
Dear audio enthusiast,

    Here you have the Nitrogen  DSP  —  a crossover fertilizer module. This
DSP designed in effort to discover true behaviour of waves in gasous media.
After almost a twelve years of development, i'm happy to report with pretty
sure, that i have studied key processes affecting final pattern of sound.
    So, you about  to  listen to latest,  exceptionally truthfull crossfeed
design generation.
    
The final notes:

    1. Activate "Crossfeed Fertilizer" from available DSP processors list.

    2. Finetune settings to your preferences.  But it's very dubtful you'll
get  better  experience  since  default  math  and  settings  are carefully
calculated.  Main  constants  finetuning made possble by my insufficient of
confidence in math precision.

    3. Since Nitrogen operates on  a  quite finer level than one considered
"enough"  by  engineers  failed  to  use  Claude Shannon findings (*), it's
made a mandatory requirement to upsample standard  definition  material  to
at  least  a  170  KHz.

    4. I'm highly recommend use Muleteer DSP just after Nitrogen output. 

    5. Enjoy.

CYa, copah (aka Vladimir Kopjov)
mailto: copah@yandex.ru

*)  Shannon-Kotelnikov  sampling theorem states: «Bandlimited analog signal
can  be perfectly reconstructed from an infinite sequence of samples if the
sampling rate exceeds 2B samples per second.»
Note: INIFINITE sequence of samples. This has very simple example. Let sine
wave  with  frequency  F at 2F samplerate have [1,-1,1,-1,1,-1,1] sequence.
So,  if  we  sample  this  sine  phase-shifted  by pi()/4 radian, we'll get
[0.7,-0.7,0.7,-0.7,0.7,-0.7,0.7].  But  we'll get same results if we sample
previous  sine  attenuated by 0.7. This evidence demonstrates impossibility
to discriminate amplitude and phase at frequency F on live (finite) signal.
For  Compact  Disc this means unavoidable decoding errors, since nobody nor
want,  nor  able play infinite note. On the other hand if samplerate equals
4F,  sampled  data  can  be  reconstructed  by  as  little  as  two samples
(equivalent  to  the  only  sample  at  2F),  since first derivative can be
calculated  as  per  2F  sample  with  perfect alignment. This way, on a 2F
samplerate, bandwidth 0F to 1/2F is error free. And bandwidth 1/2F to F has
harmonic error rate: E'f(t) = f/Fs*t if done math right… ;-)
