Hehe, taking a step further! 🙂
Originally posted by Atero
New DM: T goes to a constant of about 0.8, a better line would be something like this (this is just random-ness):
frame:
t1=t1+0.06;
t=1.1*sin(t1);
t2=t*cos(pow(t1,P=1.2));
beat:
t1=t1-1;
pixel:
d=(asin(1.05*d)-sigmoid(t2,t))/2;
Atero, I took the liberty to take a closer look on parts of the iterative function above. The variable t2 doesn't just oscilates. Because of the power term P="1.2" not being equal to "1" it has changing (over time) spectral signature. It is slightly noticible for values of "1.2", but higher values will change the spectral signature more rapidly. Try the matlab code below (if you have matlab, hehe):
--------------------------------------------------------
clear all;
t=0; t1=0; t2=0; i=1; P=1.6
t=t+0.06;
t1=1.1*sin(t);
t2=t1*cos(t^P);
while (i < 16384),
t(i+1)=t(i)+0.06;
t1(i+1)=1.1*sin(t(i));
t2(i+1)=t1(i)*cos(t(i)^P);
i=i+1; %simulating beat, mod(i,40) approximates 60 bmp
if mod(i,40)==0
t=t-1;
end
end
figure; plot(t);
figure; plot(t1);
figure; plot(t2);
f1=fft(t2(1: (end/2)));
figure; plot(abs(f1(1:end)));
f2=fft(t2(end/2:end));
figure; plot(abs(f2(1:end)));
f3=fft(t2);
figure; plot(abs(f3));
sound(t2);
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Try to set P to 1, 1.2, ..., 1.6, ..., 2.0. You can even listen to t2, if you don't want to do the fft-transform. Keep the loop spining 2^n (2^13 = 8192) times, then the fft runs faster, but you probably know that.
When P is 1, you have 1 harmonic in the t2, when t is increased t2 stops having a constant harmonic but becomes spectraly non-constant, t2 gets different spectral signatures depending when in time the spectra is estimated, just listen to t2 🙂
Originally posted by Atero
...Was that too much for everybody? 😛
My question is: ...Was THAT too much for everybody? 😛
/Spetz