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trying to calculate the central diff approximation

조회 수: 2(최근 30일)
isamh
isamh 2020년 2월 10일
댓글: Jim Riggs 2020년 2월 10일
i tried multiple ways but none of them worked. kind of stuck and not sure what to do. what i'm trying to do is find current value that is equal to one ahead minus one below all divided by .2
code is:
Phase_1 = DATA(DATA(:,4)==1,:);
Vroll_avg1_1 = movsum((1/5).*Phase_1(:,3),[2 2]);
Vdi_1 = movsum(1/5.*Vroll_avg1_1,[2 2]);
---> first try: Adi_1 = [Vdi_1(2:Vdi_1+1,:)-Vdi_1(1:Vdi_1-1,:)/(2*T)];
---> second try: Adi_1 = ((Vdi_1(2:end+1) - Vdi_1(1:end-1))/(2*T)); % error says index exceeds array bounds. i understand why but how((2:end+1)) would i get it to work?
---> third try: %K = length(Vdi_1);
%Q = length(Vti_1);
%for M = 2:(length(Vdi_1)-1)
% Adi_1 = ((Vdi_1 - Vdi_1)./(2.*T));
%end
%Adi_1 = (Vdi_1(2) - Vdi_1(1))./(2.*T);
%Adi_1(length(Vdi_1)) = (Vdi_1(K) - Vdi_1(K))./(2.*T);
really need help, please try to help me figure this out.

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Jim Riggs
Jim Riggs 2020년 2월 10일
편집: Jim Riggs 2020년 2월 10일
There is a Matlab function "diff" which will do this.
Otherwise, your subscripts must all match (2:end-1)
Adi_1 = diff(Vdi_1);
or
for i=2:numel(Vdi_1)-1
Adi_1(i) = (Vdi_1(i+1) - Vdi_1(i-1))/2/T;
end
  댓글 수: 11
Jim Riggs
Jim Riggs 2020년 2월 10일
Very good.

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