3 ms·
You are “blurring” the samples because you know your function is NOT a step-function, essentially. So instead of trying to be “super precise” (you cannot), you
by pfortuny 6y ago
You are “blurring” the samples because you know your function is NOT a step-function, essentially. So instead of trying to be “super precise” (you cannot), you “blur” your Fourier transform.
It is NOT that but my explanation is morally what you achieve.