how to plot fourier series in matlab
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how to plot fourier series in matlab
5 Kommentare
Abraham Boayue
am 16 Mär. 2018
Have you done the integrals to fined the a0, an and bn? If so, what is the expression you got for the fourier series?
Akira Agata
am 16 Mär. 2018
Bearbeitet: Akira Agata
am 16 Mär. 2018
Regarding the question (1) in the picture, I would recommend try to calculate by hand first, for your better understanding of Fourier transformation of periodic function.
Korosh Agha Mohammad Ghasemi
am 23 Sep. 2020
clear all;clc;
syms x n pi
% pi=3.14;
sum=0;
y=x %function you want
a0=(1/pi)*int(y,x,-pi,pi)
% for n=1:5
%finding the coefficients
an=(1/pi)*int(y*cos(n*x),x,-pi,pi)
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi)
sum=a0/2+(an*cos(n*x)+bn*sin(n*x))
% end
ezplot(x,y,[-pi,pi])
grid on;hold on;
ezplot(x,(sum+a0/2),[-pi,pi])
% https://www.instagram.com/koroshkorosh1/
syms x pi
F =(1/pi) * int(x^2+5*x,'Hold',true)
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi), G = bn
Gcalc = release(G)
Fcalc = int(bn)
clear all;clc;
syms x pi n
% pi=3.14;
sum=0;
y= x + (x^2) ; %function you want
a0=(1/pi)*int(y,x,-pi,pi)
% for n=1:3
%finding the coefficients
an=(1/pi)*int(y*cos(n*x),x,-pi,pi)
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi)
sum=sum+(an*cos(n*x)+bn*sin(n*x))
% end
% https://www.instagram.com/koroshkorosh1/
ezplot(x,y,[-3.14,3.14]);
grid on;hold on;
ezplot(x,(sum+a0/2),[-3.14,3.14])
% https://www.instagram.com/koroshkorosh1/
Antworten (6)
Abraham Boayue
am 18 Mär. 2018
Bearbeitet: Abraham Boayue
am 15 Jun. 2018

Here is what your Fourirer series would like if my calculations were made correctly. An attachment of the solution is also included for your reference. Take care for now.
clear variables
close all
% Fourier series of neither even nor odd function
% Decompose f(x) into even (fe) and odd (fo) functions.
% fe = (f(x) + f(-x))/2, fo = (f(x) - f(-x))/2
N = 500;
L = 4;
xd = -L:2*L/(N-1):L;
y1 = -1/8*xd.^2;
y2 = 1/8*xd.^2;
fo = y1.*(-L<=xd & xd<=0) +y2.*(0<=xd & xd<=L);
fe = 4-xd.^2/8;
f2 = fe + fo;
a0 = 10/3;
% Generate the fourier series of f(x)
y = zeros(1,N);
x = [];
K = 80;
for k = 1:K
ck = 1/(pi*k);
an = (2*L*(-1).^(k+1))*ck^2;
bn = L*(-1).^(k+1)*ck + (2*L*ck^3)*((-1)^k-1);
y = y + (an*cos(pi*k/L*xd)+ bn*sin(pi*k/L*xd)); % For fe and fo
x = [x;y];
end
y = a0 +y;
x = a0 +x;
% Color codes
s = distinguishable_colors(K); % search this function on mathworks
figure
subplot(121) % Plot f(t)
plot(xd,f2,'linewidth',2.5,'color',s(1,:))
line(xlim,[0 0],'color',s(6,:),'linewidth',3);
line([0 0],ylim,'color',s(6,:)','linewidth',3);
ylim([-.5 4]);
a= xlabel('\itt\rm (seconds)');
set(a,'fontsize',20);
a = ylabel('\itf\rm(\itt\rm)');
set(a,'fontsize',20);
a= title('f(t)');
set(a,'fontsize',14);
grid
subplot(122) % Plot fouries series of f(t);
hold on
q = length(x(:,1));
M = 1:q;
for i = 1:6:q
plot(xd,x(i,:),'linewidth',2.5,'color',s(i,:),'DisplayName',sprintf('S = %1.2f',M(i)))
end
a= title('Fourier series of f(t)');
set(a,'fontsize',14);
a= xlabel('\itt\rm (seconds)');
set(a,'fontsize',20);
a = ylabel('\itf\rm(\itt\rm)');
set(a,'fontsize',20);
line(xlim,[0 0],'color',s(6,:),'linewidth',3);
line([0 0],ylim,'color',s(6,:)','linewidth',3);
legend('-DynamicLegend','location','bestoutside');
grid
1 Kommentar
Korosh Agha Mohammad Ghasemi
am 22 Sep. 2020
clear all;clc;
syms x
pi=3.14;
sum=0;
y=exp(x); %function you want
a0=(1/pi)*int(y,x,-pi,pi);
for n=1:3
%finding the coefficients
an=(1/pi)*int(y*cos(n*x),x,-pi,pi);
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi);
sum=sum+(an*cos(n*x)+bn*sin(n*x));
end
% https://www.instagram.com/koroshkorosh1/
ezplot(x,y,[-pi,pi]);
grid on;hold on;
ezplot(x,(sum+a0/2),[-pi,pi]);
% https://www.instagram.com/koroshkorosh1/
Abraham Boayue
am 18 Mär. 2018
2 Stimmen
The is the solution file, the math is a bit messy, but I assume that you are familiar with the material that you are studying.
1 Kommentar
Rik
am 7 Mär. 2022
this guy is great
Abhishek Ballaney
am 16 Mär. 2018
0 Stimmen
https://in.mathworks.com/help/curvefit/fourier.html
2 Kommentare
Korosh Agha Mohammad Ghasemi
am 22 Sep. 2020
clear all;clc;
syms x
pi=3.14;
sum=0;
y=exp(x); %function you want
a0=(1/pi)*int(y,x,-pi,pi);
for n=1:3
%finding the coefficients
an=(1/pi)*int(y*cos(n*x),x,-pi,pi);
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi);
sum=sum+(an*cos(n*x)+bn*sin(n*x));
end
% https://www.instagram.com/koroshkorosh1/
ezplot(x,y,[-pi,pi]);
grid on;hold on;
ezplot(x,(sum+a0/2),[-pi,pi]);
% https://www.instagram.com/koroshkorosh1/
vikrant rana
am 26 Jan. 2022
hey abhishek,
what would be the changes in code if y=(pi-x)/2
and the limits are from 0 to 2 pi
like i am trying to make changes in the code by substituiting my values i not happening.
i shall be thankful to you if you resolve my doubt.
Mohamed Abugammar
am 10 Apr. 2019
0 Stimmen
clc;
close all;
clear all;
dx=0.001;
L=pi;
x=(-1+dx:dx:1)*L;
n=length(x); nquart=floor(n/4);
% define the hat function;
f=0*x;
f(nquart:2*nquart)=4*(1:nquart+1)/n;
f(2*nquart+1:3*nquart)=1-4*[1:500]/n;
plot(x, f,'r','LineWidth', 2); hold on;
%% define the coffeciet
A0=sum(f.*ones(size(x)))*dx;
fFS = A0/2;
for k=1:10
Ak=sum(f.*cos(pi*k*x/L))*dx;
Bk=sum(f.*sin(pi*k*x/L))*dx;
fFS=fFS+Ak*cos(pi*k*x/L)+Bk*sin(pi*k*x/L);
plot(x,fFS);
pause(1); drawnow;
end
1 Kommentar
Korosh Agha Mohammad Ghasemi
am 22 Sep. 2020
clear all;clc;
syms x
pi=3.14;
sum=0;
y=exp(x); %function you want
a0=(1/pi)*int(y,x,-pi,pi);
for n=1:3
%finding the coefficients
an=(1/pi)*int(y*cos(n*x),x,-pi,pi);
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi);
sum=sum+(an*cos(n*x)+bn*sin(n*x));
end
% https://www.instagram.com/koroshkorosh1/
ezplot(x,y,[-pi,pi]);
grid on;hold on;
ezplot(x,(sum+a0/2),[-pi,pi]);
% https://www.instagram.com/koroshkorosh1/
Dhiya Eid
am 20 Jul. 2020
0 Stimmen
Let f(x) be a 2π-periodic function such that f(x)=x2 for x∈[−π,π]. Find the Fourier series for the parabolic wave.
solve it in matlab
2 Kommentare
Deniz Sezen
am 28 Apr. 2023
Could you solve that question?
Rik
am 29 Apr. 2023
This is an 'answer' from almost 3 years ago. I doubt you will get a response.
Korosh Agha Mohammad Ghasemi
am 22 Sep. 2020
f=@(x)x.*(x>0 & x<-pi)-2*(x/pi+1).*(x>=-pi & x<=-pi/2);
n=50;
k=0:n;
a=1/pi*(integral(@(x)f(x).*cos(k*x),-pi,-pi/2,'ArrayValued',true)+integral(@(x)f(x).*cos(k*x),0,pi/2,'ArrayValued',true));
k=1:n;
b=1/pi*(integral(@(x)f(x).*sin(k*x),-pi,-pi/2,'ArrayValued',true)+integral(@(x)f(x).*sin(k*x),0,pi/2,'ArrayValued',true));
ffun=@(x)a(1)/2+sum(a(2:n+1).*cos((1:n)*x)+b(1:n).*sin((1:n)*x));
x=linspace(0,pi,200);
fx=arrayfun(@(x)ffun(x),x);
plot(x,fx,x,f(x))
% https://www.instagram.com/koroshkorosh1/
3 Kommentare
Korosh Agha Mohammad Ghasemi
am 22 Sep. 2020
clear all;clc;
syms x
pi=3.14;
sum=0;
y=exp(x); %function you want
a0=(1/pi)*int(y,x,-pi,pi);
for n=1:3
%finding the coefficients
an=(1/pi)*int(y*cos(n*x),x,-pi,pi);
bn=(1/pi)*int(y*sin(n*x),x,-pi,pi);
sum=sum+(an*cos(n*x)+bn*sin(n*x));
end
% https://www.instagram.com/koroshkorosh1/
ezplot(x,y,[-pi,pi]);
grid on;hold on;
ezplot(x,(sum+a0/2),[-pi,pi]);
% https://www.instagram.com/koroshkorosh1/
taha bandrawala
am 22 Sep. 2020
how to write if i want to solve f(x)=x^3 and find the fourier series of f(x)
Gülcan söm
am 30 Dez. 2020
how to write if i want to solve f(x)=cos(3x) and find the fourier series of f(x)
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