Running average after each iteration

[hiNI]

Hello,

Attached is an example of my vi file where I capture data for n iterations and then get a final average of the iterations for only the y-data values.

[hiNI]

Attached is an example of my vi file where I capture data for n iterations and then get a final average of the iterations for only the y-data values.  I then plot the final x- vs average y-data.  Is there a way to plot a graph of the average y-data after each iteration instead of just at the end?

Thanks,

hiNi.

[Pnt]

Just place a graph in the "For Loop".

You should really connect wires from left to right.

[hiNI]

Hello,

Firstly, is there a way to get the average of the y-data after each iteration?  The way of how I have it set up, the average of the y-data only comes out at the end of the iterations?  Can you assist me in getting the average of the y-data after each iteration?

Thanks,

hiNi

[Dennis_Knutson]

This is another very basic LabVIEW task. You have the orange wire going to your Y data indicator. Browse to the Mathematics>Probability and Statistics palette and select Mean. Wire it up. You could also simply drag a copy of your existing average calculation to inside the loop.

Please see the free LabVIEW training.

[hiNI]

Hello,

OK, I selected the mean function as you recommended, however, I'm having a problem with the task that I originally asked help for.  I was unclear.  I apologize.  However, this is what I would like to get some help to accomplish:

After the first iteration, the data might be:

x-data y-data (iteration #1)

1           .10

2           .30

3           .50...

The graph of the average response after the 1st iteration will have data points:

x-data y-data (Average)

1           .10

2           .30

3           .50...

For the second iteration, the raw data might be:

x-data y-data (iteration #2)

1           .30

2           .10

3           .30...

The graph of the average response after the 2nd iteration will have data points:

x-data y-data (average of iteration #1 and # 2)

1           .20  ((0.1+0.3)/2)

2           .20  ((0.3+0.1)/2)

3           .40  ((0.5+0.3)/2)...

For the third iteration, the raw data might be:

x-data y-data (iteration #3)

1           .40

2           .80

3           .10...

The graph of the average response after the 3rd iteration will have data points:

x-data y-data (average of iteration #1 and # 2 #3)

1           .30  ((0.4+0.2)/2)

2           .50  ((0.8+0.2)/2)

3           .25  ((0.4+0.1)/2)...

Thanks,

hiNi

[GerdW]

Hi hiNI,

well, what you describe isn't exactly something called a running average usually...

To get your desired result simply add the new data to theold ones and divide by 2 - it's as simple as described as LabView processes arrays the same way as scalars.

To have a "real" running average simply search this forum, you will find more than one example!

Message Edited by GerdW on 09-13-2008 05:45 PM

[altenbach]

---

hiNI wrote:

After the first iteration, the data might be:

The graph of the average response after the 1st iteration will have data points:

The graph of the average response after the 2nd iteration will have data points:

x-data y-data (average of iteration #1 and # 2)

The graph of the average response after the 3rd iteration will have data points:

x-data y-data (average of iteration #1 and # 2 #3)

---

All you need to do is sum the Y data array in a shift register and divide by the number of iteration (i+1) for display and further processing.

You code makes no sense, because you generate random x values and also average those. I assume x= constant and just the array index, else nothing would make sense. I also don't know why you generate 2D arrays with stacked FOR loops and then slice out the only data row. Way too much code!

Since X is constant and equally spaced, you can use a waveform graph. Here's a simple example, see if it makes sense. 🙂

It is not a "running average" of e.g. the last x traces, but an average of all N acquisitions as you specified

Message Edited by altenbach on 09-13-2008 10:04 AM

[hiNI]

THANK YOU!

That was exactly what I needed!!!

hiNi