Reply by robert bristow-johnson January 26, 20062006-01-26
in article p7GdnfgUx_C7r0XeRVn-qg@rcn.net, Jerry Avins at jya@ieee.org wrote
on 01/25/2006 21:21:

> robert bristow-johnson wrote: > > ... > >> as pointed out the Hamming can be complementary, too. if you scale it >> so that when it's 50% done fading out, it's at 0.5. then the Hamming >> will sum to 1. > > It adds to 1. without scaling. A rectangular windows add to 2. Of > course, any window can be scaled. For some uses, one wants the mean > ordinate rather than the peak to have a specific value. > > Jerry > > *Appendix :o)* > Hamming(i,N) = .5 + .5*cos(2pi*i/N); goes to .5 when i = +/-N/4 > Hann(i,N) = .04 + .92*Hamming(i,N); goes to .5 when i = +/-N/4
that looks backwards to me, Jerry. i would say it's Hann(i,N) = 0.50 + 0.50*cos(2*pi*i/N) Hamming(i,N) = 0.54 + 0.46*cos(2*pi*i/N) = 0.08 + 0.92*Hann(i,N) (for |i| < N/2) the appropriately scaled Hamming that would add to 1 would be (0.5/0.54)*Hamming(i,N) = 0.50 + 0.426*cos(2*pi*i/N)
> N is the total number of points; i is zero at the center.
better make N an even number (if the center is at i=0) and N-1 will be the total number of non-zero points (unless we use the two outside ones for Hamming, and then it's N+1, this mixup is evident in MATLAB's definitions). BTW, for the purposes of cross-fading (that is, if you are representing the "fade-up" or "fade-in" gain function as the rising half of the window function and "fade-down" or "fade-out" as the falling half), i find it much easier to deal with windows in continuous time even though we almost always apply them in discrete time. that way, i don't have to worry about whether or not we're including those two outside points or not. also for cross-fading, i always define the windows to be complementary: Normalized Rectangular: { 1 |u| < 0.5 w(u) = { 0.5 |u| = 0.5 { 0 |u| > 0.5 Normalized Bartlett (triangular): { 1 - |u| |u| < 1.0 w(u) = { { 0 |u| >= 1.0 Normalized Hann: { 0.5 + 0.5*cos(pi*u) |u| < 1.0 w(u) = { { 0 |u| >= 1.0 Normalized (scaled) Hamming: { 0.5 + 0.426*cos(pi*u) |u| < 1.0 w(u) = { 0.037 |u| = 1.0 { 0 |u| > 1.0 Normalized Flattened Hann: { 1/2 + (9/16)*cos(pi*u) - (1/16)*cos(3*pi*u) |u| < 1.0 w(u) = { { 0 |u| >= 1.0 for all of these windows, their leading half perfectly compliments the trailing half: +inf SUM w(u - k) = 1 k=-inf can you tell me to what derivative each is continuous to? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by Jerry Avins January 25, 20062006-01-25
robert bristow-johnson wrote:

   ...

> as pointed out the Hamming can be complementary, too. if you scale it > so that when it's 50% done fading out, it's at 0.5. then the Hamming > will sum to 1.
It adds to 1. without scaling. A rectangular windows add to 2. Of course, any window can be scaled. For some uses, one wants the mean ordinate rather than the peak to have a specific value. Jerry *Appendix :o)* Hamming(i,N) = .5 + .5*cos(2pi*i/N); goes to .5 when i = +/-N/4 Hann(i,N) = .04 + .92*Hamming(i,N); goes to .5 when i = +/-N/4 N is the total number of points; i is zero at the center. -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by robert bristow-johnson January 25, 20062006-01-25
in article 1138230108.727351.17560@g49g2000cwa.googlegroups.com, robert
bristow-johnson at rbj@audioimagination.com wrote on 01/25/2006 18:01:

> the time-scaling part of a pitch detector might be expressed as: > > > +inf > SUM{ x(t + k[n]/f0(t)) * w(t - n*T) } = x(t) > n=-inf > > where k[n] is always an integer starting with zero and f0(t) is the > fundamental frequency (determined by a simultaneous pitch detector) > around time t.
i think this is not quite right. first i shoulda said the result is the output, y(t), not "x(t)". it will sorta work for f0(t) = constant (but then why make it a function of t?) +inf SUM{ x(t + Tau[n]) * w(t - n*T) } = y(t) n=-inf where Tau[0] = 0 Tau[n] = Tau[n-1] + m[n]/f0(t + Tau[n-1]) f0(t) is the fundamental frequency around time t, and m[n] is an integer that is either -1, 0, or +1. which value m[n] takes on is whatever it has to so that t/(t + Tau[n]) = approximately the time-scaling factor. that might be a little more accurate. anyway, the undesirability of windows with discontinuities, such as the rectangular or Hamming, remains. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by robert bristow-johnson January 25, 20062006-01-25
Jerry Avins wrote:
> robert bristow-johnson wrote: > > > > it's not a bad idea to think of the window functions as the > > concatination of a fade-up and fade-down functions. the hann window is > > not bad for crossfading, but the Hamming is not so good > > (discontinuity). > > The Hann window is a raised cosine. It's performance is good, but not > outstanding; other, better windows use more terms. Hats off to Nuttall, > Blackman, Harris, Kaiser, and others. But before (he knew) their work, > Hamming had a problem with a Hann window's first sidelobe. He determined > empirically that 92% Hann and 8% rectangular knocked the first sidelobe > down as far as it would go without raising any of the others above it. > > Hamming was smart. He knew not only how to get the job done, but also > when to quit. His window is probably not optimum for much (if anything) > but it's good enough for the particular class of problems he dealt with.
...
> Despite it's relatively poor sidelobe performance, a Hann window has a > property it shares only with Bartlett (triangular) and rectangular > windows: when overlapped 50%, the sum is uniform. That property > recommends it for certain applications (NOT fast convolution).
as pointed out the Hamming can be complementary, too. if you scale it so that when it's 50% done fading out, it's at 0.5. then the Hamming will sum to 1. any of these complemetary windows satistfy: +inf SUM{ w(t - n*T) } = 1 n=-inf where T is the width between the 50% points of the window. this results with +inf SUM{ x(t) * w(t - n*T) } = x(t) n=-inf the time-scaling part of a pitch detector might be expressed as: +inf SUM{ x(t + k[n]/f0(t)) * w(t - n*T) } = x(t) n=-inf where k[n] is always an integer starting with zero and f0(t) is the fundamental frequency (determined by a simultaneous pitch detector) around time t. when no time-scaling is done, k[n] will always be zero, but when time-scaling *is* done, k[n] will increment or decrement in such a way that t/( t+ k[n]/f0(t)) is approximately the time-scaling factor. anyway, it is between these adjacent windows when k[n] jumps to another integer value that you do not want that 8% discontinuity of the Hamming window. r b-j
Reply by Jerry Avins January 25, 20062006-01-25
Ron N. wrote:
> Jerry Avins wrote: > >>The Hann window is a raised cosine. It's performance is good, but not >>outstanding; other, better windows use more terms. Hats off to Nuttall, >>Blackman, Harris, Kaiser, and others. But before (he knew) their work, >>Hamming had a problem with a Hann window's first sidelobe. He determined >>empirically that 92% Hann and 8% rectangular knocked the first sidelobe >>down as far as it would go without raising any of the others above it. >> >>Hamming was smart. He knew not only how to get the job done, but also >>when to quit. His window is probably not optimum for much (if anything) >>but it's good enough for the particular class of problems he dealt with. >>That said, it's an empirical kludge, more like what I would do than what >>we expect from people who get remembered. Hamming is remembered for many >>things besides his window. >> >>Despite it's relatively poor sidelobe performance, a Hann window has >>a property it shares only with Bartlett (triangular) and rectangular >>windows: when overlapped 50%, the sum is uniform. That property >>recommends it for certain applications (NOT fast convolution). > > > Because the Hamming window is the sum of a Hann window and a > rectangular window, it also share's this property of a uniform sum > when overlapped 50%
Yah. I wrote Hann and rectangular, assuming that we all know about superposition. Thanks for making it explicit. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Ron N. January 25, 20062006-01-25
Jerry Avins wrote:
> The Hann window is a raised cosine. It's performance is good, but not > outstanding; other, better windows use more terms. Hats off to Nuttall, > Blackman, Harris, Kaiser, and others. But before (he knew) their work, > Hamming had a problem with a Hann window's first sidelobe. He determined > empirically that 92% Hann and 8% rectangular knocked the first sidelobe > down as far as it would go without raising any of the others above it. > > Hamming was smart. He knew not only how to get the job done, but also > when to quit. His window is probably not optimum for much (if anything) > but it's good enough for the particular class of problems he dealt with. > That said, it's an empirical kludge, more like what I would do than what > we expect from people who get remembered. Hamming is remembered for many > things besides his window. > > Despite it's relatively poor sidelobe performance, a Hann window has > a property it shares only with Bartlett (triangular) and rectangular > windows: when overlapped 50%, the sum is uniform. That property > recommends it for certain applications (NOT fast convolution).
Because the Hamming window is the sum of a Hann window and a rectangular window, it also share's this property of a uniform sum when overlapped 50% IMHO. YMMV. -- rhn A.T nicholson d.0.t C-o-M
Reply by Jerry Avins January 25, 20062006-01-25
robert bristow-johnson wrote:
> Ross Clement (Email address invalid - do not use) wrote: > >>I once implemented a pitch-shifter with the ring buffer as described by >>Jani. If I remember correctly (this was a looonnngggg time ago) I used >>a cross-fade to gloss over the points where the read and write pointers >>crossed each other, which was an improvement. I've always wondered >>since then whether a better windowing scheme such as a hamming window >>would have improved things. >> >>Any comments? > > > it's not a bad idea to think of the window functions as the > concatination of a fade-up and fade-down functions. the hann window is > not bad for crossfading, but the Hamming is not so good > (discontinuity).
The Hann window is a raised cosine. It's performance is good, but not outstanding; other, better windows use more terms. Hats off to Nuttall, Blackman, Harris, Kaiser, and others. But before (he knew) their work, Hamming had a problem with a Hann window's first sidelobe. He determined empirically that 92% Hann and 8% rectangular knocked the first sidelobe down as far as it would go without raising any of the others above it. Hamming was smart. He knew not only how to get the job done, but also when to quit. His window is probably not optimum for much (if anything) but it's good enough for the particular class of problems he dealt with. That said, it's an empirical kludge, more like what I would do than what we expect from people who get remembered. Hamming is remembered for many things besides his window. Despite it's relatively poor sidelobe performance, a Hann window has a property it shares only with Bartlett (triangular) and rectangular windows: when overlapped 50%, the sum is uniform. That property recommends it for certain applications (NOT fast convolution). Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by robert bristow-johnson January 24, 20062006-01-24
Ross Clement (Email address invalid - do not use) wrote:
> I once implemented a pitch-shifter with the ring buffer as described by > Jani. If I remember correctly (this was a looonnngggg time ago) I used > a cross-fade to gloss over the points where the read and write pointers > crossed each other, which was an improvement. I've always wondered > since then whether a better windowing scheme such as a hamming window > would have improved things. > > Any comments?
it's not a bad idea to think of the window functions as the concatination of a fade-up and fade-down functions. the hann window is not bad for crossfading, but the Hamming is not so good (discontinuity). r b-j
Reply by Ross Clement (Email address invalid - do not use) January 24, 20062006-01-24
I once implemented a pitch-shifter with the ring buffer as described by
Jani. If I remember correctly (this was a looonnngggg time ago) I used
a cross-fade to gloss over the points where the read and write pointers
crossed each other, which was an improvement. I've always wondered
since then whether a better windowing scheme such as a hamming window
would have improved things.

Any comments?

Cheers,

Ross-c

Reply by robert bristow-johnson January 24, 20062006-01-24
Jani Huhtanen wrote:
> > Use ring buffer. Write at constant "speed" but read faster or slower than > you write. If you read faster pitch goes up and if you read slower pitch > goes down. Sound quality is pretty horrible due to the phase > discontinuities (i.e. phases don't match when read and write passes each > other) but it works. The length of the ring buffer is crucial parameter for > satisfactory results.
toss in a little bit of "pitch detection" (look up AMDF for "Average Magnitude Difference Function" on Google) and the splice in or out exactly one or two whole periods of the waveform. when you splice, crossfade from the old to the new. that will improve the sound a lot. r b-j