[转贴]振动加速度积分到振动位移的问题

hudaren43

[转贴]振动加速度积分到振动位移的问题
记得大家曾经在这里讨论过这个问题。在Eng-Tips上见到有人也在讨论这个问题。转过来。

suviuuno (Structural) 7 Mar 07 11:13  
Hi,

We are testing effect of accumulator in rock drill concept. We have mounted two tri-accelerometers in hydraulic hoses that feed the drill to capture the difference in acceleration/frequency, if any.
To my surprise, results were fairly same.

Now, I would be interested in position difference. Integrating signal directly gives garbage. I have tried spline interpolation with still garbage results. Signal appears to shoot way off the course. I do not have much experience in signal processing.

I have 5sec sample with 5kHz. Does someone have insight into this?

Thanks,

suviuuno






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SomptingGuy (Automotive) 7 Mar 07 12:16  
You need to low-pass filter your signal below anything of any interest.  Even then the phase response of your filter may scupper you.  I've seen this sort of (apparently good) idea end in tears many times before.

GregLocock (Automotive) 7 Mar 07 19:26  
double integration in the time domain is rarely succesful, in my experience. Even singe integration raises more problems than it solves, often. When we are interested in acceleration and displacement, we measure acceleration and displacement.
Cheers

Greg Locock

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SomptingGuy (Automotive) 8 Mar 07 5:11  
Sorry.  I meant high-pass.  But my other comments still hold and I agree with Greg - use a displacement transducer if that's what you really want.

suviuuno (Structural) 8 Mar 07 7:39  
Thanks for helpful information SomptingGuy and Greg. I will set up a small experiment, where I will measure displacement as well just to see the difference.
I filtered signal and got what might be in the right area. Quite hard to say, I think. I'll keep you posted.


Rob45 (Automotive) 8 Mar 07 9:15  
I have heard,  from an application engineer at Head Acoustics,  that you should obtain at least 100 cycles of your lowest frequency of interest.  That would suggest that 20 Hz is the lowest frequency your 5-sec sample could tell you anything meaningful about.
But why not just try manually calculating the displacement at a couple of points, based on your acceleration data,  and compare that to your integration results?

I've done double-integration on acceleration data a number of times using Head Acoustics' Artemis software,  and have always had good agreement with calculated results for individual frequencies,  and have only gotten "nonsense" results for very low frequencies.

Regards,
- R

dolsoi (Mechanical) 8 Mar 07 19:49  
It is not possible to obtain a DC displacement (at 0Hz) from an accerlation measurement, but AC displacement. (suppose that an apple drops at the accerlation of gravity, where is the position now?) In the frequency domain, it can be rather easily done by dividing the accerlation by -w^2, where w is the angular frequency. There are some commercial signal conditioner that have integrators inside, for example, Nexus (a B&K subcidiary). What it does for a single integration is, as someone said above, (a pure integrator) times (a second order high pass filter of cutoff frequency of either 0.1 Hz or 1 Hz and damping rator of about 0.7). A double integration can be done in a similar way. Hope this textbook style information helps.
Cheers, dolsoi

GregLocock (Automotive) 9 Mar 07 0:31  
...and if you try to do it in the conditioning amp, read the manual carefully. The B&K 2635 charge amps, for example, claimed to be able to integrate accelerometers, but only within strict limits.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.


MikeyP (Aerospace) 9 Mar 07 5:06  
You can do this with a reasonable amount of success in the frequency domain provided you carry out your measurement carefully.

To avoid leakage problems in the freq domain integration, you have to have a signal that begins and ends at zero. The usual approach is to apply a window (e.g. flat-top for a long data record like yours) but windowing is a non-linear process and would screw up your results big-style. The trick is to ensure that your measurement does not require windowing.

Start the da6a acquisition with the drill at rest. Start up the drill. Run it for your 5 seconds. Turn it off. Give it some time for all the vibrations to die away. Then stop the acquisition.

You can then easily fft and integrate and hp filter in the frequency domain without any phase issues and transform back to the time domain.

I have done this, even when the dominant frequency of interest was as low as 10 Hz.

M
--
Dr Michael F Platten


electricpete (Electrical) 9 Mar 07 9:28  
I have done this a few times in excel.

One rule I have is to apply the high-pass filter as mentioned by Sompting guy (to remove dc component) at each stage.  Before the first integration, after the first integration, and after the second integration.  There should be no dc in acceleration, velocity or displacement of a stationary vibrating component anyway, so you're not losing any info.  With reasonably high sample rate in relation to sampled signal and the understanding that the low freq components have been removed, I believe you can expect good result.s

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electricpete (Electrical) 9 Mar 07 9:31  
my discussion applies purely to integration in the time domain. I didn't see any mention of frequency domain or FFT in the original post. But of course integration in the frequency magnitude domain is fairly trivial.
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SomptingGuy (Automotive) 9 Mar 07 10:08  
electricpete,

How do you high pass filter in Excel?

electricpete (Electrical) 9 Mar 07 11:12  
Lots of ways.  Low pass filters are easier than high but you can often use a low-pass filter to build a high pass filter.

What's the simplest low-pass filter? Moving average.  The more points you include in the average, the lower the cutoff.

How to turn into a high-pass filter?  Just subtract the moving average from the point.  The formula for B101 might be +A101 - sum(A2:A200)/199.

Copy to B102 as +A102 - sum(A3:A201)/199
etc

You end up with the output of the filter having 100 less points on the beginning and the end than the input of the filter... not a big problem if you have a lot of data.

That's just a start.  Of course digital diltering is a well-studied field.  You can define your filter parameters and determine required FIR filter coefficients using Matlab (and then implement the filter in excel).

Of course google is your friend.  Here is a spreadsheet already built:
http://www.univie.ac.at/cga/tool ... ie.ac.at/cga/tools/

Also check out dspguide.com or google for "numerical recipes"
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SomptingGuy (Automotive) 9 Mar 07 11:39  
Digital filtering is nothing new to me.  I'm just amazed that people bother doing it in Excel.  Then again when all you have is a hammer, everything starts to look like a nail.

SomptingGuy (Automotive) 9 Mar 07 12:20  
Sorry, last post sounded somewhat snotty.  Thank you for the links (I already have dspguide bookmarked and my NR is getting really shabby these days).  However, I still HATE Excel.

electricpete (Electrical) 9 Mar 07 13:25  
You ask a question: "How do you high pass filter in Excel?"

I answered it.

Then you tell me that digital filtering is nothing new to you.  Jeesh.

Next time you want to make an editorial comment, please have the courtesy not to pose it in the form of a question so people won't waste time answering it.
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suviuuno (Structural) 12 Mar 07 18:00  
All,

Thanks for helpful comments. I am getting reasonable results now, however, I have not verified it by displacement transducer yet.
Thanks again.

suviuuno

xuezhandao

由加速度计算峰峰位移？
大侠高手，我遇到一个可能很简单的问题，想用压电式的加速度计算出峰峰位移。我在国外的一个网站上看到一个简单的正弦振动的换算公式：g=0.0511f^2D, g= acceleration, g’s peak;D= displacement, inches, peak to peak;f = frequency, Hz
我只想知道，他的这种换算关系正确吗？我大学里学的高等数学都快忘光了，单我知道从加速度到位移要积分两次。是不是对于正弦振动，这个积分结果就是上述公式？我不搞学问，只想通过测量加速度算出峰峰位移值，来估算我的设备安装环境是否满足振动的基本要求。

halisi

这个找本机械振动方面的书，我好像看到过这个公式。

guangdayou

大侠高手，我遇到一个可能很简单的问题，想用压电式的加速度计算出峰峰位移。我在国外的一个网站上看到一个简单的正弦振动的换算公式：g=0.0511f^2D, g= acceleration, g’s peak;D= displacement, inches, pea ...
实际上道理很简单。
振动加速度是A*sin(2*PI*f*t)，其中A是振动加速度的频谱峰值（peak值），f是频率；
那么位移是对加速度求两次积分，就是A*(2*PI*f)^2*sin(2*PI*f*t)，这个值也是位移的峰值（peak值），2倍的这个值就是峰峰位移了。
把你测量得到的信号，做FFT变换，得到频谱，然后导到Excel里面，各频率算完了，求个能量和（注意窗函数的修正），就是总的位移峰峰值。

李元

我也在做何楼主一样的事情，由于是初次设计这样的问题，想请问大侠fft频谱的能量和是怎样求的？

qichebaobao

加速度信号向位移的calculate，加速度信号是正弦信号，整个过程如下：
1.从加速度到速度的积分，速度信号还是对的；
2.从速度到位移的积分，结果完全失真，按高等数学上面的公式，正弦函数两次积分过后还仍然是正弦函数才对。
然而，计算结果大相径庭是一条一次函数曲线，后来认识到在信号采集的过程中，有用信号中掺杂了不少噪声信号，在计算前未加任何处理，两次积分过后，噪声干扰信号被放大从而影响了整个计算结果。