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Group 8 channel (binary) to octal/hexadecimal

Hi,
 
Let say I have 8 channel of binary data in array and plotted to the graph. If button "Grouping" pressed, then the 9th channel will be plotted, which is in octal value (Sum of binary value at the same sample).
<Please refer to the picture attach>.
I m using HSDIO 6552 to generate and acquisite the data, and I need to group some of the acquisited data.
 
Problem:
I don't know how to generate the 9th channel on the graph, as shown in the attached pic. So, I m seeking help on how to "Add" the octal-value channel. Any1.. pls help.. Thanks in advance.  
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Message 1 of 5
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Hi engwei,

do you need help on generating those "9th channel" or on how to display it?

Generate:
- Pick each bit at a given sampletime, build an array (using the bits in correct order), convert binary array to a number -> your "octal" value for given sampletime...
- Or: convert binary data to number (binary to 0/1). Then scale each "binary" array by it's bit value (1,2,4,8,...,128) and add all arrays to get an array with all numerical values.

Display:
Somewhat more tricky as you need to scale the waveform display to adapt to your array of numbers...
Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
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Message 2 of 5
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Hi GerdWm,

Thanks for your previous reply. My question is on how to display it. Ya, I may need to zoom it to see the value. If this is impossible then maybe I will display the sum of bit in table format. Thanks. . 🙂

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Message 3 of 5
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Hi engwei,
 
One way to display it programmatically is to use property nodes. They allow you to change the visibility, data type, transition type, etc. of an active plot. To access the property nodes of a graph, right-click on it and select Create >> Property Node. Attached is a screenshot of a property node I used to change one of my graph plots. Let me know if you have any further questions, thanks!
 
Cheers,
 
Jonah
Applications Engineer
National Instruments
Jonah Paul
Marketing Manager, NI Software
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Message 4 of 5
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Ok .. Thanks. .. I will try to do it by this method.. Thanks again. 🙂
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