AH 5 00 Pr ogr am m ing Ma n ua l2 - 8S E xpo ne nt Ma nti ssa8 -b it 2 3- bi tb 31 b 0S ig n b it0 : Po siti ve1 : Neg ati veEquation: ( ) 127;.121 =××− − BMBESThe single-precision floating-point numbers range from ±2 -126 to ±2+128 , and correspond to the rangefrom ±1.1755×10-38 to ±3.4028×10+38.The AH500 series PLC uses two consecutive registers to form a 32-bit floating-point number. Take(D1, D0) for example.S E7 E6 E5 E1 E0 A22 A21 A20 A6 A5 A4 A3 A2 A1 A0b0b1b2b3b4b5b6b20b21b22b23b24b28b29b30b312 2 2 2 2 2 2 2 2 2 2 2 22 2D1 (b 15 ~b 0) D0 (b 15 ~b 0)E xp on en t ( 8 b its) Ma nt issa (2 3b it s)Ma nt issa sign b it (0 : Po sit ive; 1: Ne ga tive)W hen b 0~ b3 1 a re zeros, t he con te nt is zero .T he pos ition w here the d eci mal point is hid den2.2.2.2 Double-precision Floating-point NumbersThe floating-point number is represented by the 64-bit register. The representation adopts theIEEE754 standard, and the format is as follows.S E xpo ne nt Ma nti ssa11- bi t 5 2- bi tb 63 b 0S ig n b it0 : Po siti ve1 : Neg ati veEquation: ( ) 1023;.121 =××− − BMBESThe double-precision floating-point numbers range from ±2-1022 to ±2+1024 , and correspond to therange from ±2.2250×10-308 to ±1.7976×10+308 .The AH500 series PLC uses four consecutive registers to form a 64-bit floating-point number. Take(D3, D2, D1, D0) for example.S E10 E9 E8 E1 E0 A51 A50 A49 A6 A5 A4 A3 A2 A1 A0b0b1b2b3b4b5b6b49b50b51b52b53b60b61b62b632 2 2 2 2 2 2 2 2 2 2 2 22 210 9 8 1 0 -1 -2 -3 -46 -47 -48 -49 -50 -51 -52D3(b15~b0) D2~D0Exponent(11 bits; signed value)Mantissa (52 bits)Mantissa sign bit (0: Positive; 1: Negative)When b0~b63 are zeros, the content is zero.The position where the decimal point is hiddenExample 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 4 (0111 is the mantissa, and 4 is theexponent.).Step 3: Get the value of the exponent.∵E-B=4→E-127=4 ∴E=131=10000011 2
