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 Commenti
Risposte (1)
Jan Orwat
il 27 Giu 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 Commento
Jan Orwat
il 27 Giu 2016
Modificato: Jan Orwat
il 27 Giu 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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