(x>-0.5)
So x is greater than -1/2
(x<0.5+0.5.*t)
and x < 1/2 + t/2.
If x is not in that range the the overall result will be 0. If it is in that range then the overall result is determined by the remainder:
(1.*(x>-0.5+t) + (x+0.5)./(1.0*t).*(x<-0.5+t))
That breaks into two contributions:
1.*(x>-0.5+t)
which is 1 if x > t - 1/2 and 0 otherwise
(x+0.5)./(1.0*t).*(x<-0.5+t)
which is (x + 1/2)/t if x < t - 1/2 and 0 otherwise.
Putting the parts together:
if x > -1/2 & x < (t+1)/2 & x < t - 1/2 then (x + 1/2)/t
if x > -1/2 & x < (t+1)/2 & x > t - 1/2 then 1
otherwise 0
There is probably a bug in the formula. Consider x = t - 1/2 then x > -0.5+t is false because x == -0.5+t and the > is strict not >= . So the case that contributes 1 is not invoked. And x<-0.5+t is false becuase x == -0.5+t and the < is strict not <= so the case that contributs (x+1/2)/t is not invoked. So at x = t - 1/2 neither branch is invoked and you are left with all 0. But if you examine the limit as x = t - 1/2 then if x >= t - 1/2 instead of x > t - 1/2 were coded then the contribution would be 1; if x <= t-1/2 instead of x<t-1/2 then (x+1/2)/t would be (t-1/2+1/2)/t which would be (t+0)/t which would be 1 unless t = 0 . The left and right side would both lead to 1 at that case, so the limit would be 1... but the code gives 0. Likely a bug.
The whole formula will fail for t == 0 exactly. The division by 1.0*t would give infinity and infinity times 0 gives nan.
