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Engine Test Bed Fourier Analysis

  1. Apr 16, 2014 #1
    I have done a test on a 4 piston test engine which is expected to exhibit torsional resonance at 800RPM and a vertical translational resonance at 1200RPM.

    The data we gathered from the test bed machine was as follows:

    Theta | Signal
    0 | -5
    60 | -1
    120 | 7
    180 | 4
    240 | 6
    300 | -2
    360 | -5

    We are assuming that the accelerations are periodic with a period of 2∏ (hence 0 and 360 = -5)

    I am trying to decompose the Signal using Fourier series of sin and cos, I just have no idea as how to use the data gathered in to do this, could someone please point me in the right direction?

    Regards,

    Anthony
     

    Attached Files:

  2. jcsd
  3. Apr 17, 2014 #2

    Baluncore

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    Science Advisor

    Your data is in the time domain. You must convert it to the frequency domain.
    You have 6 data items over one revolution. You can only extract 6 Fourier terms.
    For each frequency there will be a cosine term, (complex real), and a sine term, (complex imaginary).
    To identify the coefficients you must correlate your data with the functions. Correlation is multiplication.
    So for a frequency of f, tabulate your data and the function Cos(f*theta).
    Multiply each data value by it's corresponding function value and accumulate the products.
    Divide the sum of products by 6, the number of data items.
    That gives you the cosine coefficient for frequency f.
    Repeat the process with the sine function to get the sine coefficient for the same frequency.
    Those cosine and sine coefficients make a complex number, convert it from rectangular to polar to get amplitude and phase.
     
  4. Apr 17, 2014 #3

    Baluncore

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    Science Advisor

    Here is a worked example, probably with some errors to keep you awake.
    Any further questions welcome.
    Code (Text):

     Frequency = 0   ( DC component )
    theta f*theta  data  cos(f*theta) sin(f*theta)  d*cos    d*sin
       0      0   -5.00    1.000000     0.000000   -5.0000  -0.0000
      30      0   -1.00    1.000000     0.000000   -1.0000  -0.0000
      60      0    7.00    1.000000     0.000000    7.0000   0.0000
      90      0    4.00    1.000000     0.000000    4.0000   0.0000
     120      0    6.00    1.000000     0.000000    6.0000   0.0000
     150      0   -2.00    1.000000     0.000000   -2.0000  -0.0000
                                           Total    9.0000   0.0000
                            Fourier coefficients    1.5000   0.0000
                            Amplitude =   1.5000 at   0.00 deg

     Frequency = 1
    theta f*theta  data  cos(f*theta) sin(f*theta)  d*cos    d*sin
       0      0   -5.00    1.000000     0.000000   -5.0000  -0.0000
      30     30   -1.00    0.500000     0.866025   -0.5000  -0.8660
      60     60    7.00   -0.500000     0.866025   -3.5000   6.0622
      90     90    4.00   -1.000000     0.000000   -4.0000   0.0000
     120    120    6.00   -0.500000    -0.866025   -3.0000  -5.1962
     150    150   -2.00    0.500000    -0.866025   -1.0000   1.7321
                                           Total  -15.5000   1.7321
                            Fourier coefficients   -2.5833   0.2887
                            Amplitude =   2.5994 at 173.62 deg

     Frequency = 2
    theta f*theta  data  cos(f*theta) sin(f*theta)  d*cos    d*sin
       0      0   -5.00    1.000000     0.000000   -5.0000  -0.0000
      30     60   -1.00   -0.500000     0.866025    0.5000  -0.8660
      60    120    7.00   -0.500000    -0.866025   -3.5000  -6.0622
      90    180    4.00    1.000000    -0.000000    4.0000  -0.0000
     120    240    6.00   -0.500000     0.866025   -3.0000   5.1962
     150    300   -2.00   -0.500000    -0.866025    1.0000   1.7321
                                           Total   -8.5833   0.2887
                            Fourier coefficients   -1.4306   0.0481
                            Amplitude =   1.4314 at 178.07 deg
     
     
  5. Apr 19, 2014 #4
    Thats great thanks for your help, I now understand the last few steps of generating the coefficients which I was missing before! I did notice a few mistakes as you said.

    Are these last fews steps anything to do with the Riemann sum or is that something else entirely?
     
  6. Apr 19, 2014 #5

    Baluncore

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    Science Advisor

    The discrete time data you sampled, from a continuous function in the time domain, can be seen to represent an area bounded by a curve. The Riemann sum is the DC component only.

    Once transformed into the frequency domain, only fourier coefficients for discrete integer frequency harmonics are present. With the exception of the zero frequency DC component, the area of all other fourier terms over one revolution is zero since all terms represent sinusoids.

    The fourier transform does not analyse the area so much as the way area is distributed over one full rotation. Your 4 cylinder engine had a firing order that would have set up torsional vibrations along the shaft. It should be apparent that analysis over one rotation is OK for 2 stroke engines, but for four stroke, the vibration analysis should be over two full rotations of the crank shaft since the crank rotates at twice the rate of the cam shaft.
     
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