how to fill a matrix without using loop in matlab?
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I want to find the coefficient of this fourier series without using loop.I mean filling an,bn and after that is it possible to plot hplot step by step without using loop
close all; clear; clc;
N = 6;
f = @(x) rectangularPulse(-1,1,x);
x = -2:0.001:2;
%2*p is the period
p = pi;
% the main function
plot(x,f(x),'LineWidth',2);
grid;
hold on;
grid minor;
xlim([-2 2]);
ylim([-0.1 1.1]);
x = linspace(-2,2,100).';
y = linspace(0,1,0.2);
a0 = (1/(2*p))*integral(f,-p,p);
an = zeros(1,N);
bn = zeros(1,N);
% calculate an and bn till N
for n=1:N
fan = @(x) rectangularPulse(-1,1,x).*cos((n*pi/p)*x);
an(1,n) = (1/p)*integral(fan,-p,p);
fbn = @(x) rectangularPulse(-1,1,x).*sin((n*pi/p)*x);
bn(1,n) = (1/p)*integral(fbn,-p,p);
end
% create the gif
for n = 1:N
An = an(:,(1:n));
Bn = bn(:,(1:n));
fs = a0 + sum(An.*cos((1:n).*x) + Bn.*sin((1:n).*x),2);
hPlot = plot(x,fs,'color','red','LineWidth',2);
drawnow;
if(n~=N)
delete(hPlot);
end
end
0 Comments
Answers (1)
Torsten
on 10 Oct 2022
Edited: Torsten
on 10 Oct 2022
If you want to plot the partial sums of the Fourier series, you will have to keep the last loop, I guess.
close all; clear; clc;
N = 6;
f = @(x) rectangularPulse(-1,1,x);
x = -2:0.001:2;
%2*p is the period
p = pi;
% the main function
plot(x,f(x),'LineWidth',2);
grid;
hold on;
grid minor;
xlim([-2 2]);
ylim([-0.1 1.1]);
x = linspace(-2,2,100).';
y = linspace(0,1,0.2);
a0 = (1/(2*p))*integral(f,-p,p);
an = 1/p * integral(@(x) f(x).*cos((1:N)*pi/p*x),-p,p,'ArrayValued',1);
bn = 1/p * integral(@(x) f(x).*sin((1:N)*pi/p*x),-p,p,'ArrayValued',1);
fs = a0 + sum(an.*cos((1:N).*x) + bn.*sin((1:N).*x),2);
plot(x,fs,'color','red','LineWidth',2);
3 Comments
Image Analyst
on 10 Oct 2022
How many iterations do you have? Billions? If you use for n = 1 : 6, the "overhead" time to do six iterations is negligible. Your computer will do 6 iterations in nanoseconds. The bottleneck is what's happening inside the loop, not the looping itself.
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