Effect of gaussian smoothing on a circumference

Message-ID:<9287547a-c415-4157-8c57-82ea3faa73c0@b38g2000prf.googlegroups.com>
Subject:

Effect of gaussian smoothing on a circumference

Date:Wed, 21 Jan 2009 19:44:49 +0100


Consider a grey image of a circle, dark inside (say around 20) and
light outside (around 200).

The intensity across the edge is well described by G + 0.5 * L * (1 +
erf (x / S)), where G, L and S are suitable constants and x is the
abscissa normal to the edge.

Now I smooth this image by convolving it with a gaussian kernel, say k
* exp (-(x^2+y^2)/(2 s^2))

I find that the edge is still well described by the above expression
(with different constants, of course) and the radius of the
circumference is smaller.

This is not surprising, but here is my question.

Is there an analytic expression relating the two values of the radius,
before and after smoothing?

(Crosspost with comp.graphics.algorithms)

Message-ID:<6trik0FcgskvU1@mid.individual.net>
Subject:

Re: Effect of gaussian smoothing on a circumference

Date:Thu, 22 Jan 2009 16:44:57 +0100

bite wrote:

> I find that the edge is still well described by the above expression
> (with different constants, of course) and the radius of the
> circumference is smaller.

Perhaps I'm misunderstanding what you mean by "radius of the
circumference", but surely the radius of the circle (where
the edge is defined as the grey-level gradient maximum) will
remain the same?

Cheers

mark-r

Message-ID:<58016b5d-e18f-4871-9be0-73a95f72fa38@z28g2000prd.googlegroups.com>
Subject:

Re: Effect of gaussian smoothing on a circumference

Date:Thu, 22 Jan 2009 17:09:17 +0100

On 22 Gen, 16:44, Mark Robinson  wrote:
> bite wrote:
> > I find that the edge is still well described by the above expression
> > (with different constants, of course) and the radius of the
> > circumference is smaller.
>
> Perhaps I'm misunderstanding what you mean by "radius of the
> circumference", but surely the radius of the circle (where
> the edge is defined as the grey-level gradient maximum) will
> remain the same?
>
> Cheers
>
> mark-r

You are not misunderstanding.

I find edge points in the "raw" image and least squares fit a
circumference to them: I get radius r0.

Then I apply gaussian convolution and do the same with the smoothed
image: I get radius r1.

r1 < r0

This difference is significant only for small circumferences (small
when compared to the sigma of the gaussian kernel).

(By the way, I take x=0 in G + 0.5 * L * (1 + erf (x / S)) as the edge
point. It happens to be the maximum of the gradient but it is not
defined as such).

Message-ID:<6tu19eFcm6a5U1@mid.individual.net>
Subject:

Re: Effect of gaussian smoothing on a circumference

Date:Fri, 23 Jan 2009 15:07:40 +0100

bite wrote:

> Then I apply gaussian convolution and do the same with the smoothed
> image: I get radius r1.
> 
> r1 < r0
> 
> This difference is significant only for small circumferences (small
> when compared to the sigma of the gaussian kernel).

Ah, OK, I see what you mean. The result of your convolution will make
the edge slightly non-symmetrical because the edge isn't isolated.

In 1D, a model for your circle is the sum of two error functions, one
displaced and reflected relative to the other. Convolution is
distributive, so your convolved output is the sum of the convolutions
of the individual Erfs, i.e. the sum of a different pair of Erfs.

In the region of the edge, the function won't be an Erf anymore, but
an Erf with the "tail" of the other edge added to it. How this
"interference" moves the apparent edge position clearly depends on
your edge detector, but it should be possible to run this new
function through it to see what happens - a symbolic maths package
could do it for you to save some of the pencil and paper work!

Of course it gets harder in 2D, but I reckon the 1D model would be
sufficiently close for most purposes (it may even be exact?)

Cheers

mark-r