Delta AH Series Operation Manual page 111

( )
− × − ×
S
E B
1 2 . 1
Equation:
The double-precision floating-point numbers range from ±2
-308
±2.2250×10 to ±1.7976×10
The AH500 series PLC uses four consecutive registers to form a 64-bit floating-point number. Take (D3, D2, D1, D0)
for example.
D1 (b 15 ~b 0)
2 2 2
S E10 E9 E8 E1
b63 b62 b61 b60 b53
E xp on en t (11 bit s )
Ma nt is sa s ign b it (0 : Po s itiv e; 1 : Neg at iv e )
W hen b 0~ b6 3 a re z ero s, t he c on te nt is z er o.
Example 1:
23 is represented by the single-precision floating-point number.
Step 1: Convert 23 into the binary number, i.e. 23.0=10111.
Step 2: Normalize the binary number, i.e. 10111=1.0111 ×2
Step 3: Get the value of the exponent.
∵ E-B=4→E-127=4 ∴ E=131=10000011
Step 4: Combine the sign bit, the exponent, and the mantissa to form the floating-point number.
0 10000011 01110000000000000000000
23 is represented by the double-precision floating-point number.
Step 1: Convert 23 into the binary number, i.e. 23.0=10111.
Step 2: Normalize the binary number, i.e. 10111=1.0111 ×2
Step 3: Get the value of the exponent.
∵ E-B=4→E-1023=4 ∴ E=1027=10000000011
Step 4: Combine the sign bit, the exponent, and the mantissa to form the floating-point number.
0 10000000011 0111000000000000000000000000000000000000000000000000
=403700000000000
Example 2:
-23 is represented by the single-precision floating-point number.
The steps of converting -23.0 into the floating-point number are the same as those of converting 23.0 into the
floating-point number, except that the sign bit is 1.
1 10000011 01110000000000000000000
-23 is represented by the double-precision floating-point number.
The steps of converting -23.0 into the floating-point number are the same as those of converting 23.0 into the
floating-point number, except that the sign bit is 1.
Ch a pt er 5 Un d er s t a n d in g Com m on D e v ic e s
=
M ; B 1023
+308
.
2 2 2 2 2
E0 A51 A50 A49 A6
b52 b51 b50 b49
Ma nt is sa (5 2 b its )
The po si ti on w here the dec im al po int is hidd en
2
=41B80000
2
16
=C1B80000
2 16
-1022 +1024
to ±2 , and correspond to the range from
D2 ~D0
2 2 2 2 2 2 2
A5 A4 A3 A2 A1 A0
b6 b5 b4 b3 b2 b1 b0
4
(0111 is the mantissa, and 4 is the exponent.).
16
4
(0111 is the mantissa, and 4 is the exponent.).
2
2
5 - 1 3
5_