Delta DVP-ES3 Series, Руководство по эксплуатации, Страница: 114 / 411

D V P - E S 3 S e r i e s O p e r a t i o n Ma n u a l
5 - 8
_5
Example 1:
23 is represented by a single-precision floating-point number.
Step 1: Convert 23 into the binary number, 23.0=10111.
Step 2: Normalize the binary number, 10111=1.0111 ×24 (0111 is the mantissa, and 4 is the exponent.).
Step 3: Get the value of the exponent.
∵
E-
B=4→E
-127=4
∴
E=131=100000112
Step 4: Combine the sign bit, the exponent, and the mantissa to form the floating-point number.
0 100 0001 1 011 1000 0000 0000 0000 0000
2
=41B80000
16
Example 2:
-23 is represented by a single-precision floating-point number.
Converting -23.0 into the floating-point number uses the same steps as converting 23.0 into the floating-point number,
except that the sign bit is 1.
1 100 0001 1 011 1000 0000 0000 0000 0000
2
=C1B80000
16
5.2.2.2 Decimal Floating-point Numbers

Single-precision floating-point numbers and double-precision floating-point numbers can be converted into decimal
floating-point numbers so people can read them. However, internally the PLC uses single-precision floating-point
numbers and double-precision floating-point numbers.

A 32-bit decimal floating-point number is represented by two consecutive registers. The constant is stored in the first
register whose number is smaller while the exponent is stored in the register whose number is bigger. Take (D1, D0)
for example.
1 0
Decimal f loating-point number=[Constant ]*
D 0
[Exponent ]
D1
Base number D0=±1,000 to ±9,999
Exponent D1=-41 to +35
The base number 100 does not exist in D0 because 100 is represented by 1,000×10
-1
. 32-bit decimal floating-point
numbers range between ±1175×10
-41
to ±402×10
+35
.