%Project 3 problem 7: finding the value of A where sstable equilibrium becomes unstable.
%Computes approximate solution of stablized inverted pendulum
%Input: inline function f; a,b such that f(a)*f(b)<0,
%     and tolerance tol
%Output: Approximate solution xc
function xc = bisectpend2(a,b,tol)
if sign(func(a))*sign(func(b)) >= 0
  error('f(a)f(b)<0 not satisfied!') %ceases execution
end
fa=func(a);
fb=func(b);
k = 0;
while (b-a)/2>tol
  c=(a+b)/2;
  fc=func(c);
  if fc == 0              %c is a solution, done
    break
  end
  if sign(fc)*sign(fa)<0  %a and c make the new interval
    b=c;fb=fc;
  else                    %c and b make the new interval
    a=c;fa=fc;
  end
end
xc=(a+b)/2;               %new midpoint is best estimate

function pn = func(theta)

A=oscillatingpivot([0 50],[0.00000001 0],2000,theta,0,1);

[~,~,time]=size(A);
pn=1;
for i=1:time
    if A(2,4,i)==1
        pn=-1;
        break
    end
end

% The smallest value of the forcing strength A = 14.727873288228963

Error using bisectpend2 (line 7)
Not enough input arguments.

Published with MATLAB® R2013a