## 3 Proposed Fast TDE Algorithm
### Principle of Polynomial Inversion-Based Time Delay Estimation  
For ease of description, we introduce the function \(TED(\tau_{0})=\hat{\tau}_{0}\) by rewriting Equation (7) as follows:  
\[TED(\tau_{0})=\hat{\tau}_{0}=-\frac{\theta}{2}\times\frac{\hat{R}_{x_{1}x_{2}}( \theta)-\hat{R}_{x_{1}x_{2}}(-\theta)}{\hat{R}_{x_{1}x_{2}}(-\theta)+\hat{R}_{ x_{1}x_{2}}(\theta)-2\hat{R}_{x_{1}x_{2}}(0)} \tag{8}\]  
The equation above gives:  
\[\tau_{0}\approx TED^{-1}(\hat{\tau}_{0}) \tag{9}\]  
Here, \(TED^{-1}(\hat{\tau})\) is the inverse function of \(TED(\tau)\). To obtain the fractional time-delay inversion function, a set of \(P\)-order polynomial coefficients is designed to approximate \(TED^{-1}(\hat{\tau})\)--i.e., \(\sum_{p=0}^{P}a_{p}\hat{\tau}_{0}^{p}\approx TED^{-1}(\hat{\tau}_{0})\approx \tau_{0}\). To solve for the polynomial coefficients \(\{a_{0},a_{1},\cdots,a_{P}\}\), the fractional delay interval \([-0.5,+0.5]\) is divided into \(Q+1\) equal parts and \(Q\leq P\); then, \(\Delta=1/Q\), \(\tau_{q}=-0.5+q\Delta,q=0,\cdots,Q\). First, the time-delay estimate \(\hat{\tau}_{q}\) is calculated as shown in the following equation:  
\[\hat{\tau}_{q}=-\frac{\theta}{2}\frac{\hat{R}_{r}(\theta+\tau_{q})-\hat{R}_{r }(-\theta+\tau_{q})}{\hat{R}_{r}(\theta+\tau_{q})+\hat{R}_{r}(-\theta+\tau_{q })-2\hat{R}_{r}(\tau_{q})} \tag{10}\]  
here \(\hat{R}_{r}(\theta+\tau_{q})\), \(\hat{R}_{r}(-\theta+\tau_{q})\), \(\hat{R}_{r}(\tau_{q})\) are defined as:  
\[\left\{\begin{array}{rcl}\hat{R}_{r}(\theta+\tau_{q})&=&\frac{1}{N}\sum_{k= 0}^{N-1}\Re(r^{*}(kT)r(kT+\theta T+\tau_{q}T))\\ \hat{R}_{r}(\tau_{q})&=&\frac{1}{N}\sum_{k=0}^{N-1}\Re(r^{*}(kT)r(kT+\tau_{q}T ))\\ \hat{R}_{r}(-\theta+\tau_{q})&=&\frac{1}{N}\sum_{k=0}^{N-1}\Re(r^{*}(kT)r(kT- \theta T+\tau_{q}T)).\end{array}\right.\]  
where the \(r(kT)\) in the above equation is an arbitrary waveform sequence with zero noise, consistent with the spectral profile of the transmit signal \(s(kT)\). We then use the obtained \(\{\hat{\tau}_{0},\hat{\tau}_{1},\cdots,\hat{\tau}_{Q}\}\) and the polynomial to establish the equation, which can be expressed as:  
Figure 3: The first-order derivative of Equation (7) vs. bandwidth (\(\theta=1\)).  
\[\left[\begin{array}{ccccc}1&\hat{\tau}_{0}^{1}&\cdots&\hat{\tau}_{0}^{P-1}&\hat{ \tau}_{0}^{P}\\ 1&\hat{\tau}_{1}^{1}&\cdots&\hat{\tau}_{1}^{P-1}&\hat{\tau}_{1}^{P}\\ \cdots&\vdots&\cdots\\ 1&\hat{\tau}_{Q-1}^{1}&\cdots&\hat{\tau}_{Q-1}^{P-1}&\hat{\tau}_{Q-1}^{P}\\ 1&\hat{\tau}_{Q}^{1}&\cdots&\hat{\tau}_{Q}^{P-1}&\hat{\tau}_{Q}^{P}\end{array} \right]\left[\begin{array}{c}a_{0}\\ a_{1}\\ \vdots\\ a_{P-1}\\ a_{P}\end{array}\right]=\left[\begin{array}{c}\tau_{0}\\ \tau_{1}\\ \vdots\\ \tau_{Q-1}\\ \tau_{Q}\end{array}\right] \tag{11}\]  
Let \(\mathcal{A}=[a_{0},a_{1},\cdots,a_{P}]^{T}\), \(T\) denote transpose: \(\mathcal{T}=[\tau_{0},\tau_{1},\cdots,\tau_{Q}]^{T}\),  
\[\Gamma=\left[\begin{array}{ccccc}1&\hat{\tau}_{0}^{1}&\cdots&\hat{\tau}_{0}^ {P-1}&\hat{\tau}_{0}^{P}\\ 1&\hat{\tau}_{1}^{1}&\cdots&\hat{\tau}_{1}^{P-1}&\hat{\tau}_{1}^{P}\\ 1&\cdots&\vdots&\cdots\\ 1&\hat{\tau}_{Q-1}^{1}&\cdots&\hat{\tau}_{Q-1}^{P-1}&\hat{\tau}_{Q-1}^{P}\\ 1&\hat{\tau}_{Q}^{1}&\cdots&\hat{\tau}_{Q}^{P-1}&\hat{\tau}_{Q}^{P}\end{array}\right]\]  
Using the least square method, we solve for the polynomial coefficients as follows:  
\[\mathcal{A}=(\Gamma^{T}\Gamma)^{-1}\Gamma^{T}\mathcal{T}. \tag{12}\]  
Then, Equation (8) yields the fractional time-delay estimate, as follows:  
\[\tilde{\tau}_{0}=\sum_{p=0}^{P}a_{p}\tilde{\tau}_{0}^{p}. \tag{13}\]