C++ Simple Pitch Shift

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.
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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

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."

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.
�����������������������������������������������������������������������

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."