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# 2 d h 1 k m 2 c 2 2 d

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Unformatted text preview: ly damped system, ω p ≈ Ω , then expanding the single degree-of-freedom frequency response yields, 1 1 1 = k − mω 2 + jω c 2 j 2ζ k Manipulating the expression yields a form from which the half-power bandwidth points may be calculated. 1 1 = 2 2ζ 1 1− ω Ω 2 2 + jω c k 2 2 = 1− + j 2ζ 1 ω Ω 2 2 + j 2ζ ω Ω 2 2ζ = 1 − ω Ω ω Ω Squaring both sides yields ω2 ω2 ω2 ω4 ω2 8ζ = 1 − 2 + 4ζ 2 2 = 1 − 2 2 + 4 + 4ζ 2 2 Ω Ω Ω Ω Ω 2 2 ω ω 2 2 + ( 4ζ − 2 ) + (1 − 8ζ ) = 0 Ω Ω ω Solving for yields, Ω 2 − ( 4ζ 2 − 2 ) ± ω = Ω 2 2 4 2 ( 4ζ 2 − 2 ) − 4 (1 − 8ζ 2 ) 2 2 ω 2 2 = 1 − 2ζ ± 2ζ 1 + ζ Ω Assuming ζ is small (i.e. ζ approximated by, 1 ), then ζ 2 ≈ 0 , and the half-power bandwidth points are ω ≈ 1 ± 2ζ Ω 2 Lecture Notes -42- 06/16/06 12:25 PM Mechanical Vibrations I Taking the difference between the points yields, 2 2 ωH − ωL Ω 2 ω + ω L ω H − ω L ≈ H ≈ 4ζ Ω Ω Using the approximation, ω H + ω L ≈ 2Ω , allows reducing the expression to, ωH − ωL ≈ 2ζ Ω Therefore the damping factor can be approximated as, ζ ≈ ωH − ωL 2Ω ∆f BW or ζ ≈ . 2 fc 2 fc This approximation is also often written as ζ ≈ While this method appears particularly simple, in practice, there are several difficulties utilizing it effectively. Specifically, when using real sampled data, the ability to determine the actual magnitude and location of H (ω p ) is limited by the measured data. The actual sampled data will not in general include the exact frequency ( ω p ) and so both the location and magnitude will be in error. Since the magnitude of H (ω p ) is used to estimate the half-power bandwidth points, the location of ω L and ω H will be in error. In general, the errors tend to be the following: under estimating the magnitude of H (ω p ) , under estimating the magnitude of the half-power points, estimating ω L too low, and estimating ω H too high. The compound effect of these errors is that generally the damping factor ( ζ ) is estimated too high. Lecture Notes -43- 06/16/06 12:25 PM Mechanical Vibrations I FRF Example A Given a single degree-of-freedom (SDOF) system with m = 5kg , c = 20 N ⋅s m & k = 1000 N m , assuming a harmonic excitation of magnitude 20 N @ 10 Hz , determine the response of the system to the given force. Solution First, identify the actual steady-state operating frequency. ω = 10 Hz -or- 20π rad s = 62.832 rad s Then, evaluate the SDOF frequency response function. H (ω ) = H (20π rad s ) = 1 −mω + jω c + k 1 2 2 −5kg ⋅ ( 20π rad s ) + j 20π rad s ⋅ 20 N ⋅s m + 1000 N m 1 H (20π rad s ) = ( −18,739 + j1, 257 ) N m H (20π rad s ) = ( −53.125 − j 3.563) × 10−6 m N Finally, determine the response in the frequency domain. X (ω ) = H (ω ) ⋅ F (ω ) −6 m s ) = ( −53.125 − j 3.563 ) × 10 N ⋅ 20 N X (20π rad X (20π rad s ) = ( −1.063 − j 0.071) × 10−3 m -or...
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