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x=0:1:255;y=[0.266,0.379,0.336,0.106,0.689,0.118,0.587,0.715,0.509,0.814,1.902,1.495,1.389,1.377,1.263,1.4, 1.62,1.902,2.05,1.943,1.502,1.886,1.454,2.366,2.304,2.665,2.792,2.406,2.239,2.991,2.601,2.884,2.719,3.587,3.724,3.254,3.1 87,3,623,3,642,2,92,4,124 , 3.595,3.203,4.223,4.249,4.683,3.942,3.863,3.982,3.148,4.116,4.454,3.853,4.139,4.835,3.934,4.99,3.698,3.971,3.777,4.71,4. 2 86,3,946,4,545,4,153, 4.461, 3.94,3.168,3.966,3.996,4.247,3.482,4.44,3.443,3.051,3.518,3.421,2.93,3.134,2.777,1.94,1.992,2.265,2.543,1.851,2.089 ,1.455 ,1.43,1.677,0.874, , - 2.152, -2.075,-1.65,-1.692,-1.872,-2.715,-2.87,-2.626,-2.385,-2.819,-1.603,-2.425,-2.863,-1.378,-2.204,-1.629,-2.291, -1.658,-1.267,-1.454,-1.129,-0.903,-0.497,-0.52,-0.756,-0.434,0.144,-0.017,0.321,0.246,0.687,0.594,0.826,0.667,0.764,1.452 ,1,55 3 ,1.965, 1.569,1.705,2.16,2.351,1.997,1.66,1.579,1.829,1.605,1.911,1.261,1.939,1.199,0.21,1.25,0.424,0.646,0.515,0.15,-0.13 1,0,328 ,-0,439,- 1.086,- 1.006,-1.221,-0.668,-1.277,-1.58,-0.935,-1.588,-2.43,-1.087,-1.776,-1.343,-0.981,-2.258,-1.614,-1.592,-0.993, -0.808, -0.608,-0.633,-0.67,-0.514,0.178,0.179,-0.251,0.241,0.865,1.049,0.573,0.839,0.932,1.809,1.565,1.238,1.834,1.08,0.883 ,2,225 ,1,411, 1.019,1.057 ,0.28,-0.023,-0.173,0.035,-0.154,-0.663,-0.054,-0.435,-0.531,-0.497,-1.165,-0.999,-0.533,-1.38,-1.737,0.026,- 0,882,- 1,392,-0,757,-0,857,-1,212,-0,247,-0,2

84,0.592,0.26,-0.338,-0.602,-0.004,0.065,1.068,0.906,0.975,1.142,0.836,1.854,0.812,1.247,1.082,0.662,0.862,0.887,0.966,1. 431,0,482,0,475 ,- 0.538];

X=fft(y);

umbral=max(abs(X))/10;

Y=X.* (abs (X)> umbral);

yy=ifft(Y);

plot(x,y,x,yy,'r')

% YY es velocidad

%yy=delta(YY)/delta(XX)

%delta(YY)=yy*delta(XX)

%YY2 -YY1=yy(1)*(XX2-XX1)

%YY2=YY1+yy(1)*(XX2-XX1)=YY1+yy(1)

YY (1)=yy(1);

para k=1:longitud(yy)-1

YY(k+1)=YY(k)+yy( k)

final

figura

plotyy(x,yy,x,YY)

prueba=diff(YY)-yy (1 :end-1) Calcule la derivada de %diff(YY) para obtener la aceleración, réstela de yy, la diferencia es 1.0e-013*, lo que indica que el método es correcto.