How to remove a dynamic offset from an integrated signal?

[guidout]

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  

[AntonisV]

Hello guidout,

check this link below, I think it will help you 🙂

http://digital.ni.com/public.nsf/allkb/C1F39233BD94F2F08625736B00593E22?OpenDocument

[guidout]

Thanks, but I already read it and I tried to apply it to my VI but doesn't work or maybe I am not able. Are you quite sure that it is the only one solution?

[AntonisV]

You receive some kind of an error? In which way it does not work? Or you mean that you do not have the expected results?

[guidout]

Yes, I received no error but the result is not what I want. That is  the curve is the same.

[guidout]

I am attaching the VI of the link you posted and that I modified and the relative data file. So, if you have 1 minute you can see how I applied the NI solution and how it works with my data.

[Henrik_Volkers]

What and how do you measure?(physical)

What is the the value you what to know? (signal)

/cristall ball mode/

Looks like a measured vibration exited by a generator with hum

Or you have a groung loop and measure line hum

/cristall ball mode/

Depending on the decoder of the vibrometer a highpass in the Hz-mHz range might already implemented.

[guidout]

I measure vibration by a laser vibrometer (Polytec PSV-400 scanning head), the acquired signal is a velocity and, by integrating it, I want to have the displacement.

About the excitation, what is a generator with hum? Anyway I excite my test rig by a shaker but I think it is not important for this problem.

Instead, about the filter I can apply, directly from the junction box, a highpass filter (off or 100Hz). Do you have some advice?

Greetings from Germany too 🙂

Guido

LV since 4 months 🙂

[Henrik_Volkers]

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? 😉

[guidout]

First of all thanks for your time.
Maybe my problem is not clear.I am measuring hysteresis loop so I need the total signal. Moreover the offset you can see in the picture of first my post does not make sense, physically.Beacause my excitation centered around zero (shaker), of course sometimes little spikes can occur; laser or shaker spikes and I think that the offset into the integrated signal is just due to these spikes because integration "remembers" all previous hystory. Indeed, some time ago I solved this problem by decreasing the buffer per channel, in this way the offset was almost disappeard, but now I can not do a similar thing because I am performing sweep-sine tests. 

Quote:

Guess why the 100Hz HP is there?

my sweep starts from 10Hz, I do not think i can use this filter. I am not sure, I not so expert...

Guido