How to remove a dynamic offset from an integrated signal?

Question

Hello everyone, I need a big help from you.
First of all I dont know whether it is the right forums' section, in the case I am sorry.
I acquire a signal from a laser vibrometer (velocity) and, in order to obtain the displacement, I integrate the signal by using the "Integral x(t)" block.
After that operation, a dynamic offset appears and I am trying to find out the way to remove it.
The picture attached shows what I am describing.
I think a solution exists because it must be a common problem, but I am not able to find out it.
Hoping for an help.
Best regards,
Guido Ritelli  
Attachments:
vel_vs_disp.png ‏38 KB

Answer 1

Your shaker is driven by an power amplifier and the amplifier get a signal from the signalgenerator . Take a look at the powerspectrum (see examples) of your velocity voltage AND
Take your DAQ or a scope and hook it up to the shaker (output of the poweramp, but look for the rated output voltage range !) . Do you have significant 50HZ there?? (Or 60Hz ,) your main powerline frequencymn often called hum
But wait, my strategy would be to avoid the unwanted dynamic offset, you want to put grease (vaseline) on the mirror (use a filter on the signal) so you don't see the wrinkles (noise offset bias hum drift)
So you want to measure the displacement, what if the answer is, that your first post show the displacement ? Including the unwanted stuff, but hey that is a displacement!
Again: What do you want to measure?
(If your answer is: Amplitude of my sine I put on my shaker... : tryout the single tone detection vi...  and if you have a sine you don't need an integration you can divide by omega
And yes you can use the the 100Hz highpass filter ... and if you know what it will do to your signal ... even better
Quote:
I think a solution exists because it must be a common problem, but I am not able to find out it.
Guess why the 100Hz HP is there?
Greetings from Germany
Henrik
LV since v3.1
“ground” is a convenient fantasy
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