[Scilab-Dev] fftshift and ifftshift are way too different
antoine.monmayrant+scilab at laas.fr
antoine.monmayrant+scilab at laas.fr
Tue Oct 9 11:02:20 CEST 2018
Hello all,
From what I understand of fast Fourier transforms, fftshift and
ifftshift should be almost identical.
The only difference is when there are an odd number of elements in the
dimension along which the shift is performed (ie
fftshift([1:4])==ifftshift([1:4]) but fftshift([1:5])!=ifftshift([1:5])).
However, it seems that in scilab6.x fftshift and ifftshift are based on
completely different codes and do not accept the same arguments.
In particular, with fftshift, one can specify the dimension along which
to perform the shift, while it is not the case for ifftshift.
I think some update of ifftshift would be welcome.
I propose to use the code below (myifftshift), where I just changed
"ceil" by "floor" in the definition if fftshift.
What do you think, does it look right to you?
Cheers,
Antoine
function x = myifftshift(x,job)
if argn(2)<2 then job="all",end
deff("sel=fun(sk)","c=floor(sk/2);sel=[c+1:sk,1:c]")
if job=="r" then job=1,elseif job=="c" then job=2,end
ind=list()
if job=="all" then
for sk=size(x),ind($+1)=fun(sk),end
else
for sk=size(x),ind($+1)=:,end;
ind(job)=fun(size(x,job))
end
x=x(ind(:))
endfunction
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