Re: Signal to Noise Ratio again. 3d data.



On 21 Mai, 01:29, "aruzinsky" <aruzin...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
"For the noise we have no doubt an average power of \sigma."

No, sigma^2 .

"The spectral method: calculating the periodogram by fourier transforming
the signal - the squared magnitude of the fourier transform, divided by
the image size."

I'll assume your definition is correct.

"By Summing up the periodogram we should the get the average power of the
signal due to the defintion for power spectrum density. Or interpreted in
another way, this sum
should equal to R(0). R is the autocorrelation function and R(0) is
exactly the average squared amplitude."

I vaguely recall that there is no standard way of normalizing DFT
(Personally, I avoid DFT like the plague and this is not fresh in my
memory). To be consistent with your application, the orthogonal basis
should be normalized so that the conjugate product with itself equals one,
i.e., orthonormal. You can inspect the DFT source code or you can
experimentally determine the normalization constant by applying your
method to an image of a Kronecker delta (impulse).

hello aruzinsky,

you're right about the \sigma^2. I forgot the square.
As to the normalizing with the image size: I'm afraid it isn't from
the DFT.
I might have seen in the textbook like [papoulis&pillai]. in one
dimensional case the periodogram should be normalized with the time
interval 2T, if the length of the finite signal x_T(t) is [-T,T]. So I
think it's natural in this way to extend the definiton to its 2D space
domain.
But your idea to check the orthonormality of DFT is great! I have also
similar problems. Thanks for sharing your tricks. It's always a
consolation to know u r not the only sufferer.

.



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