# Passive Detection of Low-Altitude Signal Sources Using an Improved Cross-Correlation Algorithm
## 3 Proposed Algorithm
### Frequency and Direction of Arrival Estimation  
When a low-altitude signal source is present, the frequency of the received signal can be estimated using the proposed cross-correlation method. At time \(jt_{a}\), the backward cross-correlation between the template and received signal is  
\[\hat{R}_{jk}=\left\{ \begin{array}{cc}0,&k=1,2,\ldots,L\\ \frac{\sum\limits_{i=1}^{L}\left(m_{i}-\overline{m}\right)\left(\dot{y}_{j(k+i -L-1)}-\overline{z}\right)}{\sqrt{\sum\limits_{i=1}^{L}\left(m_{i}-\overline {m}\right)^{2}}\sqrt{\sum\limits_{i=1}^{L}\left(\Gamma y_{j(k+i-L-1)}- \overline{z}\right)^{2}}},&k=L+1,\ldots,(2.5-2.4)\times 10^{9}/f_{w}\end{array} \right. \tag{8}\]  
where z =\(\left[\dot{y}_{j(k-L)},\ldots,\dot{y}_{j(k-1)}\right]\) and \(\overline{z}\) is the mean of \(z\). The improved cross-correlation is given by  
\[\widetilde{R}_{jk}=\max\left(\hat{R}_{jk},\hat{R}_{jk}\right) \tag{9}\]  
where \(\widetilde{R}_{jk}\) is the \(k\)th cross-correlation coefficient at time \(jt_{a}\), the cross-correlation array at time \(jt_{a}\) is \(\widetilde{R}_{j}=\left[\widetilde{R}_{j1},\ldots,\widetilde{R}_{jk},\ldots,\widetilde{R}_{j((2.5-2.4)\times 10^{9}/f_{w})}\right]\), \(k=1,2,\ldots,(2.5-2.4)\times 10^{9}/f_{w}\), and the improved cross-correlation matrix in the time-frequency domain is \(\widetilde{R}=\left[\widetilde{R}_{1},\ldots,\widetilde{R}_{j},\ldots, \widetilde{R}_{8/t_{a}}\right]^{H}\), \(j=1,\ldots,8/t_{a}\). In order to reduce the cross-correlation value of the clutter signal and enhance the cross-correlation value of the signal source simultaneously, a time-frequency domain accumulation method is employed. The improved cross-correlation matrix after three rounds of accumulation can be expressed as  
\[R_{jk}=\sum\limits_{l=1}^{3}\widetilde{R}_{jk}^{l} \tag{10}\]  
Figure 5a shows the improved cross-correlation matrix of the signal in Figure 3b after accumulation, and Figure 5b shows the matrix of the signal in Figure 3d after accumulation. The low-altitude signal shown in Figure 5c has a bandwidth of 9.47 MHz between 2.401758 GHz and 2.411228 GHz. The start of this signal is  
\[f_{start}=\left\{\begin{array}{cc}K_{b}f_{w}+2.4\times 10^{9}&s.t. (K_{b+1}-K_{ b})\geq 8\times 10^{6}/f_{w},\\ K_{b}\in K,K_{1}\leq\ldots\leq K_{b}\leq\ldots\leq K_{B},b=1,\ldots,B,\\ K=k\Big{|}_{(\exists\;R_{jk}\geq 0.7)},\\ 1\leq j\leq 8/t_{a},\\ 1\leq k\leq(2.5-2.4)\times 10^{9}/f_{w}-L+1,\\ 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\where \(K\) is the array of frequency positions that exceed the cross-correlation threshold and \(B\) is the length of \(K\). The elements in \(K\) are sorted in ascending order. When the difference between adjacent elements exceeds \(8\times 10^{6}/f_{w}\), the start frequency is found, and vice versa. The end position of the frequency range is found in the same way and is given by  
\[f_{end}=\left\{ \begin{array}{ll}K_{b+1}f_{w}+2.4\times 10^{9}&s.t.(K_{b+1}-K_{ b})\geq 8\times 10^{6}/f_{w},\\ &K_{b}\in K,K_{1}\leq\ldots\leq K_{b}\leq\ldots\leq K_{B},b=1,\ldots,B,\\ &K=k\Big{|}_{(\exists\ R_{jk}\geq 0.7)},\\ &1\leq j\leq 8/t_{a},\\ &1\leq k\leq(2.5-2.4)\times 10^{9}/f_{w}-L+1,\\ 0&\text{otherwise}\end{array} \right. \tag{12}\]  
The frequency range is then  
\[f_{range}=\left\{ \begin{array}{ll}f_{end}-f_{start}&s.t.f_{end}\
eq 0,\\ &f_{start}\
eq 0,\\ 0&\text{otherwise}\end{array} \right. \tag{13}\]  
From the rotational speed and location of the antenna, the correspondence between the time and direction can be expressed as  
\[p=\frac{180t}{8} \tag{14}\]  
Note that a directional antenna is employed. The DOA of the low-altitude signal source can be obtained using frequency \(f_{start}\) as shown in Figure 5d. The time of the maximum power can be used as an estimate of the time when the source appears which is given by  
\[ \begin{array}{ll}\dot{t}=Jt_{a}\\ s.t.\ J=j\Big{|}_{(\forall\max(R_{jk}))},\\ 1\leq j\leq 16/t_{a}\\ k=\big{(}f_{start}-2.4\times 10^{9}\big{)}/f_{w}\end{array}