So I:1/3 will become 1 when the pressure is above 118 I:1/3 will become 0 when the pressure is below 98 (=118-20) I:1/3 will not change otherwise. I thought of one more thing: a few posts ago you showed PE2 (I:1/3) parameters of high limit = 118 and span = 20. Ill post again if i have any more questions. Ill start the Part 4 and see how it goes. Oh okay! thank you and yes you can add comments in every rung! That would not change the behavior, but I think it will be a bit more clear that said, the inverted logic on the inputs might be confusing, so maybe change the names of I:1/2 and I:1/3 to something like and, respectively. Putting O:0/2 at the start of rungs 003 and 004,.You can get rid of Rungs 001 and 002, and also intermediate bits B3:0/2 and B3:0/4 by I'm partial to the top two rungs of the image below. You might want to ask The Google about and see if there are any arrangements that are more clear. I would do the flip-flop (rung 005) a little differently, but that's just a style issue. I did some changes to the program and its working just fine, as it supposed to according to the exercise, is there anything wrong where you can help me to improve it?
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