Matlab graph doesnt look right...not a sinusoid...

Ok so I had to plot a graph that fulfills those conditions in the function.. But when I plot the graph version using fplot(@qtsolver,[0 6*pi]) it plots out a diagonal line with a kink at pi and 2 pi, but shouldnt it print a sinusoid or at least something curvy? I feel like my codes all correct but I dont feel like the graph is :(

function qt = qtsolver(t)

if (0<=t) && (t<pi) qt= 2*t - 0.8*sin(2.5)*t;

else if (pi<=t) && (t<2*pi) qt = 4*pi - 2*t -0.8*sin(2.5)*t-1.6*cos(2.5)*t;

else if (t>= 2*pi) qt = -1.6*cos(2.5)*t;

end

end

end

[Jan]

Of course the graph looks exactly as the code forces it to do. As long as we see the code only, how could we suggest modifications?

Please learn how to format code in the forum. Follow the "? Help" link.

[Andrei Bobrov]

sin(2.5*t) instead of sin(2.5)*t

Answer 1

[Jan]

Your function calculates the points separately. But as far as I can see, fplot expects a vector output for a vector input:

function qt = qtsolver(t)
qt = zeros(size(t));   % Pre-allocate
index     = (0<=t) && (t<pi);
qt(index) = 2*t(index) - 0.8*sin(2.5)*t(index);
index     = (pi<=t) && (t<2*pi);
qt(index) = 4*pi - 2*t(index) -0.8*sin(2.5)*t(index)-1.6*cos(2.5)*t(index);
index     = (t>= 2*pi)
qt(index) = -1.6*cos(2.5)*t(index);

[EDITED], t -> t(index)

Answer 2

[Andrei Bobrov]

function out = qtsolver(x)
    f = {@(t)2*t-.8*sin(2.5*t),...
         @(t)4*pi-2*t-0.8*sin(2.5*t)-1.6*cos(2.5*t),...
         @(t)-1.6*cos(2.5*t)};
    [~,iii] = histc(x,[0 pi 2*pi inf]);
    out = arrayfun(@(y,z)y{:}(z),f(iii),x);
end