Find velocity from accelerometer without initial velocity? 3

[Popawheely]

I am having trouble figuring out the velocity from an accelerometer data set. My problems are two ... how do you determine the initial velocity and what about the constant from integration?

V(T) = V(o) + A(o)T + C

I deployed an accelerometer on a pole. I then later turned on the datalogeer to record the accelerations at a frequency of 10 Hz. I must assume that the pole was initially moving and hence V(o) is not zero or known.

Also, what about the constant of integration?

 

[electricpete]

It seems to me you are only missing one parameter... the initial velocity v(0) which is the same thing as the integration constant.

v(t)=v(0)+ Integral{a(tau)}dtau from tau=0 to t

I would select the integration constant to make the average value of v(t) be zero. Either over the entire sample or over as long an interval as possible.

For example start by arbitrarily assuming v(0)=0, then compute a first estimate v1(t). Then your final estimate
v(t) = v1(t)-average(v1(t)) will have an average value of zero.

 

[anands78]

Hello
One rule of thumb to get velocities from accelerometer readings is to divide it by the frequency(displacements are obtained by dividing the square of the frequency).
So if you have a set of acceleration data over a range of frequencies, divide the values at each and every freq by the correspondig freq to get the velocity.
Hope this helps.
Anand

 

[MikeyP]

The constant of integration implies a constant velocity component which in turn implies that the test structure is flying across the room! It is a safe bet to assume that the integration constant is zero.

You do raise a serious issue however. Real world noisy measurements may contain spurious non-zero constant components even for a constrained structure. When you integrate an acceleration signal in the time domain you have to filter out the the lowest frequency components to avoid this drift. Doing this with a digital filter can introduce unwanted phase shifts so an alternative approach is to do a "trend removal" on the integrated velocity data. ie. remove the straight line (and possible quadratic and cubic) trends. This has its own set of problems.

The best way to deal with this is to take an acceleration measurement where the vibrating test structure STARTS from a state of rest and is allowed to DECAY to a state of rest (ie displacement, velocity and acceleration start and end at zero). You can then use a Fourier transform to convert the measured signal to the frequency domain (without any problems from leakage effects) and, as anands78 suggests divide by i*omega to obtain the velocity spectrum (obeying the rules of complex arithmetic). You can then simply set the amplitudes of the first few frequency lines to zero before performing the inverse Fourier transform back to the time domain.

As anands78 says, we can take this a stage further an integrate again to obtain displacement (FFT, divide velocity spectrum by i*omega, inverse FFT). However we must be careful. It IS possible for the DISPLACEMENT to have a constant term associated with it (for example think of buckled beam vibrating in the post-buckled state).

M

 

[skfmike]

This is why I majored in Mechanical Engineering Technology as opposed to ME. Just show us where to place the accelerometers, collect the data, and analyze the data.

 

[jjmoore]

I agree with the second posting. For steady state vibration, just divide the acceleration by omega (rad/s)with the appropriate conversions. For steady state vibration, the initial velocity is meaningless.

 

[electricpete]

I also agree IF the objective is simply to find the spectrum. In that case, calculate the acceleration spectrum and divide it by 2*pi*frequency.

IF you were interested in a velocity time waveform, then you would need to simply subtract out the average value of the acceleration to ensure you don't have an unwanted dc offset arising from initial conditions.

 

[anands78]

Electripete
IF you configure the setting of the spectrum analyzer to AC coupling, do you think it will still gather the DC offset?

 

[electricpete]

Yes, an ac coupling feature will remove the dc.

It is still good practice to remove the dc during processing both before and after integration.

 

[anands78]

Electripete
I still dont get you. When we have not at all gathered a DC component(i.e, the time domain data is oscillating about a zero value; say the readings are from an accelerometer) then why would we have to remove it? And another question that i have is how can we do a conversion from acc to vel in the time domain? ie, u said that "IF you were interested in a velocity time waveform"; my ques is how do you post process the acceleratioin data into velocity in the time domain?
Thanks

 

[MikeyP]

anands78. Re ac coupling.

A little thought experiment. I have an accelerometer which is attached to something that is vibrating it sinusoidally at 50 Hz. My measurement equipment has a DC decoupled input. If I measure 0.03 seconds of data, I will have measured 3 1/2 sine waves. This most definitely has a DC value associated with it. Of course if I had measured a whole number of cycles then I would have zero DC component.

The DC decoupling on the input does not necessarily mean that there is no DC component in the measured data. Plus of course there will be all kinds of other errors such as quantisation noise on the ADC. It only takes a very small bias on the digital representation of the signal to make your time domain integration fly off towards infinity. As I said before, If you need to do the integration in the time domain then the signal must be detrended afterward (remove DC, linear and possibly higher order trends)

M

 

[GregLocock]

""IF you were interested in a velocity time waveform"; my ques is how do you post process the acceleratioin data into velocity in the time domain?"

You integrate it. Either with an electronic integration circuit, as seen in a B&K 2635 charge amp, or digitally, by using V(t+dt)=V(t)+dt*A(t).


Cheers

Greg Locock

 

[pwtem]

anands78
"I still dont get you. When we have not at all gathered a DC component(i.e, the time domain data is oscillating about a zero value; say the readings are from an accelerometer) then why would we have to remove it?"

Yes, if neither your sensor nor amplifiers (if any) or digitizers have any DC offset, then you probably don't have to worry. I still agree with eletricpete and MikeyP, its a good idea to remove linear trends before and after integration.

MikeyP's posts agree with my experience processing earthquake data. Initial values make integration very difficult. In the old days, before we had pre-event memory in the accelerographs, we had to pad the beginning and ending of the waveforms with zeros in order to get a stable integration. Then we had to use a special Ormsby band-pass filter... I forget all the steps now but it was not trivial. Main problem was, earthquake engineers were really interested in the low frequency motions. If not, we could have just filtered them out. Problem is, buildings respond at frequencies less than 10Hz, so we needed to try and recover them.

These steps are only necessary for transient signals. If you have a steady state signal, the easiest way to integrate in the time domain is to throw in a single pole (-6dB/octave) low-pass (hi-cut) filter with a 1 Hz corner frequency. This is equivalent to a 1/s kind of integration. (Boy, I haven't thought about this stuff in years, hope I remembered right)