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 件のコメント
回答 (1 件)
Torsten
2022 年 10 月 10 日
編集済み: Torsten
2022 年 10 月 10 日
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 件のコメント
Image Analyst
2022 年 10 月 10 日
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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