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J
JBlackhole16-Pearl
16-Pearl
March 30, 2016
Question

Freq. Resp. Assurance Criteria (FRAC)

  • March 30, 2016
  • 21 replies
  • 12979 views

to all

I am playing around with what is known as 'Frequency Response Assurance Criteria' (FRAC). I have attached an example  with some ref

I understand that the Freq. Response Function (FRF), noted H(w) is a complex function containing magnitude and phase information

  1. If one has got magnitude and phase info (at each frequency) (from a physical test) does one need to create a complex equivalent of the data to apply the criteria (A*exp(iw+phase)) ?
    1. In the case of the analytical data (FRF) I am not sure I can get the phase (but will need to check) as it is from a random analysis
  2. I found an example of a FRAC equation involving sum, see attached, assuming it is correct (not sure it is!), how does one "code" the sum? I did it but it look wrong. I interpreted the equation as calculating a FRAC value in a range w1 to w2. I f guess if one plots it one can see the "evolution" of the FRAC

The example attached contains dummy vectors. I'll need to see if I can put my hand on "real" data

Thanks

Regards

21 replies

F
Fred_Kohlhepp23-Emerald I
23-Emerald I
March 30, 2016

The CFFT function in Mathcad creates a vector of complex values that could be used for your test.

A place to start:

R
RichardJ19-Tanzanite
19-Tanzanite
March 30, 2016

1) Yes, you will have to convert the magnitude and phase to a vector of complex numbers.

2) The equation doesn't look right to me either. I can't figure out any way to get it to give the same number as your other function. Note however that in the numerator you have the square in the wrong place: it is the square of the sum, not the sum of the squares.

J
JBlackhole16-PearlAuthor
16-Pearl
March 30, 2016

Thanks for the input/helps. Much appreciated

found another version of the paper and as suspected the eq. was not correct. See attached for updated file

Fred

Really stupid question (but I don't ask i'll never know): Why doe one need the CFFT function tp create the complex vector?

If I have a vector or magnitude and a vector of phase (for each frequency) is not possible to "combine" both to create the complex vector?



F
Fred_Kohlhepp23-Emerald I
23-Emerald I
March 30, 2016

J B,

The FFT functions in Mathcad (there are at least 4 of them) all take a vector of sampled values (assumed acquired at fixed sampling frequency) and move directly to a complex number value that describes the amplitude and phase at that frequency.  (Amplitude is simply the magnitude, phase is recovered with the function "arg" (I believe.)  There are subtle differences in how each of the functions provides their answer,  how the amplitudes are normalized, some by how many samples they need to have (lower case FFT requires number of samples be a power of 2, I think.)  Details can be found in help and examples are many.

So, if you're end goal is the complex vector and you have a time history these are a direct path.  Now if you have amplitude and phase, then you need to construct the vector; and how you would do that becomes another discussion.

Note that the FFT functions return a vector the same length as your time history sample.  And sample size determines your frequency resolution (along with sampling frequency.)  So if you had a real "experimental" distribution, in my experience what you have is a digitally sampled time history file.  My example had three frequencies, amplitudes, phases, and "experimental deviations."  (Because I was lazy.)  Assume for a minute that we have A(t) as the analytical description, the way we think the system responds.  By measurement (shaker and accelerometer) we acquire eSt, the experimental time history of the response function.  That file is 5000 samples long, the FFT (eHf) will be 5000 frequencies long.  How will you create a 5000 frequency complex vector from three?  Or, conversely, how will you strip eSt down to three frequencies?

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