Talk:Wiener deconvolution

message: It is valuable to give a derivation of this useful expression.
However, in this part, the expression that is defined as "the power spectral density"
is actually the "energy spectral density" expectation.
There is much confusion about this in Wikipedia related articles, especially in the time domain,
but litterature generally agrees about the following definitions in the frequency domain:
If we consider a real signal x(t) in the time domain,
its Fourier transform is X(f) = FT{x(t)};
then, its "energy spectral density" (ESD) is :
Sxx(f) = |X(f)|2
(Some people use another letter than S).
It is easy to check that this expression has the dimension of an energy (x squared) divided by a frequency, as a "spectral density" of energy.
The integral of Sxx(f) = |X(f)|2 over all the frequencies is indeed the total energy of the signal, (i.e. the integral of |x(t)|2 over time) according to the Parseval-Plancherel theorem.
Taking its expectation E{Sxx(f)} (i.e. its statistical average over all the possibilities of the signal x(t)) does make it a power (i.e. an energy per time unit).
The "power spectral density" (PSD) is :
Pxx(f) = FT{Cxx(T)}
with :
Cxx(T) = COVARIANCE{x(t), x(t+T)} = E{[x(t)-mx ] . [x(t+T)-mx]}
where:
mx = E{x}
E is the expectation operator.
I assume that x(t) is stationary up to the second order.
(Many -but not all- call the expression Cxx(T) above the autocorrelation function).
This expression Pxx(f) has the dimension of a power (energy per unit of time) divided by a frequency, as a "spectral density" of power.
The usual expression with the PSD looks the same, (SNR is a dimensionless ratio between -usually- powers or -here- energies) but is more difficult to demonstrate.
This is why more accessible litterature uses the energy rather than the power, but this should be clearly stated.
Almipa (talk) 15:45, 6 May 2010 (UTC)
thread_title: Energy spectral density

message: the wiener filter expression is incorrect ... ref. gonzalez and woods book p263 —The preceding unsigned comment was added by 67.87.222.13 (talk • contribs) 03:48, 19 March 2007  (UTC)
:In my copy of this book it appears on page 170, and while it does not have the same symbols exactly, it is mathematically equivalent. Just below it is written that |H(u,v)^2| is H*(u,v)H(u,v), which allows an H(u,v) to cancel on the top and bottom. There are many subtly different ways to write this expression, but I believe the one we have is correct. (Test it out on a deconvolution problem and see, if you're not sure.) - Rainwarrior 04:27, 19 March 2007 (UTC)
thread_title: Expression incorrect

message: * The sentence "...As this is a complex value, \ G^*(f) acts as a constant..." does not make any sense and it is misleading at best, if not completely wrong. This last step involving the derivation of the quantity \ G(f) G^*(f) should be corrected and made clear, since the argumentation used now is not valid, although the resulting expression for G is correct.
* As a counterexample to the current argument, just suppose that both the image \ x and the convolution kernel \ h are centrally symmetric, and noise is white. Then we know that the Fourier transforms \ H and \ X will be real (imaginary part equal to zero), thus also \ G will be centrally symmetric and real. In such a case the quantity \ G(f) G^*(f) would reduce to \ G(f)^2, whose derivative is \ 2G(f), and clearly in this case \ G^*(f) does not act as constant, as claimed by the author.
* A more reasonable argument could be that, if we assume that the Wiener filter \ g in time domain is real, then we have that in Fourier domain \ G^*(f) = G(-f), which would indeed justify treating \ G^*(f) "as a constant" when differentiating w.r.t. \ G(f). However I don't know if the Wiener filter \ g must be strictly real (if not, then the above argument is not valid either).
* PS: furthermore, all the instances of \ V  should be replaced with \ N for consistency, and it should be specified that noise is assumed to have zero mean, otherwise \ \mathbb{E} \{ X(f)N^*(f) \} \neq 0.
 91.145.120.246 (talk) 18:14, 2 September 2013 (UTC)
thread_title: Mistake in final step of Derivation

message: SNR, as written |S(f)/N(f)|, is not a unitless quantity and therefore rather misleading because that is not what is normally understood as signal to noise ratio.  Depending on the units of input and output signal (x and y, or alternatively S and N), the ratio can be anything.  This can be also seen from the expression |H(f)|^2 + 1/SNR(f)|.  Since H is generally not unitless either, neither can be |S(f)/N(f)|.
46.39.166.145 (talk) 20:17, 10 May 2018 (UTC)
thread_title: SNR is not Dimensionless