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The Gaussian filter can be applied with a 2d mask, or two 1d masks in sequence. Is "Linearly Separable" really the proper term for this? "Sequenced Convolution" perhaps? —Preceding unsigned comment added by 82.46.170.107 ( talk) 21:09, 25 April 2008 (UTC)
I spent some time looking through the article to find what the term "Gaussian" means, and didn't find any answers. After some further googling I found that it is named after a man named Gauss( http://www.webmonkey.com/webmonkey/glossary/gaussian.html). I am surprised the introduction doesn't mention this, and assumes the reader knows (or doesn't care) what Gaussian actually means -- I looked up this article to discover that, and the entire Wiki article didn't help.
Additionally, does Gaussian always necessarily mean it relates in some way to Karl Friedrich Gauss? Perhaps that clarification is beyond the scope of this article, but surely the fact that Gaussion in this case refers to Karl Friedrich Gauss is not? —Preceding unsigned comment added by Abeall ( talk • contribs) 18:55, 3 January 2008 (UTC)
The page claims most programs use a radius of 6σ+1. Won't this depend entirely on the bit depth? That is, to get "perfect" results you need a radius such that the value of the convolution kernel at that radius would round to zero in that bit depth? Actually, you might need to go further... I think the worst case is bluring a black circle with a white background. To find the result corresponding to the center of the circle you want to make sure your filter is large enough to get the correct value at the center of the circle in the current bit depth. I could work out the math, but either way, the optimum radius of the kernel will certainly be a function of bit depth. —Ben FrantzDale 15:14, 2 May 2006 (UTC)
Image:Josefina with Bokeh.jpg - Try doing this in the GIMP.
Shouldn't it rather say something like "viewing through a translucent screen". According to my understanding the blur you get with an out of focus lens has a distribution that's close to even inside a projection of the iris and zero outside it. That doesn't look a lot like the bell curve. See bokeh. — Home Row Keysplurge 14:50, 10 May 2006 (UTC)
#!/usr/bin/python import math size = 41 mid = size/2 for x in range(0,size): for y in range(0, size): dist = math.sqrt(pow(x-mid,2)+pow(y-mid,2)) if dist < mid-0.5: print 1, elif dist > mid+0.5: print 0, else: print dist-(mid-0.5), print ""
or
Why has the square root disappeared from the second formula, and why has a appeared?
We're told that so why wasn't that substitution the only difference?
I ask not because I think it's wrong, but because I'm trying to understand the formula before I use it in a program I'm writing.
Edited to add: for example see
http://www.maths.abdn.ac.uk/~igc/tch/mx4002/notes/node99.html - this page shows this formula:
It looks to me as if the square root should be in the second formula above, and that it should extend over the sigma squared (and perhaps the first formula should extend it too and square the sigma for consistency?).
Can anyone clear this up for me?
I modified the equations to the equations that actually fit the example matrix. However after some more research I found that both the current equations and the equations I chagned to are used commonly. Therefore I reverted my changes so now the example matrix doesn't match up with the presented equations. Rune Hunter ( talk) 05:27, 11 June 2008 (UTC)
With no doubt, the convolution of the signal by a gaussian kernel should preserve the total energy of the signal which is only possible if the integral of the function on the whole domain is equal to one. Hence, as the 2D integral of the non-normalized gaussian kernel is expressed as
, we have , therefore .
As the integral of the non-normalized 1D gaussian is , the 2D integral is equal to . Therefore the normalized 2D gaussian equation must be . This is of course generalizable to nD dimensions: where is a vector of dimensions. —Preceding
unsigned comment added by
130.104.224.111 (
talk)
09:59, 11 June 2008 (UTC)
Yes, this is similar to what was previously in this article and the formula should probably be reverted. Side note... The Fourier transform of a Gaussian is another Gaussian, but I have not been able to find a good resource that defines what this transformed Gaussian is in terms of n dimensions, and the constant needed to preserve the signal properly; if anyone feels up to adding that, it would be appreciated. The relevance is the topic of how Gaussian blurs work in reciprocal space. —Preceding unsigned comment added by 137.131.204.136 ( talk) 00:31, 21 January 2009 (UTC)
The article mentions the radius as if it was a well-known property, but nowhere is it explained what the blur radius is. Sam Hocevar ( talk) 09:12, 28 July 2008 (UTC)
Something should probably done to connect this article to the gaussian filter one, especially since the subjects are so closely related and there's already some duplication of content. -Roger ( talk) 18:36, 24 February 2009 (UTC)
I've come to the discussion page after seeing the merge proposal with Gaussian filter and reviewing the two articles. I would vote against, pretty strongly, for these reasons:
That's my feedback, anyway. 88.96.159.86 ( talk) 10:50, 18 February 2011 (UTC)
Would anyone address those issues using Gaussaian Filters? -- Royi A ( talk) 12:39, 16 October 2009 (UTC)
"Applying multiple, successive gaussian blurs to an image has the same effect as applying a single, larger gaussian blur, whose radius is the square root of the sum of the squares of the blur radii that were actually applied. For example, applying successive gaussian blurs with radii of 6 and 8 gives the same results as applying a single gaussian blur of radius 10, since . Because of this relationship, processing time cannot be saved by simulating a gaussian blur with successive, smaller blurs — the time required will be at least as great as performing the single large blur."
I don't believe this is accurate, referencing gpu massively parallel applications and smaller mask windows — Preceding unsigned comment added by 70.208.73.107 ( talk) 17:42, 16 November 2013 (UTC)
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While the Gaussian function is very important in statistics, does the same hold true for optics? The Gaussian blur seems to be the go-to blur method, preferred over more naïve digital methods for its supposedly appealing retention of edges, but this alone doesn't say much about its objective basis. I find it interesting that the similar Cauchy function never gets used for this purpose, despite it also offering seemingly natural-looking blurs with long tails yielding an oneric aesthetic. I guess the question boils down to how close the Gaussian function resembles the behaviour of blurring incurred by the interaction of light with real materials and the human eye. A notable case is how it poorly represents the bokeh blur of out-of-focus camera shots. 212.60.107.203 ( talk) 09:21, 13 June 2019 (UTC)
This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
The Gaussian filter can be applied with a 2d mask, or two 1d masks in sequence. Is "Linearly Separable" really the proper term for this? "Sequenced Convolution" perhaps? —Preceding unsigned comment added by 82.46.170.107 ( talk) 21:09, 25 April 2008 (UTC)
I spent some time looking through the article to find what the term "Gaussian" means, and didn't find any answers. After some further googling I found that it is named after a man named Gauss( http://www.webmonkey.com/webmonkey/glossary/gaussian.html). I am surprised the introduction doesn't mention this, and assumes the reader knows (or doesn't care) what Gaussian actually means -- I looked up this article to discover that, and the entire Wiki article didn't help.
Additionally, does Gaussian always necessarily mean it relates in some way to Karl Friedrich Gauss? Perhaps that clarification is beyond the scope of this article, but surely the fact that Gaussion in this case refers to Karl Friedrich Gauss is not? —Preceding unsigned comment added by Abeall ( talk • contribs) 18:55, 3 January 2008 (UTC)
The page claims most programs use a radius of 6σ+1. Won't this depend entirely on the bit depth? That is, to get "perfect" results you need a radius such that the value of the convolution kernel at that radius would round to zero in that bit depth? Actually, you might need to go further... I think the worst case is bluring a black circle with a white background. To find the result corresponding to the center of the circle you want to make sure your filter is large enough to get the correct value at the center of the circle in the current bit depth. I could work out the math, but either way, the optimum radius of the kernel will certainly be a function of bit depth. —Ben FrantzDale 15:14, 2 May 2006 (UTC)
Image:Josefina with Bokeh.jpg - Try doing this in the GIMP.
Shouldn't it rather say something like "viewing through a translucent screen". According to my understanding the blur you get with an out of focus lens has a distribution that's close to even inside a projection of the iris and zero outside it. That doesn't look a lot like the bell curve. See bokeh. — Home Row Keysplurge 14:50, 10 May 2006 (UTC)
#!/usr/bin/python import math size = 41 mid = size/2 for x in range(0,size): for y in range(0, size): dist = math.sqrt(pow(x-mid,2)+pow(y-mid,2)) if dist < mid-0.5: print 1, elif dist > mid+0.5: print 0, else: print dist-(mid-0.5), print ""
or
Why has the square root disappeared from the second formula, and why has a appeared?
We're told that so why wasn't that substitution the only difference?
I ask not because I think it's wrong, but because I'm trying to understand the formula before I use it in a program I'm writing.
Edited to add: for example see
http://www.maths.abdn.ac.uk/~igc/tch/mx4002/notes/node99.html - this page shows this formula:
It looks to me as if the square root should be in the second formula above, and that it should extend over the sigma squared (and perhaps the first formula should extend it too and square the sigma for consistency?).
Can anyone clear this up for me?
I modified the equations to the equations that actually fit the example matrix. However after some more research I found that both the current equations and the equations I chagned to are used commonly. Therefore I reverted my changes so now the example matrix doesn't match up with the presented equations. Rune Hunter ( talk) 05:27, 11 June 2008 (UTC)
With no doubt, the convolution of the signal by a gaussian kernel should preserve the total energy of the signal which is only possible if the integral of the function on the whole domain is equal to one. Hence, as the 2D integral of the non-normalized gaussian kernel is expressed as
, we have , therefore .
As the integral of the non-normalized 1D gaussian is , the 2D integral is equal to . Therefore the normalized 2D gaussian equation must be . This is of course generalizable to nD dimensions: where is a vector of dimensions. —Preceding
unsigned comment added by
130.104.224.111 (
talk)
09:59, 11 June 2008 (UTC)
Yes, this is similar to what was previously in this article and the formula should probably be reverted. Side note... The Fourier transform of a Gaussian is another Gaussian, but I have not been able to find a good resource that defines what this transformed Gaussian is in terms of n dimensions, and the constant needed to preserve the signal properly; if anyone feels up to adding that, it would be appreciated. The relevance is the topic of how Gaussian blurs work in reciprocal space. —Preceding unsigned comment added by 137.131.204.136 ( talk) 00:31, 21 January 2009 (UTC)
The article mentions the radius as if it was a well-known property, but nowhere is it explained what the blur radius is. Sam Hocevar ( talk) 09:12, 28 July 2008 (UTC)
Something should probably done to connect this article to the gaussian filter one, especially since the subjects are so closely related and there's already some duplication of content. -Roger ( talk) 18:36, 24 February 2009 (UTC)
I've come to the discussion page after seeing the merge proposal with Gaussian filter and reviewing the two articles. I would vote against, pretty strongly, for these reasons:
That's my feedback, anyway. 88.96.159.86 ( talk) 10:50, 18 February 2011 (UTC)
Would anyone address those issues using Gaussaian Filters? -- Royi A ( talk) 12:39, 16 October 2009 (UTC)
"Applying multiple, successive gaussian blurs to an image has the same effect as applying a single, larger gaussian blur, whose radius is the square root of the sum of the squares of the blur radii that were actually applied. For example, applying successive gaussian blurs with radii of 6 and 8 gives the same results as applying a single gaussian blur of radius 10, since . Because of this relationship, processing time cannot be saved by simulating a gaussian blur with successive, smaller blurs — the time required will be at least as great as performing the single large blur."
I don't believe this is accurate, referencing gpu massively parallel applications and smaller mask windows — Preceding unsigned comment added by 70.208.73.107 ( talk) 17:42, 16 November 2013 (UTC)
Hello fellow Wikipedians,
I have just modified one external link on Gaussian blur. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.
This message was posted before February 2018.
After February 2018, "External links modified" talk page sections are no longer generated or monitored by InternetArchiveBot. No special action is required regarding these talk page notices, other than
regular verification using the archive tool instructions below. Editors
have permission to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the
RfC before doing mass systematic removals. This message is updated dynamically through the template {{
source check}}
(last update: 5 June 2024).
Cheers.— InternetArchiveBot ( Report bug) 21:07, 11 October 2017 (UTC)
While the Gaussian function is very important in statistics, does the same hold true for optics? The Gaussian blur seems to be the go-to blur method, preferred over more naïve digital methods for its supposedly appealing retention of edges, but this alone doesn't say much about its objective basis. I find it interesting that the similar Cauchy function never gets used for this purpose, despite it also offering seemingly natural-looking blurs with long tails yielding an oneric aesthetic. I guess the question boils down to how close the Gaussian function resembles the behaviour of blurring incurred by the interaction of light with real materials and the human eye. A notable case is how it poorly represents the bokeh blur of out-of-focus camera shots. 212.60.107.203 ( talk) 09:21, 13 June 2019 (UTC)