Hioki MEMORY HiCORDER 8855, Инструкция по эксплуатации

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APPENDIX 15
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Appendix 3 Reference
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(8) Differentiation once (DIF)
(9) Differentiation twice (DIF2)
1st and 2nd differential are calculated using the 5th-order Lagrange
interpolation equation, whereby data from a range of five surrounding points
are used to determine the value of the current point.
Data corresponding to sample time
t
1
- t
n
are taken as d
1
- d
n
and used for
calculating the differential.
When the input voltage becomes small, processing results will show little
variation. In such a case, apply the MOV operator.
1st differential
Point t
1
b
1
= (-25d
1
+48d
2
-36d
3
+16d
4
-3d
5
) /12h
Point t
2
b
2
= (-3d
1
-10d
2
+18d
3
-6d
4
+d
5
) /12h
Point t
3
b
3
= (d
1
- 8d
2
+ 8d
4
- d
5
) /12h
Point t
i
b
i
= (d
i-2
-8d
i-1
+8d
i+1
-d
i+2
) /12h
Point t
n-2
b
n-2
= (d
n-4
-8d
n-3
+8d
n-1
-d
n
)/12h
Point t
n-1
b
n-1
= (-d
n-4
+6d
n-3
-18d
n-2
+10d
n-1
+3d
n
)/12h
Point t
n
b
n
= (3d
n-4
-16d
n-3
+36d
n-2
-48d
n-1
+25d
n
)/12h
b
1
to b
n:
data of calculation result
h =
∆
t : sampling period
2st differential
Point t
1
b
1
= (35d
1
-104d
2
+114d
3
-56d
4
+11d
5
)/12h
2
Point t
2
b
2
= (11d
1
-20d
2
+6d
3
+4d
4
-d
5
)/12h
2
Point t
3
b
3
= (-d
1
+16d
2
-30d
3
+16d
4
-d
5
)/12h
2
Point t
i
b
i
= (-d
i-2
+16d
i-1
-30d
i
+16d
i+1
-d
i+2
)/12h
2
Point t
n-2
b
n-2
= (-d
n-4
+16d
n-3
-30d
n-2
+16d
n-1
-d
n
)/12h
2
Point t
n-1
b
n-1
= (-d
n-4
+4d
n-3
+6d
n-2
-20d
n-1
+11d
n
)/12h
2
Point t
n
b
n
= (11d
n-4
-56d
n-3
+114d
n-2
-104d
n-1
+35d
n
)/12h
2
(10) 1st integral (INT)
(11) 2nd integral (INT2)
The 1st and 2nd integral calculation uses the trapezoidal rule.
Data corresponding to sample time
t
1
- t
n
are taken as d
1
- d
n
and used for
calculating the integral.
1st integral
Point t
1
I
1
= 0
Point t
2
I
2
= (d
1
+d
2
)h/2
Point t
3
I
3
= (d
1
+d
2
)h/2 + (d
2
+d
3
)h/2 = I
2
+ (d
2
+d
3
)h/2
Point t
n
I
n
= I
n-1
+ (d
n-1
+d
n
)h/2
I
1
to I
n:
processing result data
h =
∆
t: sampling period