
Re: interpolation on geometric progression data
Posted:
Jan 15, 2013 5:19 PM


On Wednesday, January 16, 2013 9:45:15 AM UTC+13, Anil Kumar Palaparthi wrote: > From Geometric Progression, what I mean is that the data is not sampled uniformly or in linear sequence. It is sampled in geometric sequence. > > For example, y = f(x) where 'x' is sampled in geometric sequence not in linear sequence and I can't fit any polynomial to 'f'. > > > > Anil Palaparthi. > > > > "Barry Williams" <barry.r.williamsnospam@saic.com> wrote in message <kd4d6v$pv0$1@newscl01ah.mathworks.com>... > > > "Anil Kumar Palaparthi" wrote in message <kd4acs$eja$1@newscl01ah.mathworks.com>... > > > > Hi, > > > > > > > > I need to interpolate my data whose 'x' values are in geometric progression rather than linear. For example, x = [0.005,0.01,0.02,0.04,0.08] and 'y' can be anything. > > > > Can anyone suggest me how I can interpolate this kind of data? > > > > Is there a specific algorithm that can interpolate geometric progression data? > > > > > > > > Best Regards, > > > > Anil Palaparthi. > > > > > > What I prefer to do whenever possible is to interpolate from the underlying function. If by *geometric progression* you are referring to a polynomial of form: > > > y = a0 + a1(x) + a2(x^2 + a3(x^3) ... > > > Then you could fit the data to the polynomial, and then use polyval to evaluate it at the needed values of x. > > > If you have no idea of the for f(x) takes, then dpb is right. The method you use with interp1 is irrelevant. > > > Barry
Perhaps what you need is: yint=interp1(log10(x),y,log10(xint)); in other words, linearise the independent variable by taking logs.
BTW, don't top post: follow to hard thread the makes it. Put your reply UNDERNEATH.

