Re: interpolation w/ cubic convolution kernel: boundary treatment?
- From: Markus <iandjohn@xxxxxxxxx>
- Date: 26 Apr 2007 15:30:35 -0700
This same topic has been posted just before in another thread.
I am sorry for this, buit it happened due to the very slow response of
Google: It took Google 20 hours to display my message after I had
posted it.
If you want to reply, please reply to the other thread "cubic
convolution interpolation at boundaries?"
Markus
On Apr 26, 2:18 am, Thomas Kluge <thomas.kl...@xxxxxxx> wrote:
Hi,
I am using the symmetric cubic convolution kernel
("Catmull-Rom splines") to interpolate data over a limited
range in a variable x. For the interpolation I am using
typically 10 nodes which are equidistant in x.
Example: The interpolated function between nodes 4 and 5 is
computed based on the data points at nodes 3,4,5,6
A nice property of the Catmull-Rom splines is that I get a
continuous 1st derivative everywhere.
My question is now: how should I treat the ranges near the
boundary?
With 10 nodes, how should I interpolate the data between
node 9 and 10? So far I am using linear interpolation here -
but this is conceptually ugly (and it's not precise,
although the latter is not my biggest problem since I want a
_nice_ solution).
I am especially worried that in my current approach the
interpolated function has no continuous 1st derivative at
node 9.
Is there a solution, in which I could use e.g. a
non-symmetric convolution kernel to interpolate between
nodes 9 and 10, based on the nodes 8,9,10 or maybe 7,8,9,10
- in a way that I get a continuous 1st derivative everywhere?
I have not found any discussion about the boundary-treatment
in the literature (and neither on the Google-wide web). In
image processing sometimes people mirror the image at the
boundaries (i.e. they would introduce a hypothetical 11th
node for which the data value is set equal to the data at
the 9th node and then interpolate using the data points at
nodes 8,9,10,11) - but this would not work in my case.
.
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- From: Thomas Kluge
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