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Re: fftshift
Posted:
Dec 27, 2012 7:09 PM
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"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
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