how to plot fourier series in matlab

how to plot fourier series in matlab

5 commentaires

Abraham Boayue
Abraham Boayue le 16 Mar 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
Akira Agata le 16 Mar 2018
Modifié(e) : Akira Agata le 16 Mar 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.
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/
Salihath
Salihath le 15 Déc 2022
Déplacé(e) : Rik le 16 Déc 2022
how to write matlab code for time modulated array
Rik
Rik le 16 Déc 2022
@Salihath Since your post was not an answer, I moved it to the comment section. Have a read here and here. It will greatly improve your chances of getting an answer.

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Réponses (6)

Abraham Boayue
Abraham Boayue le 18 Mar 2018
Modifié(e) : Abraham Boayue le 15 Juin 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 commentaire

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/

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Abraham Boayue
Abraham Boayue le 18 Mar 2018

2 votes

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.
Abhishek Ballaney
Abhishek Ballaney le 16 Mar 2018

0 votes

https://in.mathworks.com/help/curvefit/fourier.html

2 commentaires

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
vikrant rana le 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.

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Mohamed Abugammar
Mohamed Abugammar le 10 Avr 2019

0 votes

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 commentaire

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/

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Dhiya Eid
Dhiya Eid le 20 Juil 2020

0 votes

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 commentaires

Deniz Sezen
Deniz Sezen le 28 Avr 2023
Could you solve that question?
Rik
Rik le 29 Avr 2023
This is an 'answer' from almost 3 years ago. I doubt you will get a response.

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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 commentaires

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
taha bandrawala le 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
Gülcan söm le 30 Déc 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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Rik
le 29 Avr 2023

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