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

*From*: ernie <Binjian.Xin@xxxxxxxxx>*Date*: 23 May 2007 07:57:37 -0700

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.

.

**References**:**Signal to Noise Ratio again. 3d data.***From:*ernie

**Re: Signal to Noise Ratio again. 3d data.***From:*aruzinsky

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