Subscript indices must either be real positive integers or logicals.

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Yintong Zheng
Yintong Zheng el 27 de Feb. de 2016
Respondida: Image Analyst el 27 de Feb. de 2016
function T=baryeval2(tau,s,c,belta,x)
u=zeros(size(x))
d=zeros(size(x))
T=zeros(size(x))
for k=1:5
u=u+s(k).*belta(k).*(x-tau(k)).^(-1);
d=d+c(k).*belta(k).*(x-tau(k)).^(-1);
end
T=u./d;
end

Respuestas (4)

Azzi Abdelmalek
Azzi Abdelmalek el 27 de Feb. de 2016
Editada: Azzi Abdelmalek el 27 de Feb. de 2016
You cant use an index equal to zero, Matlab allows only positive integer indices, or logical indices
  4 comentarios
Yintong Zheng
Yintong Zheng el 27 de Feb. de 2016
I changed my code and now I get answers. But apparently something is wrong! Please help me to fix it.
Yintong Zheng
Yintong Zheng el 27 de Feb. de 2016
I insert all the values of tau, s, c, belta at command window. And all those '-Inf' are what I got.

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the cyclist
the cyclist el 27 de Feb. de 2016
MATLAB has 1-based indexing. When you enter
tau(k)
for k = 0, there is no such element.

Walter Roberson
Walter Roberson el 27 de Feb. de 2016
You have, in part, (x-tau(k)).^(-1) . If there is any x equal to tau(k) then the subtraction would give 0 and raising that to power -1 would give infinity. Infinity times any non-zero value gives infinity (infinity times 0 gives nan), so you would be adding infinity to the value, resulting in infinity. And then since you are effectively totaling values, you are going to get an infinite result.
  2 comentarios
Yintong Zheng
Yintong Zheng el 27 de Feb. de 2016
can you advice me how to fix this?
Walter Roberson
Walter Roberson el 27 de Feb. de 2016
If you do have an x equal to some tau, then inf or nan is the correct answer for the formulae you have written. The situation is exactly like asking to total 1/x when x is allowed to be 0.

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Image Analyst
Image Analyst el 27 de Feb. de 2016

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