# Power System State Estimation Approach Considering Transmission Line Temperature
## 3 Augmented Jacobian Etc-Se Approach
### Augmented Jacobian Matrix Incorporating Line Temperature  
With the augment of measurement and state variables, the Jacobian matrix \(\mathbf{H}\) is also augmented. In order to determine the elements related to line temperatures, the augmented Jacobian matrix \(\mathbf{H}_{aug}\) is partitioned, which can be represented as,  
\[\mathbf{H}_{aug}=\frac{\partial\mathbf{h}(\mathbf{x})}{\partial\mathbf{x}}=\left[\begin{array} []{cccc}\mathbf{H}_{11}&\mathbf{H}_{12}&\mathbf{H}_{13}\\ \mathbf{H}_{21}&\mathbf{H}_{22}&\mathbf{H}_{23}\end{array}\right]=\left[\begin{array}{ ccc}\partial\mathbf{h}_{RTU}/\partial\mathbf{U}&\partial\mathbf{h}_{RTU}/\partial\mathbf{0}& \partial\mathbf{h}_{RTU}/\partial\mathbf{T}\\ \partial\mathbf{h}_{HBE}/\partial\mathbf{U}&\partial\mathbf{h}_{HBE}/\partial\mathbf{T}\end{array}\right] \tag{19}\]  
Moreover, the submatrices \(\mathbf{H}_{11}\) and \(\mathbf{H}_{12}\) are identical to the Jacobian matrix of WLS-SE. Taking \(\mathbf{H}_{23,i}\) as an example, the magnitude of a branch current flowing through the line \(l\) can be obtained directly via a current magnitude meter. Further, the partial differential coefficients corresponding to the line temperature \(T_{l}\) can be expressed as,  
\[\mathbf{H}_{23,i}=\begin{bmatrix}&&&0&0&\\ \cdots&\vdots&\vdots&\vdots&\vdots&\cdots\\ &\frac{\partial\mathbf{U}_{l}}{\partial T_{l}}&\frac{\partial\mathbf{U}_{i}}{\partial T _{l}}&\frac{\partial\mathbf{U}_{i,j,mpl}}{\partial T_{l}}&\frac{\partial\mathbf{U}_ {i,j,mpl}}{\partial T_{l}}\\ \cdots&\vdots&\vdots&\vdots&\vdots&\cdots\\ &&&0&0\end{bmatrix}^{T} \tag{20}\]with  
\[\frac{\partial U_{i}}{\partial T_{l}}=0 \tag{21}\]  
\[\frac{\partial\theta_{i}}{\partial T_{l}}=0 \tag{22}\]  
\[\frac{\partial I_{ij,real}}{\partial T_{l}}=-(U_{i}\cos\theta_{i}-U_{j}\cos \theta_{j})\frac{\partial G_{ij}(T_{l})}{\partial T_{l}}+(U_{i}\sin\theta_{i}- U_{j}\sin\theta_{j})\frac{\partial B_{ij}(T_{l})}{\partial T_{l}} \tag{23}\]  
\[\frac{\partial I_{ij,imag}}{\partial T_{l}}=-(U_{i}\cos\theta_{i}-U_{j}\cos \theta_{j})\frac{\partial B_{ij}(T_{l})}{\partial T_{l}}-(U_{i}\sin\theta_{i}- U_{j}\sin\theta_{j})\frac{\partial G_{ij}(T_{l})}{\partial T_{l}} \tag{24}\]  
It should be noted that only when temperature \(T_{l}\) corresponds to transmission line \(l\), is the partial differential coefficients not equal to 0, i.e., merely one element in each row of \(\partial_{ij,real}/\partial T\) and \(\partial I_{ij,imag}/\partial T\) is not equal to 0.  
Moreover, regarding the submatrix \(\mathbf{H}_{31}\), at most two elements in each row are not equal to 0. Then, the submatrix \(\mathbf{H}_{31}\) can be expressed as,  
\[\mathbf{H}_{31,J}=\begin{bmatrix}0&\cdots&\frac{\partial h_{HBEJ}}{\partial U_{i} }&\cdots&0&\cdots&\frac{\partial h_{HBEJ}}{\partial U_{j}}&\cdots&0\end{bmatrix} \tag{25}\]  
In addition, the corresponding partial differential coefficient is calculated as,  
\[\frac{\partial h_{HBEJ}}{\partial U_{i}}=(G_{ij}^{2}+B_{ij}^{2})(2U_{i}-U_{j} \cos\theta_{ij})R \tag{26}\]  
Similarly, in the submatrix \(H_{32}\), at most two elements in each row are not equal to 0, the submatrix \(\mathbf{H}_{32}\) can be represented as,  
\[\mathbf{H}_{32,J}=\begin{bmatrix}0&\cdots&\frac{\partial h_{HBEJ}}{\partial\theta _{i}}&\cdots&0&\cdots&\frac{\partial h_{HBEJ}}{\partial\theta_{j}}&\cdots&0 \end{bmatrix} \tag{27}\]  
Then, the corresponding partial differential coefficient is calculated as,  
\[\frac{\partial h_{HBEJ}}{\partial\theta_{i}}=(G_{ij}^{2}+B_{ij}^{2})(2U_{i}U_ {j}\sin\theta_{ij})R \tag{28}\]  
Since the HBE of the transmission line \(l\) is merely relevant to the corresponding line temperature \(T_{l}\), the submatrix \(\mathbf{H}_{33}\) is a diagonal matrix and represented as,  
\[\mathbf{H}_{33}=\begin{bmatrix}\ddots&&0\\ 0&\frac{\partial h_{HBEJ}}{\partial T_{l}}&\cdots&0\\ 0&\cdots&\frac{\partial h_{HBEJ}}{\partial T_{m}}&0\\ 0&&\ddots\end{bmatrix} \tag{29}\]  
with  
\[\begin{array}{l}\frac{\partial h_{HBEJ}}{\partial T_{l}}=(2G_{ij}\frac{ \partial G_{ij}}