Re: [Fwd: table_model constraints]

I see that all results obtained here are based on the assupmtion of
cartesain grid (M points in x direction & N points in y direction)
What is treated in the LRM as I saw through the example posted is that
data are scattered.
So I believe that the results will differ in that case because there
aren't any specified N & M.

Also basing some limitations on the cartesian case may not support the
general scattered one.

Regards,
Tarek

Ahmad Khaled Al-Kashef wrote:

>
>
> -------- Original Message --------
> Subject: table_model constraints
> Date: Tue, 10 Aug 2004 16:04:58 -0400
> From: "Geoffrey.Coram" <Geoffrey.Coram@analog.com>
> To: verilog-ams@eda.org, Jonathan Sanders <jons@cadence.com>
> CC: Jim Barby <jabarby@sun5.vlsi.uwaterloo.ca>, david_root@agilent.com
>
>
>
>I've been having a look at the numbers of equations and unknowns
>for a 2-d interpolation, 2nd and 3rd order.
>
>**** 2nd order splines
>
>If one has a polynomial Cij x^i y^j for i=0..2, j=0..2, there are
>9 coefficients per region.
>
>If we assume a Cartesian grid, with M points in the x direction
>and N points in the y direction, that gives (M-1)*(N-1) regions.
>
>Thus, we have a total of 9MN-9M-9N+9 unknowns.
>
>
>In each region, the polynomial must fit the data given for the four
>corners. This is 4MN-4M-4N+4 equations/constraints.
>
>The derivative df/dx must be continuous at the boundaries. I believe
>we want the derivative to be continuous for all y along the edge,
>not just at the corner points. df/dx will have terms that are
>order y^0,y^1, and y^2; the coefficients of each power must be
>equal on either side of the boundary, so this means we have 3
>equations for each boundary. There are (M-2) boundaries per "row"
>and (N-1) rows. This makes 3MN-3M-6N+6 equations.
>
>The derivative df/dy gives us 3MN-6M-3N+6 for similar reasons.
>
>Thus, we have a total of 10MN-13M-13N+16 equations, but only
>9MN-9M-9N+9 unknowns. The system is *overconstrained* once we
>get bigger than 7x7.
>
>I don't see that a 2nd order spline is useful in 2d (or 3d),
>since we have to give up some constraints, either fitting
>the data or continuity of derivatives.
>
>
>**** 3rd order splines
>
>The book "Numerical Recipes in C" indicates that a bicubic spline can
>be set up by specifying f, df/dx, df/dy, and d2f/dxdy on the Cartesian
>grid. The interpolation function will have these specified derivatives
>continuous across the region boundaries. Note that, even though it is
>cubic, we don't necessarily get 2nd order derivatives continuous.
>The bicubic spline is of the form
> (sum on i=0..3)(sum on j=0..3) Cij x^i y^j
>which is 16 terms per region.
>
>Since users generally will not have derivative information, let's see
>how many constraints we need beyond the function values and the
>continuity of derivatives.
>
>The values provide 4 constraints per region.
>
>The derivative df/dx will contain terms of order y^0, y^1, y^2, y^3,
>and these must vanish independently; that means we have 4 equations
>per boundary, and (M-2)*(N-1) boundaries. We get the same number of
>constraints for df/dy.
>
>It turns out that we do not get any further constraints by requiring
>continuity of d2f/dxdy, because these constraints are linearly related
>to the constraints enforced by the first derivatives.
>
>So, we have 16MN-16M-16N+16 unknowns, and 12MN-16M-16N+20 equations,
>so we are missing 4MN-4 constraints.
>
>Anyone one want to take a stab at defining those constraints?
>
>
>-Geoffrey
>
>
Received on Thu Aug 12 00:45:36 2004