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20mA
32,764 10V
0 0V
(Scaled Max.)
(Scaled Min.)
(Shifted Input Min.) (Shifted Input Max.)
Calculating the Shifted Linear Relationship
Use the following equations to calculate the scaled units:
Scaled value = (input value x rate) + offset
Rate = (scaled max. – scaled min.) / (input max. – input min.)
 (32,764 − 0) / (13,107 − 0) = 2.4997 (or 24,997/10000)
Offset = scaled min. – (input min. x rate)
 0 – (0 × 2.4997) = 0
PrefaceInstruction Set Reference Manual
3–22
In this example, the SCL instruction is entered in the ladder logic program as
follows:
SCALE
Source N7:0
Rate [/10000] 24997
Offset 0
Dest O:2.0
SUBTRACT
Source A I:1.0
Source B 3277
Dest N7:0
Analog Input
Analog Output
Apply the Shift
Scale Shifted Analog Value
SUB
SCL
Math Instructions
3–23
Application Example 3 – Convert Voltage Input to Percent (MicroLogix)
The following example takes a 0V to 10V analog input from a MicroLogix 1000
analog controller and scales the raw input data to a value between 0 and 100%. The
input value range is 0V to 10V which corresponds to 0 to 31,207 counts. The scaled
value range is 0 to 100%.
Scaled Value
(percent)
Input Value
0V 31,207 10V
100
0
(Scaled Max.)
(Scaled Min.)
(Input Min.) (Input Max.)
0
Calculating the Linear Relationship
Use the following equations to calculate the scaled units:
Scaled value = (input value x rate) + offset
Rate = (scaled max. – scaled min.) / (input max. – input min.)
 = (100 – 0) / (31,207 – 0)
= .00320 (or 320/10000)
Offset = scaled min. – (input min. x rate)
 = 0 − (0 × .00320) = 0
PrefaceInstruction Set Reference Manual
3–24
Absolute (ABS)
Use the ABS instruction to calculate the absolute value of the Source and place the
result in the Destination. This instruction supports integer and floating point values.
Use this instruction with SLC 5/03 (OS302), SLC 5/04 (OS401), and SLC 5/05
processors.
Entering Parameters
Enter the following parameters when programming this instruction:
• Source can be a word address, an integer constant, floating point data element,
or a floating point constant.
• Destination can only be a word address or a floating point data element.
Updates to Arithmetic Status Bits
With this Bit: The Processor:
Carry (C) always resets.
Overflow (V) always resets with a floating point value; sets if the input is –32,768
(integer value).
Zero (Z) sets when destination value is zero; otherwise resets.
Sign (S) always resets.
333
ABSOLUTE VALUE
Source
Dest
ABS
Output Instruction
Math Instructions
3–25
Compute (CPT)
The CPT instruction performs copy, arithmetic, logical, and conversion operations.
You define the operation in the Expression and the result is written in the
Destination. The CPT uses functions to operate on one or more values in the
Expression to perform operations such as:
• converting from one number format to another
• manipulating numbers
• performing trigonometric functions
Use this instruction with SLC 5/03 (OS302), SLC 5/04 (OS401), and SLC 5/05
processors.
Instructions that can be used in the Expression include:
 +, –, *, | (DIV), SQR, – (NEG), NOT, XOR, OR, AND, TOD, FRD, LN,
TAN, ABS, DEG, RAD, SIN, COS, ATN, ASN, ACS, LOG, and ** (XPY).
Note The execution time of a CPT instruction is longer than a single arithmetic
operation and uses more instruction words.
Entering Parameters
Enter the following parameters when programming this instruction:
• Destination can be a word address or the address of a floating-point data
element.
• Expression is zero or more lines, with up to 28 characters per line, up to 255
characters.
Updates to Arithmetic Status Bits
With this Bit: The Processor:
Carry (C) sets based on the result of the last instruction in the Expression.
Overflow (V) sets any time an overflow occurs during the evaluation of the Expression.
Zero (Z) sets based on the result of the last instruction in the Expression.
Sign (S) sets based on the result of the last instruction in the Expression.
The above bits are cleared at the start of the CPT instruction. See S:34/2 for special
handling of the math status bits when using floating point.
333
COMPUTE
Dest
Expression
CPT
Output Instruction
PrefaceInstruction Set Reference Manual
3–26
Application Example
This application example uses Pythagorean’s theorem to find the length of the long
leg of a triangle, knowing the two other leg lengths. Use the following equation:
c2 = a2 + b2
Rung 2:0 uses standard math instructions to implement Pythagorean’s theorem.
Rung 2:1 uses the CPT instruction to obtain the same calculation.
| Rung 2:0 +XPY–––––––––––––––+ ||–––––––––––––––––––––––––––––––––––––––––––––––––––––+–+X TO POWER OF Y +–+–|| | |Source A N7:1| | || | | 3| | || | |Source B 2| | || | | | | || | |Dest N7:3| | || | | 0| | || | +––––––––––––––––––+ | || | +XPY–––––––––––––––+ | || +–+X TO POWER OF Y +–+ || | |Source A N7:2| | || | | 4| | || | |Source B 2| | || | | | | || | |Dest N7:4| | || | | 0| | || | +––––––––––––––––––+ | || | +ADD–––––––––––––––+ | || +–+ADD +–+ || | |Source A N7:3| | || | | 0| | || | |Source B N7:4| | || | | 0| | || | |Dest N7:5| | || | | 0| | || | +––––––––––––––––––+ | || | +SQR–––––––––––––––+ | || +–+SQUARE ROOT +–+ || |Source N7:5| || | 0| || |Dest N7:0| || | 0| || +––––––––––––––––––+ || Rung 2:1 +CPT––––––––––––––––––––––––+ ||––––––––––––––––––––––––––––––––––––––––––––––––+COMPUTE +–|| |Dest