This
level-5 vital article is rated C-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
This page has archives. Sections older than 365 days may be automatically archived by Lowercase sigmabot III when more than 5 sections are present. |
The article states, "It is important to note that the Kronecker delta is not the result of sampling the Dirac delta function." Isn't this incorrect? If we use the ideal lowpass to limit the bandwidth for sampling, then we will be sampling a sinc function at the middle peak and then zero-crossings. I won't change anything until I cook up a proof or find a ref to add here, or until someone (myself included) shows I'm wrong. Any thoughts? Herr Lip ( talk) 19:50, 31 March 2011 (UTC)
I went ahead and put something in, albeit without a proof or ref, for now. I also changed "It is important to note that the Kronecker delta is not the result of sampling the Dirac delta function." to "It is important to note that the Kronecker delta is not the result of directly sampling the Dirac delta function." Adding the "directly" to clarify (what I assume is) the point of the original statement, and help compare to my addition. Herr Lip ( talk) 01:16, 30 April 2011 (UTC)
I think what this statement means is that sampling a Dirac delta function at x=0 (for a sufficiently short sample "time" around x=0) must yield a value greater than 1, the expected discrete Kronecker delta value for [0,0]. The reason is that the value of a Dirac delta at exactly x=0 (with infinitely small sample time) must be infinite for the total integral over the number line to be 1. Does anyone agree? David Spector ( talk) 22:01, 23 July 2013 (UTC)
what is the discrete convolution of sampled Dirac delta function with discrete sinc function? 1*delta(x)= delta(x), but what about 0*delta(x)=?, since 0*infinity is undetermined or undefined, so the lowpass filter idea is questionable. the idea indeed seems to implicitly take continuous convolution of sampled Dirac delta function with discrete sinc function first and then sample the convolution and get the Kronecker delta, that's no problem — Preceding unsigned comment added by 123.119.83.143 ( talk)
I agree that it may look like original research, but it is nonetheless a true algebraic representation, as long as it is used only on integers. I have my doubts about how useful it is, though, as all practical programming languages interprets boolean truth as 1 if converted to integer, which means that the Kronecker delta function can simply be represented as something like: int(i==j) —Preceding unsigned comment added by 77.40.128.194 ( talk) 14:27, 6 December 2009 (UTC)
Isn't the section on the unit Impusle Misleading? Shouldn't the value at 0 for the impulse be '+inf', not '1'? (I know thats not actually correct either, but I'm saying its not right as is) -- 143.107.106.100 ( talk) 22:50, 14 April 2011 (UTC)
I notice that in this article, the relationship with the Levi-Civita symbol differs by a factor of n!. Here it is given as
whereas in Gamma matrices#The fifth gamma matrix, γ5 (open the "Proof" box to see it) it is given as
This discrepancy is possibly a matter of definition of the generalized Kronecker delta, and should be clarified. — Quondum 04:52, 29 October 2012 (UTC)
The definition of Dirac delta should not be that it is infinity at x=0, but just that its area is one. That is, it is a special infinity, or a limit value that tends to infinity. Gah4 ( talk) 07:01, 20 February 2022 (UTC)