plotting step function with variable step length

Question

sasha 2014-12-15

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https://ww2.mathworks.cn/matlabcentral/answers/166741-plotting-step-function-with-variable-step-length

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.

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Henrik 2014-12-15

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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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Answer 1

Guillaume 2014-12-15

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https://ww2.mathworks.cn/matlabcentral/answers/166741-plotting-step-function-with-variable-step-length#answer_162454

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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 2014-12-15

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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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