creating Lagrange interpolation w.r.t. two arrays

Asked by sermet on 25 Dec 2017

Latest activity Edited by sermet on 26 Dec 2017

6 views (last 30 days)

time=[t1 t2 t3];
data=[y1 y2 y3];
specific_time=tt;
% for specific time tt, computation of data as follows;
data(tt)=((tt-t2)*(tt-t3))/((t1-t2)*(t1-t3))*y1 + ((tt-t1)*(tt-t3))/((t2-t1)*(t2-t3))*y2 + ((tt-t1)*(tt-t2))/((t3-t1)*(t3-t2))*y3

How can I create data(tt) w.r.t. arbitrary number of time and data array?

0 Comments

1 Answer

Link

Answer by ANKUR KUMAR on 25 Dec 2017

time=[t1 t2 t3];
data=[y1 y2 y3];
tt=[tt1 tt2 tt3 tt4 tt5 tt6 tt7];
for i=1:length(tt)
data(i)=((tt(i)-t2)*(tt(i)-t3))/((t1-t2)*(t1-t3))*y1 +...
    ((tt(i)-t1)*(tt(i)-t3))/((t2-t1)*(t2-t3))*y2 + ((tt(i)-t1)*(tt(i)-t2))/((t3-t1)*(t3-t2))*y3;
end

You can even replace t1 by time(1) and so on.

3 Comments

sermet on 26 Dec 2017

t1,t2,t3 and y1,y2,y3 should be given as loop. Because their's number are variable.

ANKUR KUMAR on 26 Dec 2017

Is your problem resolved?

sermet on 26 Dec 2017

No. Your codes are just working for above example. When t and y vectors' length are taken into consideration (they are variable, i.e, changes w.r.t. example), they should be represented as t(i), y(i) within loop.

You are now following this question

sermet (view profile)