band-pass filter does nothing to the data...

[yolobia]

Hi everyone!

I intend on applying a band-pass filter (Butterworth filter, 3rd degre) to my data to cut highy and low frequencies (from 1.2 MHz to 2.2 MHz). I attached a picture of my vi. But what I get after filtering is the same as the raw data.

When am I wrong??

Thaks for any help.

yolobia

[johnsold · Accepted solution]

It is difficult to tell from a picture.

Most likely the problem is due to not providing timing information to the filter VI.  When you convert the 2D array to dynamic data, nothing specifies the sampling rate.

Lynn

[yolobia]

Hi Lynn,

Thank you for your answer.

I didn't kow we had to be careful about the samplig rate. My data are already sampled though. It is a time trace with 17 ps time step.

I did not find yet how to provide timing iformation to the filter vi.

Laurence

[Coq_Rouge]

The sample rate is indeed important information. In order to to get your filter to work properly. I guess the Express VI concept has deluded you to think not. The best thing is to avoid use of the Express VI concept and use proper Labview instead. Also for the beginner. NI used to have have a lot of good application notes about digital  signal processing. Also filtering. But I could not find anyone. But I did find this http://my.safaribooksonline.com/9780132442534

I will also say with a rate equal to 17 sample pr second, you will not be able to do much filtering. You should use a higher samplerate say 1Khz if possible. You should tell us (a lot) more about what you plan to do. And also post your data and code. Remember good question draw good answers

[yolobia]

Hi!

Thank you for your advices and links!

I solved the timing problem by creating a wave form (using Build Waveform) and feeding it with my Y data and my time step.

Thank you very much!

Greetings from France 🙂

Yolobia