Sum of fourier series:

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Real Name
Real Name on 6 Jan 2017
Commented: Ale gutierrez on 21 Jun 2018
I just have a quick question. How do I enter the formula shown above? I can't seem to get n = 1, 3, 5, ... I tried different code along the line of this:
syms k x
x = 1/2 + symsum(k^2, k, [1:2:15])
But there is an error message.
Also, on a side note, what's the purpose of syms and the "x" variable shown in the matlab documentation example where I got this code?

Accepted Answer

John BG
John BG on 6 Jan 2017
Edited: John BG on 6 Jan 2017
the 1st part of Walter's answer has syntax error:
  • it's most likely that t will have a higher amount of elements than vector n
  • therefore sin(n.*pi.*t) will always find one or more terms (of n) with a numel(n) href = ""</a> numel(t) mismatch
returning error, from the attempted d.*t
Let me suggest the following:
for n = 1:2:15
x=x+ sin(n*pi*t)/n;
plot(t,x);grid on
Increase L to improve signal time resolution.
if you find these lines useful would you please mark my answer as Accepted Answer?
To any other reader, if you find this answer of any help please click on the thumbs-up vote link,
thanks in advance for time and attention
John BG
Ale gutierrez
Ale gutierrez on 21 Jun 2018
I don't understand why you divide x by./2

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More Answers (1)

Walter Roberson
Walter Roberson on 6 Jan 2017
Easiest is to use a definite sum:
n = 1:2:15;
x = 1/2 + sum( 2./(pi*n) .* sin(n.*pi.*t) );
But you could use
syms n t
x = 1/2 + symsum( 2./(pi*(2*n-1)) .* sin((2*n-1).*pi.*t), n, 1, 8 );
Walter Roberson
Walter Roberson on 7 Jan 2017
For vector t, if you are using R2016b or later,
n = (1:2:15).';
x = 1/2 + sum( 2./(pi*n) .* sin(n.*pi.*t) );
For previous versions,
n = (1:2:15).';
x = 1/2 + sum( bsxfun(@times, 2./(pi*n), sin( bsxfun(@times, n.*pi, t) ) ) );

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