Evaluating the integral of a function?
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
ROSHAN SINGH
le 18 Août 2015
Commenté : ROSHAN SINGH
le 19 Août 2015
I want to evaluate integral of (Hr(w)*cos(n*w*t)-Hi(w)sin(n*w*t))dw with limits 1 to 2 where t=0.2 and T=2000. where Hr(w) is the real part of a function H(w)=10*(((i*w)/0.006)-1)*((-(w^2)/0.64)+(((2*(i*w))/0.8)+1))*(((i*w)/10)-1)*((-(w^2)/2250000)+(((2*(i*w))/1500)+1))/((((w^2)/0.000004)+(((2*(i*w))/0.002)+1))*((-(w^2)/49)+(((2*(i*w))/7)+1))*((-(w^2)/400000000)+(((2*(i*w))/20000)+1))*(((i*w)/100)-1))? And w is a vecor from [0.0005 to 5].
I am trying the following codes but the integral is not happening and instead it shows the error Undefined function 'int' for input arguments of type 'double
w=[0.0005:0.1:5]; n=1:length(w); t=0.1; T=2000;
for j=1:length(w) y(j)=10*(((i*w(j))./0.006)-1)*((-(w(j).^2)./0.64)+(((2*(j.*w(j)))./0.8)+1))*(((j.*w(j))./10)-1)*((-(w(j).^2)./2250000)+(((2*(j.*w(j)))./1500)+1))/((((w(j).^2)./0.000004)+(((2*(j.*w(j)))./0.002)+1))*((-(w(j).^2)./49)+(((2*(j.*w(j)))./7)+1))*((-(w(j).^2)./400000000)+(((2*(j.*w(j)))./20000)+1))*(((j.*w(j))./100)-1)); rr(j)=real(y(j)) ii(j)=imag(y(j)) xc(j)=cos(n(j)*w(j)*t)*rr(j) xs(j)=sin(n(j)*w(j)*t)*ii(j) ad(j)=xc(j)-xs(j) ri(j)=int(ad(j),w,1,2) end
Please help me and even let me know is my method of finding the integral of the function correct?
0 commentaires
Réponse acceptée
Steven Lord
le 18 Août 2015
Use INT to integrate an expression written using symbolic variables created using SYM or SYMS from Symbolic Math Toolbox.
Use INTEGRAL to integrate a function whose handle (which could be an anonymous function) you specify as the first input.
Use TRAPZ to integrate a function represented by a set of (X, Y) data.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Numerical Integration and Differentiation dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!