Separating linear and nonlinear parts in PhyLSTM

2024-04-26: This has become obsolete

2023-08-15: not yet complete

For the PhyLSTM, we want to separate the modeling of the linear part of the transfer function $I\to B$ from the nonlinearities, while still retaining the mathematical correctness of the solution of the Bouc-Wen Model.

Conceptually, the model maps $I\to \hat{B},\hat{\dot{B}},\hat{r}$, however in reality we are mapping $I_{\mathrm{s}}\to {\hat{B}}_{\mathrm{s}},{\hat{\dot{B}}}_{\mathrm{s}},\hat{r}$, where $\mathrm{s}$ stands for scaled. This is because we input a scaled $I$ to the PhyLSTM model, and output scaled variables. The variables are scaled using standard scaling, so for example $I_s=\frac{I-{\mu }_I}{{\sigma }_I}$ .

However when we split the output into a linear and nonlinear part, the output can be written as

$$\{\begin{array}{l}{\hat{B}}_{\mathrm{s}}={\hat{B}}_{\ln ,\mathrm{s}}+{\hat{B}}_{\mathrm{nl},\mathrm{s}}\\ {\hat{\dot{B}}}_{\mathrm{s}}={\hat{\dot{B}}}_{\ln ,\mathrm{s}}+{\widehat{\hat{\dot{B}}}}_{\mathrm{nl},\mathrm{s}}\end{array}$$

Where the nonlinear parts are modeled as

$$\{{\hat{B}}_{\mathrm{nl},\mathrm{s}},{\hat{\dot{B}}}_{\mathrm{nl},\mathrm{s}}\}={\mathrm{PhyLSTM}}_1(I_{\mathrm{s}})$$

For the linear parts, we define

$$\begin{array}{rl}{\hat{B}}_{\ln }& :=kI+m\\ {\hat{\dot{B}}}_{\ln }& =\frac{\mathrm{d}{B}_{\ln }}{\mathrm{dt}}:=k\dot{I}\end{array}$$

And then we express the same with its scaled versions

\begin{align} \hat{B}_{\mathrm{ln}} &= \sigma_B \hat{B}_{\mathrm{ln}, \mathrm{s}} + \mu_B \\ \hat{\dot{B}}_{\mathrm{ln}} & = \sigma_\dot{B} \hat{\dot{B}}_{\mathrm{ln}, \mathrm{s}} + \mu_\dot{B} \end{align}

Similarly for the scaled $B$ we define

$$\begin{array}{rl}{\hat{B}}_{\ln ,\mathrm{s}}& :=k^{\prime }{I}_{\mathrm{s}}+m^{\prime }\\ {\hat{\dot{B}}}_{\ln ,\mathrm{s}}& :=\frac{\mathrm{d}{\hat{B}}_{\ln ,\mathrm{s}}}{\mathrm{dt}}=k^{\prime }{\dot{I}}_s\text{\# can we do this?}\end{array}$$

Linear part of B

So when we factor in

$$\begin{array}{rl}{\hat{B}}_{\ln }& =kI+m={\sigma }_B{\hat{B}}_{\ln ,\mathrm{s}}+{\mu }_B\\ & ={\sigma }_B(k^{\prime }{I}_{\mathrm{s}}+m^{\prime })+{\mu }_B\\ & ={\sigma }_B\left(k^{\prime }\left(\frac{I-{\mu }_I}{{\sigma }_I}\right)+m^{\prime }\right)+{\mu }_B\\ & =k^{\prime }\frac{{\sigma }_B}{{\sigma }_I}I-k^{\prime }\frac{{\sigma }_B}{{\sigma }_I}{\mu }_I+{\sigma }_B{m}^{\prime }+{\mu }_B\end{array}$$

by identifying terms we get

$$\{\begin{array}{l}{k}^{\prime }=\frac{{\sigma }_I}{{\sigma }_B}k\\ m^{\prime }=\frac{k^{\prime }}{{\sigma }_I}{\mu }_I+\frac{1}{{\sigma }_B}\left[m-{\mu }_B\right]=\frac{1}{{\sigma }_B}\left[k{\mu }_I+m-{\mu }_B\right]\end{array}$$

Linear part of $\dot{B}$

Similarly for its derivative

$$\begin{array}{rl}{\hat{\dot{B}}}_{\ln }& =k\dot{I}\\ & =k\frac{\mathrm{d}I}{\mathrm{d}t}\\ & =k\frac{\mathrm{d}}{\mathrm{d}t}\left({\sigma }_I{I}_{\mathrm{s}}+{\mu }_I\right)\\ & =k{\sigma }_I\frac{\mathrm{d}{I}_s}{\mathrm{d}t}\\ & =k{\sigma }_I{\dot{I}}_s\end{array}$$

However,

\begin{align} \hat{\dot{B}}_{\mathrm{ln}} & = \sigma_\dot{B} \hat{\dot{B}}_{\mathrm{ln}, \mathrm{s}} + \mu_{\dot{B}} \\ & = \sigma_{\dot{B}} k^{\prime} \dot{I}_{s} + \mu_{\dot{B}} \end{align}

And therefore

$$k^{\prime }=k\frac{{\sigma }_I}{{\sigma }_B}$$

as before. For the constant term we might add a bias for good measure.