Generate unevenly-spaced synthetic time series

Hello,

I have created a synthetic time series (x) over time (tsyn), but I need to remake this so that the spacing between points in tsyn is both uneven and random between 1 and 5.

tsyn = 1:1600;
x = 2*sin(2*pi*tsyn/100)+1*sin(2*pi*tsyn/41)+0.5*sin(2*pi*tsyn/21);
figure;
plot(tsyn,x);

I thought it would be something like the below code, but that didn't work.

tsyn = 1:randn(1:5):1600; 

Any advice?

[Adam Danz]

tsyn is only integers. Is that a requirement?

[John D'Errico]

I assumed non-integer, because randn is floating point, and that is what was attempted for use. But integer increments are a possibility. You never know. :)

[Adam Danz]

Ah, I just saw this. I updated my solution to cover both possibilities.

Answer 1

[Adam Danz]

If your samples are not integers

You can set the maximum sample, the [min, max] of your bounds, and the starting number. tsyn is a monotonically increasing vector with random intervals from a bounded, uniform distribution starting at your specified starting value and ending at or just before your maximum sample.

maxSample = 1600;       %maximum value allowed
bounds = [1,5];         %bounds to the uniformly distribution random intervals between samples (inclusive)
startNum = 1;           %first sample (starting point)

% Create samples at random intervals
minSamples = floor(maxSample / bounds(1));   %you'll end up with at least this many samples
randIntervals = (rand(minSamples,1) * range(bounds)) + bounds(1);
tsyn = [startNum; cumsum(randIntervals)+startNum]; 
tsyn(tsyn > maxSample) = []; %get rid of extra values

figure
 histogram(diff(tsyn), bounds(1):bounds(2))
 xlabel('Intervals')
 ylabel('Frequency')
 set(gca, 'Xtick', bounds(1):bounds(2))

If your samples should be intergers

Replace the "randIntervals =" line above with the one below

randIntervals = randi(range(bounds)+1, minSamples, 1)+bounds(1)-1;

Test

And if you need convinced that the intervals are randomly distributed between your bounds:

figure
 histogram(diff(tsyn), bounds(1):bounds(2)+1)
 xlabel('Intervals')
 ylabel('Frequency')
 set(gca, 'Xtick', bounds(1):bounds(2))

Answer 2

[John D'Errico]

It failed, because the colon operator does not accept a random increment. Anyway, randn(1:5) does not generate a random number between 1 and 5 anyway. I'd suggest you need to read the getting started tutorials.

Doing what you have asked is not trivial however. One approach would use Poisson arrivals. But I recall that would imply an exponential interarrival time. And you are asking for a uniformly distributed interarrival time.

Simplest would be to use Roger Stafford's randfixedsum, available for free download from the File Exchange.

If the spacing is to be from 1 to 5, then the average spacing is 3. Over the interval [1,1600], that means you would expect

(1600-1)/3
ans =
   533

So you want to find a set of 533 random numbers, uniformly distributed on the interval [1,5] that sum to 1599.

delta = randfixedsum(533,1,1599,1,5)';
tsyn = cumsum([1,delta]);

As you can see, this set has the desired properties. It runs from 1 to 1600.

tsyn(1)
ans =
     1
     
tsyn(end)
ans =
                      1600

As well, it is uniformly distributed in terms of the inter-arrivals. (Well approximately so. With a larger sample size, it will be closer to uniform.)

min(diff(tsyn))
ans =
          1.00672171800682
          
max(diff(tsyn))
ans =
          4.99817125723121
          
hist(diff(tsyn),10)          

You can find randfixedsum here:

https://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum