# Key Action Mechanisms of Intentional Mistuning
## 2 AMM Formulation  
The AMM for the case of forcing a blade dominated modal family can be written as (see [11; 12; 13] for a detailed explanation):  
\[\left[ \begin{array}{cccc}\Delta\omega_{1}-\Delta\omega+\mathrm{i}\xi&\delta_ {N-1}\dots&\dots\delta_{1}\\ \delta_{1}&\Delta\omega_{2}-\Delta\omega+\mathrm{i}\xi&\dots\delta_{2}\\ \ddots&\ddots&\ddots\\ \delta_{N-1}\dots&\dots\delta_{1}&\Delta\omega_{N}-\Delta\omega+\mathrm{i} \xi\end{array} \right]\left[ \begin{array}{c}A_{1}\\ \vdots\\ A_{N}\end{array} \right]=\left[ \begin{array}{c}0\\ \vdots\\ \mathrm{i}\xi\\ 0\end{array} \right],\]  
where \(N\) is the number of blades of the rotor, and \((A_{1},\dots,A_{N})\) are the amplitudes of the travelling wave (TW) modes of the family. TW \(A_{j}\) has wavenumber \(j\) and (due to the invariance of the rotor with respect to a complete turn) all indexes have to be considered modulo N, that is, \(A_{j+N}\equiv A_{j}\). The diagonal contains the material damping of the TW modes \(\xi\), the relative deviation of the forcing frequency (\(\omega\)) with respect to the modal family mean frequency (\(\omega_{0}\)), \(\omega=\omega_{0}(1+\Delta\omega)\), and the relative deviations of the modal frequencies \(\omega_{j}=\omega_{0}(1+\Delta\omega_{j})\) (note that the \(\Delta\omega_{j}\) have zero mean value). The mistuning is contained in the off-diagonal term (coupling different tuned TW modes): \(\delta_{j}\) is the Fourier mode with wavenumber \(j\) of the mistuning distribution. The mistuning coefficients verify \(\delta_{j}=\delta_{-j}\) (the mistuning is a real distribution), and \(\delta_{N}=\delta_{0}=0\) (the tuned rotor is the average of all sectors). Finally, the forcing takes the form of a TW with engine order \(r\), which has been rescaled to have maximum tuned response equal to 1. The AMM formulation is, in this case of a nearly flat modal family, completely similar to that of the fundamental mistuning model [14]. The effect of the intentional mistuning is better analysed if we change the AMM from the TW basis \(\mathbf{A}=(A_{1},\dots,A_{N})\) back to the blade displacement basis \(\mathbf{X}=(x_{1},\dots,x_{N})\), with the transformation  
\[\mathbf{X}=\left[ \begin{array}{c}x_{1}\\ \vdots\\ x_{N}\end{array} \right]=\mathbf{PA},\text{ with }\mathbf{P}=\left[ \begin{array}{ cccc}\mathrm{e}^{\mathrm{i}(\frac{2\pi l}{N})1}&\dots&\mathrm{e}^{ \mathrm{i}(\frac{2\pi l}{N})1}\\ \vdots&&\vdots\\ \mathrm{e}^{\mathrm{i}(\frac{2\pi l}{N})j}&\dots&\mathrm{e}^{\mathrm{i}(\frac {2\pi l}{N})j}\\ \vdots&&\vdots\\ \mathrm{e}^{\mathrm{i}(\frac{2\pi l}{N})N}&\dots&\mathrm{e}^{\mathrm{i}(\frac {2\pi l}{N})N}\end{array} \right]\]  
The resulting equations for the AMM in terms of the blade displacements take the form:  
\[\left[ \begin{array}{cccc}\mathrm{i}\xi-\Delta\omega+d_{1}&a_{1}&\dots&a_{N- 1}\\ a_{N-1}&\mathrm{i}\xi-\Delta\omega+d_{2}&\dots&a_{N-2}\\ \vdots&\ddots&\ddots&\vdots\\ a_{1}&\dots&a_{N-1}&\mathrm{i}\xi-\Delta\omega+d_{N}\end{array} \right]\left[ \begin{array}{c}x_{1}\\ \vdots\\ x_{N}\end{array} \right]=\mathrm{i}\xi\left[ \begin{array}{c}\mathrm{e}^{ \mathrm{i}(\frac{2\pi l}{N})1}\\ \vdots\\ \mathrm{e}^{\mathrm{i}(\frac{2\pi l}{N})N}\end{array} \right]\]  
where the coefficients \(d_{1},\dots,d_{N}\) are the inverse Fourier transform of the mistuning coefficients \(\delta_{1},\dots,\delta_{N}\) (and they now correspond to the physical frequency mistuning of the blades), and the off-diagonal coefficients \(a_{1},\dots,a_{N}\) to the Fourier transform of the frequency corrections of the modal family: \(\Delta\omega_{1},\dots,\Delta\omega_{N}\), and can be regarded as the influence coefficients of the elastic coupling through the disk. The coupling coefficients are all real and symmetric (\(a_{j}=a_{N-j}\)), because the distribution of frequencies \(\Delta\omega_{1},\dots,\Delta\omega_{N}\)is real and symmetric, and, since the average of the modal frequency deviations is 0, \(a_{N}=a_{0}=0\).