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Topic: fftshift
Replies: 4   Last Post: Jan 5, 2013 12:08 AM

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maryam

Posts: 17
Registered: 12/25/12
Re: fftshift
Posted: Dec 28, 2012 11:58 AM
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"Greg Heath" <heath@alumni.brown.edu> wrote in message <kbinv7$4ev$1@newscl01ah.mathworks.com>...
> "maryam" wrote in message <kbie5v$29n$1@newscl01ah.mathworks.com>...
> > Hi dear friends, thanks for your reading
> > I would like to know when a vector is transformed to Fourier domain using fftshift(fft(fftshift(s))),
> > S=[t1 t2 t3 t4 ?. tn] & sampling frequency=Fr
> > result vector fftshift(fft(fftshift(s))) is equal to [0 ?.. Fr] & sampling=Fr/n or [-Fr/2 ?.. Fr/2]?

>
> If time is bipolar and xb is defined over
>
> tb = dt*[ -(N-1)/2 : (N-1)/2 ] for N odd
> tb = dt*[ -N/2 : N/2-1 ] for N even
>
> then
>
> x = ifftshift(xb);
>
> is defined over unipolar time
>
> t = dt*[ 0 : N-1];
>
> Furthermore,
>
> X = fft(x);
>
> is defined over unipolar frequency
>
> f = df * [ 0 : N-1 ]:
>
> and
>
> Xb = fftshift(X);
>
> is defined over bipolar frequency
>
> fb = df*[ -(N-1)/2 : (N-1)/2 ]; for N odd
> fb = df*[ -N/2 : N/2-1 ]; for N even
>
> When N is even, fftshift and ifftshift are equal. In general, however, they are inverses.
> Therefore, in general,
>
> Xb = fftshift(fft(ifftshift(xb));
>
> and
>
> xb = fftshift(ifft(ifftshift(Xb)));
>
> Alternate forms of the time and frequency intervals can be obtained by using the
> the following dual relationships of time period T and sampling frequency Fs:
>
> Fs = 1/dt, T = 1/df
>
> dt = T/N, df = Fs/N
>
> t = 0: dt : T - dt;
>
> f = 0: df : Fs - df
>
> I will let you obtain the corresponding forms for tb and fb;
>
> Hope this helps.
>
> Greg

----------------------------------------------------------------
Hi thanks a lot for your reply
To express clearly what I mean, I would explain my question by an example;
Assume that we have S matrix as bellow:

ti=1.44e-4; to=1.48e-04; dt=1e-9;
m=2*ceil(.5*(to-ti)/dt);
n=linspace(0,2,600);
t=ti+(0:m-1)*dt;

s=zeros(600,m);
for k=1:600
s(k,:)=exp(i*2*pi*10e5*sqrt(3*n(k))+i*pi*t.^2);
end
F=zeros(600,m); F2=F;
for k=1:600
F(k,:)=fftshift(fft(fftshift(s(k,:).'))).';
end
for l=1:m
F2(:,1)=fftshift(fft(fftshift(F(:,l))));
end

and S matrix is transformed to Fourier domain independent twice
what would be the values of frequencies on vertical & horizontal axis(from 0 to sampling frequency(fs) or from ?fs/2 to fs/2) when we want to show F & F2 matrix by ?imagesc? command?

Thanks & Best Regards


Date Subject Author
12/27/12
Read fftshift
maryam
12/27/12
Read Re: fftshift
ImageAnalyst
12/27/12
Read Re: fftshift
Greg Heath
12/28/12
Read Re: fftshift
maryam
1/5/13
Read Re: fftshift
Greg Heath

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