how can i perform a Fourier series on this function?

조회 수: 22 (최근 30일)
Mohammad Adeeb
Mohammad Adeeb 2018년 3월 17일
댓글: Mohammad Adeeb 2018년 3월 20일
clear all;
close all;
clc;
syms n t1 t2 T f0 t;%t2=T;
func=(f0/t2-t1)*(t-t1);
n=1:10;
syms t t1 T;%T=t2;
w = (2*pi)/T;
a0 = (2/T)*int(func,t,0,T);
an = (2/T)*int(func*cos(n*w*t),t,0, T);
bn = (2/T)*int(func*sin(n*w*t),t, 0,T);
f = a0/2 + dot(an,cos(n*w*t)) + dot (bn, sin(n*w*t));
ezplot(func);
hold on;
grid on;
plot(f);

채택된 답변

Abraham Boayue
Abraham Boayue 2018년 3월 20일
Here is the full solution to your problem. Good luck.

추가 답변 (2개)

Abraham Boayue
Abraham Boayue 2018년 3월 19일
Hey Mahammed, instead of calculating the Fourier series like you did in matlab, why don't you calculate the coefficients by hand? What is the expression for f(x) and the values of f0 and t1?
  댓글 수: 1
Mohammad Adeeb
Mohammad Adeeb 2018년 3월 19일
편집: Mohammad Adeeb 2018년 3월 19일
F(x)=(fo/t2-t1)*(t-t1)
F0=20
t1=2
T=t2=4

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Abraham Boayue
Abraham Boayue 2018년 3월 20일
Mohammed, if you calculate the coefficients of the Fourier series of f(t), you will get a0 and an equal to zero. This is because f(t) is an odd function, only bn has value. The following code implements your equation.
clear variables
close all
N = 1000;
%%1. Generate and plot f(x)
T = 4;
t1 = 2;
t2 = T;
fo = 20;
t = -T:2*T/(N-1):T;
f = (fo/(t2-t1))*(t-t1);
figure
plot(t,f,'linewidth',1.5,'color','m')
grid;
a = title('f(x)');
set(a,'fontsize',14);
a = ylabel('y');
set(a,'Fontsize',14);
a = xlabel('x (-4 4]');
xlim([-4 4]);
set(a,'Fontsize',14);
%%Define variable for F series
to = -16;
tf = 16;
t2 = to:(tf-to)/(N-1):tf;
F = zeros(1,N);
F1 = F;
F2 = F1;
% This is just a vectorized version of the for loop. It's a faster way
% of coding the F-series.
m = 1:N;
bn = (T^2./(pi*m)).*(-1).^(m+1);
F1000 = bn*sin((pi/T)*m'*t2); % 1000th Harmonic
% Gererate the F-series with a for loop, flexible.
% 5th Harmonic
for n = 1:5
an = 0;
bn = (T^2/(n*pi))*(-1)^(n+1);
w = (pi*n)/T;
F = F + bn*sin(w*t2);
end
% 15th Harmonic
for n = 1:15
an = 0;
bn = (T^2/(n*pi))*(-1)^(n+1);
w = (pi*n)/T;
F1 = F1 +bn*sin(w*t2);
end
% 25th Harmonic
for n = 1:25
bn = (T^2/(n*pi))*(-1)^(n+1);
w = (pi*n)/T;
F2 = F2 + bn*sin(w*t2);
end
figure
plot(t2,F,'linewidth',1.5,'color','r')
hold on
plot(t2,F1,'linewidth',1.9,'color','g')
plot(t2,F2,'linewidth',1.5,'color','m')
plot(t2,F1000,'linewidth',1.8,'color','b')
grid;
a = title('Fourier Series of f(x) 5th, 15th , 25th and 1000th Harmonics');
legend('F5','F15','F25','F1000');
set(a,'fontsize',14);
a = ylabel('y');
set(a,'Fontsize',14);
a = xlabel('t [-16 16]');
xlim([to tf]);
set(a,'Fontsize',14);

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