plotting step function with variable step length

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sasha
sasha 2014-12-15
评论： sasha 2014-12-15
I want to do a plot of y as a function of time. Where the values of y are either zero or one and they alternate. The time that y is either zero or one is given by t(n) where I solved for t(n) (a vector of different times). Basically I want to have y(t=0 -> t=t(1)) = 0, then y(t=t(1) -> t = t(2)) = 1, then y(t = t(2) -> t(3)) = 0, and so on. I have already solved for a distribution of the t's. I just don't know how to plot this.
  1 个评论
Henrik
Henrik 2014-12-15
Your notation is difficult to understand, so I don't know exactly what you're asking.
From what I understood, you could make a vector of the t-values, another vector with the y-values, and plot them using
plot(t,y)
example:
t=[1 2 4 6 7 9];
y=[1 1 -1 0 -1 1];
plot(t,y)

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回答（1 个）

Guillaume
Guillaume 2014-12-15
One possible way:
t=[2 7 10 15 20]; %for example
x = [0 reshape(repmat(t(1:end-1), 2, 1), 1, []) t(end)];
y = repmat([1 1 0 0], 1, ceil(numel(x)/4));
y = y(1:numel(x));
plot (x, y)
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sasha
sasha 2014-12-15
I actually messed around with it and it appears that defining
n = 1;
n1 = 1;
while n< nmax
x(n+1) = x(n) + t(n);
n = n+1;
end
while n1<nmax/2
t_odd = 2*n1-1;
t_even = 2*n1;
y(t_odd) = 0;
y(t_even) = 1;
n1 = n1+1
end
because t is just the time step, while now x is the absolute time. And should have the first value be at 0 because we start with a zero by the definition of my plot.
And then just using
stairs(x,y)
Gives the correct plot

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