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creating Lagrange interpolation w.r.t. two arrays

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sermet
sermet 2017 年 12 月 25 日
編集済み: sermet 2017 年 12 月 26 日
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?

回答 (1 件)

ANKUR KUMAR
ANKUR KUMAR 2017 年 12 月 25 日
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 件のコメント
sermet
sermet 2017 年 12 月 26 日
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.

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