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

2013 4 10
2 个回答
3 次查看（30 天）

0 个投票

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 个评论
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Jan
Jan 2013-4-10
编辑：Jan 2013-4-10
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
Andrei Bobrov 2013-4-10
sin(2.5*t) instead of sin(2.5)*t

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回答（2 个）

Jan
Jan 2013-4-10
编辑：Jan 2013-4-10

0 个投票

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)
Andrei Bobrov
Andrei Bobrov 2013-4-10

0 个投票

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

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mike
mike
2013-4-10

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