Hello Samuel,
Sorry I might not have made myself clear: fft and fftshift provide the
ability to perform transform along only one of the dimensions of a
multidimensional array.
Something like S(x,y,z) --[FFT along 3rd dim]--> ffthift(ffft(S(x,y,z),
-1,3),3)=Ŝ(x,y,kz).
In that case, you need to perform fft and eventually fftshift along only
the dimension of the transform.
ifftshift should also provide the same possibility to perform the
inverse transform: Ŝ(x,y,kz) --[IFFT along 3rd dim]-->
iffthift(ffft(Ŝ(x,y,kz), +1,3),3)=S(x,y,z).
This is a basic signal processing requirement in my field.
Cheers,
Antoine
Le 09/10/2018 à 16:05, [email protected] a écrit :
Hello Antoine,
fttshift() switches all halves along all dimensions. This is mandatory.
So for instance in 2D, opposite quadrants (wrt the center) are switched.
In 3D, opposite cubes are switched. etc.
To me, the current implementation is right:
--> m=grand(3,5,"uin",0,9)
m =
2. 5. 9. 3. 0.
2. 4. 5. 5. 6.
4. 1. 8. 5. 9.
--> fftshift(m)
ans =
5. 9. 4. 1. 8.
3. 0. 2. 5. 9.
5. 6. 2. 4. 5.
--> ifftshift(fftshift(m))
ans =
2. 5. 9. 3. 0.
2. 4. 5. 5. 6.
4. 1. 8. 5. 9.
--> and(ifftshift(fftshift(m))==m)
ans =
T
BR
Samuel
----- Mail d'origine -----
De: antoine monmayrant+scilab <[email protected]>
À: International users mailing list for Scilab. <[email protected]>, List
dedicated to development questions <[email protected]>
Envoyé: Tue, 09 Oct 2018 11:02:20 +0200 (CEST)
Objet: [Scilab-users] fftshift and ifftshift are way too different
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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