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mike

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

Asked by mike
on 10 Apr 2013

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

  2 Comments

Jan Simon
on 10 Apr 2013

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
on 10 Apr 2013

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

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2 Answers

Answer by Jan Simon
on 10 Apr 2013
Edited by Jan Simon
on 10 Apr 2013

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)

  0 Comments


Answer by Andrei Bobrov
on 10 Apr 2013
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

  0 Comments


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