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Re: fftshift
Posted:
Jan 5, 2013 12:08 AM
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"maryam" wrote in message <kbkj30$hc3$1@newscl01ah.mathworks.com>...
> "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;
WARNING!
FFT is only defined from t =[ 0 : dt : max(t) ] to f = [ 0 : df : max(f) ]
IFFT is only defined from f = [ 0 : df : max(f) ] to t = [ 0 : dt : max(t) ]
Use IFFTSHIFT to shift from tb to t or from fb to f
Use FFTSHIFT to shift from f to fb or t to tb
Therefore, you should redefine your t to ti+t before transforming, but you cannot
use MATLAB's shift functions unless yout original t was tb.
> s=zeros(600,m);
> for k=1:600
> s(k,:)=exp(i*2*pi*10e5*sqrt(3*n(k))+i*pi*t.^2);
> end
What is the physical interpretation of this??
Is this a 2-D problem in (n,t) space where n is a normalized space variable?
If so, why the square root?
Why are you trying to use for loops instead of vectorizing???
Why didn't you use FFT2?
help FFT2
doc FFT2
> F=zeros(600,m); F2=F;
> for k=1:600
> F(k,:)=fftshift(fft(fftshift(s(k,:).'))).';
> end
NO! See my post. Even if s were properly defined over tb, you would
use ifftshift, not its inverse fftshift in the time domain.
> for l=1:m
> F2(:,1)=fftshift(fft(fftshift(F(:,l))));
> end
>
No. Use ifftshift in the inner parentheses.
> and S matrix is transformed to Fourier domain independent twice
No. The two dimensional space-time function is transformed into a two dimensional wavelength-frequency space.
> 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?
Reread my previous post.
Hope this helps.
Greg
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