Using ifft to get the Fourier Coefficient

What exactly does the ifft() gives me?

I have a real data in 'x' where,

f=summation over -N to N-1 [C(n)exp(2*pi*1i*x/L)]

So, here f is known at every point (2N points in total). fftshift(ifft(f)) also gives an array of 2N size. So, does it gives me the coefficients C(n). If so, then can you please check the following code.

N=256;
X=2*N;
L=2*pi;
x=linspace(-pi,pi,X);
c=0;
for n=1:2*N
      k(n)=2*pi*(n-N-1)/L;        
end
y=x; 
z=fftshift(ifft(y));
for i=1:2*N
  c=c+z(i)*exp(1i*k(i)*x);
end
plot(x,y);hold on;plot(x,c);

Here, if ifft() gave the coefficients, then shouldn't the plots have matched?

[Jan Orwat]

What is this and where is c?

y=x/pi;
...
plot(x,y);hold on;plot(x,c);

Why/How do you use loop here? i? n?

for i=1:2*N
F=F+z(n)*exp(1i*k(n)*x);
end

Why like that?

z=fftshift(ifft(y));

[Jan Orwat]

Documentation

[Raunak Raj]

Hi, I am sorry, that wasn't the intended code. I have edited the code correctly. fftshift is just to shift the values from 0 to 2N to -N to N-1 frequencies (actually wave numbers). Moreover, the code works as intended for y=sin(x) but for other functions there appears a shift in the graphs.

Answer 1

[Jan Orwat]

N=256;
X=2*N;
L=2*pi;
x=linspace(-pi,pi,X);
c=0;
k = 2*pi*((1:2*N)-N-1)/L; % vectorised
y = sin(x); % don't understand why it is here, why not defined earlier
z = ifftshift(ifft(y)); % would be more logical to use fft here
for i=1:2*N
  c=c+z(i)*exp(1i*k(i)*(pi-x));
end
plot(x,y);hold on;plot(x,real(c));

[Jan Orwat]

I'm still not sure why you calculate ifft of signal, then dft of ifft and compare with original signal. From mathematical point of view it makes no difference, because y, ifft(fft(y)) and fft(ifft(y)) are equal (within numerical precision), but it's logically weak.