Studio 5000: Working on Structured Text Commands, and...

[AutomationTechBrian]

I'm working on learning Structured Text, and I'm stuck on something so simple:

I have an If-Then statement using Bool tags, and I'm having a hard time understanding why this doesn't work...

If Homing_Process and NOT Homing_Complete then Machine_State_Homing := 1;

NOT is not working. Both Homing_Complete and Machine_State_Homing are 1. Is there another way to write the not-bit?

 

[drbitboy]

Perhaps the syntax is

NOT(Homing_Complete)
​

?

 

[AutomationTechBrian]

No, that didn't work either. I'm checking out the possibility of the result being latched on.

 

[L D[AR2P#0.0]]

How do you know that Homing_Complete and Machine_State_Homing were both 0?

 

[AutomationTechBrian]

I'm hovering the cursor over the tags, which shows me the current value. Everything starts out 0. After homing, all three tags are 1s. I could do this simply in ladder logic, but I can see ST will be a great tool, once I learn it better. So I'm working through the learning curve.

 

[AutomationTechBrian]

I'm using my Tech Connect... I'm on hold. He's having trouble, too.

 

[bb76]

If it's the same as Siemens you have set the bit but have no reset.

You can use

If Homing_Process and NOT Homing_Complete
then Machine_State_Homing := 1
else Machine_State_Homing := 0;

or (in Siemens at least) you could simply use

Machine_State_Homing :=
Homing_Process and NOT Homing_Complete;

 

[Snyd1]

No, that didn't work either. I'm checking out the possibility of the result being latched on.

I know nothing about structured text but do spend a too much time on PLCtalk.

maybe this

If Homing_Process and NOT Homing_Complete then Machine_State_Homing [:= 1]

See http://www.plctalk.net/qanda/showthread.php?t=78088

 

[nehpets]

Have you checked your syntax?



If Homing_Process and not Homing_Complete then
Machine_State_Homing :=1 ;
End_if;


This works for me.


Steve

 

[_Dock_]

I know nothing about structured text but do spend a too much time on PLCtalk.

maybe this

If Homing_Process and NOT Homing_Complete then Machine_State_Homing [:= 1]

See http://www.plctalk.net/qanda/showthread.php?t=78088

This is correct. If your familiar with ladder, think of :=1 as a OTL. You need a condition of := 0 to “unlatch” your expression.

Think of [:= 1] as an OTE, the bit will be a 1 if the expression is true, in other words, not retained.

 

[_Dock_]

Edit: Just read the link, there are better explanations than mine but the same apply.

 

[drbitboy]

I'm hovering the cursor over the tags, which shows me the current value. Everything starts out 0. After homing, all three tags are 1s. I could do this simply in ladder logic, but I can see ST will be a great tool, once I learn it better. So I'm working through the learning curve.

The logic in the OP

If Homing_Process and NOT Homing_Complete then Machine_State_Homing := 1;

is equivalent to an OTL (latch), so if ever Homing_Process was 1 and Homing_Complete was 0 on the same scan, then Machine_State_Homing would have been assigned a 1, and would remain 1 after either of the inputs changed, no?

 

[AutomationTechBrian]

Just got off the phone. I haven't read your responses yet because I'm on-site. I found a solution before he did. I switched it around:

Machine_State_Homing := Homing_Process and NOT(Homing_Complete);

That worked. I don't know why the other didn't, but I'll look at that later on when I'm home.

ST will be a big deal for me. The math is so much easier. This is changing the way I think about solutions.

 

[drbitboy]

Just got off the phone. I haven't read your responses yet because I'm on-site. I found a solution before he did. I switched it around:

Machine_State_Homing := Homing_Process and NOT(Homing_Complete);

That worked. I don't know why the other didn't, but I'll look at that later on when I'm home.

ST will be a big deal for me. The math is so much easier. This is changing the way I think about solutions.

Yah, that changes the logic from an OTL to an OTE i.e. Machine_State_Homing is assigned a value on every scan; in the OP logic that is not true.




Specifically:


This:

If Homing_Process and NOT Homing_Complete then Machine_State_Homing := 1
​

is equivalent to this:

   Homing_Process  Homing_Complete  Machine_State_Homing
---] [-------------]/[--------------(L)--------------------

and this:

Machine_State_Homing := Homing_Process and NOT(Homing_Complete);

(and probably also this:
Machine_State_Homing := Homing_Process and NOT Homing_Complete
)
​

is equivalent to this:

   Homing_Process  Homing_Complete  Machine_State_Homing
---] [-------------]/[--------------( )--------------------

Sorry I didn't see it sooner.

 

[kamenges]

I personally like the "equation" form (A = B AND C). It is just more clear to me. However, if you are more of an IF-THEN type person, the OTE equivalent requires an ELSE:

IF (B AND C) THEN
A := 1
ELSE
A :=0
END_IF

Keith

P.S: Sorry, missed the post by bb76 which says exactly the same thing. My eye skipped right over that.