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How to plot a random impulse train with random height and space? From the following code, I was able to get a random height graph and a random interval space graph.

randHeights = Mod[{RandomVariate[NormalDistribution[5, 2], 20]}, 1];
randSpace = RandomVariate[NormalDistribution[5, 2]];
f[t_] := 1;

GraphicsRow[{
  ListPlot[randHeights, Filling -> Axis, PlotStyle -> Red, 
   PlotRange -> All, AspectRatio -> .5],
  
  ListPlot[Table[{t, f[t]}, {t, 0, 50, randSpace}], Filling -> Axis, 
   AspectRatio -> .5]
  }]

You may get a different result from the one I have here, and I'm still unsure how to combine these two concepts.

I'm looking for a single graph that has different intervals & heights (with a maximum of 1) on every run. Any idea, anyone? Thanks for the help!
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  • $\begingroup$ randSpace = Table[RandomVariate[NormalDistribution[5, 2]], 1000]; and Sort[randSpace][[1 ;; 10]] can easily generate negative numbers. Usually random delay is specified around a base delay and is a fraction of a base delay. $\endgroup$
    – Syed
    Commented Dec 6, 2021 at 16:54
  • $\begingroup$ @Syed I tried to use this code, but it's not generating any graph. Is it because of the Sort command? $\endgroup$
    – nightcape
    Commented Dec 7, 2021 at 12:29
  • $\begingroup$ This was to show that delays could be negative too (not just zero) if a UniformDistribution were chosen. $\endgroup$
    – Syed
    Commented Dec 7, 2021 at 14:03

1 Answer 1

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Something like this?

n = 20;
t1 = RandomReal[{0, 1}, n];
t2 = RandomReal[{0, 1}, n];
t1 = Accumulate[t1];
dat = Transpose[{t1, t2}];
ListPlot[dat, Filling -> Axis]

enter image description here

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  • $\begingroup$ Almost, but I don't want to make it overlap or have a close interval from one to another, that is why I use NormalDistribution $\endgroup$
    – nightcape
    Commented Dec 6, 2021 at 11:12
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    $\begingroup$ I think you can adapt this yourself $\endgroup$
    – Daniel Huber
    Commented Dec 6, 2021 at 11:33
  • $\begingroup$ Got it, I use RandomVariate[NormalDistribution[10, 2], n] for one of the t. Thanks! $\endgroup$
    – nightcape
    Commented Dec 7, 2021 at 2:37
  • 1
    $\begingroup$ 👍 Glad to hear this. You learn more if you do it yourself. $\endgroup$
    – Daniel Huber
    Commented Dec 7, 2021 at 9:25

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