possible situation is using linear/hermite interpolation as an upsampling filter in a 2^N oversampler. It was only specified 18 posts later, that the interpolation is used for variable pitch playback. These three situations all are different, and different formulas apply... And the combination oversampling and linear/hermite interpolation can also be meant in multiple ways. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Here's a graph of performance in mflops of varying length FFT transforms from the fftw.org benchmark page, for Intel Pentium 4: http://morpheus.spectralhead.com/img/fftw_benchmark_pentium4.png Afaik Pentium 4 has 16 KB of L1 data cache. If you check the graph, around 8-16k the performance starts to drop drastically. I believe the main reason for this is that the data doesn't fit into the L1 data cache any more, which is 16 KB. You'll see similar graphs for most other CPU types as well, there's a dropoff near the L1 cache size. So using more memory is only free(ish) until a certain point - if your data doesn't fit into the L1 cache any more, it will cause cache misses and give you a performance penalty, because the CPU needs to fetch the data from the L2 cache, which is several times slower. In the graph, you can see ~3-4x performance difference between transforms that fit into the L1 cache, and transforms that don't. For this reason, very large filters have a notable performance penalty. The coefficients for a FIR filter of length 20,000 will certainly not fit into a 16 KB L1 data cache. Here's the memory topology of AMD Bulldozer server microarchitecture: https://upload.wikimedia.org/wikipedia/commons/9/95/Hwloc.png Each core has a 16 KB L1 data cache. The further away you go from the CPU core, the slower the memory access gets. L2 cache is 2 MB, and there's a shared 8 MB L3 cache across cores. There's a 64 KB instruction cache per two cores. Similar cache architectures are common among computer processors (sometimes without L3 cache). There's a document that discusses this in depth: Ulrich Drepper, What Every Programmer Should Know About Memory http://morpheus.spectralhead.com/pdf/cpumemory.pdf This document gives the following memory access times for Intel Pentium M: Register: = 1 cycle L1 data cache: ~3 cycles L2 cache: ~14 cycles RAM: ~240 cycles So this means, on a Pentium M, accessing data in the L1 cache is ~3x slower, accessing data in the the L2 cache is ~14x slower, and accessing data in the RAM is ~240x slower than accessing data in a register. (Earlier I wrongly said RAM is about 10x slower, rather that's about the L2 cache speed.) So if the data doesn't fit into the L1 cache and needs to be fetched from the L2 cache, that's nearly 5x slower on the Pentium M. A notable part of the L2 cache penalty is caused by the physical limits of the universe - data travels in wires at the speed of light, which is about 1 foot (30 cm) per nanosecond. The larger the cache, the longer the wires, hence the longer the data access delay. For further details and detailed performance analysis, see the above paper. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 overall accuracy of the interpolator. At low oversampling ratios, the sinc^2 rolloff of the linear interpolator response isn't effective at squashing the signal images, so you end up with aliasing corrupting your results. Hence the need for higher order interpolation at lower oversampling ratios, as described in Ollie's paper. If you want to get high SNR out of linear interpolation, you need to crank up the oversampling considerably - far beyond what is needed just to avoid the attenuation of high frequencies of the in-band component, in order to sufficiently squash the images. E On Thu, Aug 20, 2015 at 12:18 PM, Chris Santoro chris.sant...@gmail.com wrote: As far as the oversampling + linear interpolation approach goes, I have to ask... why oversample so much (512x)? Purely from a rolloff perspective, it seems you can figure out what your returns are going to be by calculating sinc^2 at (1/upsample_ratio) for a variety of oversampling ratios. Here's the python code to run the numbers... #- import numpy as np #normalized frequency points X = [1.0/512.0, 1.0/256.0, 1.0/128.0, 1.0/64.0, 1.0/32.0, 1.0/16.0, 1.0/8.0, 1.0/4.0] #find attenuation at frequency points due to linear interpolation worst case (halfway in between) S = np.sinc(X) S = 20*np.log10(S*S) print S #--- and here's what it spits out for various attenuation values at what would be nyquist in the baseband: 2X: -7.8 dB 4X: -1.8 dB 8X: -0.44 dB 16X: -0.11 dB 32X: -0.027 dB 64X: -0.0069 dB 128X: -0.0017 dB 256X: -0.00043 dB 512X: -0.00010 dB If all you're trying to do is mitigate the rolloff of linear interp, it looks like there's diminishing returns beyond 16X or 32X, where you're talking about a tenth of a dB or less at nyquist, which most people can't even hear in that range. Your anti-aliasing properties are going to be determined by your choice of upsampling/windowed-sync/anti-imaging filter and how long you want to let that be. Or am I missing something? It just doesn't seem worth it go to that high. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Let me just add, that in case of having a non-oversampled linearly
interpolated fractional delay line with exactly 0.5 sample delay (most
high-frequency roll-off position), the frequency response formula is
not sinc^2, but rather, sin(2*PI*f)/(2*sin(PI*f)), as I discussed
earlier.
In that case, the results are slightly different:
# Tcl code -
set pi 3.141592653589793238
set freqs {1/4. 1/8. 1/16. 1/32. 1/64. 1/128. 1/256. 1/512. 1/1024.}
set amt 2
foreach freq $freqs {
set amp [expr sin(2*$pi*$freq)/(2*sin($pi*$freq))]
set db [expr 20.0 * log($amp)/log(10)]
puts [format %-8s ${amt}X:][format %f $db] dB
set amt [expr $amt*2]
}
# End of code --
Results:
2X: -3.01 dB
4X: -0.688 dB
8X: -0.169 dB
16X:-0.0419 dB
32X:-0.0105 dB
64X:-0.00262 dB
128X: -0.00065 dB
256X: -0.000164 dB
512X: -0.41 dB
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Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 20/08/2015, Ethan Duni ethan.d...@gmail.com wrote: But I'm on the fence about whether it's the tightest use of resources (for whatever constraints). Then try and measure it yourself - you don't believe my words anyways. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Comparison of the two formulas from previous post: (1) in blue, sinc^2 (2) in red: http://morpheus.spectralhead.com/img/sinc.png sin(pi*x*2) -(1) 2*sin(pi*x) (Formula from Steven W. Smith, absolute value taken on graph) sin(pi*x) --- ^2 = sinc^2(x)(2) pi*x (Formula from JOS, Nigel R.) (1) and (2) (blue and red curve) are quite different. Let's test how equation (1) compares against measured frequency response of a LTI filter with coeffs [0.5, 0.5]: http://morpheus.spectralhead.com/img/halfsample_delay_response.png The maximum error between formula (1) and the measured frequency response of the filter (a0=0.5, a1=0.5) is 3.3307e-16, or -310 dB, which about equals the limits of the floating point precision at 64 bits. The frequency response was measured using Octave's freqz() function, using 512 points. Conclusion: Steven W. Smith's formula (1) seems correct. Frequency response of the same filter in decibel scale: http://morpheus.spectralhead.com/img/halfsample_delay_response2.png (this graph is normalized to 0..1 rad, not 0..0.5) The pole-zero plot was shown earlier, having a zero at z=-1, meaning -Inf gain at Nyquist. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 19/08/2015, Ethan Duni ethan.d...@gmail.com wrote: But why would you constrain yourself to use first-order linear interpolation? Because it's computationally very cheap? The oversampler itself is going to be a much higher order linear interpolator. So it seems strange to pour resources into that Linear interpolation needs very little computation, compared to most other types of interpolation. So I do not consider the idea of using linear interpolation for higher stages of oversampling strange at all. The higher the oversampling, the more optimal it is to use linear in the higher stages. So heavy oversampling seems strange, unless there's some hard constraint forcing you to use a first-order interpolator. The hard constraint is CPU usage, which is higher in all other types of interpolators. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
i would say way more than 2x if you're using linear in between. if memory is cheap, i might oversample by perhaps as much as 512x and then use linear to get in between the subsamples (this will get you 120 dB S/N). But why would you constrain yourself to use first-order linear interpolation? The oversampler itself is going to be a much higher order linear interpolator. So it seems strange to pour resources into that, just so you can avoid putting them into the final fractional interpolator. Is the justification that the oversampler is a fixed interpolator, whereas the final stage is variable (so we don't want to muck around with anything too complex there)? I've seen it claimed (by Julius Smith IIRC) that oversampling by as little as 10% cuts the interpolation filter requirements by over 50%. So heavy oversampling seems strange, unless there's some hard constraint forcing you to use 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 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., Lagrange interpolation filter, windowed sinc, etc). It's not an artifact of the simple linear approach, at precisely Nyquist, you're right. as you approach Nyquist, linear interpolation is worser than cubic Hermite but better than cubic B-spline (better in terms of less roll-off, worser in terms of killing images). it's a feature of the symmetric, finite nature of the fractional interpolator. Since there are good reasons for the symmetry constraint, we are left to trade off oversampling and filter order/design to get the final passband as flat as we need. My view is that if you are serious about maintaining fidelity across the full bandwidth, you need to oversample by at least 2x. i would say way more than 2x if you're using linear in between. if memory is cheap, i might oversample by perhaps as much as 512x and then use linear to get in between the subsamples (this will get you 120 dB S/N). That way you can fit the transition band of your interpolation filter above the signal band. In applications where you are less concerned about full bandwidth fidelity, oversampling isn't required. Some argue that 48kHz sample rate is already effectively oversampled for lots of natural recordings, for example. If it's already at 96kHz or higher I would not bother oversampling further. i might **if** i want to resample by an arbitrary ratio and i am doing linear interpolation between the new over-sampled samples. remember, when we oversample for the purpose of resampling, if the prototype LPF is FIR (you know, the polyphase thingie), then you need not calculate all of the new over-sampled samples. only the two you need to linear interpolate between. so oversampling by a large factor only costs more in terms of memory for the coefficient storage. not in computational effort. Also this is recommended reading for this thread: https://ccrma.stanford.edu/~jos/Interpolation/ https://ccrma.stanford.edu/%7Ejos/Interpolation/ quite familiar with it. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
SOmetimes I feel the personal integrity about these undergrad level scientific quests is nowhere to be found with some people, and that's a shame. Working on a decent subject like these mathematical approximations in the digital signal processing should be accompanied with at least some self-respect in the treatment of subjects one involves oneself in, obviously apart from chatter and stories and so on, because otherwise people might feel hurt to be contributing only as it were to feed da Man or something of that nature, and that's not cool in my opinion. T.V. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/19/15 1:43 PM, Peter S wrote: On 19/08/2015, Ethan Duniethan.d...@gmail.com wrote: But why would you constrain yourself to use first-order linear interpolation? Because it's computationally very cheap? and it doesn't require a table of coefficients, like doing higher-order Lagrange or Hermite would. The oversampler itself is going to be a much higher order linear interpolator. So it seems strange to pour resources into that Linear interpolation needs very little computation, compared to most other types of interpolation. So I do not consider the idea of using linear interpolation for higher stages of oversampling strange at all. The higher the oversampling, the more optimal it is to use linear in the higher stages. here, again, is where Peter and i are on the same page. So heavy oversampling seems strange, unless there's some hard constraint forcing you to use a first-order interpolator. The hard constraint is CPU usage, which is higher in all other types of interpolators. for plugins or embedded systems with a CPU-like core, computation burden is more of a cost issue than memory used. but there are other embedded DSP situations where we are counting every word used. 8 years ago, i was working with a chip that offered for each processing block 8 instructions (there were multiple moves, 1 multiply, and 1 addition that could be done in a single instruction), 1 state (or 2 states, if you count the output as a state) and 4 scratch registers. that's all i had. ain't no table of coefficients to look up. in that case memory is way more important than wasting a few instructions recomputing numbers that you might otherwise just look up. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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. Meanwhile, the oversampler itself needs a table of coefficients. Assuming we're talking about FIR interpolation, to avoid phase distortion. But that's a single fixed table for supporting a single oversampling ratio, so I can see how it would add up to a memory savings compared to a bank of tables for different fractional interpolation points, if you're looking for really fine/arbitrary granularity. If we're talking about a fixed fractional delay, I'm not really seeing the advantage. Obviously it will depend on the details of the application, it just seems kind of unbalanced on its face to use heavy oversampling and then the lightest possible fractional interpolator. It's not clear to me that a moderate oversampling combined with a fractional interpolator of modestly high order wouldn't be a better use of resources. So it doesn't make a lot of sense to me to point to the low resource costs of the first-order linear interpolator, when you're already devoting resources to heavy oversampling in order to use it. They need to be considered together and balanced, no? Your point about computing only the subset of oversamples needed to drive the final fractional interpolator is well-taken, but I think I need to see a more detailed accounting of that to be convinced. E On Wed, Aug 19, 2015 at 1:00 PM, robert bristow-johnson r...@audioimagination.com wrote: On 8/19/15 1:43 PM, Peter S wrote: On 19/08/2015, Ethan Duniethan.d...@gmail.com wrote: But why would you constrain yourself to use first-order linear interpolation? Because it's computationally very cheap? and it doesn't require a table of coefficients, like doing higher-order Lagrange or Hermite would. The oversampler itself is going to be a much higher order linear interpolator. So it seems strange to pour resources into that Linear interpolation needs very little computation, compared to most other types of interpolation. So I do not consider the idea of using linear interpolation for higher stages of oversampling strange at all. The higher the oversampling, the more optimal it is to use linear in the higher stages. here, again, is where Peter and i are on the same page. So heavy oversampling seems strange, unless there's some hard constraint forcing you to use a first-order interpolator. The hard constraint is CPU usage, which is higher in all other types of interpolators. for plugins or embedded systems with a CPU-like core, computation burden is more of a cost issue than memory used. but there are other embedded DSP situations where we are counting every word used. 8 years ago, i was working with a chip that offered for each processing block 8 instructions (there were multiple moves, 1 multiply, and 1 addition that could be done in a single instruction), 1 state (or 2 states, if you count the output as a state) and 4 scratch registers. that's all i had. ain't no table of coefficients to look up. in that case memory is way more important than wasting a few instructions recomputing numbers that you might otherwise just look up. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
3.2 Multistage 3.2.1 Can I interpolate in multiple stages? Yes, so long as the interpolation ratio, L, is not a prime number. For example, to interpolate by a factor of 15, you could interpolate by 3 then interpolate by 5. The more factors L has, the more choices you have. For example you could interpolate by 16 in: - one stage: 16 - two stages: 4 and 4 - three stages: 2, 2, and 4 - four stages: 2, 2, 2, and 2 3.2.2 Cool. But why bother with all that? Just as with decimation, the computational and memory requirements of interpolation filtering can often be reduced by using multiple stages. 3.2.3 OK, so how do I figure out the optimum number of stages, and the interpolation ratio at each stage? There isn't a simple answer to this one: the answer varies depending on many things. However, here are a couple of rules of thumb: - Using two or three stages is usually optimal or near-optimal. - Interpolate in order of the smallest to largest factors. For example, when interpolating by a factor of 60 in three stages, interpolate by 3, then by 4, then by 5. (Use the largest ratio on the highest rate.) http://dspguru.com/dsp/faqs/multirate/interpolation ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Nope. Ever heard of multistage interpolation? I'm well aware that multistage interpolation gives cost savings relative to single-stage interpolation, generally. That is beside the point: the costs of interpolation all still scale with oversampling ratio and quality requirements, just like in single stage interpolation. There's no magic to multi-stage interpolation that avoids that relationship. that's just plain wrong and stupid, and that's what all advanced multirate books will also tell you. You've been told repeatedly that this kind of abusive, condescending behavior is not welcome here, and you need to cut it out immediately. Tell me, you don't have an extra half kilobyte of memory in a typical computer? There are lots of dsp applications that don't run on personal computers, but rather on very lightweight embedded targets. Memory tends to be at a premium on those platforms. 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 is the right balance overall, when considering the combined costs of the oversampler and the fractional interpolator. It's hard to tell in general. It depends on various factors, including: - your desired/available CPU usage - your desired/available memory usage and cache size - the available instruction set of your CPU - your desired antialias filter steepness - your desired stopband attenuation ...and possibly other factors. Since these may vary largely, I think it is impossible to tell in general. What I read in multirate literature, and what is also my own experience, is that - when using a relatively large oversampling ratio - then it's more cost-effective to use linear interpolation at the higher stages (and that's Olli's conclusion as well). You can leverage any finite interpolator to skip computations in an FIR oversampler, not just linear. You get the most skipping in the case of high oversampling ratio and linear interpolation, but the same trick still works any time your oversampling ratio is greater than your interpolator order. But to a varying degree. A FIR interpolator is still heavy if you skip samples where the coefficient is zero, compared to linear interpolation (but it is also higher quality). The flipside is that the higher the oversampling ratio, the longer the FIR oversampling filter needs to be in the first place. Nope. Ever heard of multistage interpolation? You may do a small FIR stage (say, 2x or 4x), and then a linear stage (or another, low-complexity FIR stage according to your desired specifications, or even further stages). Seems you still don't understand that you can oversample in multiple stages, and use a linear interpolator for the higher stages of oversampling... Which is almost always optimal than using a single costy FIR filter to do the interpolation. You don't need to use a 512x FIR at 100 dB stopband attentuation, that's just plain wrong and stupid, and that's what all advanced multirate books will also tell you. Same for IIR case. Since memory is usually not an issue, There are lots of dsp applications where memory is very much the main constraint. Tell me, you don't have an extra half kilobyte of memory in a typical computer? I hear, those have 8-32 GB of RAM nowadays, and CPU cache sizes are like 32-128 KiB. The performance of your oversampler will be garbage if you do that. And so there will be no point in worrying about the quality of fractional interpolation after that point, since the signal you'll be interpolating will be full of aliasing to begin with. Exactly. But it won't be heavy! So it's not the oversampling what makes the process heavy, but rather, the interpolation / anti-aliasing filter!! And that means it needs lots of resources, especially as the oversampling ratio gets large. It's the required quality that drives the oversampler costs (and filter design choices). Which is exactly what I said. If your specification is low, you can have a 128x oversampler that is (relatively) low-cost. It's not the oversampling ratio what matters most. If you are willing to accept low quality in order to save on CPU (or maybe there's nothing in the upper frequencies that you're worried about), then there's no point in resampling at all. Just use a low order fractional interpolator directly on the signal. Seems you still miss the whole point of multistage interpolation. I recommend you read some books / papers on multirate processing. It should also be noted that the linear interpolation can be used for the upsampling itself as well, reducing the cost of your oversampling, Again, that would add up to a very low quality upsampler. You're wrong. Read Olli Niemitalo's paper again
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 19/08/2015, Ethan Duni ethan.d...@gmail.com wrote: Obviously it will depend on the details of the application, it just seems kind of unbalanced on its face to use heavy oversampling and then the lightest possible fractional interpolator. It should also be noted that the linear interpolation can be used for the upsampling itself as well, reducing the cost of your oversampling, not just as your fractional delay. A potential method to do fractional delay is to upsample by a large factor, then delay by an integer number of samples, and then downsample, without the use of an actual fractional delay. Say, if the fraction is 0.37, then you may upsamle by 512x, then delay the upsampled signal by round(512*0.37) = 189 samples, then downsample back. So you did a fractional delay without using actual fractional interpolation for the delay - you delayed by an integer number of samples. You'll also have a little error - your delay is 0.369140625 instead of the desired 0.37, since it's quantized to 512 steps, so the error is -0.000859375. I'm not saying this is ideal, I'm just saying this is one possible way of doing a fractional delay. This is discussed by JOS[1]: In discrete time processing, the operation Eq.(4.5) can be approximated arbitrarily closely by digital upsampling by a large integer factor M, delaying by L samples (an integer), then finally downsampling by M, as depicted in Fig.4.7 [96]. The integers L and M are chosen so that eta ~= L/M, where eta the desired fractional delay. [1] Julius O. Smith, Physical Audio Signal Processing https://ccrma.stanford.edu/~jos/pasp/Convolution_Interpretation.html Ref. [96] is: R. Crochiere, L. Rabiner, and R. Shively, ``A novel implementation of digital phase shifters,'' Bell System Technical Journal, vol. 65, pp. 1497-1502, Oct. 1975. Abstract: A novel technique is presented for implementing a variable digital phase shifter which is capable of realizing noninteger delays. The theory behind the technique is based on the idea of first interpolating the signal to a high sampling rate, then using an integer delay, and finally decimating the signal back to the original sampling rate. Efficient methods for performing these processes are discussed in this paper. In particular, it is shown that the digital phase shifter can be implemented by means of a simple convolution at the sampling rate of the original signal. In short, there are a zillion ways of implementing both oversampling and fractional delays, and they can be combined arbitrarily. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 20/08/2015, Ethan Duni ethan.d...@gmail.com wrote: Ugh, I suppose this is what I get for attempting to engage with Peter S again. Not sure what I was thinking... Well, you asked, why use linear interpolation at all? We told you the advantages - fast computation, no coefficient table needed, and (nearly) optimal for high oversampling ratios, and you were given some literature. If you don't believe it - well, not my problem... it's still true. #notmyloss ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 19/08/2015, Peter S peter.schoffhau...@gmail.com wrote: Another way to show that half-sample delay has -Inf gain at Nyquist: see the pole-zero plot of the equivalent LTI filter a0=0.5, a1=0.5. It will have a zero at z=-1. A zero on the unit circle means -Inf gain, and z=-1 means Nyquist frequency. Therefore, a half-sample delay has -Inf gain at Nyquist frequency. It looks like this: http://morpheus.spectralhead.com/img/halfsample_delay_zplane.png -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Another way to show that half-sample delay has -Inf gain at Nyquist: see the pole-zero plot of the equivalent LTI filter a0=0.5, a1=0.5. It will have a zero at z=-1. A zero on the unit circle means -Inf gain, and z=-1 means Nyquist frequency. Therefore, a half-sample delay has -Inf gain at Nyquist frequency. It would be ill-advised to dismiss Nyquist frequency because it may alias to DC signal when sampling. The zero on the unit circle is at Nyquist (z=-1), not at DC (z=1). Frequency response graphs of linear interpolation, according to JOS: https://ccrma.stanford.edu/~jos/Interpolation/Frequency_Responses_Linear_Interpolation.html ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 table space for the fractional interpolator? I wonder if the salient design concern here is less about balancing resources, and more about isolating and simplifying the portions of the system needed to support arbitrary (as opposed to just very-high-but-fixed) precision. I like the modularity of the high oversampling/linear interp approach, since that it supports arbitrary precision with a minimum of fussy variable components or arcane coefficient calculations. It's got a lot going for it in software engineering terms. But I'm on the fence about whether it's the tightest use of resources (for whatever constraints). Typically those are the arcane ones that take a ton of debugging and optimization :P E On Wed, Aug 19, 2015 at 1:00 PM, robert bristow-johnson r...@audioimagination.com wrote: On 8/19/15 1:43 PM, Peter S wrote: On 19/08/2015, Ethan Duniethan.d...@gmail.com wrote: But why would you constrain yourself to use first-order linear interpolation? Because it's computationally very cheap? and it doesn't require a table of coefficients, like doing higher-order Lagrange or Hermite would. The oversampler itself is going to be a much higher order linear interpolator. So it seems strange to pour resources into that Linear interpolation needs very little computation, compared to most other types of interpolation. So I do not consider the idea of using linear interpolation for higher stages of oversampling strange at all. The higher the oversampling, the more optimal it is to use linear in the higher stages. here, again, is where Peter and i are on the same page. So heavy oversampling seems strange, unless there's some hard constraint forcing you to use a first-order interpolator. The hard constraint is CPU usage, which is higher in all other types of interpolators. for plugins or embedded systems with a CPU-like core, computation burden is more of a cost issue than memory used. but there are other embedded DSP situations where we are counting every word used. 8 years ago, i was working with a chip that offered for each processing block 8 instructions (there were multiple moves, 1 multiply, and 1 addition that could be done in a single instruction), 1 state (or 2 states, if you count the output as a state) and 4 scratch registers. that's all i had. ain't no table of coefficients to look up. in that case memory is way more important than wasting a few instructions recomputing numbers that you might otherwise just look up. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, Nigel Redmon earle...@earlevel.com wrote: well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? -Inf at Nyquist when you're halfway between two samples. Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1, -1... After interpolating with fraction=0.5, it becomes a constant signal 0,0,0,0,0,0,0... (because (-1+1)/2 = 0) ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 expectation that digital processing of such a signal is going to work properly in general. But even given that, the interpolator outputting the zero signal in that case is exactly correct. That's what you would have gotten if you'd sampled the same sine wave (*not* square wave - that would imply frequencies above Nyquist) with a half-sample offset from the 1, -1, 1, -1, ... case. The incorrect behavior arises when you try to go in the other direction (i.e., apply a second half-sample delay), and you still get only DC. But, again, that doesn't really say anything about interpolation. It just says that you sampled the signal improperly in the first place, and so digital processing can't be relied upon to work appropriately. E On Tue, Aug 18, 2015 at 1:40 AM, Peter S peter.schoffhau...@gmail.com wrote: On 18/08/2015, Nigel Redmon earle...@earlevel.com wrote: well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? -Inf at Nyquist when you're halfway between two samples. Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1, -1... After interpolating with fraction=0.5, it becomes a constant signal 0,0,0,0,0,0,0... (because (-1+1)/2 = 0) ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/18/15 4:28 PM, Peter S wrote: 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. well Peter, here again is where you overreach. assuming, without loss of generality that the sampling period is 1, the continuous-time signals x(t) = 1/cos(theta) * cos(pi*t + theta) are all aliases for the signal described above (and incorrectly as contain[ing] no aliasing). -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
What's causing you to be unable to reconstruct the waveform? There are an infinite number of different nyquist-frequency sinusoids that, when sampled, will all give the same ...,1, -1, 1, -1, ... sequence of samples. The sampling is a many-to-one mapping in that case, and so cannot be inverted. See here: https://en.wikipedia.org/wiki/Nyquist–Shannon_sampling_theorem#Critical_frequency Or consider what happens if you shift a nyquist-frequency sinusoid by half a period before sampling it. You get ..., 0, 0, 0, 0, ... - which is quite obviously the zero signal. It is not going to reproduce 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 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 expectation that digital processing of such a signal is going to work properly in general. Not necessarily, at least in theory. In practice, an anti-alias filter will filter out a signal exactly at Nyquist freq, both when sampling it (A/D conversion), and both when reconstructing it (D/A conversion). But that doesn't mean that a half-sample delay doesn't have -Inf dB gain at Nyquist frequency. It's another thing that the anti-alias filter of a converter will typically filter it out anyways when reconstructing - but we weren't talking about reconstruction, so that is irrelevant here. A Nyquist frequency signal (1, -1, 1, -1, ...) is a perfectly valid bandlimited signal. But even given that, the interpolator outputting the zero signal in that case is exactly correct. That's what you would have gotten if you'd sampled the same sine wave (*not* square wave - that would imply frequencies above Nyquist) with a half-sample offset from the 1, -1, 1, -1, ... case. More precisely: a bandlimited Nyquist frequency square wave *equals* a Nyquist frequency sine wave. Or any other harmonic waveform for that matter (triangle, saw, etc.) In all cases, only the fundamental partial is there (1, -1, 1, -1, ... = Nyquist frequency sine), all the other partials are filtered out from the bandlimiting. So the signal 1, -1, 1, -1, *is* a Nyquist frequency bandlimited square wave, and also a sine-wave as well. They're identical. It *is* a bandlimited square wave - that's what you get when you take a Nyquist frequency square wave, and bandlimit it by removing all partials above Nyquist freq (say, via DFT). You may call it a square, a sine, saw, doesn't matter - when bandlimited, they're identical. The incorrect behavior arises when you try to go in the other direction (i.e., apply a second half-sample delay), and you still get only DC. What would be incorrect about it? I'm not sure what is your assumption. Of course if you apply any kind of filtering to a zero DC signal, you'll still have a zero DC signal. -Inf + -Inf = -Inf... Not sure what you're trying to achieve by applying a second half-sample delay... That also has -Inf dB gain at Nyquist, so you'll still have a zero DC signal after that. Since a half-sample delay has -Inf gain at Nyquist, you cannot undo it by applying another half-sample delay... But, again, that doesn't really say anything about interpolation.It just says that you sampled the signal improperly in the first place, and so digital processing can't be relied upon to work appropriately. That's false. 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. It is a bandlimited Nyquist frequency square wave (which is equivalent to a Nyquist frequency sine wave). From that, you can reconstruct a perfect alias-free sinusoid of frequency SR/2. What's causing you to be unable to reconstruct the waveform? -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/18/15 4:50 PM, Nigel Redmon wrote: I’m sorry, I’m missing your point here, Peter (and perhaps I missed Roberts, hence the “No?” in my reply to him). The frequency response of linear interpolation is (sin(pi*x)/(pi*x))^2, -7.8 dB at 0.5 of the sample rate... i will try to spell out my point. there are probably a zillion applications of fractional-sample interpolation. Vesa Valimaki's IEEE article Splitting the Unit Delay from the 90s is sorta equivalently seminal as fred harris's classic windowing paper about this. but, within the zillion of applications, i can think of two classes of application in which all applications will fall into one or the other: 1. slowly-varying (or constant) delay with a fractional component. this would be a precision delay we might use to time-align things (like speakers) or for effects like flanging or to compensate for the delay of some other component like a filter. 2. rapidly-varying delay (again with a fractional component). this would be sample-rate-conversion (SRC), resampling sound files, pitch shifting (either the splicing thing a Harmonizer might do or the sample-playback and looping a sampler might do), and wild-assed delay effects. it's only in the second class of application that i think the sinc^2 frequency rolloff (assuming linear interpolation) is a valid model (or hand-wavy approximation of a model). in the first class of application, i think the model should be what you get if the delay was constant. 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. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, Ethan Duni ethan.d...@gmail.com wrote: 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 sampling, there's no way to reconstruct without some additional side information. You cannot calculate 1/x when x=0, can you? Since that's division by zero. Yet you'll know when x tends to zero from right towards left, then 1/x will tend to +infinity. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Okay, I get what you mean. But that doesn't change the frequency response of a half-sample delay, or doesn't mean that a half-sample delay doesn't have a specific gain at Nyquist. Never said that it did. In fact, I explicitly said that this issue of sampling of Nyquist frequency sinusoids has 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 of signals is band limited to SR/2. The aliasing is in the amplitude/phase offset, not the frequency. Okay, I get what you mean. But that doesn't change the frequency response of a half-sample delay, or doesn't mean that a half-sample delay doesn't have a specific gain at Nyquist. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
In order to reconstruct that sinusoid, you'll need a filter with an infinitely steep transition band. You've demonstrated that SR/2 aliases to 0Hz, i.e. DC. That digital stream of samples is not reconstructable. On 8/18/2015 1:28 PM, Peter S wrote: That's false. 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. It is a bandlimited Nyquist frequency square wave (which is equivalent to a Nyquist frequency sine wave). From that, you can reconstruct a perfect alias-free sinusoid of frequency SR/2. NOTICE: This electronic mail message and its contents, including any attachments hereto (collectively, this e-mail), is hereby designated as confidential and proprietary. This e-mail may be viewed and used only by the person to whom it has been sent and his/her employer solely for the express purpose for which it has been disclosed and only in accordance with any confidentiality or non-disclosure (or similar) agreement between TEAC Corporation or its affiliates and said employer, and may not be disclosed to any other person or entity. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/18/15 5:01 PM, Emily Litella wrote: ... Never mind. too late. :-) -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
You cannot calculate 1/x when x=0, can you? Since that's division by zero. Yet you'll know when x tends to zero from right towards left, then 1/x will tend to +infinity. Not sure what that is supposed to have to do with the present subject. If you want to put it in terms of simple arithmetic, 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...@gmail.com wrote: 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 sampling, there's no way to reconstruct without some additional side information. You cannot calculate 1/x when x=0, can you? Since that's division by zero. Yet you'll know when x tends to zero from right towards left, then 1/x will tend to +infinity. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/18/15 3:44 PM, Ethan Duni wrote: 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 expectation that digital processing of such a signal is going to work properly in general. But even given that, the interpolator outputting the zero signal in that case is exactly correct. That's what you would have gotten if you'd sampled the same sine wave (*not* square wave - that would imply frequencies above Nyquist) with a half-sample offset from the 1, -1, 1, -1, ... case. The incorrect behavior arises when you try to go in the other direction (i.e., apply a second half-sample delay), and you still get only DC. But, again, that doesn't really say anything about interpolation. It just says that you sampled the signal improperly in the first place, and so digital processing can't be relied upon to work appropriately. as suprizing as it may first appear, i think Peter S and me are totally on the same page here. regarding *linear* interpolation, *if* you use linear interpolation in a precision delay (an LTI thingie, or at least quasi-time-invariant) and you delay by some integer + 1/2 sample, the filter you get has coefficients and transfer function H(z) = (1/2)*(1 + z^-1)*z^-N (where N is the integer part of the delay). the gain of that filter, as you approach Nyquist, approaches -inf dB. *my* point is that as the delay slowly slides from a integer number of samples, where the transfer function is H(z) = z^-N to the integer + 1/2 sample (with gain above), this linear but time-variant system is going to sound like there is a LPF getting segued in. this, for me, is enough to decide never to use solely linear interpolation for a modulateable delay widget. if i vary delay, i want only the delay to change. and i would prefer if the delay was the same for all frequencies, which makes the APF fractional delay thingie problematic. bestest, r b-j On Tue, Aug 18, 2015 at 1:40 AM, Peter S peter.schoffhau...@gmail.com mailto:peter.schoffhau...@gmail.com wrote: On 18/08/2015, Nigel Redmon earle...@earlevel.com mailto:earle...@earlevel.com wrote: well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? -Inf at Nyquist when you're halfway between two samples. Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1, -1... After interpolating with fraction=0.5, it becomes a constant signal 0,0,0,0,0,0,0... (because (-1+1)/2 = 0) -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
*my* point is that as the delay slowly slides from a integer number of samples, where the transfer function is H(z) = z^-N to the integer + 1/2 sample (with gain above), this linear but time-variant system is going to sound like there is a LPF getting segued in. this, for me, is enough to decide never to use solely linear interpolation for a modulateable delay widget. if i vary delay, i want only the delay to change. Yeah, absolutely. The variable suppression of high frequencies when fractional delay changes is undesirable, and indicates that better interpolation schemes should 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 wrote: 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 expectation that digital processing of such a signal is going to work properly in general. But even given that, the interpolator outputting the zero signal in that case is exactly correct. That's what you would have gotten if you'd sampled the same sine wave (*not* square wave - that would imply frequencies above Nyquist) with a half-sample offset from the 1, -1, 1, -1, ... case. The incorrect behavior arises when you try to go in the other direction (i.e., apply a second half-sample delay), and you still get only DC. But, again, that doesn't really say anything about interpolation. It just says that you sampled the signal improperly in the first place, and so digital processing can't be relied upon to work appropriately. as suprizing as it may first appear, i think Peter S and me are totally on the same page here. regarding *linear* interpolation, *if* you use linear interpolation in a precision delay (an LTI thingie, or at least quasi-time-invariant) and you delay by some integer + 1/2 sample, the filter you get has coefficients and transfer function H(z) = (1/2)*(1 + z^-1)*z^-N (where N is the integer part of the delay). the gain of that filter, as you approach Nyquist, approaches -inf dB. *my* point is that as the delay slowly slides from a integer number of samples, where the transfer function is H(z) = z^-N to the integer + 1/2 sample (with gain above), this linear but time-variant system is going to sound like there is a LPF getting segued in. this, for me, is enough to decide never to use solely linear interpolation for a modulateable delay widget. if i vary delay, i want only the delay to change. and i would prefer if the delay was the same for all frequencies, which makes the APF fractional delay thingie problematic. bestest, r b-j On Tue, Aug 18, 2015 at 1:40 AM, Peter S peter.schoffhau...@gmail.com mailto:peter.schoffhau...@gmail.com wrote: On 18/08/2015, Nigel Redmon earle...@earlevel.com mailto:earle...@earlevel.com wrote: well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? -Inf at Nyquist when you're halfway between two samples. Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1, -1... After interpolating with fraction=0.5, it becomes a constant signal 0,0,0,0,0,0,0... (because (-1+1)/2 = 0) -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, Ethan Duni ethan.d...@gmail.com wrote: 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. That's not weird, and that's *exactly* what you have in the highest bin of an FFT. The signal 1, -1, 1, -1, 1, -1 ... is the highest frequency basis function of the DFT: http://www.dspguide.com/graphics/F_8_5.gif If you think that's weird, then I guess you think that the Fourier transformation is weird. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, Tom Duffy tdu...@tascam.com wrote: In order to reconstruct that sinusoid, you'll need a filter with an infinitely steep transition band. I can use an arbitrarily long sinc kernel to reconstruct / interpolate it. Therefore, for any desired precision, you can find an appropriate sinc kernel length. Where's the problem? I can also oversample the signal arbitrarily, using an arbitrarily long sinc kernel. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 sampling, there's no way to reconstruct without some additional side information. E On Tue, Aug 18, 2015 at 1:45 PM, Tom Duffy tdu...@tascam.com wrote: In order to reconstruct that sinusoid, you'll need a filter with an infinitely steep transition band. You've demonstrated that SR/2 aliases to 0Hz, i.e. DC. That digital stream of samples is not reconstructable. On 8/18/2015 1:28 PM, Peter S wrote: That's false. 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. It is a bandlimited Nyquist frequency square wave (which is equivalent to a Nyquist frequency sine wave). From that, you can reconstruct a perfect alias-free sinusoid of frequency SR/2. NOTICE: This electronic mail message and its contents, including any attachments hereto (collectively, this e-mail), is hereby designated as confidential and proprietary. This e-mail may be viewed and used only by the person to whom it has been sent and his/her employer solely for the express purpose for which it has been disclosed and only in accordance with any confidentiality or non-disclosure (or similar) agreement between TEAC Corporation or its affiliates and said employer, and may not be disclosed to any other person or entity. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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., Lagrange interpolation filter, windowed sinc, etc). It's not an artifact of the simple linear approach, it's a feature of the symmetric, finite nature of the fractional interpolator. Since there are good reasons for the symmetry constraint, we are left to trade off oversampling and filter order/design to get the final passband as flat as we need. My view is that if you are serious about maintaining fidelity across the full bandwidth, you need to oversample by at least 2x. That way you can fit the transition band of your interpolation filter above the signal band. In applications where you are less concerned about full bandwidth fidelity, oversampling isn't required. Some argue that 48kHz sample rate is already effectively oversampled for lots of natural recordings, for example. If it's already at 96kHz or higher I would not bother oversampling further. Also this is recommended reading for this thread: https://ccrma.stanford.edu/~jos/Interpolation/ E On Tue, Aug 18, 2015 at 1:45 PM, Tom Duffy tdu...@tascam.com wrote: In order to reconstruct that sinusoid, you'll need a filter with an infinitely steep transition band. You've demonstrated that SR/2 aliases to 0Hz, i.e. DC. That digital stream of samples is not reconstructable. On 8/18/2015 1:28 PM, Peter S wrote: That's false. 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. It is a bandlimited Nyquist frequency square wave (which is equivalent to a Nyquist frequency sine wave). From that, you can reconstruct a perfect alias-free sinusoid of frequency SR/2. NOTICE: This electronic mail message and its contents, including any attachments hereto (collectively, this e-mail), is hereby designated as confidential and proprietary. This e-mail may be viewed and used only by the person to whom it has been sent and his/her employer solely for the express purpose for which it has been disclosed and only in accordance with any confidentiality or non-disclosure (or similar) agreement between TEAC Corporation or its affiliates and said employer, and may not be disclosed to any other person or entity. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 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 expectation that digital processing of such a signal is going to work properly in general. Not necessarily, at least in theory. In practice, an anti-alias filter will filter out a signal exactly at Nyquist freq, both when sampling it (A/D conversion), and both when reconstructing it (D/A conversion). But that doesn't mean that a half-sample delay doesn't have -Inf dB gain at Nyquist frequency. It's another thing that the anti-alias filter of a converter will typically filter it out anyways when reconstructing - but we weren't talking about reconstruction, so that is irrelevant here. A Nyquist frequency signal (1, -1, 1, -1, ...) is a perfectly valid bandlimited signal. But even given that, the interpolator outputting the zero signal in that case is exactly correct. That's what you would have gotten if you'd sampled the same sine wave (*not* square wave - that would imply frequencies above Nyquist) with a half-sample offset from the 1, -1, 1, -1, ... case. More precisely: a bandlimited Nyquist frequency square wave *equals* a Nyquist frequency sine wave. Or any other harmonic waveform for that matter (triangle, saw, etc.) In all cases, only the fundamental partial is there (1, -1, 1, -1, ... = Nyquist frequency sine), all the other partials are filtered out from the bandlimiting. So the signal 1, -1, 1, -1, *is* a Nyquist frequency bandlimited square wave, and also a sine-wave as well. They're identical. It *is* a bandlimited square wave - that's what you get when you take a Nyquist frequency square wave, and bandlimit it by removing all partials above Nyquist freq (say, via DFT). You may call it a square, a sine, saw, doesn't matter - when bandlimited, they're identical. The incorrect behavior arises when you try to go in the other direction (i.e., apply a second half-sample delay), and you still get only DC. What would be incorrect about it? I'm not sure what is your assumption. Of course if you apply any kind of filtering to a zero DC signal, you'll still have a zero DC signal. -Inf + -Inf = -Inf... Not sure what you're trying to achieve by applying a second half-sample delay... That also has -Inf dB gain at Nyquist, so you'll still have a zero DC signal after that. Since a half-sample delay has -Inf gain at Nyquist, you cannot undo it by applying another half-sample delay... But, again, that doesn't really say anything about interpolation.It just says that you sampled the signal improperly in the first place, and so digital processing can't be relied upon to work appropriately. That's false. 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. It is a bandlimited Nyquist frequency square wave (which is equivalent to a Nyquist frequency sine wave). From that, you can reconstruct a perfect alias-free sinusoid of frequency SR/2. What's causing you to be unable to reconstruct the waveform? -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, Ethan Duni ethan.d...@gmail.com wrote: You cannot calculate 1/x when x=0, can you? Since that's division by zero. Yet you'll know when x tends to zero from right towards left, then 1/x will tend to +infinity. Not sure what that is supposed to have to do with the present subject. You cannot calculate 1/x when x=0, because that's division by zero, yet you can calculate the limit of 1/x as x tends towards zero. Meaning that you can approach zero arbitrarily, and 1/x will approach +infinity arbitrarily. Similarly, even if frequency f=0.5 may be considered ill-specified (because it's critical frequency), you can still approach it to arbitrary precision, and the gain will approach -infinity. So f=0.4 f=0.49 f=0.499 f=0.4999 f=0.49 f=0.499 etc. The more you approach f=0.5, the more the gain will approach -infinity. Even if f=0.5 is a critical frequency. f=0.4 isn't, and it's quite close to f=0.5. That's what I mean. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On Aug 17, 2015, at 9:38 AM, Esteban Maestre este...@ccrma.stanford.edu wrote: No experience with compensation filters here. But if you can afford to use a higher order interpolation scheme, I'd go for that. Using Newton's Backward Difference Formula, one can construct time-varying, table-free, efficient Lagrange interpolation schemes of arbitrary order (up to 30-th or 40-th order) which stay within linear complexity while allowing for run-time modulation of the interpolation order. https://ccrma.stanford.edu/~jos/Interpolation/Lagrange_Interpolation.html Cheers, Esteban I would think that polynomial interpolators of order 30 or 40 would provide no end of unpleasant surprises due to the behavior of high-order polynomials. I'm thinking of weird spikes, etc. Have you actually used polynomial interpolators of this order? Jerry On 8/17/2015 12:07 PM, STEFFAN DIEDRICHSEN wrote: I could write a few lines over the topic as well, since I made such a compensation filter about 17 years ago. So, there are people, that do care about that topic, but there are only some, that do find time to write up something. ;-) Steffan On 17.08.2015|KW34, at 17:50, Theo Verelst theo...@theover.org wrote: However, no one here besides RBJ and a few brave souls seems to even care much about real subjects. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp -- Esteban Maestre CIRMMT/CAML - McGill Univ MTG - Univ Pompeu Fabra http://ccrma.stanford.edu/~esteban ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/18/2015 6:41 AM, Jerry wrote: I would think that polynomial interpolators of order 30 or 40 would provide no end of unpleasant surprises due to the behavior of high-order polynomials. I'm thinking of weird spikes, etc. Have you actually used polynomial interpolators of this order? I remember going even above 40-th order with no problems. But I also remember having problems with 80-th order interpolation. I think it's called the /Runge phenomenon/. Esteban -- Esteban Maestre CIRMMT/CAML - McGill Univ MTG - Univ Pompeu Fabra http://ccrma.stanford.edu/~esteban ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, robert bristow-johnson r...@audioimagination.com wrote: *my* point is that as the delay slowly slides from a integer number of samples [...] to the integer + 1/2 sample (with gain above), this linear but time-variant system is going to sound like there is a LPF getting segued in. Exactly. As the fractional delay varies between 0..1, it will sound like a fluttering LP filter that closes and opens as the delay varies, having the most 'muffled' (LPF'ed) sound when fraction = 1/2. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
I’m sorry, I’m missing your point here, Peter (and perhaps I missed Roberts, hence the “No?” in my reply to him). The frequency response of linear interpolation is (sin(pi*x)/(pi*x))^2, -7.8 dB at 0.5 of the sample rate... On Aug 18, 2015, at 1:40 AM, Peter S peter.schoffhau...@gmail.com wrote: On 18/08/2015, Nigel Redmon earle...@earlevel.com wrote: well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? -Inf at Nyquist when you're halfway between two samples. Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1, -1... After interpolating with fraction=0.5, it becomes a constant signal 0,0,0,0,0,0,0... (because (-1+1)/2 = 0) __ ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, robert bristow-johnson r...@audioimagination.com wrote: On 8/18/15 4:28 PM, Peter S wrote: 1, -1, 1, -1, 1, -1 ... is a proper bandlimited signal, and contains no aliasing. That's the maximal allowed frequency without any aliasing. well Peter, here again is where you overreach. assuming, without loss of generality that the sampling period is 1, the continuous-time signals x(t) = 1/cos(theta) * cos(pi*t + theta) are all aliases for the signal described above (and incorrectly as contain[ing] no aliasing). Well, strictly speaking, that is true. But I assumed the signal to be bandlimited to 0..SR/2. In that case, you can perfectly reconstruct it, as you have no other alias between 0..SR/2. -P ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
OK, I looked back at Robert’s post, and see that the fact his reply was broken up into segments (as he replied to segments of Peter’s comment) made me miss his point. At first he was talking general (pitch shifting), but at that point he was talking about strictly sliding into halfway between samples in the interpolation. Never mind. On Aug 18, 2015, at 1:50 PM, Nigel Redmon earle...@earlevel.com wrote: I’m sorry, I’m missing your point here, Peter (and perhaps I missed Roberts, hence the “No?” in my reply to him). The frequency response of linear interpolation is (sin(pi*x)/(pi*x))^2, -7.8 dB at 0.5 of the sample rate... On Aug 18, 2015, at 1:40 AM, Peter S peter.schoffhau...@gmail.com wrote: On 18/08/2015, Nigel Redmon earle...@earlevel.com wrote: well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? -Inf at Nyquist when you're halfway between two samples. Assume you have a Nyquist frequency square wave: 1, -1, 1, -1, 1, -1, 1, -1... After interpolating with fraction=0.5, it becomes a constant signal 0,0,0,0,0,0,0... (because (-1+1)/2 = 0) __ ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 18/08/2015, Ethan Duni ethan.d...@gmail.com wrote: That class of signals is band limited to SR/2. The aliasing is in the amplitude/phase offset, not the frequency. Okay, I get what you mean. But that doesn't change the frequency response of a half-sample delay, or doesn't mean that a half-sample delay doesn't have a specific gain at Nyquist. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
And to add to what Robert said about “write code and sell it”, sometimes it’s more comfortable to make general but helpful comments here, and stop short of detailing the code that someone paid you a bunch of money for and might not want to be generally known. And before people assume that I mean strictly “keep the secret sauce secret”, there’s also the fact that marketing might not want it known that every detail of their expensive plug-in is not 256x oversampled, 128-bit floating point data path throughout, dithered every stage. :-D On Aug 17, 2015, at 1:46 PM, robert bristow-johnson r...@audioimagination.com wrote: On 8/17/15 12:07 PM, STEFFAN DIEDRICHSEN wrote: I could write a few lines over the topic as well, since I made such a compensation filter about 17 years ago. So, there are people, that do care about that topic, but there are only some, that do find time to write up something. ;-) Steffan On 17.08.2015|KW34, at 17:50, Theo Verelst theo...@theover.org mailto:theo...@theover.org wrote: However, no one here besides RBJ and a few brave souls seems to even care much about real subjects. Theo, there are a lotta heavyweights here (like Steffan). if you want a 3-letter acronym to toss around, try JOS. i think there are plenty on this list that care deeply about reality because they write code and sell it. my soul is chicken-shit in the context. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
OK, Robert, I did consider the slow versus fast issue. But there have been few caveats posted in this thread, so I thought it might be misleading to some to not be specific about context. The worst case would be a precision delay of an arbitrary constant. (For example, at 44.1 kHz SR, there would be a significant frequency response difference between 250 ms and 250.01 ms, despite no perceptible difference in time. Of course, in some cases, even when using interpolated delays, you can quantize the delay time to a sample boundary—say if modulation is transient and the steady state is the main concern.) So, yes, the context means a lot, so we should be clear. (And can you tell I’m doing something with delays right now?) Personally, I’m a fan of upsampling, when needed. On Aug 17, 2015, at 1:55 PM, robert bristow-johnson r...@audioimagination.com wrote: On 8/17/15 2:39 PM, Nigel Redmon wrote: Since compensation filtering has been mentioned by a few, can I ask if someone could get specific on an implementation (including a description of constraints under which it operates)? I’d prefer keeping it simple by restricting to linear interpolation, where it’s most needed, and perhaps these comments will make clearer what I’m after: As I noted in the first reply to this thread, while it’s temping to look at the sinc^2 rolloff of a linear interpolator, for example, and think that compensation would be to boost the highs to undo the rolloff, that won’t work in the general case. Even in Olli Niemitalo’s most excellent paper on interpolation methods (http://yehar.com/blog/wp-content/uploads/2009/08/deip.pdf), he seems to suggest doing this with pre-emphasis, which seems to be a mistake, unless I misunderstood his intentions. In Olli’s case, he specifically recommended pre-emphasis (which I believe will not work except for special cases of resampling at fixed fractions between real samples) over post, as post becomes more complicated. (It seems that you could do it post, taking into account the fractional part of a particular lookup and avoiding the use of recursive filters—personally I’d just upsample to begin with.) to me, it really depends on if you're doing a slowly-varying precision delay in which the pre-emphasis might also be slowly varying. but if the application is sample-rate conversion or similar (like pitch shifting) where the fractional delay is varying all over the place, i think a fixed compensation for sinc^2 might be a good idea. i don't see how it would hurt. especially for the over-sampled case. i like Olli's pink-elephant paper, too. and i think (since he was picking up on Duane's and my old and incomplete paper) it was more about the fast-varying fractional delay. and i think that the Zölzer/Bolze paper suggested the same thing (since it was even worse than linear interp). -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Thanks for the suggestions and discussion. In my application I'm playing back 44.1khz wavefiles with variable pitch envelopes. I'm currently using hermite interpolation and the quality seems fine for playback. It's only after resampling and running through the audio engine multiple times does the high frequency roll off become a problem. I'll try adding in some oversampling. Cheers Shannon ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 2015-08-17, robert bristow-johnson wrote: As I noted in the first reply to this thread, while it’s temping to look at the sinc^2 rolloff of a linear interpolator, for example, and think that compensation would be to boost the highs to undo the rolloff, that won’t work in the general case. Even in Olli Niemitalo’s most excellent paper on interpolation methods (http://yehar.com/blog/wp-content/uploads/2009/08/deip.pdf), he seems to suggest doing this with pre-emphasis, which seems to be a mistake, unless I misunderstood his intentions. Actually it's not that simple. Substandard interpolation methods do lead to high frequency rolloff, which can be corrected to a degree with a complementary filter. But the trouble is, at the same time they lead to aliasing and even nonlinear artifacts, whose high frequency content will be amplified by the compensatory filter as well. As such, that approach is basically sound...but at the same time only within a very narrowly parametrized envelope. to me, it really depends on if you're doing a slowly-varying precision delay in which the pre-emphasis might also be slowly varying. In slowly varying delay it ought to work no matter what. but if the application is sample-rate conversion or similar (like pitch shifting) where the fractional delay is varying all over the place, i think a fixed compensation for sinc^2 might be a good idea. i don't see how it would hurt. especially for the over-sampled case. It doesn't necessarily hurt, but here it isn't guaranteed to do any good either. And it's close to doing something bad instead. -- Sampo Syreeni, aka decoy - de...@iki.fi, http://decoy.iki.fi/front +358-40-3255353, 025E D175 ABE5 027C 9494 EEB0 E090 8BA9 0509 85C2___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On Aug 17, 2015, at 7:23 PM, robert bristow-johnson r...@audioimagination.com wrote: On 8/17/15 7:29 PM, Sampo Syreeni wrote: to me, it really depends on if you're doing a slowly-varying precision delay in which the pre-emphasis might also be slowly varying. In slowly varying delay it ought to work no matter what. well, if it's linear interpolation and your fractional delay slowly sweeps from 0 to 1/2 sample, i think you may very well hear a LPF start to kick in. something like -7.8 dB at Nyquist. no, that's not right. it's -inf dB at Nyquist. pretty serious LPF to just slide into. Right the first time, -7.8 dB at the Nyquist frequency, -inf at the sampling frequency. No? ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
No experience with compensation filters here. But if you can afford to use a higher order interpolation scheme, I'd go for that. Using Newton's Backward Difference Formula, one can construct time-varying, table-free, efficient Lagrange interpolation schemes of arbitrary order (up to 30-th or 40-th order) which stay within linear complexity while allowing for run-time modulation of the interpolation order. https://ccrma.stanford.edu/~jos/Interpolation/Lagrange_Interpolation.html Cheers, Esteban On 8/17/2015 12:07 PM, STEFFAN DIEDRICHSEN wrote: I could write a few lines over the topic as well, since I made such a compensation filter about 17 years ago. So, there are people, that do care about that topic, but there are only some, that do find time to write up something. ;-) Steffan On 17.08.2015|KW34, at 17:50, Theo Verelst theo...@theover.org mailto:theo...@theover.org wrote: However, no one here besides RBJ and a few brave souls seems to even care much about real subjects. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp -- Esteban Maestre CIRMMT/CAML - McGill Univ MTG - Univ Pompeu Fabra http://ccrma.stanford.edu/~esteban ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
For people including scientific oriented it always surprises me how little actual science is involved in this talk about tradeoffs. First, what it is you want to achieve by preserving high frequencies (which of course I'm all for)? Second, is it really only at the level of first order interpolations ? And if so, isn't the compensation interpolation much more expensive than a solution that tries to qualify, and preferably quantify the errors involved. Using least squares and error estimates is a bit too easy for our sampling issues because of at least the mid and high frequencies getting interpreted but the DAC reconstruction filter, subsequent digital signal processing or as I prefer: the perfect reconstruction interpretation of the resulting digital signal streams. IMO, high frequencies will be most served by leaving them alone as much as possible, and honoring the studio and post processing that has checked them out and pre-averaged them for normal sound reproduction. However, no one here besides RBJ and a few brave souls seems to even care much about real subjects. Now I get it, everyone has a sound card and endless supplies of digital materials, and from my student efforts I recall it is fun to understand the theoretics of interpolation curves (and (hyper-) surfaces), but unfortunately they correlate only very very loosely with useful sampled signal theories, unless you want an effort for a particular niche. T.V. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
I could write a few lines over the topic as well, since I made such a compensation filter about 17 years ago. So, there are people, that do care about that topic, but there are only some, that do find time to write up something. ;-) Steffan On 17.08.2015|KW34, at 17:50, Theo Verelst theo...@theover.org wrote: However, no one here besides RBJ and a few brave souls seems to even care much about real subjects. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Since compensation filtering has been mentioned by a few, can I ask if someone could get specific on an implementation (including a description of constraints under which it operates)? I’d prefer keeping it simple by restricting to linear interpolation, where it’s most needed, and perhaps these comments will make clearer what I’m after: As I noted in the first reply to this thread, while it’s temping to look at the sinc^2 rolloff of a linear interpolator, for example, and think that compensation would be to boost the highs to undo the rolloff, that won’t work in the general case. Even in Olli Niemitalo’s most excellent paper on interpolation methods (http://yehar.com/blog/wp-content/uploads/2009/08/deip.pdf), he seems to suggest doing this with pre-emphasis, which seems to be a mistake, unless I misunderstood his intentions. In Olli’s case, he specifically recommended pre-emphasis (which I believe will not work except for special cases of resampling at fixed fractions between real samples) over post, as post becomes more complicated. (It seems that you could do it post, taking into account the fractional part of a particular lookup and avoiding the use of recursive filters—personally I’d just upsample to begin with.) It just occurred to me that perhaps one possible implementation is to cross-fade between a pre-emphasized and normal delay line, depending on the fractional position (0.5 gets all pre-emph, 0.0 gets all normal). This sort of thing didn’t seem to be what Olli was getting at, since he only gave the worst-case rolloff curve and didn’t discuss it further. I also think about the possibility of crossfading between emphasis and none, depending on the fractional position (full emphasis for I’m not asking because I need to do this—I’m asking for the sake of the thread, where people are talking about compensation, but not explaining the implementation they have in mind, and not necessarily explaining the conditions under which it works. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
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 filtering has been mentioned by a few, can I ask if someone could get specific on an implementation (including a description of constraints under which it operates)? ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/17/15 12:07 PM, STEFFAN DIEDRICHSEN wrote: I could write a few lines over the topic as well, since I made such a compensation filter about 17 years ago. So, there are people, that do care about that topic, but there are only some, that do find time to write up something. ;-) Steffan On 17.08.2015|KW34, at 17:50, Theo Verelst theo...@theover.org mailto:theo...@theover.org wrote: However, no one here besides RBJ and a few brave souls seems to even care much about real subjects. Theo, there are a lotta heavyweights here (like Steffan). if you want a 3-letter acronym to toss around, try JOS. i think there are plenty on this list that care deeply about reality because they write code and sell it. my soul is chicken-shit in the context. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/17/15 2:39 PM, Nigel Redmon wrote: Since compensation filtering has been mentioned by a few, can I ask if someone could get specific on an implementation (including a description of constraints under which it operates)? I’d prefer keeping it simple by restricting to linear interpolation, where it’s most needed, and perhaps these comments will make clearer what I’m after: As I noted in the first reply to this thread, while it’s temping to look at the sinc^2 rolloff of a linear interpolator, for example, and think that compensation would be to boost the highs to undo the rolloff, that won’t work in the general case. Even in Olli Niemitalo’s most excellent paper on interpolation methods (http://yehar.com/blog/wp-content/uploads/2009/08/deip.pdf), he seems to suggest doing this with pre-emphasis, which seems to be a mistake, unless I misunderstood his intentions. In Olli’s case, he specifically recommended pre-emphasis (which I believe will not work except for special cases of resampling at fixed fractions between real samples) over post, as post becomes more complicated. (It seems that you could do it post, taking into account the fractional part of a particular lookup and avoiding the use of recursive filters—personally I’d just upsample to begin with.) to me, it really depends on if you're doing a slowly-varying precision delay in which the pre-emphasis might also be slowly varying. but if the application is sample-rate conversion or similar (like pitch shifting) where the fractional delay is varying all over the place, i think a fixed compensation for sinc^2 might be a good idea. i don't see how it would hurt. especially for the over-sampled case. i like Olli's pink-elephant paper, too. and i think (since he was picking up on Duane's and my old and incomplete paper) it was more about the fast-varying fractional delay. and i think that the Zölzer/Bolze paper suggested the same thing (since it was even worse than linear interp). -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 17/08/2015, STEFFAN DIEDRICHSEN sdiedrich...@me.com wrote: I could write a few lines over the topic as well, since I made such a compensation filter about 17 years ago. So, there are people, that do care about that topic, but there are only some, that do find time to write up something. I also made a compensation filter for linear interpolation. Definitely doable. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
[music-dsp] Compensate for interpolation high frequency signal loss
Hi, Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Are there any papers that detail what such a filter might look like? Thanks Shannon ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 2015-08-16, Sham Beam wrote: Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Are there any papers that detail what such a filter might look like? Look at Vesa Välimäki's work, and his students'. They did fractional delay delay lines, which had just this problem in the high end. Also, Julius O. Smith's work with waveguides bumped into this very same thing, because they're implemented as (fractional) delay lines as well. Beyond that, most reverb designers could tell you about this sort of thing, only they tend to keep their secret sauce *most* secret. ;) The usual thing you do is to go for higher order interpolation, with the interpolating polynomial being designed for flatter performance over the utility band than the linear spline. It's already very much better at 3rd order, and if you do something like 4th to 5th order with 2x oversampling, it's essentially perfect. -- Sampo Syreeni, aka decoy - de...@iki.fi, http://decoy.iki.fi/front +358-40-3255353, 025E D175 ABE5 027C 9494 EEB0 E090 8BA9 0509 85C2___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
As far as compensation: Taking linear as an example, we know that the response rolls off (“sinc^2). Would you compensate by boosting the highs? Consider that for a linearly interpolated delay line, a delay of an integer number of samples, i, has no high frequency loss at all. But that the error is max it you need a delay of i + 0.5 samples. More difficult to compensate for, would be such a delay line where the delay time is modulated. A well-published way of getting around the fractional problem is allpass compensation. But a lot of people seem to miss that this method doesn’t lend itself to modulation—it’s ideally suited for a fixed fractional delay. Here’s a paper that shows one possible solution, crossfading two allpass filters: http://scandalis.com/jarrah/Documents/DelayLine.pdf Obviously, the most straight-forward way to avoid the problem is to convert to a higher sample rate going into the delay line (using windowed sinc, etc.), then use linear, hermite, etc. On Aug 16, 2015, at 1:09 AM, Sham Beam sham.b...@gmail.com wrote: Hi, Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Are there any papers that detail what such a filter might look like? Thanks Shannon ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Hi Shannon, If the number of reads from the delay line per sample cycle is high enough, as a less expensive alternative to the most obvious solution (higher order interpolation based on multiple samples before and after, with some fancy set of coefficients calculated on the spot, or looked up from a table of sufficiently high resolution, depending on the fraction of a sample delay involved) you might like to consider upsampling the input signal to twice the normal rate, and then doing simple linear interpolation. This would not be mathematically perfect, since the high frequency response would be slightly reduced if the delay fraction was 0.25 or 0.75, whereas it would be flat for 0, and as flat as the upsampling algorithm for 0.5 (I recall that the upsampling algorithm produces the odd numbered samples in the final output, with the even ones being the input samples). However, assuming the sampling rate is 44.1kHz, 48kHz or higher, I think this slight variation is unlikely to be perceptible to human ears. Robin On 2015-08-16 6:09 PM, Sham Beam wrote: Hi, Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Are there any papers that detail what such a filter might look like? Thanks Shannon ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
Is this Robin Whittle of Devilfish fame? I bought a Devilfish from you back in the mid-1990s. Best mod ever! On Aug 16, 2015, at 8:07 PM, Robin Whittle r...@firstpr.com.au wrote: Hi Shannon, If the number of reads from the delay line per sample cycle is high enough, as a less expensive alternative to the most obvious solution (higher order interpolation based on multiple samples before and after, with some fancy set of coefficients calculated on the spot, or looked up from a table of sufficiently high resolution, depending on the fraction of a sample delay involved) you might like to consider upsampling the input signal to twice the normal rate, and then doing simple linear interpolation. This would not be mathematically perfect, since the high frequency response would be slightly reduced if the delay fraction was 0.25 or 0.75, whereas it would be flat for 0, and as flat as the upsampling algorithm for 0.5 (I recall that the upsampling algorithm produces the odd numbered samples in the final output, with the even ones being the input samples). However, assuming the sampling rate is 44.1kHz, 48kHz or higher, I think this slight variation is unlikely to be perceptible to human ears. Robin On 2015-08-16 6:09 PM, Sham Beam wrote: Hi, Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Are there any papers that detail what such a filter might look like? Thanks Shannon ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp signature.asc Description: Message signed with OpenPGP using GPGMail ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Compensate for interpolation high frequency signal loss
On 8/16/15 4:09 AM, Sham Beam wrote: Hi, Is it possible to use a filter to compensate for high frequency signal loss due to interpolation? For example linear or hermite interpolation. Are there any papers that detail what such a filter might look like? besides the well-known sinc^2 rolloff that comes with linear interpolation, this paper discussed interpolation using higher-order B-splines. Udo Zölzer and Thomas Bolze, Interpolation Algorithms: Theory and Application, http://www.aes.org/e-lib/browse.cfm?elib=6334 . i thought i had a copy of it, Duane Wise and i referenced the paper in our 2-decade-old paper about different polynomial interpolation effects and which were better for what. (you can have a .pdf of that paper if you want.) but Zölzer and Bolze (they might be hanging out on this list, i was pleasantly surprized to see JOS post here recently) *do* discuss pre-compensation of high-frequency rolloff due to interpolation polynomials that cause such rolloff. you just design a filter (using MATLAB or whatever) that has magnitude response that is, in the frequency band of interest, approximately the reciprocal of the rolloff effect from the interpolation. Zölzer and Bolze suggested Nth-order B-spline without really justifying why that is better than other polynomial kernels such as Lagrange or Hermite. the Nth-order B-spline (at least how it was shown in their paper), is what you get when you convolve N unit-rectangular functions with each other (or N zero-order holds). the frequency response of a Nth-order B-spline is sinc^(N+1). this puts really deep and wide notches at integer multiples of the original sampling frequency (other than the integer 0) which is where all those images are that you want to kill. linear interpolation is all of a 1st-order Lagrange, a 1st-order Hermite, and a 1st-order B-spline (and the ZOH or drop-sample interpolation is a 0th-order realization of those three). an Nth-order polynomial interpolator will have, somewhere in the frequency response, a H(f) = (sinc(f/Fs))^(N+1) in there, but if it's not the simple B-spline, there will be lower order terms of sinc() that will add to (contaminate) the highest order sinc^(N+1) and make those notches less wide. any other polynomial interpolation (at higher order than 1), will have at least one sinc() term with lower order than N+1. so the cool thing about interpolating with B-splines is that it kills the images (which become aliases when you resample) the most, but it also has wicked LPFing that needs to be compensated unless your sampling frequency is *much* higher than twice the bandwidth (oversampled big-time). but if you *are* experiencing that LPFing, as you have suspected, you can design a filter to undo that for much of the baseband. not all of it. -- r b-j r...@audioimagination.com Imagination is more important than knowledge. ___ music-dsp mailing list music-dsp@music.columbia.edu https://lists.columbia.edu/mailman/listinfo/music-dsp
