Convolution, integration

[Jbryant61]

Hi. I am revisiting some earlier posts and trying to combine them. Jean had shown me how to perform a convolution to allow me to estimate the image of a bead represented by a gaussian if the psf is known (i.e. how a single point appears in the image). I now need to take this convolved image and calculate the diameter that contains half the total intensity. Again, Richard/Tom have kindly helped before, but I don't know how to do it on the output of a convolution as I don't have a function to integrate except just a matrix.

Thanks
Jason

[ptc-1368288]

Jason,

Will look at it again in case something comes back to mind.

Jean

[ptc-1368288]

On 6/19/2009 11:00:02 AM, Jbryant61 wrote:
>Hi. I am revisiting some
>earlier posts and trying to
>combine them. ....
>but I don't know how to do it on
>the output of a convolution as
>I don't have a function to
>integrate except just a
>matrix.
>
>Thanks
>Jason
_________________________________

That's your question (last green):

"How to calculate the radius that contains 50% of the total intensity for the image of a point and, Image of a bead profile as I don't have a functional form but instead a matrix of data ?".

Simple:
Fit a 2d Gaussian to the convolved result, integrate triple, solve for the circle of � intensity. Something to come hopefully by tomorrow or near future.

Jean

[RichardJ]

I can think of two approaches.

One is to find the center point (which in general will not be exactly on one pixel). Then create a function that sums all the pixels within a given radius from the center point. To make that a smooth function you will need to weight the pixels close to the boundary, so that they are not just counted as 1 or 0.

The second is to create x and y vectors, and then use a 2D spline to create a function. Then integrate the spline (this is the approach I would choose).

Richard

[ptc-1368288]

Interesting ,

We did it in the past and got the radius , but for that particular one it fails. A generalised method has to be found... RemToDo.

jmG

[ptc-1368288]

... that's what it looks like in term of spatial pixels.

Jean

[Jbryant61]

Hi Jean, thanks, its very nice. To answer your question, yes by 1/2 volume I mean 1/2 intensity.

I think I am seeing something strange. When I change the size of the bead and hence the end result due to the convolution, the 1/2 intensity and hence radius do not change. Could there be an error somewhere?

thanks
Jason