Using ifft to get the Fourier Coefficient
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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?
3 comentarios
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Jan Orwat
el 27 de Jun. de 2016
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));
1 comentario
Jan Orwat
el 27 de Jun. de 2016
Editada: Jan Orwat
el 27 de Jun. de 2016
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.
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