Talk:Integrator

Attention: Make sure the sum of your inputs does not exceed the machine unit (±10V), otherwise an overload will occur

I don't think this is correct. An overload occurs when the output exceeds the machine unit. However it should be no problem if the sum of the inputs exceeds the machine unit. — Preceding unsigned comment added by MKoch (talk • contribs)

Well, if the sum of the inputs exceeds the machine unit, the output will also do so, isn't it? You could think of cancellation of overloads (like in a sum with three inputs, two in overload but opposite sign, like +15V-15V+5V = +5V). This is a border case similar to saying +infinity - infinity + 5 = 5, because actually an overload value is typically an illegal value on the computer and the upcoming computation is most likely wrong. --Sven (talk) 13:51, 1 October 2021 (CEST)

No, if the sum if the inputs exceeds the machine unit, that doesn't mean the integrator's output will exceed the machine unit. At least not immediately. Your example with +-infinity isn't applicable, because every signal comes from another output and is always in the -1 to +1 range (or slightly above).--MKoch (talk) 14:04, 1 October 2021 (CEST)