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why analog input task shows different signal shape than DAQmx Test Panel?

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I have a PXIe-6363multifunction  DAQ device. I need to visualize CHA and CHB of an encoder. I had connected CHA to ai4+ and CHB to ai5+ through a SCC-68 enclosure in differential mode. I am providing +5VDC from the PXI-4110 powersupply for the encoder. I connected ai4- and ai5- to the GND of the encoder power on the SCC-68 screw terminals.

PROBLEM: When I create a TASK acquiring from ai4 and ai5 simultaneously, the signal shape is distorted. See attached image.

If I look at the signals with an external scope touching the screw terminals, the signals have the correct, undistorted shape.

Also, if i look at one of the signals with the DAQmx Test Panel i n MAX, I see the correct, undistorted shape.

I also included a snippet of my code.

 

Is something wrong with my PXIe-6363?

Scientia est potentia!
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I can get the actual shape of the two signals if I lower the sample rate to 40kS/sec.

This is much much lower than the expected 1MS/s. Isn't it true that, if I can do 2MS on one channel I should be able to get 1MS/s acquiring two channels at a time?  Or somewhat lower than that. Certainly not 40kS/s?

Scientia est potentia!
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Solution
Accepted by magyar

Acquiring from a single channel at a time on the 6363 would give you the 2 MHz sample rate.  However, the aggregate sample rate is only specced at 1 MHz due to settling time considerations of the multiplexer used. 

 

What you are seeing is likely ghosting, and it should occur in the test panels as well if you measure from both channels at a time.  This occurs if you don't allow enough time for the channels to settle after multiplexing.  Here's a chart from the 6363 specifications that shows the settling time for various source impedances:

 

        2011-02-16_172609.png

 

In your first attached image, the error is about 500 mV on a 5V step.  This would be 10%, or 100 000 ppm.

 

From the same image, I see you are setting a sample clock rate of 200 kHz on 2 channels.  So, the aggregate of 400 kHz would allow for 2.5 us of settling between channels.

 

Looking at the graph above, (2.5 us, 100 k ppm) is off the chart, but if you extrapolate the curves we might expect that your source impedance is somewhere around 5-10 kOhms.  Is this correct?  If you have a link to the spec sheet for your encoder I'd be happy to take a look.

 

 

You don't see the error at 40 kHz on 2 channels because this allows up to 12.5 us (1/80 kHz) for settling--actually, DAQmx defaults to an 11 us maximum convert period on the 6363 when you are acquiring slowly enough to allow for it so the allowed settling time in this case would be 11 us.  If you needed to, you can reduce the convert clock rate manually through a DAQmx Timing Property Node if the 11 us default does not allow for enough time to settle for your application.  The maximum convert period that DAQmx will default to is the inverse of the maximum aggregate sample rate (in your case, 1 / 1MHz = 1 us) + 10 us.

 

 

The ghosting article linked above has some suggestions for eliminating ghosting, the most likely solution for you would be to implement a voltage follower if you need to acquire at higher rates than the settling of the multiplexer allows given your signal source.  The voltage follower serves to dramatically reduce the impedance as seen at the input of the DAQ card.

 

 

Best Regards,

John Passiak
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Thank you for your excellent explanation.

I see ghosting only with our HEM-series encoders, the oldest design we still produce.

In this design, the encoder pulse is produced on the collector of a transistor. The emitter is tied to GND and the collector is connected to the Vcc through a resistor.

 

You wrote: "your source impedance is somewhere around 5-10 kOhms.  Is this correct?"

You are absolutely correct, we have 10K resistor between the Vcc and the encoder output!

 

Thank you for the fast response.

Scientia est potentia!
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That explains it, thanks for the info.  It also explains why you only see ghosting on the channel measuring 5V since the line is either driven low (low output impedance) or pulled-up through the 10k resistor (high output impedance):

 

        2011-02-17_140710.png

 

 

Best Regards,

John Passiak
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