@@ -32,9 +32,7 @@ The goal of WLS state estimation is to evaluate the state variable with the high
3232measurement input, by solving:
3333
3434$$
35- \begin{eqnarray}
36- \min r(\underline{U}) = \dfrac{1}{2} (f(\underline{U}) - \underline{z})^H W (f(\underline{U}) - \underline{z})
37- \end{eqnarray}
35+ \min r(\underline{U}) = \dfrac{1}{2} (f(\underline{U}) - \underline{z})^H W (f(\underline{U}) - \underline{z})
3836$$
3937
4038Where:
@@ -82,23 +80,21 @@ For example, there can be multiple voltage sensors on the same bus.
8280The measurement data can be merged into one virtual measurement using a Kalman filter:
8381
8482$$
85- \begin{eqnarray}
86- z = \dfrac{\sum_{k=1}^{N_{sensor}} z_k \sigma_k^{-2}}{\sum_{k=1}^{N_{sensor}} \sigma_k^{-2}}
87- \end{eqnarray}
83+ z = \dfrac{\sum_{k=1}^{N_{sensor}} z_k \sigma_k^{-2}}{\sum_{k=1}^{N_{sensor}} \sigma_k^{-2}}
8884$$
8985
9086Where $z_k$ and $\sigma_k$ are the measured value and standard deviation of individual measurements.
9187
9288Multiple appliance measurements (power measurements) on one bus are aggregated as the total injection at the bus:
9389
9490$$
95- \begin{eqnarray }
96- \underline{S} = \sum_{k=1}^{N_{appliance}} \underline{S}_k
97- \quad\text{and}\quad
98- \sigma_P^2 = \sum_{k=1}^{N_{appliance}} \sigma_{P,k}^2
99- \quad\text{and}\quad
100- \sigma_Q^2 = \sum_{k=1}^{N_{appliance}} \sigma_{Q,k}^2
101- \end{eqnarray }
91+ \begin{aligned }
92+ \underline{S} = \sum_{k=1}^{N_{appliance}} \underline{S}_k
93+ \quad\text{and}\quad
94+ \sigma_P^2 = \sum_{k=1}^{N_{appliance}} \sigma_{P,k}^2
95+ \quad\text{and}\quad
96+ \sigma_Q^2 = \sum_{k=1}^{N_{appliance}} \sigma_{Q,k}^2
97+ \end{aligned }
10298$$
10399
104100Where $S_k$ and $\sigma_ {P,k}$ and $\sigma_ {Q,k}$ are the measured value and the standard deviation of the individual
@@ -150,26 +146,14 @@ The following illustrates how this works for `sym_current_sensor`s in symmetric
150146See also [ the full mathematical workout] ( https://github.com/PowerGridModel/power-grid-model/issues/547 ) .
151147
152148$$
153- \begin{eqnarray}
154- & \mathrm{Re}\left\{I\right\} = I \cos\theta
155- \end{eqnarray}
156- $$
157- $$
158- \begin{eqnarray}
159- & \mathrm{Im}\left\{I\right\} = I \sin\theta
160- \end{eqnarray}
161- $$
162- $$
163- \begin{eqnarray}
164- & \text{Var}\left(\mathrm{Re}\left\{I\right\}\right) =
165- \sigma_i^2 \cos^2\theta + I^2 \sigma_{\theta}^2\sin^2\theta
166- \end{eqnarray}
167- $$
168- $$
169- \begin{eqnarray}
170- & \text{Var}\left(\mathrm{Im}\left\{I\right\}\right) =
171- \sigma_i^2 \sin^2\theta + I^2 \sigma_{\theta}^2\cos^2\theta
172- \end{eqnarray}
149+ \begin{aligned}
150+ & \mathrm{Re}\left\{I\right\} = I \cos\theta \\
151+ & \mathrm{Im}\left\{I\right\} = I \sin\theta \\
152+ & \text{Var}\left(\mathrm{Re}\left\{I\right\}\right) =
153+ \sigma_i^2 \cos^2\theta + I^2 \sigma_{\theta}^2\sin^2\theta \\
154+ & \text{Var}\left(\mathrm{Im}\left\{I\right\}\right) =
155+ \sigma_i^2 \sin^2\theta + I^2 \sigma_{\theta}^2\cos^2\theta
156+ \end{aligned}
173157$$
174158
175159## Iterative linear state estimation
@@ -188,9 +172,7 @@ Therefore, traditional measurements are linearized prior to running the algorith
188172 $\theta_i$ is the intrinsic transformer phase shift:
189173
190174$$
191- \begin{eqnarray}
192- \underline{U}_i = U_i \cdot e^{j \theta_i}
193- \end{eqnarray}
175+ \underline{U}_i = U_i \cdot e^{j \theta_i}
194176$$
195177
196178- Branch current with global angle: The global angle current measurement captures the phase offset relative to the same
201183 angle).
202184
203185$$
204- \begin{eqnarray}
205- \underline{I} = I_i \cdot e^{j \theta_i}
206- \end{eqnarray}
186+ \underline{I} = I_i \cdot e^{j \theta_i}
207187$$
208188
209189- Branch current with local angle: Sometimes, (accurate) voltage measurements are not available for a branch, which
217197 Given the measured (linearized) voltage phasor, the current phasor is calculated as follows:
218198
219199$$
220- \begin{equation}
221- \underline{I} = \underline{I}_{\text{local}}^{*} \frac{\underline{U}}{|\underline{U}|}
222- = \underline{I}_{\text{local}}^{*} \cdot e^{j \theta}
223- \end{equation}
200+ \underline{I} = \underline{I}_{\text{local}}^{*} \frac{\underline{U}}{|\underline{U}|}
201+ = \underline{I}_{\text{local}}^{*} \cdot e^{j \theta}
224202$$
225203
226204where $\underline{U}$ is either measured voltage magnitude at the bus or assumed unity magnitude, with the intrinsic
@@ -232,19 +210,15 @@ $\theta$ is the phase angle of the voltage.
232210 Given the measured (linearized) voltage phasor, the current phasor is calculated as follows:
233211
234212$$
235- \begin{eqnarray}
236- \underline{I} = (\underline{S}/\underline{U})^*
237- \end{eqnarray}
213+ \underline{I} = (\underline{S}/\underline{U})^*
238214$$
239215
240216- Bus power injection: Similar as above, to translate the power flow to a complex current phasor, if the bus voltage is
241217 not measured, the nominal voltage with zero angle will be used as an estimation.
242218 The current phasor is calculated as follows:
243219
244220$$
245- \begin{eqnarray}
246- \underline{I} = (\underline{S}/\underline{U})^*
247- \end{eqnarray}
221+ \underline{I} = (\underline{S}/\underline{U})^*
248222$$
249223
250224The aggregated apparent power flow is considered as a single measurement, with variance
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