
Appendix B. Processing Time Limitations
If the T
t
for Output Option XXXX is greater than the specified interval, the
extra time will be added to the execution time of Instruction 101.
Example 1
Given
:
Output Option = XXXX
Specified interval = S = 500 ms
Function = frequency (Function 2)
Number of channels measured = N = 6
Average input frequency = F = 1kHz
Find
: Total Processing Time, T
t
Solution
:
T
t
= T
o
+ΣT
i
= T
o
+T
2
= 0.511*S+E
t
*(.034+.010*N) +0.015*E
2
= 583 ms
S = 500 ms
N = 6
E
t
= N * S * F = 3000
E
2
= E
t
= 3000
In this case 83 ms is added to the execution time of Instruction 101.
Example 2
Given
: N = number of channels measured
F = Frequency of input signal, kHz
Function = frequency (Function 2)
Output Option = 0 or 0--
Find
: Maximum average frequency, F, at which the processing time is ≤ the
measuring/storing time.
Solution
:
T
t
must be ≤ to S;
T
o
+T
2
≤S
0.511*S +E
t
*(.034+.010*N) + 0.015*E
2
≤ S
Note: E
t
= E
2
= N*F*S
0.511*S+N*F*S*(.034+.010*N)+
0.015*N*F*S≤S
F<.489/(N*(.049+.01*N))
Results for the above example using all Functions is presented in TABLE 2,
Section 5.4.1
B-2
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