0
$\begingroup$

The Connes Window function is defined as :

w(f)=$(1-(\frac{f}{ \Delta f})^2)^2$ for $f<|\Delta f| $

w(f)=0 otherwise

The inverse fourier transform of this function can be analytically calculated to be a purely real valued function.

However performing IFFT using matlab gives me a complex function whith real and imaginary parts. The code I used in the following:

delta_f=10;

fs=300; 

nCyl = 5;

t=0:1/fs:nCyl*1/f;

x=(1-(t/delta_f).^2).^2;

plot(t,x);

NFFT=1024;  

X=fftshift(ifft(x,NFFT));       

fVals=fs*(-NFFT/2:NFFT/2-1)/NFFT;   

plot(fVals,real(X),'b'); 

I think this discrepancy has something to do with the function being piecewise. Is there something which I am missing?

$\endgroup$
  • $\begingroup$ is the last sample equal to the first sample? it shouldn't be. it should be cyclic. ifft([1,2,3,4,3,2,1]) is complex but ifft([1,2,3,4,3,2]) is real $\endgroup$ – endolith Aug 9 '15 at 5:35
  • $\begingroup$ no it still gives imaginary values.but why does that happen to the example you gave? shouldnt the same function same inverse fourier transform? $\endgroup$ – rsujatha Aug 9 '15 at 5:47
  • $\begingroup$ [1,2,3,4,3,2] is symmetrical, with the first sample occurring right after the last sample. So is [1,0,0,2,0,0], for instance. Anything of the form [a,b,c,d,e,d,c,b] will have a real transform. $\endgroup$ – endolith Aug 19 '15 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.