calculating fourier series for the given functions and drawing the curve?
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
hey guys, one of my friends has just given me a question.I wrote the code for it but don't know how to draw the chart or curve for that,would you please help me out...??thanks
here's the code: clc syms x f=exp(-abs(x)); N=15; a_0=(1/pi)*int(f*cos(0*x),-pi,pi) for n=1:N a_n(n)=(1/pi)*int(f*cos(n*x),-pi,pi); b_n(n)=(1/pi)*int(f*sin(n*x),-pi,pi); end a_n b_n f_new=a_0/2; for n=1:N f_new=f_new+a_n(n)*cos(n*x)+b_n(n)*sin(n*x); end
0 Kommentare
Antworten (1)
VBBV
am 17 Sep. 2024
clear
syms x
N=[5 15 30];
f1=exp(-abs(x));
a_01=(1/pi)*int(f1.*cos(0*x),x,-pi,pi);
f_new10=vpa(a_01/2)
f2=exp(-(x));
a_02=(1/pi)*int(f2.*cos(0*x),x,-2,1);
f_new20=vpa(a_02/2)
f3 = -2*x.^2;
a_03=(1/pi)*int(f3.*cos(0*x),x,1,2);
f_new30=vpa(a_03/2)
f4 = 4;
a_04=(1/pi)*int(f4.*cos(0*x),x,2,pi);
f_new40=vpa(a_04/2)
X = linspace(-pi,pi,100);
for k = 1:length(X)
for n=1:N(2)
a_n1(n)=(1/pi)*int(f1.*cos(n*x),x,-pi,pi);
b_n1(n)=(1/pi)*int(f1.*sin(n*x),x,-pi,pi);
f_new1(n,k)=f_new10+a_n1(n).*cos(n*X(k))+b_n1(n).*sin(n*X(k));
a_n2(n)=(1/pi)*int(f2.*cos(n*x),x,-2,1);
b_n2(n)=(1/pi)*int(f2.*sin(n*x),x,-2,1);
f_new2(n,k)=f_new20+a_n2(n).*cos(n*X(k))+b_n2(n).*sin(n*X(k));
a_n3(n)=(1/pi)*int(f3.*cos(n*x),x,1,2);
b_n3(n)=(1/pi)*int(f3.*sin(n*x),x,1,2);
f_new3(n,k)=f_new30+a_n3(n).*cos(n*X(k))+b_n3(n).*sin(n*X(k));
a_n4(n)=(1/pi)*int(f4.*cos(n*x),x,2,pi);
b_n4(n)=(1/pi)*int(f4.*sin(n*x),x,2,pi);
f_new4(n,k)=f_new40+a_n4(n).*cos(n*X(k))+b_n4(n).*sin(n*X(k));
end
F_new(k) = sum(f_new1(:,k)); % similarly do it for remaining functions
end
plot(F_new)
0 Kommentare
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!