## 3 Discussion of Results
### Low Damping Case  
The first case considered is that with low damping (\(\zeta_{1,2}=0.0001\)) and well-separated modes. The vibrating mode is the one corresponding to the first family at ND = 5. Due to mistuning the EO = 5 excitation produces two frequency response peaks at two different rotational speeds. Figure 7 shows the response computed mathematically (continuous line) overlapped on the reconstructed response curves (dotted lines) using both fitting methods (2 DOF fit and Sine fitting). The two figures correspond to the two different random mass distribution patterns. It can be seen that there is an excellent match between the curve obtained by the 2 DOF fitting and the original response. In comparison to the Sine fit method, the 2 DOF method better capture the amplitude of the two resonance peaks. The parameters directly obtained by the fitting procedure with the 2 DOF model are the resonance frequencies and the damping ratios. The obtained values can be compared to the real ones set in the mathematical model. The frequency and damping values identified by the 2 DOF fitting method are listed in Table 2 and compared to the exact values of the mathematical model. The comparison of the values confirms the good estimation obtained by the 2 DOF model, the value of residual close to zero confirms the good quality of fitting. In a second step of computation, modes corresponding to higher nodal diameters are considered in particular modes corresponding to ND = 6, 7, 10 excited by the corresponding EO. As in the previous case, for each nodal diameter there is a splitting of the two resonance frequencies excited at two different rotational speeds. The response curves (the mathematical one and the two curves obtained by fitting) are shown in Figures 8-10. For all the cases it is confirmed that the 2 DOF method predicts the amplitude of the response at resonance better than the Sine fitting method. This difference can be attributed to the fact that the Sine fitting method is influenced by the sampling frequency of the signal. Since the BTT signals are sampled once per revolution, for the Sine fit method more samples are needed to successfully cover the whole range of the resonance. Particularly for sharp peaks (small damping) close to the resonance a higher sampling rate would be required. This can be achieved by reducing the acceleration/deceleration rate which is considered here with a realistic value of 0.25 Hz/s. On the contrary the 2 DOF method is not so sensitive to the BTT sampling frequency since it uses the data of the entire resonance region to be fitted by using the mathematical formula of Equation (6). In the particular cases, as the proposed examples, of sharp amplitude peaks with low damping values, applying only the Sine fitting method might be inadequate with the risk of underestimating the amplitude of the peaks. In these cases the 2 DOF method proved to be a good numerical tool to verify and to correct the results of the Sine fitting method. \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline \multirow{2}{*}{**Parameters**} & \multicolumn{3}{c}{**SD = 0.05**} & \multicolumn{3}{c}{**SD = 0.10**} \\ \cline{2-7}  & **Exact Value** & **Calculated** & **Difference (\%)** & **Exact Value** & **Calculated** & **Difference (\%)** \\ \hline Frequency 1 & 157.1322 & 157.1322 & 0.00 & 157.4456 & 157.4456 & 0.00 \\ Frequency 2 & 157.6046 & 157.6046 & 0.00 & 158.1039 & 158.1039 & 0.00 \\ Damping 1 & 0.00009985 & 0.00009987 & 0.02 & 0.00009979 & 0.00009981 & 0.02 \\ Damping 2 & 0.0001015 & 0.00010000 & 0.15 & 0.00010021 & 0.00010017 & 0.04 \\ \hline Residual & & \multicolumn{3}{c}{\(5.3537\times 10^{-6}\)} & \multicolumn{3}{c}{\(1.6128\times 10^{-6}\)} \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Comparison of vibration parameters identified by the 2 degrees of freedom (2 DOF) fit model with the exact values for Engine Order (EO) = 5. Figure 8: Comparison of non-dimensional amplitudes obtained by different fitting methods for EO = 6 and \(\zeta_{1,2}=0.0001\), (**a**) SD = 0.05, (**b**) SD = 0.10.