## 3. Planar Modal Characteristic of Heavy Load Radial Tire
### Modal Resonance Parameters Identification  
The least square complex exponential method is utilized for the modal parameters identification, including the modal resonance frequency, modal damping, and modal shape. Figures 7 and 8 present the frequency domain vibration characteristic, meanwhile the impulse is needed. Utilizing the inverse FFT method, the impulse response function \(h(t)\) is given as:  
\[h(t)=\text{Re}\left(\sum\limits_{r=1}^{2N}A_{r}e^{s_{r}t}\right) \tag{18}\]  
where Re is the real part. The impulse response function of the finite time form \(t_{k}=k\Delta t\) transforming from the impulse response function is given as:  
\[h(k\Delta t)=\sum\limits_{r=1}^{2N}A_{r}e^{s_{r}k\Delta t}=\sum\limits_{r=1}^ {2N}A_{r}V_{r}^{k} \tag{19}\]  
where \(V_{r}=e^{s_{r}\Delta t}k=0,1,\cdots,L\). Impulse response functions of the different time can be concluded with the finite time form:  
\[\left\{ \begin{array}{l}h_{0}=\sum\limits_{r=1}^{2N}A_{r}V_{r}^{0}=A_{1}+A_{2 }+\cdots+A_{2N}\\ h_{1}=\sum\limits_{r=1}^{2N}A_{r}V_{r}^{1}=A_{1}V_{1}+A_{2}V_{2}+\cdots+A_{2N} V_{2N}\\ h_{2}=\sum\limits_{r=1}^{2N}A_{r}V_{r}^{2}=A_{1}V_{1}^{2}+A_{2}V_{2}^{2}+ \cdots+A_{2N}V_{2N}^{2}\\ \vdots\\ h_{L}=\sum\limits_{r=1}^{2N}A_{r}V_{r}^{L}=A_{1}V_{1}^{L}+A_{2}V_{2}^{L}+ \cdots+A_{2N}V_{2N}^{L}\end{array} \right. \tag{20}\]  
where \(\beta_{2N}=1\). \[\sum\limits_{r=1}^{2N}\beta_{k}V^{k}=\prod\limits_{r=1}^{N}(V-V_{k})(V-V_{k}^ {*})=0 \tag{21}\]  
Multiplying Equation (21) by \(\beta_{k}\), the calculated equations are summed up:  
\[\sum\limits_{k=1}^{2N}\beta_{k}h_{k}=\sum\limits_{k=1}^{2N}\beta_{k}\sum \limits_{r=1}^{2N}A_{r}V_{r}^{k}=\sum\limits_{r=1}^{2N}A_{r}\sum\limits_{k=1} ^{2N}\beta_{k}V_{r}^{k} \tag{22}\]  
Substituting \(\sum\limits_{k=0}^{2N}\beta_{k}V_{r}^{k}=0\) and \(\beta_{2N}=1\), Equation (22) is transformed as:  
\[\sum\limits_{k=1}^{2N-1}\beta_{k}h_{k}=-h_{2N} \tag{23}\]Constructing Equation (23) with the time domain result to solve the autoregressive coefficient \(\beta_{k}\), the equation is set with the same time interval \(\Delta t\):  
\[\left\{ \begin{array}{l}\sum\limits_{k=1}^{2N-1}\beta_{k}h_{k}=\beta_{0}h_{0}+ \beta_{1}h_{1}+\cdots+\beta_{2N-1}h_{2N-1}=-h_{2N}\\ \sum\limits_{k=1}^{2N-1}\beta_{k}h_{k+1}=\beta_{0}h_{1}+\beta_{1}h_{2}+\cdots+ \beta_{2N-1}h_{2N}=-h_{2N+1}\\ \\ \sum\limits_{k=1}^{2N-1}\beta_{k}h_{k+M-1}=\beta_{0}h_{M-1}+\beta_{1}h_{M}+ \cdots+\beta_{2N-1}h_{L-1}=-h_{L}\end{array} \right. \tag{24}\]  
in which, \(M=L-2N\). Transforming Equation (24) in the matrix form, the equation set is given as Equation (25):  
\[\left[ \begin{array}{ccccc}h_{0}&h_{1}&h_{2}&\cdots&h_{2N-1}\\ h_{1}&h_{2}&h_{3}&\cdots&h_{2N}\\ \vdots&\vdots&\vdots&&\vdots\\ h_{M-1}&h_{M}&h_{M+1}&\cdots&h_{L-1}\end{array} \right]\left[ \begin{array}{ c}\beta_{0}\\ \beta_{1}\\ \vdots\\ \beta_{2N-1}\end{array} \right]=\left[ \begin{array}{c}h_{2N}\\ h_{2N+1}\\ \vdots\\ h_{L}\end{array} \right] \tag{25}\]  
Utilizing the least square method, the autoregressive coefficient \(\beta_{k}\) is gained:  
\[\left\{\beta\right\}=\left([h]^{T}[h]\right)^{-1}\left([h]\left\{h^{\prime} \right\}\right) \tag{26}\]  
\[\sum\limits_{r=1}^{2N}\beta_{k}V^{k}=\beta_{0}+\beta_{1}V+\beta_{2}V^{2}+ \cdots+\beta_{2N-1}V^{2N-1}+V^{2N}=0 \tag{27}\]  
In which \(V_{r}\) can be given by solving the polynomial equation formed with the autoregressive coefficient \(\beta_{k}\). The modal parameters, including modal resonance frequency \(\omega_{r}\) and modal damping \(\xi_{r}\) are calculated in Equation (28) and modal shape coefficient \(A_{r}\) is obtained in Equation (29):  
\[\left\{ \begin{array}{ccccc}R_{r}=\ln V_{r}=s_{r}\Delta t\\ \omega_{r}=\left|R_{r}\right|/\Delta t\\ \xi_{r}=\sqrt{\frac{1}{1+\left(\Im(R_{r})/\Re(R_{r})\right)^{2}}}\end{array} \right. \tag{28}\]  
\[\left[ \begin{array}{ccccc}1&1&1&\cdots&1\\ V_{1}&V_{2}&V_{3}&\cdots&V_{2N}\\ \vdots&\vdots&\vdots&&\vdots\\ V_{1}^{L}&V_{2}^{L}&V_{3}^{L}&\cdots&V_{2N}^{L}\end{array} \right]\left[ \begin{array} []{c}A_{1}\\ A_{2}\\ \vdots\\ A_{2N}\end{array} \right]=\left[