I have a mathematical problem with calculating the average value of the analog current/Kilowatt signals received from the RTU over one full minute.

I have used the normal average calulation method
below which is given the wrong result.

For example

time Value
s.ms ( A )
03.600 10
10.200 12
11.200 20
59.900 15
Avr = (10+12+20+15)/4 = 14.25 A

I know that this is wrong but I do not know where can I find the other way.

I have heard that integate calculation is need to be used but I do not know the fulmular.

So please advise me how and what is the right method to calculate the time oriented average calculation for analog value.

 

In your calculation, sampling rate should be same. In the other word you need sampling current with constant time. if not you have to include variable time in your formula.

Regards,
Udin

 

Responding to Tong's query:

If the measurements are instantaneous values taken at the 4 times noted and the values at 0.0 and 60.0 ms are zero, then you calculate the RMS equivalent or Root-Mean-Square value. That is, the square-root of the mean-value of the square. The formula is:

A(rms) = SQRT [ (A1^2+A2^2+A3^2+A4^2)/4 ]

For your case the value is 14.74 !

However, if the values at 0 and 60 ms are not zero then you should consider another method. Contact me if you can obtain y-values (ordinates) corresponding to the time-base (abscissa) divided into 6 equal parts.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

the sampling rate is not the same since the values will be sent from the RTU to the control center once the process valuse being changed.

I need the average value of the received value from the RTU over one full minute.

So could you please tell me what should be the suitable formula in this case?

 

Time-weight the numbers. It would be best if you use midpoints but end-points will probably produce similar results. If T=time, the algorithm might look like this:

Avg=[(T1-T0)*(A1-A0)/2]+[(T2-T1)*(A2-A1)/2]+...[(Tn-Tn-1)*(An-An-1)/2]

Joe Hohn

 

Responding to Joe Hohn's Thu, Jul 14, 5:09pm comments:

A) I want to apologize to Tong for my earlier reply. It was way, way off base.

B) Joe your formula has several errors:

1. It addresses only the triangle segment of each interval area.

2. The rectangular segment (if any) is omitted!

3) The calculation produces a dimension in kW-msec or Amp-ms. The total area must be divided by the time of one period (60 msec) in order to yield the average kW or Amp.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Sorry Tong. Thanks Phil.

 

Responding to Joe Hohn's Jul 16 reply... all is not lost:

Change the sign in the height terms for each interval from (-) to (+) Then, divide the total sum of intervals by Tn.

Changing subtraction to addition yields the area of a trapezoid for each interval. It also gives the triangular area of the first and last intervals.

The corrected formula will give 15.73 as the average kW or Amps.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Dear Phil,

Thanks Phil for your suggestion. Your suggestion have solved by problem.

By finding the total area of the ploted curved and divided it by the time is the right way to get the average value.

Regards,
Tong