| Re: IDL 5.5, 2D FFT indexing confusion. [message #44912 is a reply to message #44831] |
Thu, 21 July 2005 11:47  |
R.G. Stockwell
Messages: 363
Registered: July 1999
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Senior Member |
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"Pitufa" <c.c.calderon@gmail.com> wrote in message
news:1121966932.337140.282750@g49g2000cwa.googlegroups.com.. .
> but then it only works if the fourier array is real, and 2-5 don't make
> any sense physically! My conclusion from this is that you are better
> off using an odd N if you are doing this sort of calculation.
>
> Agree? Disagree?
If speed of calculation is a concern, "even N" is good (tends to be
more factorable), and powers of two are especially nice.
If you happen across a large prime number for N, then idl routines
are going to crawl. (and I have actually done that in the past).
Of course, actual measured time series are often even in N,
because the number of hours in day is even, minutes in hour, seconds in
minutes, etc all even
Fractions of a time unit are often in halfs or in tens, etc.
But I don't see any great compelling reason not use (nicely factorable) odd
numbers for N.
> Why the nyquist rows/columns need to be positive?
They don't. They (the symmetric nyquist pairs) just need to be the same
sign (and you had positive, so I said positive). Sorry, should have been
more clear.
As to why they must be the same, if you look at the math for a particular
nyquist pair,
(apply discrete inverse dft to just those two spectral components)
you basically get the following
using
image[x,y] = sum_u sum_v F[u,v] exp(i 2 pi (ux/M + vy/N))
input in the symmetric pair ( A1 at (u,N/2) and A2 at (-u, N/2)
then you get
image(x,y) = (-1)^y * [A1*(cos(p1) + i sin(p1)) + A2*(cos(p1) - i
sin(p1)) ]
where the * are just normal multiplication, i've chosen to do the nyquist
frequency in
the y direction which is where the (-1)^y comes from.
and p1 is shorthand for the argument p1 = i 2 pi ux/M.
So, requiring image to be real means that the sin components must cancel out
therefore
A1sin(p1) = A2 sin(p1)
A1 = A2
[if p1 ~=0 of course. But that is your DC Nyquist points, and they do not
have to be symmetric]
Cheers,
bob
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