[Scilab-users] fourier series and fft

[Samuel Gougeon]

Le 22/09/2016 23:09, paul.carrico at free.fr a écrit :
> dear all
>
> I'm novice in Fourier series and other and my question is probably 
> naive (sorry for this) => I'm wondering if scilab can directly 
> calculate the Fourier coefficient a0, a_k and b_k ?
>
> I'm currently doing it "by hand" is order to familiarise myself with 
> it (and I'm looking at the same time to documents on  FFT use and 
> rules to refind the 2 natural frequencies of the example here bellow), 
> but it seems I'll need to code the coefficient calculations ... Am I 
> right ?
.
Doing that is the main purpose of fft().
fft() returns the (bilateral) complex coefficients c_k of the series.
a_k and b_k are easily computed from them.

// Example #1: n  =  100;
x  =  linspace(0,  1,  n+1);
x  =  x(1:n);
y  =  sin(2*%pi*x);
ft  =  fft(y);
ft  =  clean(ft);
c  =  ft/n;
c0  =  c(1)
// c(k) = ft(k+1)
// c(-k) = ft(n+1-k)

a0  =  c0 // average
a  =  (c(2:(n/2))  +  c(n:-1:(n/2+2)))      // a_k = c_k + c_(-k)
b  =  %i*(c(2:n/2)  -  c(n:-1:n/2+2)) // b_k = %i*(c_k - c_(-k))
// Example #2:
n = 100; x = linspace(0, 2, n+1); x = x(1:n); y  =  2*sin(2  *  %pi  *  x)  -  3*cos(%pi  *  x);
ft  =  fft(y);
ft  =  clean(ft);
c  =  ft/n;
c0  =  c(1)
a0  =  c0
a  =  (c(2:(n/2))  +  c(n:-1:(n/2+2)))
b  =  %i*(c(2:n/2)  -  c(n:-1:n/2+2))

--> a = (c(2:(n/2)) + c(n:-1:(n/2+2)))
  a  =
          column 1 to 22
   -3.   0.   0.   0.   0.   0.   0.  ...
  // here is the "-3" coeff of the first even cos() harmonic

--> b = %i*(c(2:n/2) - c(n:-1:n/2+2))
  b  =
          column 1 to 22
    0.   2.   0.   0.   0.   0.   0.  ...
  // here is the "3" coeff of the second odd sin() harmonic

Anyway, the full x-range is considered as the period.
If it is not actually the case, some artefacts appear in the results.
To avoid them, it is mandatory to sample /x/ over a multiple of its 
"natural" period.
Example #1: The period of sin(2*%pi*x) is 2*pi, this is why x must be 
sampled from 0 to 1.
Example #2: The period of 2*sin(2 * %pi * x) - 3*cos(%pi * x)is also 2*pi,
                       this is why x must be sampled from 0 to 2 (in the 
cos() argument).

HTH
Samuel

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