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16bit counter in S7?

Created by: sky59 at: 4/27/2012 2:17 PM (12 Replies)

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4/27/2012 2:17 PM	Rate (0)
sky59 Posts: 30 Rating: (1)	I still learn S7.... I programmed a lof of microcontrollers and with surprise I can not find the way how to do 16bit software counter in s7 something like this: variable type word iii iii = iii + 1 so iii would count from 0 to FFFF then 0 and again to FFFF..... so transition from FFFF to 0 is natural the benefit is to measure intervals when I can simply substract two samples like this: iii(momentA) - iii(momentB) gives interval from A to B moments, of course it works around FFFF and 0 naturally any explanation would be very helpful for me so far I realize only signed mathematics with INT and DINT exist for S7 or am I wrong? is there any instruction for sum of thwo words?
sky59 Posts: 30 Rating: (1)
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4/27/2012 2:55 PM	Rate (1)
MM1234 Posts: 270 Rating: (52)	Hello sky59, as far as I know, it doesn't really matter if the variable is WORD or INT, if you add two WORDs with the +I instruction, it will computethe result like it would do it with two INTs. Now to your application. In the incrementing part you will simply do what you said, with the exception that first you will check if iii = FFFF and if it does, then you will set it to zero. So it will be something like (take it as an inspiration, I haven't programmed in stl for a while and I don't give you proper code like with included initialization of variables etc.): [code]in pseudocode (shall you make it in other than STL): if (iii >= FFFF) iii = 0 else iii = 1 end if [/code] [code]L #iii; L w#16#FFFF; >=I; JCN _res; L 1; L #iii; +I; T #iii; JU _end; _res: L 0; T #iii; _end: nop 0; [/code] Now to the rest, I don't understand this part: sky59 iii(momentA) - iii(momentB) gives interval from A to B moments, of course it works around FFFF and 0 naturally Do you want to assign some certain values to each of the counter values to create some kind of a map (because that's what it looks like to me)? That would be a bit more complicated. But if you just want to substract two different numbers and take into consideration the fact, that FFFF+1 = 0, then just do this: [code]pseudocode (iii1 is the earlier value, iii2 is the later value, result is the difference): if (iii1 - iii2 < 0) result = iii2 - iii1; else result = FFFF - (iii1 - iii2) end if [/code] Problem I see with this is, that you have no clue how many times have you "overflown" the counter, if it was just once, twice or thousand times. Also in this case, if both values are same, it assumes you didn't start and press the counter without a single tick (which I believe is the right behavior). So the code would be something like: [code] L #iii1; L #iii2; -I; L 0; <I; JCU _ovr; T #result; JU _end; _ovr: L W#16#FFFF; L #iii1; -I; L #iii2; +I; T #result; _end: nop 0; [/code] I hope this clears more confusion than it creates and there's not many mistakes, as mentioned, it's a while since I did something with STL. :-]
MM1234 Posts: 270 Rating: (52)	Last edited by: MM1234 at: 4/27/2012 3:03 PM This obviously also doesn''''t ensure saving of your variables in the counter during power shortage, so you might wanna consider using DB for the counter value.
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4/27/2012 3:25 PM	Rate (0)
sky59 Posts: 30 Rating: (1)	thanx for the help pseudocode would be even simpler than you think: iii = iii +1 you do not fhave to care for FFFF it overflovs naturally also calculation is simple iii1 - iii2 and you get the result I am interested only in WORD+1 and WORD-WORD :-) regarding WORD and INT I read somewhere the automatic type checking would be unhappy
sky59 Posts: 30 Rating: (1)
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4/27/2012 3:50 PM	Rate (0)
MM1234 Posts: 270 Rating: (52)	Silly me,I must say,you got me there. Somehow, I expected you asking about the two things, rather than stating them, as not rarely you can see people pulling out some incredible constructs when trying to write something in english and I failed terribly. :-]]] I doubt you would get the right result with your "simple computation" though, assuming that your counter can overflow. If you will have the value of iii1 bigger than the value of iii2, if you will simply substract them, you will get the distance literaly between them, not the amount of ticks between each of the moments. Example to ilustrate: You have counter from 0 to 1000 with overflow. Measured values are iii1 = 500, iii2 = 300. The "gap" between those two (iii1-iii2) is 200. But the amount of ticks it took between them is 800, because it had to count all the way to 1000 and then start from 0 again. Well, asuming the way I suggeste to do the "manual overflow", it would be 801, but that's not the point. :-] I don't think I've ever run into problems with using the +I/-I with WORD, but I can imagine there could be some troubles when you will try to display the value in regard of possibility of signed integer?
MM1234 Posts: 270 Rating: (52)	Last edited by: MM1234 at: 4/27/2012 3:55 PM
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4/27/2012 9:44 PM	Rate (0)
sky59 Posts: 30 Rating: (1)	hmmm, it seems the counter finishes at 999 for step7.... for me it is very strange, it seems no othrer option for me just to accept it I have always been used to have 16 fingers (hexadecimal) now I do not know what to do with 999 :-))
sky59 Posts: 30 Rating: (1)
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4/28/2012 8:34 AM	Rate (0)
Aret Posts: 2348 Rating: (259)	Counters are sign BCD-coded, so unsurprisingly range for standard is +- 999. Although you can use DINT BCD-coded and get +- 999 9999.
Aret Posts: 2348 Rating: (259)
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5/5/2012 4:28 PM	Rate (0)
Marciano Bronze Member Joined: 4/30/2007 Last visit: 3/9/2023 Posts: 231 Rating: (23)	sky59 I is there any instruction for sum of thwo words? Yes, It is possible to increment words L MW10 LB#16#1 +I TMW 10 However you may face problem for your subtraction over 32K, since S7 will see it as a negative number

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5/5/2012 4:53 PM	Rate (0)
SolutionFinder Posts: 559 Rating: (87)	Marciano sky59 I is there any instruction for sum of thwo words? Yes, It is possible to increment words L MW10 LB#16#1 +I TMW 10 However you may face problem for your subtraction over 32K, since S7 will see it as a negative number Dear Marciano, It is possible to add 1 in words. But please take care of Positive edge for the relevant input for whom u want to calculate the counts. Otherwise your addition will be done for every cycle.
SolutionFinder Posts: 559 Rating: (87)
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5/5/2012 6:55 PM	Rate (0)
Marciano Bronze Member Joined: 4/30/2007 Last visit: 3/9/2023 Posts: 231 Rating: (23)	Of course SolutionFinder ...

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5/6/2012 10:24 AM	Rate (0)
Marko Bursic Platinum Member Joined: 9/8/2009 Last visit: 3/15/2023 Posts: 1410 Rating: (148)	In SCL is this solution: Count2:=Count; Count:=DWORD_TO_DINT( DINT_TO_DWORD(Count+1) AND DW#16#FFFF); Diff:=DWORD_TO_DINT( DINT_TO_DWORD(Count-Count2) AND DW#16#FFFF); DW#16#FFFF, can be replaced to any number that is 2^N-1 if you want a ring counter This converts in STL: L #Count L L#1 +D L DW#16#FFFF AD T #Count L#Count L #Count2 -D L DW#16#FFFFF AD T Diff I do use this kind of calculation for ring buffers, fifos, measuring distances on conveyor with encoder,...number must roll-over 2^n-1 -> 0 -> 2^n-1 -> 0 .....

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