3 ms·
Discreetization (what you are referring to) always involves loss, and, rather than instantly getting the "correct" result from your calculations, you will need
by anonymousiam 4y ago
Discreetization (what you are referring to) always involves loss, and, rather than instantly getting the "correct" result from your calculations, you will need to wait for enough clock cycles to progress until the remaining error value is less than the (ever present) noise.