Spatially Varying Blur - Revision history

Jaspervdg at 16:26, 26 April 2010

2010-04-26T16:26:36Z

← Older revision		Revision as of 16:26, 26 April 2010
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	One possibility is to do a simple linear interpolation between two blurs, but this tends to give an effective kernel which is only a very poor fit to the true kernel desired, it tends to give a much too sharp result.		One possibility is to do a simple linear interpolation between two blurs, but this tends to give an effective kernel which is only a very poor fit to the true kernel desired, it tends to give a much too sharp result.

	This ~~leads to wanting to create a better fitting interpolation between the two blurs~~.		This page looks at possibilities for creating an improved spatially varying blur.

	== L2-minimized fit ==		== L2-minimized fit ==

Jaspervdg at 11:32, 26 April 2010

2010-04-26T11:32:57Z

← Older revision		Revision as of 11:32, 26 April 2010
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	0&=2\left[G(\sigma,\sigma_1)-\,G(\sigma,\sigma_2)\right]-2(1-a)G(\sigma_1,\sigma_1)+2(1-2a)G(\sigma_1,\sigma_2)+2a\,G(\sigma_2,\sigma_2)		0&=2\left[G(\sigma,\sigma_1)-\,G(\sigma,\sigma_2)\right]-2(1-a)G(\sigma_1,\sigma_1)+2(1-2a)G(\sigma_1,\sigma_2)+2a\,G(\sigma_2,\sigma_2)
	\\0&=G(\sigma,\sigma_1)-\,G(\sigma,\sigma_2)-G(\sigma_1,\sigma_1)+a\,G(\sigma_1,\sigma_1)+G(\sigma_1,\sigma_2)-2a\,G(\sigma_1,\sigma_2)+a\,G(\sigma_2,\sigma_2)		\\0&=G(\sigma,\sigma_1)-\,G(\sigma,\sigma_2)-G(\sigma_1,\sigma_1)+a\,G(\sigma_1,\sigma_1)+G(\sigma_1,\sigma_2)-2a\,G(\sigma_1,\sigma_2)+a\,G(\sigma_2,\sigma_2)
	\\a&=\frac{G(\~~sigma_1~~,\~~sigma_1~~)-G(\~~sigma_1~~,\~~sigma_2~~)-(G(\~~sigma~~,\sigma_1)-G(\~~sigma~~,\sigma_2))}{G(\sigma_1,\sigma_1)-2\,G(\sigma_1,\sigma_2)+G(\sigma_2,\sigma_2)}		\\a&=\frac{G(\sigma,\sigma_2)-G(\sigma,\sigma_1)+G(\sigma_1,\sigma_1)-G(\sigma_1,\sigma_2)}{G(\sigma_1,\sigma_1)-2\,G(\sigma_1,\sigma_2)+G(\sigma_2,\sigma_2)}
	.\end{align}</math><!-- Verified numerically for s,s1,s2=2,1,3 -->		.\end{align}</math><!-- Verified numerically for s,s1,s2=2,1,3 -->

	~~For reference:~~		This uses

	<math>g(\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/(2\sigma^2)}</math>		<math>\begin{align}
			g(\sigma)&=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/(2\sigma^2)}
			\\G(\sigma_a,\sigma_b)&=\frac{1}{\sqrt{2\pi(\sigma_a^2+\sigma_b^2)}}
			.\end{align}</math>

	<math>G(\~~sigma_a~~,\~~sigma_b~~)=\frac{1}{\sqrt{2\pi(\~~sigma_a~~^2+\~~sigma_b~~^2)}}</math>		<!-- I think it's probably just best to use feComponentTransfer with type=table

			Filling in and simplifying we get

			<math>\begin{align}
			a&=\frac{G(\sigma,\sigma_2)-G(\sigma,\sigma_1)+G(\sigma_1,\sigma_1)-G(\sigma_1,\sigma_2)}{G(\sigma_1,\sigma_1)-2\,G(\sigma_1,\sigma_2)+G(\sigma_2,\sigma_2)}
			\\&=\frac{\frac{1}{\sqrt{2\pi(\sigma^2+\sigma_2^2)}}-\frac{1}{\sqrt{2\pi(\sigma^2+\sigma_1^2)}}+\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_1^2)}}-\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_2^2)}}}{\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_1^2)}}-2\,\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_2^2)}}+\frac{1}{\sqrt{2\pi(\sigma_2^2+\sigma_2^2)}}}
			\\&=\frac{\frac{1}{\sqrt{\sigma^2+\sigma_2^2}}-\frac{1}{\sqrt{\sigma^2+\sigma_1^2}}+\frac{1}{\sqrt{\sigma_1^2+\sigma_1^2}}-\frac{1}{\sqrt{\sigma_1^2+\sigma_2^2}}}{\frac{1}{\sqrt{\sigma_1^2+\sigma_1^2}}-2\,\frac{1}{\sqrt{\sigma_1^2+\sigma_2^2}}+\frac{1}{\sqrt{\sigma_2^2+\sigma_2^2}}}
			.\end{align}</math>
			-->

Jaspervdg: Start

2010-04-23T12:25:53Z

Start

New page

Spatially varying blur can be important for a number of effects, not the least of which is depth-of-field. Unfortunately SVG does not support this directly and we need to make some approximation of the effect.

One possibility is to do a simple linear interpolation between two blurs, but this tends to give an effective kernel which is only a very poor fit to the true kernel desired, it tends to give a much too sharp result.

This leads to wanting to create a better fitting interpolation between the two blurs.

== L2-minimized fit ==
In general the easiest global error metric to minimize is the l2-norm. This is simply (the root of) the integral of the squared error. Which in this case can be formulated as (<math>g(\sigma)</math> is the kernel of a Gaussian blur with standard deviation <math>\sigma</math>, <math>G(\sigma_a,\sigma_b)</math> is the integral of the product of two kernels, <math>\sigma_1<\sigma<\sigma_2</math> and <math>0<a<1</math>)

<math>\begin{align}
e(\sigma,\sigma_1,\sigma_2,a)&=\int_{-\infty}^{\infty}\left[g(\sigma)-\left((1-a)g(\sigma_1)+a\,g(\sigma_2)\right)\right]^2dx
\\&=\int_{-\infty}^{\infty}g(\sigma)^2-2g(\sigma)\left((1-a)g(\sigma_1)+a\,g(\sigma_2)\right)+\left((1-a)g(\sigma_1)+a\,g(\sigma_2)\right)^2dx
\\&=G(\sigma,\sigma)-2\left[(1-a)G(\sigma,\sigma_1)+a\,G(\sigma,\sigma_2)\right]+(1-a)^2G(\sigma_1,\sigma_1)+2a(1-a)G(\sigma_1,\sigma_2)+a^2\,G(\sigma_2,\sigma_2)
.\end{align}</math>

Equating the derivative to zero (in order to minimize the quadratic equation in <math>a</math>) we get

<math>\begin{align}
0&=2\left[G(\sigma,\sigma_1)-\,G(\sigma,\sigma_2)\right]-2(1-a)G(\sigma_1,\sigma_1)+2(1-2a)G(\sigma_1,\sigma_2)+2a\,G(\sigma_2,\sigma_2)
\\0&=G(\sigma,\sigma_1)-\,G(\sigma,\sigma_2)-G(\sigma_1,\sigma_1)+a\,G(\sigma_1,\sigma_1)+G(\sigma_1,\sigma_2)-2a\,G(\sigma_1,\sigma_2)+a\,G(\sigma_2,\sigma_2)
\\a&=\frac{G(\sigma_1,\sigma_1)-G(\sigma_1,\sigma_2)-(G(\sigma,\sigma_1)-G(\sigma,\sigma_2))}{G(\sigma_1,\sigma_1)-2\,G(\sigma_1,\sigma_2)+G(\sigma_2,\sigma_2)}
.\end{align}</math>

For reference:

<math>g(\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/(2\sigma^2)}</math>

<math>G(\sigma_a,\sigma_b)=\frac{1}{\sqrt{2\pi(\sigma_a^2+\sigma_b^2)}}</math>