# Algorithms Utilized for Creep Analysis in Torque Transducers for Wind Turbines
## 11 Signal Variance Value Estimators
### Variance Estimator for Integral Mean  
This estimator is equal to the square of uncertainty determined by the mean uncertainty corridor around the line \(y=\theta_{1}\xi+\theta_{0}\), where the variance for a line is given by \(\sigma^{2}=u^{2}(\theta_{1})\,\xi^{2}\) \begin{table}
\begin{tabular}{c c c c c} \hline \hline  & \multicolumn{4}{c}{**(a)**} \\ \hline
**Level/Level Difference** & \(\overline{y}\) & \(y_{M}\) & \(y_{R}\) & \(\overline{y}_{\rm W}\) \\ \hline Level 0 kN\(\cdot\)m (mV/V) & 0.000234 & 0.000188 & 0.000292 & 0.000225 \\ Level 2000 N\(\cdot\)m (mV/V) & 1.046705 & 1.046708 & 1.046090 & 1.046729 \\ Level 2000–Level 0 (mV/V) & 1.046471 & 1.046520 & 1.045798 & 1.046504 \\ \hline  & \multicolumn{4}{c}{**(b)**} \\ \hline  & \multicolumn{4}{c}{\(<y>_{\rm tp1}\)} & \multicolumn{2}{c}{\(<y>_{\rm tp2}\)} & \multicolumn{2}{c}{\(<y>_{\rm tp3}\)} \\ \cline{2-6}
**Level/Level Difference** & \(u(y_{i})=0.0001\) & \(u(t_{i})=0.3\) & \(u(y_{i})=0.0001\) & \(u(t_{i})=0.3\) & \(u(y_{i})=0.0001\) & \(u(t_{i})=0.3\) \\  & \multicolumn{4}{c}{\(u(y_{i})=0.0001\)} & \multicolumn{2}{c}{\(u(y_{i})=0.0001\)} & \(u(y_{i})=0.0001\) & \(u(y_{i})=0.0001\) \\ \hline Level 0 kN\(\cdot\)m (mV/V) & 0.000246 & 0.000316 & 0.000343 & 0.000246 & 0.000594 & 0.000246 \\ Level 2000 N\(\cdot\)m (mV/V) & 1.047195 & 1.046692 & 1.046693 & 1.046700 & 1.046688 & 1.046690 \\ Level 2000–Level 0 (mV/V) & 1.046949 & 1.046376 & 1.046350 & 1.046454 & 1.046094 & 1.046444 \\ \hline \end{tabular}
\end{table} 
Table 3: Values of the mean signal level and difference estimators. \(+2\rho_{0100}\)\(\xi\)\(u(\theta_{0})\)\(u(\theta_{1})+u^{2}(\theta_{0})\). This estimator, fitted to, in general, a nonlinear curve and averaged over time is equal to:  
\[<u^{2}>_{t}=\frac{1}{t_{n}-t_{1}}\int_{t_{1}}^{t_{n}}\left[u^{2}(\theta_{1})\xi^ {2}(t)+2\rho_{\theta_{0}\theta_{1}}\xi(t)u(\theta_{1})u(\theta_{0})+u^{2}( \theta_{0})\right]\,\mathrm{d}t \tag{22}\]  
Case tp1. Straight line fit  
For the equation of a line \(y=at+b\), function \(\xi(t)=t\) and parameters \(\theta_{1}=a,\ \theta_{0}=b\). Then, the result from Formula (22) is as follows:  
\[<u_{p1}^{2}>_{t}=u^{2}(\theta_{1})(t_{1}^{2}+t_{1}t_{n}+t_{n}^{2})\,/3+\rho_{ \theta_{0}\theta_{1}}u(\theta_{1})u(\theta_{0})(t_{n}+t_{1})+u^{2}(\theta_{0}) \tag{23}\]  
Variance values \(u^{2}(\theta_{1})\) i \(u^{2}(\theta_{0})\) and covariance element \(\rho_{\theta 100}\)\(u(\theta_{0})u(\theta_{1})\) for variables \(\theta_{1}\) and \(\theta_{0}\) are determined numerically from the law of uncertainty propagation using the matrix \(\boldsymbol{L}\)\(\boldsymbol{U}_{\text{TS}}\boldsymbol{L}\) after adjusting the curve parameters and obtaining the global minimum of the criterion function. After linearization of the model, the slope \(\theta_{1}\) and intercept \(\theta_{0}\) are linearly dependent on the time and signal coordinates. It is necessary to determine the matrix of the sensitivity coefficients by numerically differentiating the increments for both parameters and calculating appropriate differential quotients with the use of appropriately selected increments of time and signal coordinates. A bilaterally multiplied matched covariance matrix by a sensitivity coefficient matrix (right-hand transposed) leads to the determination of the variances and covariances contained in the formula (23). Case tp2. Fitting a Third-Order Polynomial Curve  
For a polynomial \(y=b_{2}\)\(t^{3}+b_{1}\)\(t+b_{0}\), the function in (22) is determined by the \(\xi(t)=t^{3}+\beta_{1}t\), and parameter \(\beta_{1}=b_{1}/b_{2}\), and the parameters of a straight line \(\theta_{1}=b_{2}\), \(\theta_{0}=b_{0}\). The time-average value of the variance after integration is:  
\[<u_{p2}^{2}>_{t}=u^{2}(\theta_{1})\Big{(}W_{7}/7+(2b_{1}/b_{2})W_{5}/5+(b_{1} /b_{2})^{2}W_{3}/3\Big{)}+\rho_{\theta_{0}\theta_{1}}u(\theta_{1})u(\theta_{ 0})(W_{4}/2+b_{1}W_{2}/b_{2})+u^{2}(\theta_{0}) \tag{24}\]  
where this is denoted by  
\[W_{k}=\frac{(t_{n})^{k}-(t_{1})^{k}}{t_{n}-t_{1}}=\sum_{j=0}^{k}(t_{1})^{j}(t _{n})^{k-j} \tag{25}\]  
Case tp3.