Fergal Cassidy
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John
Are you absolutely sure that a USB 1808 has 8M internal memory to allow for a load of 2 outputs x 2,147,483,647 points?
Regards
Fergal
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int cbAOutScan(int BoardNum, int LowChan, int HighChan, long NumPoints, long *Rate, int Range, int MemHandle, int Options)
Using the cbAOutScan call for the USB 1808 and assuming a 100kHZ output rate, channels 0 to 1 and CONTINUOUS option what is the maximum number of NumPoints that can be loaded.
Maximum Number of Points in the cbAoutScan call
in Measurement Computing (MCC)
Posted
John
Sorry that should read 8GB of memory? Is it that the Universal library co-ordinates the transfer and uses 8GB Windows memory and there is no concern for the memory capacity of the USB 1808.
Regards
Fergal