why chart buffer doesn accept cluster of two arrays?

[looser]

Hi all..
In the attached vi, i m trzing ti ouse xy chart vi in order to plot real time measurement. Well, the problem is one of mz output is scalar tzpe whereas other 1d array. In order to make cluster which is needed for buuerf chart vi., i first convert the scalar to array, and later bundle two arrays whose out put cluster feeds the input of buffer chart vi. But it doesnt accept wiring..""You have connected scalar type to an array of that type"" error given. But I dont get it, i converted the scalar already array, then why do i get this error message

[Dennis_Knutson]

There is nothing in this VI related to a chart or graph.

If you are using the XY Chart Buffer from the XY Chart example, then you do not use the bundle function. The input is either a cluster array or a cluster. Look at XY Chart again. Do you see how it's done? There is a XY Chart Buffer for each plot. Then the outputs are combined with a Build Cluster Array function before being wired to the graph.

[looser]

I am completely confused now. I thought that the best way to draw my x and corresponding values in reat time with history is xy chart.

it s like  x=0     y=5.3

             x=1.3   y= 2.3

                .            .

                .            .

so each x and y value is supposed to be one point in the graph and there ll be line spectrum binding them. Now what I understand from XY chart is that i need to sprovide 2 values to chart by clustering them. If i dont use the chart buffer then how can i determine the history length?

[Dennis_Knutson]

Here is another example of using the XY Buffer. Since you have not modified your VI, I don't know what you are trying to plot and how. What is the x axis and what is the y? You've got three different measurements.

Message Edited by Dennis Knutson on 03-27-2007 08:03 AM

[looser]

exactly here is the my problem i can not explain. you can think my x and y values each of output of random number generator. so my point in the graph will be at intersection of x and corresponding y value.

I have updated vi, u see that there are 2 agilent 34411A reading voltage, and one power supply reading current.

my y value it sgoing to be ratio of those 2 multimeter signal whereas corresponding x value coming from power supply. So after each iteartion i ll have 3 values, i want to plot  a graph which has a point at where Agilent1/Agilent 2  value versus Kepco value. I want to repeat at iteration with new values as keeping history.

Now the problem is that, first the out put of kepco is 1D array whereas that of kepco is 64-bt real scalar. so in order to make them as cluster i need bundle right? after that this bundle should feed the buffer chart where i can set the history length, right? and then i can connect its output to xy chart right? if rioght, i dont know what i dont understand?

i m trapped.please help me

[looser]

in order to clarify i attach also the type of graph that i want to plot!

[Dennis_Knutson]

I will repeat, if you use the XY Chart Buffer, you need a cluster array and not a bundle if you want to plot a whole curve at once. This would be done with something like the attached. If you want to plot a single point with each iteration, then you need to reduce the data in the array to a scalar (i.e. with the Mean function). The you could bundle two scalars together.

[AlexxxR]

Hi Looser,

I´m an AE in Switzerland and worked also in Germany. As an AE i give a lot of Trainingcourses and especially in the Basic 1 Course which is offered from NI, it will be discussed how to wire a XY-Chart or why I need to use a cluster, etc.! From the threads before, I think you should consider to visit one of our training courses.

Best regards,

Alex R

NI Switzerland

PS: You should start with a small example, like Dennis mentioned. Just take the outputs of your Example, fill them with some expected values, take them to a new VI and convert them from indicators to controls, so you can try how to connect it to a XY-Chart!

[looser]

thanks.

I ll consider that

[looser]

Thanks Dennis..

It was the terms that i have misunderstood.Now, i know what to do.

thanks