Problem with fzero function

15 views (last 30 days)
OldCar
OldCar on 22 Feb 2016
Commented: Star Strider on 23 Feb 2016
I had some troubles in using fzero.
Here is my script: lambda=0.23; T_orb=86400; D=130*1000; omega=2*pi/T_orb; distance=1000*[37700;38400;39000;39600]; for q=1:4 APS.tint(q)=fzero(@(T_int)[distance(q)*lambda/(2*D/2*-(cos(omega*T_int+omega*(T_orb/4-T_int/2))-cos(omega*(T_orb/4-T_int/2))))-2000],100);
end
and it gives me as result tint=[1.527152190882263e-12 1.527032893943011e-12 1.526723991273891e-12 1.527144367230523e-12], but using this values the function has not a zero! Using a bigger first guess (like 1000) it finds the correct first solution (the function is periodical). Why it gives me this result of magnitude 10^-12?
  1 Comment
jgg
jgg on 22 Feb 2016
If your function is non-linear, fzero can find local near-zeroes. You will want to trial it from several points.
I can't debug your code further because you didn't include lambda.

Sign in to comment.

Accepted Answer

Star Strider
Star Strider on 22 Feb 2016
‘Why it gives me this result of magnitude 10^-12?’
That means that the value of ‘T_int’ it estimates is close to zero.
You left out ‘lambda’, so I created a value for it. I have no idea what its true value should be, so if I chose the wrong value, your function could have true zeros, but it does not for the ‘lambda’ I chose.
However, your function does not have any true zero-crossings in the sense that fzero can locate them. Instead, it has discontinuities with sign-reversals, so you need to search for them. Define the ‘T_int’ vector here to have the range and resolution (number of points) you need to define the sign-reversal points with acceptable accuracy.
The code:
lambda = 1000; % <— Insert Correct Value Here
T_orb=86400;
D=130*1000;
omega=2*pi/T_orb;
distance=1000*[37700;38400;39000;39600];
fcn = @(T_int)[distance(q).*lambda./(2*D/2*-(cos(omega*T_int+omega*(T_orb/4-T_int/2))-cos(omega*(T_orb/4-T_int/2))))-2000];
T_int = linspace(0, 5E5, 100);
y = fcn(T_int);
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Zero-Crossing Indices Of Argument Vector
zero_cross_idx = zci(y); % Approximate Zero-Crossing Indices Of Your Function
figure(1)
plot(T_int, y)
hold on
plot(T_int(zero_cross_idx), y(zero_cross_idx), '*r')
hold off
grid
  2 Comments
OldCar
OldCar on 23 Feb 2016
I have wrote the lambda value. I have not understood your answer. Why fzero gives me a value that drives to a big value of the fuction?
Star Strider
Star Strider on 23 Feb 2016
Even with ‘lambda’ defined, my Answer remains the same. Plot it and you will see the reason.
Your function is periodic, however it periodically goes to +Inf then goes to -Inf and does not cross y=0 between them, so your function has no true zero crossings in the sense that fzero can determine them.
Testing for the +Inf to -Inf sign changes is the only way to detect the periodicity. The plot function connects the +Inf and -Inf values with a line, but that line is not an actual zero-crossing. Your function is not defined between the +Inf and -Inf transitions, so it does not cross zero there.

Sign in to comment.

More Answers (0)

Products

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!