回复： Pcie6537and start digital pattern

[ylongwu]

Hello:

I have some problems with the pcie6537card.

I want to continuously collect data, if encounted a specified 32bit tag i needs to save the next 80bit data .then goon to wait the next tag. 

Will the "Start digital pattern "be able to detect the tags?

Another problem is that if using "digital 2D U16 Nsample Nchannel" to obtain the data, and if there are 32 channel, if it means that  in an instance there will be given 32 * 16bit data,or only 32bit? what is the U16 means?

 following are the explains of digital u8 1chan 1Samp:

Digital U8 1Chan 1Samp 

Reads a single 8-bit unsigned integer sample from a task that contains a single digital input channel. Use an instance that reads 8-bit unsigned integers for devices with up to eight lines per port.what is the lines mean?  i thought it was the same mean as channel ,but it seems wrong

thanks !

[yq_LabVIEW]

hi

1 “Start Digital Pattern” will be able to detect the tag and then trigger a finite acquisition.

2 When using "digital 2D U16 Nsample Nchannel", and you set "one channel for one line", then you will acquire 32*16bit Data for each instance.

   The data on a single digital line will be represented as 16bit(U16).

   As for the 80bit Data you specified, the total result after one tag will be 32*16bit*80bit.

AE

NISH

[ylongwu]

Hi yq_LabVIEW , thanks for your help!

1.  you say "The data on a single digital line will be represented as 16bit(U16)" do you meas that it will get 16 bit (are they represent  in a array which has a length of 16 or in a single U16 numeric ?)data in one line during one instance?

then if there are only ten bit data in every lines,and all together there are 32 lines, what will i get during one instance? 

2.to pcie6537,  if i  using "digital 1D U132 Nsample 1channel", and set "one channel for all lines", then what will i acquire for each instance?

  3  you say " As for the 80bit Data you specified, the total result after one tag will be 32*16bit*80bit" 

I mean all the lines' data add together 80bit.  32*16 already larger then 80bits, what do you mean by 32*16bit*80bit?

hope for your help.

 thanks!

[yq_LabVIEW]

Hi

I think you just need a try.

AE

NISH