Siemens 6ED1050-1AA00-0BE8 manual Possible errors Zero division and overflow, Examples, Result

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4.4 Special functions list - SF

Functional Description

The mathematic instruction function combines the four operands and three operators to form an equation. The operator can be any one of the four standard operators: +, -, *, or /. For each operator, you must set a unique priority of High (H), Medium (M), or Low (L). The high operation will be performed first, followed by the medium operation, and then by the low operation. You must have exactly one operation of each priority. The operand values can reference another previously-defined function to provide the value. The mathematic instruction function rounds the result to the nearest integer value.

The number of operand values is fixed at four and the number of operators is fixed at 3. If you need to use fewer operands, use constructions such as + 0 or * 1 to fill the remaining parameters.

You can also configure the behavior of the function when the Enable parameter En=0. The function block can either retain its last value, or be set to 0. If the parameter Qen → 0 = 0, then the function sets AQ to 0 when En=0. If the parameter Qen → 0 = 1, then the function leaves AQ at its last value when En=0.

Possible errors: Zero division and overflow

If the mathematic instruction function block execution results in zero division or overflow, it sets internal bits that indicate the type of error that occurred. You can program a mathematic instruction error detection function block in your circuit program to detect these errors, and to control the program behavior as needed. You program one mathematic instruction error detection function block to reference one specific mathematic instruction function block.

Examples

The following tables show some simple example mathematic instruction block parameters, and the resulting equations and output values:

V1	Op1	V2	Op2	V3	Op3	V4
	(Pr1)		(Pr2)		(Pr3)
12	+ (M)	6	/ (H)	3	- (L)	1

Equation: (12 + (6 / 3)) - 1

Result: 13

V1	Op1	V2	Op2	V3	Op3	V4
	(Pr1)		(Pr2)		(Pr3)
2	+ (L)	3	* (M)	1	+ (H)	4

Equation: 2 + (3 * (1 + 4))

Result: 17

V1	Op1	V2	Op2	V3	Op3	V4
	(Pr1)		(Pr2)		(Pr3)
100	- (H)	25	/ (L)	2	+ (M)	1

Equation: (100 - 25) / (2 + 1)

Result: 25

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Manual, 04/2011, A5E03556174-01	267