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本帖最后由 meiyongyuandeze 于 2011-4-13 10:42 编辑
下面是一个求解系数为常数的分数阶微分方程的程序,仅供你参考,希望对你有帮助!
建议斑竹将该帖移☞matlab讨论区!
% This function solves linear fractional-order differential equation (fode)
% with constant coefficients of the form:
%
% n/v (n-1)/v 1/v
% c1(1)*D y(t)+c1(2)*D y(t)+...+c1(n)*D y(t)+c1(n+1)*y(t) =
%
% m/v (m-1)/v 1/v
% c2(1)*D r(t)+c2(2)*D r(t)+...+c2(m)*D r(t)+c2(m+1)*r(t)
%
%
% Inputs:
%
% v : common denominator
% c1 : vector of output coefficients (1x(n+1))
% c2 : vector of input coefficients (1x(m+1))
% r : samples of the input signal r(t)
% h : sampling period
%
% Outputs:
% y : vector of output samples
% t : time vector (corresponding to y)
%
%
%
% Note: this function calls the function "fderiv.m" which is also
% downloadable from MathWorks-File Exchange. The parameter "h" can
% easily be tuned; it must be as small as approximate the input signal
% r(t). The short memory principle has not neen used here, so the
% length of input signal is limited to few hundred samples.
%
% Copyright (c), 2007.
%
function [t,y] = fode2(v,c1,c2,r,h)
n = length(c1)-1;
% r = k*ones(1,100); % if the input signal is unit step
temp1 = zeros(size(r));
for i=1:length(c2)
r_new = fderiv((length(c2)-i)/v,r,h);
temp1 = temp1+c2(i)*r_new;
end
r = temp1;
t = [0:1:length(r)-1]*h;
y = zeros(1,length(r));
temp = zeros(1,n);
a = 0;
for i=1:length(c1)
a = a+c1(i)/h^((n-i+1)/v);
end
for i=1:length(r)
for k=n:-1:1
for j=1:i-1
temp(n-k+1) = temp(n-k+1)+(-1)^j*gamma(k/v+1)*y(i-j)/(gamma(j+1)*gamma(k/v-j+1));
end
temp(n-k+1) = -c1(n-k+1)*temp(n-k+1)/h^(k/v);
end
y(i) = (sum(temp)+r(i))/a;
temp = zeros(1,n);
end
y = [0 y(1:length(y)-1)];
%plot(t,y)
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