Impossible to get a sufficient number of samples per second

Similar Questions

• How to abort Continuous Acquisition (which is set to acquire a constant number of samples per channel each iteration)?

  Hi all

  I would like to continuously acquire an input channel analog and, if necessary, be able to stop the acquisition as quickly as possible.

  I use a card NI USB-6289. I implemented a DAQmx (voltage) analog input task, set the clock sampling DAQmx 'continuous samples' and the rate required (IE 1000, which is 1 kHz). For the acquisition, I used a while loop with inside the DAQmx Read with Terminal 'number of samples per channel', wired (IE 1000). The while loop waits until 1000 samples are acquired and therefore continuously reads the channel in uniform batches of 1000 samples every second ticking. The while loop can be easily interrupted when the DAQmx has completed the acquisition of the lot being 1000 samples, but it takes the DAQmx finish their current task.

  My problem is how to stop the loop while the DAQmx Read lies in the middle to get 1000 samples WITHOUT delay all 1000 samples are received? Is it possible to interrupt the DAQmx Read?

  I could reduce "the number of samples per second", increasing the responsiveness of the while loop, but this is not the solution I prefer. I tried to destroy the task (outside the while loop), but that does not stop immediately the acquisition, Read DAQmx still finish acquire 1000 samples. I've included an example Subvi. When I was using a PCI card, I used a timed while loop with inside the DAQmx Read with Terminal 'number of samples per channel' wired-1 (= which means ' read everything in the buffer "). In this case, it was easy to stop the acquisition at any time: the timed, while the loop is abandoned and the acquisition has ceased immediately. But it does not work with an NI USB-6289 map (see thread http://forums.ni.com/ni/board/message?board.id=170&message.id=386509&query.id=438879#M386509) because of the different way the data are transferred to the PC.

  Thank you very much for your help!

  Have a great day,

  LucaQ

  Hi LucaQ,

  Your solutions are decrease of the number of samples, or record the time you want to stop and remove samples that have been acquired beyond this time. There is no other way to stop the actual reading out of the hardware store.

  Flash

• Why do I get a huge number of samples even with small flow and execution time

  I use the express Signal as my data acquisition software. NI9239 Daq is data acquisition system is my data acquisition equipment. Problem I have is, regardless of the data rate and the number of samples, I put, I get a lot of samples. for example I put the flow to 10 samples per second, and the number of samples to 600 (in continuous mode) I get more than 64000 samples where I should only get 600 samples total.  I tried to adjust the flow to 10 Hz and number of sasmples at 10, it gives me an error indicating that the data buffer overrun. Does anyone know a way around it?

  all responses are much appreciated.

  Hi odessy27,

  Maybe you can use the > Analog > step subset and resampling to resample the data at a lower sampling rate.

  Brad

• Total number of samples per channel

  Hello

  I was wondering if you can call a function that returns the total number of samples per channel read. I try to get a precise timestamp.

  I use DAQmx ANSI C developer environment.

  According to NOR:

  Option #1:

  Try and do exactly what the driver is doing.  This will require you to do exactly what you do in the displayed example.  Call time current system immediately before calling the DAQmx Read and subtract dt * x where x is the number of samples already acquired.  It will take you to know exactly how many samples were acquired.  It can be found by calling the Total property of the samples by chain acquired immediately before the DAQmx Read.  This is to introduce some points for innaccuracy.  For example, you're time system is already innaccurate to a certain amount.  In addition, it takes some time between the system times call, calling the total number of samples acquired and calling the DAQmx reading.  If 2 samples are acquired between the call to the time system and the total number of samples acquired, you could be shifted a few samples.  For the slower clock rates, you will have more precision.

  I try to attempet to program this solution, but cannot find the total number of samples per channel acquired property. If anyone can help me, your help will be very appreciated.

  Thank you for your help,

  Vladimir

  Hi Vladimir,.

  Here is the property you are looking for (as the C reference help):

  

  You will be able to get accurate relative timestamps, since we know that our acquisition is based on a sample with a certain dt clock.  The value of initial time must still come from the OS.

  Best regards

  John

• samples per channel and the number of samples per channel

  in my DAQ mode samples finished program, there are two screws: timing and read.vi DAQmx DAQmx.

  I have to set the parameter to "samples per channel" DAQmx timing.vi and 'number of samples per channel' on DAQmx read.vi... Is there a relationship between these two?

  My laser runs at 1 K Hz. I want to go to the wavelength, wait for a number of shooting lasers, read the data and move on to the next page...

  Thank you

  Lei

  In your case, the VI will acquire the lesser of either:

  The "samples per channel" that you have defined on the timing DAQmx VI

  -OR-

  The number of iterations of your for loop (N) times the 'number of samples per channel"that you have defined on the DAQmx read VI

  The "samples per Channel" VI DAQmx of timing for a finite acquisition dictates how many samples the DAQ hardware should acquire in it's onboard buffer before indicating that the acquisition is complete.  "The number of samples per Channel" on the read DAQmx VI dictates how many samples the DAQmx driver must return buffer on board the aircraft to your application.

  Let's say the "samples per channel" on the calendar DAQmx VI is set to 50.  Thus, the card will acquire 50 samples and place them in the edge of the buffer, then stops.  Suppose we have the 'number of samples per Channel"on the DAQmx reading VI the value 3 and what we call the VI in a loop For which runs 10 times.  Thus, every time the DAQmx lu VI is called, it will wait until there are at least 3 samples in the buffer, and then return these three.  We call the VI a total of 10 times, then we will answer 30 total samples.  Thus, the last 20 samples acquired the card remains in the buffer and are destroyed when the task is disabled.

  Now let's say that we increase the "number of samples per Channel" on our DAQmx Read VI at 10.  VI Read will wait until 10 or more samples are in the buffer, and then return these 10. Thus, we will be back all 50 samples map acquired by the 5th iteration of the loop For.  The 6th time we call him VI DAQmx Read it expires, because there will never be another 10 samples in the buffer, and the VI returns a warning.

  This clarifies things?

  The purpose of this behavior is to allow you to both set the total number of samples that the DAQ hardware will acquire and also control how much of these samples is returned whenever you call the DAQmx Read VI.

  Kind regards

• How can I throw 100 samples per second on a graph?

  Hello

  I take a CANBus 100 samples per second, and I would like to present these data in a real-time graph. I want to see all 100 samples and the graph refresh every 1 dry, I use a waveform table.

  I would like to than my table updated with new data every one second (1 SEC 100 samples, 200 samples of 2 sec, etc.) and be able to see all 100 samples we take in real-time in the chart. My question is it possible? I think that some how I need to change the x in the graph scale? I did a little code in which I change the x of the chart scale and put in real time, but in the graph shows only a sample of 100 samples that I take every second, can you please suggest a solution for this?

  
  

• 10 samples per second 9229

  Looking to capture the 10 samples per second in a file by using a NI9229.

  I think I have the output to work correctly using a backup in format ascii/lvm.

  My problem is that I can't know the ratio to capture who gives me 10 samples (10 rows of data) per second using a 'samples of N' 'samples to read' and 'rate (Hz).

  My goal is to have 10 samples per second and the first column correctly lable time since when I click 'run' until I hit the stop and tell me how many seconds it failed because I stopped before her.

  Thank you

  Matt

  Hi Matt,

  I didn't know that you use SignalExpress.

  This is what you want?

  • DAQmx acquire:
  • • Acquisition mode = continuous samples
    • Samples to read = 200 (or 2000, if you want to see 1 second at a time on the graph)
    • Rate (Hz) = 2 k
  • Subset and resampling:
  • • Subset of return = false
    • Resample = true
    • DT = 100 m

  Brad

• I need to improve the performance of a Web site and I want to track the total number of requests per second...

  I need to improve the performance of a Web site and I want to track the total number of requests per second... we are currently hitting...

  Please I need to hit the current with 1200 queries/second application, please help in this regard.

  Hi Nivask,

  If you created the Web site, the question would be better suited in the MSDN Forums. I would recommend posting your query in the MSDN Forums for assistance:

  Internet Explorer Web development

• Is there a way to get the total number of samples to a TDMS file?

  I'm looking at the documentation for the function of properties Get TDMS and I see that it has a property: wf_samples which "represents the number of samples in the first data segment. Is it possible to get the number of samples in, say, the second segment of data? My goal is to get the number of samples in all segments of data for a single channel.

  To retrieve the total number of samples, I use the properties of 'NI_ChannelLength' to be linked to the function of TDMS properties get.

  The I64 outbut gives you the total number of samples of the selected channel.

  See attachment so useful

  Marco

• What is the relationship with the sampling frequency and number of samples per channel?

  In my world, if I wanted to taste 10 seconds 10 Hz (100 s/ch), specify a rate of 10 and a number of samples of 100.  This would take 10 seconds to return data.  The task does not appear to behave this way.  No matter what rate and the number of samples, I chose, I spammed with data at 1 Hz or more.

  What I am doing wrong?

  This problem is resolved by making a request for telephone assistance.  It turns out that the minimum sampling frequency of the NI 9239 is 14xx s/s.  I don't know why there is a minimum sampling frequency, but now I have to go to the next question discussed at this link:

  http://forums.NI.com/T5/Multifunction-DAQ/how-can-I-read-gt-10x-samples-at-a-sample-rate-of-X/TD-p/2...

• Impossible to get a serial number #-purchased from Best Buy

  I bought the 10 version of Best Buy.  Under the sticker, it says to go to www.adobe.com/go/getserial to register the product.  This is an invalid address.  Now what?

  I tried this address and it worked for me. This link will take you to: https://na1p.licenses.adobe.com/getserial/

  You are able to hit that URL directly? If so, this is where you can enter your redemption code.

• The configuration step of the calendar after the call to DAQm angular encoder, takes a lot of time vey even to 1000 samples per second. No idea why?

  I use an encoder in quadrature with DAQmx 6259, the call to read the outputs of the meter is slow, much slower than the sampling rate? No idea why?
  Thank you

  Hello Roxanne,

  In the title of this thread, you mention that the calendar configuration step takes a long time.  Is now called node configuration several times in your code?  You may find that your code runs faster if the channel configurations and the task doesn't occur once.

  Our example of the expedition, ' counting digital Events.vi ' may be of interest to you.

  Can be found by browsing our example search tool:

  'Help' examples

  Once opened, the example Finder you can find this example by browsing:
  "" material input and output"DAQmx" counter measures "County Digital Events.vi

  Best regards.

• How do I see the number of events per second received by MARCH

  Hello

  I have a MARCH installed in a customer demonstration porpuse. Y at - there a way for me to see how many events per second has the MARCH has received?

  In the newspapers of the access list when even if she cannot connect due to the limit of the flow it is said 'missed 20 bags due to pulsations.

  Then maybe the MARCH cannot process all the events, but he can tell how many events, he received per second on average.

  access to the admin-> maintenance-> view system logs and files search messages rates

• Change the number of frames per second

  It's very simple stuff - so I hope I'm missing an obvious answer... that is.

  I have an animation that goes 15 seconds and 16 fps. Total of 240 frames.

  I want to change to 15 seconds at 30 frames per second. A total of 450 executives.

  Is it possible to do - that will keep all the keyframes in relative place? Or I have to recreate the test specimen at 450 and pass each unique to his new position keyframe?

  Thank you for your help.

  T

  Unfortunately, no, you can't do that and keep the same length. You CAN change the frame rate, but increase the rate of 30 frames per second will make the scene twice as long because are presented twice as many images per second.

• 9239 number of samples compared to rates

  Hi all

  I'm running a VI for the 9239.

  In it, I updated the sample clcok 2048.

  I then put the number of samples to 20480, IE rotate for 10 seconds.

  However, instead of the final point to be measured to 10s it measure to the 9.82992

  If it is set to 2048 samples the final time is 0.98256

  I realize that say, if I had a rate of 5 samples per second and one took five samples, I'd get the next 0,.2,.4,.6.8

  But if I had to wear it to 10 samples, I'd get

  0,.2,.4,.6,.8, 1, 1.2, 1.3, 1.4, 1.6, 1.8

  The important point here is that the offset remains the same

  In addition 1-1/2048 =.99951171875 pas.98256.

  Anyone know what is happening here?

  Sean

  Add to that,

  You made a very good point about the first sample being taken at t = 0.  Accounting for this, we get:

  20480 samples

  (20480 / 2083.33333) * (1-1/20480) = 9.82992

  2048 samples

  (2048 / 2083.33333) * (1-1/2048) = 0.98256

  What exactly gives the number you reported.

  Best regards