Convolution Problem

[Jbryant61]

Hi, can anyone point out any obvious mistakes with my convolution - it apppears to create a convolved object smaller than the original object itself which can't be right.

[PhilipOakley]

Isn't it a visualisation artefact due to the normalisation when being displayed.

That is, the graph shows that the psf has spread the bead shape (the slopes are shallower), which is to be expected.

When they are viewed as grey scale then the auto adjust of brightness makes the eye percieve just the central part of the top.

It hasn't narrowed the top, rather it has shaved away the corners leaving less top.

Try with a larger bead to see if this is the effect?

Philip Oakley

[Jbryant61]

I see it even on the centroid line scan towards the end.


Thanks
Jason

[IRstuff]

On 5/6/2010 8:25:28 AM, Jbryant61 wrote:
>Hi, can anyone point out any
>obvious mistakes with my
>convolution - it apppears to
>create a convolved object
>smaller than the original
>object itself which can't be
>right.

I think you're not considering "width" correctly. The usual definitions, FWHM, 1/e, 1/e², etc, would show that your resultant is larger than the original.

However, that specifically does not mean that input curve will be smaller at EVERY point. Just consider the RECT function you applied to the pixel; it's definitely not smaller at the top, compared to the resultant.

TTFN,
Eden

[Jbryant61]

Where I perform the convolution, for the bead and pixel function, should I normalise these to unity (I haven't done yet, only have done it for the graphic at the top). I think ther eis no need to normalise the psf, as this can have an influence of less than unity.

Thanks
Jason

[PhilipOakley]

rescaling doesn't really affect the result.

Rather it is the perception of the result that causes the issue. This is partly a optical illusion effects (for the greyscale image of the spot) and partly a misunderstanding of what the effect should be (which affects how we see the graph.

To take an analogy, the effect of the psf is not to pile dirt against the side of the kerb(solid step), rather, it needs to erode the top of the curb so as to create the dirt that it can then pile up at the bottom.

The problem comes when we have the classic explanation of the point source. for the point source there is no edge to erode back, merely a narrow tall spire with the top eroded away. we never really see the height of the original spire.

For example, think of starting with a butte that is 1000m high that is then eroded to dust; see Black Butte or Signal Butte in http://en.wikipedia.org/wiki/Butte rather than thinking Merrick's Butte. In these case we "think" that the psf has widened the original spike. However if we apply that to an edge....

Philip Oakley

[Jbryant61]

Thanks Philip, I thought your analogy with the kerb (or as we call it in the UK - curb) was very nice.

I was anxious since when I changed the NA to a lower NA, the convolved object appeared to get smaller, but as you have already pointed out, this is probably due to my definition of size being the half intensity diameter.

Thanks
Jason

[RichardJ]

On 5/7/2010 7:21:03 AM, Jbryant61 wrote:
>Thanks Philip, I thought your
>analogy with the kerb (or as
>we call it in the UK - curb)
>was very nice.

Yes, it is a good analogy. Remember that convolution with a spread function preserves area if the spread function has an area of 1 (which it should do if represents any sort of instrumental broadening effect). So what you gain at one height must be lost at another height.

>I was anxious since when I
>changed the NA to a lower NA,
>the convolved object appeared
>to get smaller, but as you
>have already pointed out, this
>is probably due to my
>definition of size being the
>half intensity diameter.

I'm not sure what you mean there. The width at half height of the convolved function is greater that the width at half height of the bead.

Richard

[Jbryant61]

Hi Richard.

I actually use the width at 73% of the peak intensity as my measure of width as this represents the Rayleigh criteria (27% dip). I found if I want to work out the imaged size of say a 1um bead using a NA 0.5 and an NA 0.8, then the NA 0.5 gave a narrower (73% peak intensity) width. I was immediately concerned, but am now reassured that this actually can be the case. But this would then assume the Resolution (via Raylegh Criteria) can be better..???

Jason

[RichardJ]

On 5/7/2010 6:21:24 PM, Jbryant61 wrote:
>But this
>would then assume the
>Resolution (via Raylegh
>Criteria) can be better..???

Better than what?

The problem you are having stems form the fact that you are trying to use the width of the convolved function as a measure of resolution. The shape of the convolved function changes depending on the relative widths of the two underlying functions (bead and psf) though, so you get different answers depending on how you define the width.

Richard

[IRstuff]

From a more pragmatic perspective, the changing of the relative shape of the point spread function is not necessarily unusual. It is something that many edge enhancement filters do.


TTFN,
Eden

[Jbryant61]

Hi,

Yes I have been using the convolved size to define "pitch" rather then "size". I assumed that a dip of 27% between two objects is the limit of resolution according to the Rayleigh Criteria. By measuring this width (i.e. at 73% peak intensity), this could then misleadingly sometimes allows for a closer pitch for a lower NA (due to thinner top, but wider bottom) and (hence resolution in this regard would seem to be improved), even though the classical 0.61 lambda / NA would suggest otherwise.

I guess I am simplifying Rayleigh's Criteria somewhat.

[RichardJ]

Rayleigh's criterion is not defined in terms of the dip between the two objects. It is defined based on the maximum of a particular psf for one object coinciding with the first minimum of the psf of another object. It's mathematically convenient to define it that way, and it happens to result in a dip between the two objects that is large enough that you could resolve them by eye. You could measure the width of the psf anywhere, although the most common are at the base (i.e. to the first minimum) or at half height. Wherever you measure it, a broader psf will give a larger width, because the shape does not change. But if you convolve the psf with a function representing the object, then the shape can change and you cannot define resolution in terms of the width. As a simple example, take a rectangle of with W and a triangle of width W at half height (which happens to be the convolution of two rectangles of width W). Which is wider? It depends entirely on how you choose to define "width". If your definition of width is based on some arbitrary (but convenient) criterion then any such comparison is meaningless.

Richard

[Jbryant61]

Thanks Richard (and everyone else) for your comments.

Jason

[PhilipOakley]

On 5/7/2010 7:21:03 AM, Jbryant61 wrote:
>Thanks Philip, I thought your
>analogy with the kerb (or as
>we call it in the UK - curb)
>was very nice.
>
>Thanks
>Jason

I thought you were in USofA with the .com address...

I decided not to talk about pavement edges (an even
deeper hole) .....

I note that our UK election resulted in no change
here in Scotland, yet London is jumping up and down
in excitement. It's a funny old world.

Philip Oakley

[Jbryant61]

English company (spun out of Cambridge Uni) brought out by American company!! I thought you were based in US, in fact Im not sure I know anyone on the forum from UK, glad to hear at least you are.

Yes its all fun and games with the election.

[PhilipOakley]

There are a reasonable number of UK folks, and ex-
UK. Have a look under "more .." for the member info.

Stuart Bruff is here in UK, and Richard J is
originally from these shores.

The math(s) is still international though 😉

Philip