Borrar filtros
Borrar filtros

Integrate a polyfit-polyval in matlab, I would appreciate any input. I want to calculate the integral of the sine function from a trend li

5 visualizaciones (últimos 30 días)
X=[ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 ];
Y=[ 0 0.017451121 0.034896927 0.052332105 0.069751344 0.08714934 0.104520792 0.121860412 0.139162917 0.156423038 0.173635518 0.190795114 0.207896601 0.224934771 0.241904433 0.258800419 0.275617584 0.292350805 0.308994987 0.32554506 0.341995983 0.358342746 0.374580371 0.390703911 0.406708457 0.422589134 0.438341105 0.453959573 0.469439781 0.484777014 0.4999666 0.515003915 0.529884377 0.544603456 0.559156667 0.573539579 0.587747811 0.601777035 0.61562298 0.629281427 0.642748218 0.65601925 0.669090481 0.681957931 0.694617681 0.707065874 0.719298721 0.731312494 0.743103535 0.754668253 0.766003125 0.7771047 0.787969596 0.798594504 0.808976189 0.819111488 0.828997314 0.838630657 0.848008582 0.857128234 0.865986835 0.874581687 0.882910172 0.890969753 0.898757976 0.90627247 0.913510945 0.920471196 0.927151104 0.933548635 0.939661839 0.945488856 0.95102791 0.956277314 0.96123547 0.965900868 0.970272086 0.974347793 0.978126748 0.9816078 0.984789889 0.987672046 0.990253392 0.992533143 0.994510602 0.996185169 0.997556332 0.998623675 0.999386873 0.999845692 0.999999993 ];
plot(X,Y)
p2=polyfit(X,Y,4);
p3=polyfit(X,Y,3);
hold on
c=polyval(p2,X);
plot(X,c)
FS=7;
leg = legend(["Seno","Aproximacion Seno"],'Location','northwest','Orientation','horizontal','FontSize',FS,"NumColumns",1);
x1 = 12;
x2 = 12;
leg.ItemTokenSize = [x1, x2];
legend('boxoff');
hold off

Respuesta aceptada

KSSV
KSSV el 11 de Ag. de 2022
%% Integration
p2=polyfit(X,Y,4);
fun = @(x) p2(1)*x.^4+p2(2)*x.^3+p2(3)*x.^2+p2(4)*x+p2(5) ; % using p2 from polyfit
val = integral(fun,X(1),X(end))
  1 comentario
Juan David Parra Quintero
Juan David Parra Quintero el 11 de Ag. de 2022
Note that the integral must give 1 between 0 and 90 degrees of the sine function, but with the code it gives 57.2. Do you know what error is happening?
X=[ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 ];
Y=[ 0 0.017451121 0.034896927 0.052332105 0.069751344 0.08714934 0.104520792 0.121860412 0.139162917 0.156423038 0.173635518 0.190795114 0.207896601 0.224934771 0.241904433 0.258800419 0.275617584 0.292350805 0.308994987 0.32554506 0.341995983 0.358342746 0.374580371 0.390703911 0.406708457 0.422589134 0.438341105 0.453959573 0.469439781 0.484777014 0.4999666 0.515003915 0.529884377 0.544603456 0.559156667 0.573539579 0.587747811 0.601777035 0.61562298 0.629281427 0.642748218 0.65601925 0.669090481 0.681957931 0.694617681 0.707065874 0.719298721 0.731312494 0.743103535 0.754668253 0.766003125 0.7771047 0.787969596 0.798594504 0.808976189 0.819111488 0.828997314 0.838630657 0.848008582 0.857128234 0.865986835 0.874581687 0.882910172 0.890969753 0.898757976 0.90627247 0.913510945 0.920471196 0.927151104 0.933548635 0.939661839 0.945488856 0.95102791 0.956277314 0.96123547 0.965900868 0.970272086 0.974347793 0.978126748 0.9816078 0.984789889 0.987672046 0.990253392 0.992533143 0.994510602 0.996185169 0.997556332 0.998623675 0.999386873 0.999845692 0.999999993 ];
plot(X,Y)
p2=polyfit(X,Y,4);
hold on
time= linspace(0,90,90);
c=polyval(p2,time);
plot(time,c)
FS=7;
leg = legend(["Seno","Aproximacion Seno"],'Location','northwest','Orientation','horizontal','FontSize',FS,"NumColumns",1);
T1 = 12;
T2 = 12;
leg.ItemTokenSize = [T1, T2];
legend('boxoff');
hold off
fun = @(x) p2(1)*x.^4+p2(2)*x.^3+p2(3)*x.^2+p2(4)*x+p2(5) ; % using p2 from polyfit
val = integral(fun,X(1),X(end));

Iniciar sesión para comentar.

Más respuestas (1)

Walter Roberson
Walter Roberson el 11 de Ag. de 2022
https://www.mathworks.com/help/matlab/ref/polyint.html
polyint() the component vector to get the integral of the polynomial, and polyval to evaluate.
  1 comentario
Walter Roberson
Walter Roberson el 12 de Ag. de 2022
You are not integrating sin(x), you are integrating sind(x). sind(x) = sin(alpha*x) for alpha = π/180
syms alpha x
int(sin(alpha*x), x, 0, sym(pi)/2/alpha)
ans = 1/alpha
with alpha being π/180, then 1/alpha is 180/π which is 57 point something

Iniciar sesión para comentar.

Categorías

Más información sobre Mathematics en Help Center y File Exchange.

Productos


Versión

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by