Order power spectrum - order resoluiton

[charlie87]

Hi,

I used NI example from Sound and Vibration library (attached below) and I tried to apply that VI for my input data and computing order power spectrum. However I cannot reach better order resolution than 0.1. When I set required resolution e.g. 0.01, no spectrum is computed. 

It is the same problem as when you set in the attached VI the order resolution 0.001. (no spectrum is computed).

In my example I let the item MAX ORDER TO ANALYZE (block OAT converter to even angle signal) empty = auto. So the output ACTUAL MAX ORDER is 480. This value is used in the block to compute the order resolution. However I do not understand why it is not working and no spectrum is displayed...

[preston_johnson]

As in any spectrum calculation, when you reduce the delta frequency (in this case order), you need more time series data.  So as an illustration, lets say 10 seconds of time series data gets you to 0.1 resolution in the spectrum.  If you want 0.05 resolution in the spectrum, you will need 20 seconds of data.  The general rule of thumb is:

FMAX (usable maximum frequency in the spectrum) is = sampling rate / 2.56

Lines of resolution (the number of frequency bins from 0 to FMAX) is equal to # samples in the block / 2.56

Recommendation - take a longer time series reading, and you will have more frequency bins in your FFT, and you will have better resolution. 

Hope this helps. 

[charlie87]

Thank you for your quick reply. I have found something about the stuff you mentioned and think about it. Now I would like to be more specific.

I have about 210 seconds of data (vibration) sampled by cca 5008 Hz. It is about 1,050,000 samples. The corresponding speed signal is 6269 samples, 120 smp/rev. The rotational speed is nearly constant in this case about 0.25 RPS => 4 Hz.

Acc. to the standard formulas:

FMAX = 5008/2.56 = 1956.25 Hz -> 1956.25/4 = 489 max order ( = cca 480 computed by the VI mentioned in my previous message , OK)

But I am not sure how to substitute the variable in the formula for resolution you mentioned. # samples in the block ? I do not specifie any data block...

[preston_johnson]

The calculations are similar, just a bit more. 

I trust you are using the resampling function to get the data into the angular domain.  When the VI reports it has max order of 480, I believe that is just under 400 samples per revolution.  We would want to check. 

Now we are in the revolution domain, as compared to the time domain. 

So we have lets say 360 points per revolution in the resampled data (you will need to check this). The FFT will produce 180 FFT bins for one revolution and spread those out to max order of 480. Two revolutions will create 360 bins, and so on.  The more revolutions we feed to the Order Spectrum function, the finer the order resolution.  Similar math applies as in the time domain.   

Does this make sense

[charlie87]

I tried to do some research acc. to your instructions.

Yes, I am using the VI OAT Convert to even-angle signal. The output even angle signal has 49920 samples in 210 seconds. When the average speed is 0.25 RPS than it is about 52.5 revolutions in 210 seconds. Than 49920/52,5 = 950 even angle samples per rev....

does it make sense ?

[charlie87]

I realised I was wrong in my previous message. 0.25 RPS is not 4 Hz but just 0.25 Hz. Now I am totaly lost in this theory ............

[preston_johnson]

Would you be able to post your code, and perhaps a data file (TDMS for example).  I'd be happy to take a look. 

Regards,

[charlie87]

I decided to start from the beginning and apply the basic formula to the NI library example. However the results are not acc. to my (your) formula...

I attached the modified VI (just add some indicators). The signal rotatin speed is constant 3000 RPM, so it is 50 RPS = 50Hz. Than the first order (harmonic) 1X corresponds to 50 Hz. ( am I right ? I hope so...)

Sampling freq. is 10kHz and sampling time 2 sec, so final samples count is 20000.

The max freq. Fmax = fs/2.56 ............... so 10kHz/2.56 = 3906.25 Hz. This value transformed to order (devided by 50) is 78.125X. But the max order acc. to the VI OAT Convert to even angle signal is 84....

I was looking for the relationship among the order resolution and samples counts and the other values which are available but I failed.

I believe when I understand this I can solve my problem I mentioned above...