Difference between revisions of "Talk:Integrator"
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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. {{Unsigned|MKoch}} | 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. {{Unsigned|MKoch}} | ||
: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 <code>+15V-15V+5V = +5V</code>). This is a border case similar to saying <code>+infinity - infinity + 5 = 5</code>, because actually an overload value is typically an illegal value on the computer and the upcoming computation is most likely wrong. --[[User:Sven|Sven]] ([[User talk:Sven|talk]]) 13:51, 1 October 2021 (CEST) | :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 <code>+15V-15V+5V = +5V</code>). This is a border case similar to saying <code>+infinity - infinity + 5 = 5</code>, because actually an overload value is typically an illegal value on the computer and the upcoming computation is most likely wrong. --[[User:Sven|Sven]] ([[User talk: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).--[[User:MKoch|MKoch]] ([[User talk:MKoch|talk]]) 14:04, 1 October 2021 (CEST) | ||
Revision as of 13:04, 1 October 2021
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)
