I have a camera from Imaging Source and it captures just one image for every external trigger pulse. But I want it to capture 3 images. Is there any idea of how to do it ?

The datasheetfor the camera DMK 23UV024

Steps executed so far:

  1. Generate trigger pulse from a datagenerator and trigger the camera
  2. Developed a LabVIEW program that can capture images on trigger
  3. I have two different images to be captured for one single trigger, since the camera captures only one image for each trigger, it will create a confusion as to which trigger image belongs to which one. How can I solve this problem.
  4. The idea is to capture 2 images based on 1 trigger pulse without any interference between the two images
  • \$\begingroup\$ You link to the datasheets, explain what you have already done etc. \$\endgroup\$ Oct 21, 2013 at 20:14
  • \$\begingroup\$ Edited the question with details. \$\endgroup\$ Oct 21, 2013 at 20:54
  • \$\begingroup\$ I don't understand the question. What is the problem with sending n triggers for n images? And how would an image interfere with another? \$\endgroup\$ Oct 21, 2013 at 21:14

1 Answer 1


I think you need to develop a circuit to create multiple pulses on a single pulse input. The timing of your circuit will need to coordinate to the timing of your program so when your program executes ICImagingControl1.MemorySaveImage "Triggered.bmp" it will have enough time to respond to each pulse. Your program can utilize a string variable to relate the images to each other such as "Triggered1.bmp", "Triggered2.bmp", "Triggered2.bmp". There is some good documentation in your link under the Using the trigger section.

As far as your circuit to create the 3 pulses you can use some cascading 555 timers. You would need 3 timers for the on pulses and 3 timers for the off pulses. I am sure there are better ways of generating the pulses but that is the way I have done this in the past.

  • \$\begingroup\$ You'd only need two 555 timers (or one 556). See this answer to a recent question. \$\endgroup\$
    – Samuel
    Oct 21, 2013 at 22:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.