CD600 Plus - User's Manual
4.48
EXAMPLE 2:
Using the Taylor Series, the 4th-order polynomial can be used to represent functions as:
24
x
+
6
x
+
2
x
+ x + 1 =
e
432
x
24
a) (x.
4
+
6
a) (x.
3
+
a)
2
lnln
2
. (x
+ a x. + 1 =
a
x
ln
ln
24
1)-(x
4
-
3
1)-(x
3
+
2
1)-(x
2
- 1)-(x = x
ln
The coefficients must be adjusted keeping in mind that they will be multiplied by 100. For example, if
the polynomial is used to represent
e
x
, "x" would be given by the input varying from -100 to +100%.
Therefore
-1 ≤ x ≤ 1 and 0.368 ≤ 'e
x
≤ 2.718.
If the coefficients are used like in the Taylor Series, the output would vary between 36.89% and
271.8%. In order to avoid this, the coefficients must be divided by 2.718:
K
0
= 36.79%
K
1
= 0.3679
K
2
= 0.1839
K
3
= 0.06131
K
4
= 0.01533
Gives:
13.5%
≤ output ≤ 100%
If input represents other values than -1 to 1 an output of 0-100% is desired, other coefficients must
be calculated.
TYPE MNEM DESCRIPTION RANGE DEFAULT
I LIA Input A 0
I LIB Input B 0
I LIC Input C
Addresses
0 to 170/225 to 240
0
0 - A-B difference (F
0
)
P CTYP Type of desired equation
1 - 4th-order polynomial (F
1
) 2 - 3-input
sum (F
2
)
0
R A-K0 Coefficient K
0
-300.00% to 300.00% 0.00%
R A-K1 Coefficient K
1
-10 E 37 to 10 E 37 0
R A-K2 Coefficient K
2
-10 E 37 to 10 E 37 0
R A-K3 Coefficient K
3
-10 E 37 to 10 E 37 0
R A-K4 Coefficient K
4
-10 E 37 to 10 E 37 0
Number of Bytes per Type of Parameter: A = 20 C = 2 L = 6