Omron CQM1H - PROGRAM Programming Manual page 374

Floating-point Math Instructions
Numbers Expressed as Floating-point Values
Normal Numbers
Non-normal Numbers
Zero
Infinity
The following types of floating-point numbers can be used.
Mantissa (f)
0
Not 0
Note A non-normal number is one whose absolute value is too small to be
expressed as a normal number. Non-normal numbers have fewer significant
digits. If the result of calculations is a non-normal number (including interme-
diate results), the number of significant digits will be reduced.
Normal numbers express real numbers. The sign bit will be 0 for a positive
number and 1 for a negative number.
The exponent (e) will be expressed from 1 to 254, and the real exponent will
be 127 less, i.e., –126 to 127.
The mantissa (f) will be expressed from 0 to 2
the real mantissa, bit 2
Normal numbers are expressed as follows:
(sign s) (exponent e)–127
(–1) x 2
Example
31 30
1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Sign: –
Exponent: 128 – 127 = 1
Mantissa: 1 + (2
Value: –1.75 x 2
Non-normal numbers express real numbers with very small absolute values.
The sign bit will be 0 for a positive number and 1 for a negative number.
The exponent (e) will be 0, and the real exponent will be –126.
The mantissa (f) will be expressed from 1 to 2
the real mantissa, bit 2
Non-normal numbers are expressed as follows:
(sign s) –126
(–1) x 2
Example
3130
0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Sign: –
Exponent: –126
Mantissa: 0 + (2
Value: –0.75 x 2
Values of +0.0 and –0.0 can be expressed by setting the sign to 0 for positive
or 1 for negative. The exponent and mantissa will both be 0. Both +0.0 and –
0.0 are equivalent to 0.0. Refer to Floating-point Arithmetic Results, below, for
differences produced by the sign of 0.0.
Values of + ∞ and – ∞ can be expressed by setting the sign to 0 for positive or 1
for negative. The exponent will be 255 (2
0
0 Normal number
Non-normal
number
33
is 1 and the binary point follows immediately after it.
x (1 + mantissa x 2
23 22
22 21 –23
+ 2 ) x 2 = 1 + (2
1
= –3.5
33
is 0 and the binary point follows immediately after it.
–23
x (mantissa x 2 )
23 22
22 21 –23
+ 2 ) x 2 = 0 + (2
–126
8
Section 5-24
Exponent (e)
Not 0 and All 1's (255)
not all 1's
Infinity
NaN
33
– 1, and it is assume that, in
–23
)
0
–1 –2
+ 2 ) = 1 + 0.75 = 1.75
33
– 1, and it is assume that, in
0
–1 –2
+ 2 ) = 0 + 0.75 = 0.75
– 1) and the mantissa will be 0.
349