Here, I am attaching the complete works on how do I find the Fourier series expansion just in case it would help anyone who want to help me to make the subplot (see in problem 1).
Plotting Fourier Series Expansion
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I was given a problem of the Fourier series expansion of with a periodicity interval . I found that the Fourier series expansion of such function with given periodicity interval is . However, I am new to MATLAB and my knowledge is limited to simple plotting. Suppose that my answer is correct, how to subplot and its Fourier series expansion using loop that generates the n-order of the expansion? I searched in this community and actually find a really good plotting, but I don't understand how it works. Thanks in advance!
Accepted Answer
Paul
on 9 Mar 2023
Hi Hanif,
Maybe this will help.
First, the problem statement says that one period spans -1 < x <= 1 (I don't think it matter if using < or <=)
x = linspace(-1,1,100);
Instead of using your problem, let's do something simpler, like 1 - 2*sum(n=1:5,n*x)
fsum = 0*x; % initialize
for n = 1:5
fsum = fsum + n*x;
end
plot(x,1-2*fsum)
I think you can adapt that to your problem. Of course we can't compute an infinite sum, so you'll have to pick an upper bound for the sum over n.
3 Comments
Paul
on 9 Mar 2023
Edited: Paul
on 9 Mar 2023
Use plot inside the loop, with hold set to on. Also, the initial value for fsum was incorrect. And still need to define x over the correct interval
% x = linspace(-10,10,100); should be from -1 to 1
% fsum = 4*cos((pi*x)/(pi^2)); % initialize, should be 4*cos(pi*x)/pi^2;
% for n = 2:10
% fsum = fsum + 4*cos((2*n-1)*pi*x)/(pi^2*(2*n-1)^2);
% end
% y = 1-fsum;
% plot(x,y,'r-')
x = linspace(-1,1,100); % changed this line
fsum = 0*x; % initialize
figure;
hold on;
for n = 1:10 % start loop at 1
fsum = fsum + 4*cos((2*n-1)*pi*x)/(pi^2*(2*n-1)^2);
plot(x,1-fsum)
end
The sum seems like it's biased up by 0.5. Also, as your code showed, the reconstruction is not periodic with period = 2. Recheck derivation?
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