STEP 5 Operations S5-115F Manual
015
ExplanationSTL
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L DW 2
SLW 3
T DW 3
Load the contents of data word DW 2 into ACCUM 1.
Shift the bit pattern in ACCUM 1 three positions to the left.
Transfer the result (contents of ACCUM 1) to data word DW 3.
Numeric Example
The value 464
10
is stored in data
word DW 2. Multiply this value by
2
3
=8. Do so by shifting the bit
pattern of DW 2 in ACCUM 1 three
positions to the left.
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0 0 0 0 1 1 1 0 1 0 0 0 0 0 0 0
15 0
0 0 0 0 0 0 0 1
3712
10
ACCUM 1
ACCUM 1
464
10
(DW 2)
1 1 0 1 0 0 0 0
SLW 3
015
ExplanationSTL
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L IW 128
SRW 4
T QW 160
Load the value of input word IW 128 into ACCUM 1.
Shift the bit pattern in ACCUM 1 four positions to the right.
Transfer the result (contents of ACCUM 1) to output word QW 160.
Numeric Example
The value 352
10
is stored in IW 128.
Shift the corresponding bit pattern
in ACCUM 1 four positions to the
right to divide the value 352
10
by
2
4
= 16.
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0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0
15 0
0 0 0 0 0 0 0 1
22
10
ACCUM 1
ACCUM 1
352
10
(IW 128)
0 1 1 0 0 0 0 0
SRW 4
3-50
EWA 4NEB 811 6149-02
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