Input argument "alpha" is undefined.

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naeem
naeem 2014-11-27
评论: Torsten 2014-12-1
i get this error. how to solve this error?????
""""??? Input argument "alpha" is undefined.
Error in ==> newvod3RL at 6 alphaINT = (real(alpha)>0) .* ceil(real(alpha)) + ... """
function out = vod3RL(t, in, alpha)
% function out = vod3RL(t, in, alpha)
% Variable order derivative, Riemann-Liouville definition, variant 3.
% Duarte Valrio, 2009.
alphaINT = (real(alpha)>0) .* ceil(real(alpha)) + ...
( (real(alpha)==0) & (imag(alpha)~=0) ); % number of differentiations to perform
alphaRES = alpha - alphaINT; % residue of alphaINT with negative real part only
% negative differentiation (i.e. integration) performed
out = zeros(size(t));
for k = 2 : length(t) % the first integral is always 0, since no time elapsed, hence no need to make k = 1...
tau = t(1:k);
integrand = (t(k) - tau).^(-alphaRES(k - (1:k) + 1) - 1) ./ gamma(-alphaRES(k - (1:k) + 1)) .* in(1:k);
if any(~isfinite(integrand)) % eliminate eventual inf and nan points
integrand(find(isinf(integrand))) = 0;
integrand(find(isnan(integrand))) = 0;
end
integrand(1) = 2*integrand(1); % this is to handle steps properly with Caputo's definition
% (otherwise the first point only has half the influence and the result comes halved all the way to the end)
out(k) = sum( (integrand(1:end-1)+integrand(2:end))/2 .* (tau(2:end)-tau(1:end-1)) );
end
% necessary differentiations performed
while 1
if sum(alphaINT) == 0, break, end % exit condition
temp = diff([0; out]) ./ [t(2)-t(1); diff(t)]; % this is to handle steps properly
out = out.*(alphaINT==0) + temp.*(alphaINT~=0);
alphaINT = (alphaINT-1) .* (alphaINT~=0); % take away one integration
end
alphaINT = (real(alpha)>0) .* ceil(real(alpha)) + ...
  1 个评论
Torsten
Torsten 2014-11-27
The error addresses a function called "newvod3RL", but you posted "vod3RL".
I suspect that a variable "alpha" does not appear in the list of parameters for function "newvod3RL".
Best wishes
Torsten.

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回答(1 个)

Stalin Samuel
Stalin Samuel 2014-11-27
编辑:Stalin Samuel 2014-11-27
before using a variable alpha in function you must use that variable alpha.It may be initialization or some other value
  2 个评论
naeem
naeem 2014-11-30
i am supplying three inputs as
alpha=0.9 in=0.5 t=1 and now i am getting an error of
??? Attempted to access t(2); index out of bounds because numel(t)=1.
Error in ==> vod1RL at 30
temp = diff([0; out]) ./ [t(2)-t(1); diff(t)]; % this is to handle steps properly
Torsten
Torsten 2014-12-1
But you write yourself that you supply t=1 to vod3RL.
This is a scalar value - consequently t(2) (which you address to in the above expression) does not exist.
Best wishes
Torsten.

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