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

Asked by mike

### mike (view profile)

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

Jan Simon

### Jan Simon (view profile)

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

### Andrei Bobrov (view profile)

on 10 Apr 2013

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

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## 2 Answers

### Jan Simon (view profile)

Answer by Jan Simon

### Jan Simon (view profile)

on 10 Apr 2013
Edited by Jan Simon

### Jan Simon (view profile)

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)

### Andrei Bobrov (view profile)

Answer by Andrei Bobrov

### Andrei Bobrov (view profile)

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
```

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