Daggos Posted February 24 Share Posted February 24 Hi, I'm trying to get the average of 5 channels as 1 average for all. How would i go about that? Quote Link to comment Share on other sites More sharing options...
AzeoTech Posted February 24 Share Posted February 24 For the current average across 5 channels, it is simply: (ChanA[0] + chanB[0] + ChanC[0] + ChanD[0] + ChanE[0]) / 5 Quote Link to comment Share on other sites More sharing options...
Daggos Posted February 25 Author Share Posted February 25 Sorry was totally overthinking that cheers. Quote Link to comment Share on other sites More sharing options...
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