﻿ 木材含水率的神经网络模型构建

# 木材含水率的神经网络模型构建Neural Network Model for Wood Moisture Content

Abstract: Wood moisture content is an important technical parameter in the wood drying process. In order to deal with data of the wood drying test more accurately and effectively, we decided to use the advantages of RMA neural network model to design an algorithm for wood moisture content in-fluencing factors and relationship of time. Firstly, we select a kind of neural network named Re-current Neural Network to set up a model algorithm. And next thing is building an operation pro-gram which can express the RMA neural network model well. Finally, we combine real empirical data with the program, and we get functional statements of systemic wood moisture content.

1. 引言

2. 实时建模算法RMA的理论建立

2.1. 实时建模算法RMA的基本思想

2.1.1. 非线性连续系统离散化

$\frac{\partial x\left(t,k\right)}{\partial t}=-\alpha x\left(t,k\right)+f\left({W}_{1}\left(t,k\right),x\left(t,k\right)\right)$ (1-1)

${W}_{1}\left(t,k+1\right)={W}_{1}\left(t,k\right)+\Delta {W}_{1}\left(t,k\right)$ (1-2)

2.1.2. 连接权矩阵的训练法

$e\left(t,k\right)=y\left(t\right)-x\left(t,k\right),t\in \left[0,\infty \right)$ (1-3)

$x\left(0,k\right)=x\left(0\right)=y\left(0\right),k=0,1,2,\cdots$ (1-4)

$\eta \left(t,k\right)={\int }_{0}^{t}\left[x\left(\tau ,k+1\right)-x\left(\tau ,k\right)\right]\text{d}\tau$ (1-5)

$\eta \left(0,k\right)=0$ ，由此可得

$\begin{array}{c}\frac{\partial \eta \left(t,k\right)}{\partial t}=x\left(t,k+1\right)-x\left(t,k\right)\\ =x\left(t,k+1\right)-x\left(0,k+1\right)-x\left(t,k\right)+x\left(0,k\right)\\ ={\int }_{0}^{t}\frac{\partial x\left(\tau ,k+1\right)}{\partial \tau }\text{d}\tau -{\int }_{0}^{t}\frac{\partial x\left(\tau ,k\right)}{\partial \tau }\text{d}\tau \\ =-\alpha {\int }_{0}^{t}x\left(\tau ,k+1\right)\text{d}\tau +{\int }_{0}^{t}f\left({W}_{1}\left(\tau ,k+1\right),x\left(\tau ,k+1\right)\right)\text{d}\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\alpha {\int }_{\text{0}}^{t}x\left(\tau ,k\right)\text{d}\tau -{\int }_{0}^{t}f\left({W}_{1}\left(\tau ,k\right),x\left(\tau ,k\right)\right)\text{d}\tau \\ =-\alpha \eta \left(t,k\right)+{\int }_{0}^{t}\left[f\left({W}_{1}\left(\tau ,k+1\right),x\left(\tau ,k+1\right)\right)-f\left({W}_{1}\left(\tau ,k\right),x\left(\tau ,k\right)\right)\right]\text{d}\tau \end{array}$

$\begin{array}{l}e\left(t,k+1\right)-e\left(t,k\right)\\ =x\left(t,k\right)-x\left(t,k+1\right)\\ =\alpha \eta \left(t,k\right)-{\int }_{0}^{t}\left[f\left({W}_{1}\left(\tau ,k+1\right),x\left(\tau ,k+1\right)\right)-f\left({W}_{1}\left(\tau ,k\right),x\left(\tau ,k\right)\right)\right]\text{d}\tau \end{array}$ (1-6)

$\begin{array}{l}f\left({W}_{1}\left(t,k+1\right),x\left(t,k+1\right)\right)-f\left({W}_{1}\left(t,k\right),x\left(t,k\right)\right)\\ ={K}_{1}\frac{\partial \eta \left(t,k\right)}{\partial t}+{K}_{2}\frac{\partial e\left(t,k\right)}{\partial t}\end{array}$ (1-7)

$\left[\begin{array}{c}\frac{\partial \eta \left(t,k\right)}{\partial t}\\ e\left(t,k+1\right)\end{array}\right]=\left[\begin{array}{cc}-\alpha I+{K}_{1}& {K}_{2}\\ \alpha I-{K}_{1}& I-{K}_{2}\end{array}\right]\left[\begin{array}{c}\eta \left(t,k\right)\\ e\left(t,k\right)\end{array}\right]+\left[\begin{array}{cc}-{K}_{1}& -{K}_{2}\\ {K}_{1}& {K}_{2}\end{array}\right]\left[\begin{array}{c}\eta \left(0,k\right)\\ e\left(0,k\right)\end{array}\right]$ (1-8)

$\left[\begin{array}{c}\frac{\partial \eta \left(t,k\right)}{\partial t}\\ e\left(t,k+1\right)\end{array}\right]=\left[\begin{array}{cc}-\alpha I+{K}_{1}& {K}_{2}\\ \alpha I-{K}_{1}& I-{K}_{2}\end{array}\right]\left[\begin{array}{c}\eta \left(t,k\right)\\ e\left(t,k\right)\end{array}\right]$ (1-9)

$\left[\begin{array}{c}\frac{\partial x\left(t,k\right)}{\partial t}\\ y\left(t,k+1\right)\end{array}\right]=\left[\begin{array}{cc}{A}_{1}& {A}_{2}\\ {A}_{3}& {A}_{4}\end{array}\right]\left[\begin{array}{c}x\left(t,k\right)\\ y\left(t,k\right)\end{array}\right]$

$\underset{k\to \infty }{\mathrm{lim}}\left[\begin{array}{c}x\left(t,k\right)\\ y\left(t,k\right)\end{array}\right]=0,t\in \left[0,T\right]$

2.2. 非线性连续系统的建模算法

$f\left({W}_{1}\left(t\right),x\left(t\right)\right)={W}_{1}\sigma \left(x\left(t\right)\right)，f\left({W}_{1}\left(t\right),x\left(t\right)\right)=\sigma \left({W}_{1}x\left( t \right) \right)$

$f\left({W}_{1}\left(t\right),x\left(t\right)\right)={W}_{1}\left(t\right)\sigma \left(x\left(t\right)\right)$ (1-10)

$\begin{array}{l}{W}_{1}\left(t,k\right)\sigma \left(x\left(t,k+1\right)\right)+\Delta {W}_{1}\left(t,k\right)\sigma \left(x\left(t,k+1\right)\right)-{W}_{1}\left(t,k\right)\sigma \left(x\left(t,k\right)\right)\\ ={K}_{1}\frac{\partial \eta \left(t,k\right)}{\partial t}+{K}_{2}\frac{\partial e\left(t,k\right)}{\partial t}\end{array}$ (1-11)

$\begin{array}{l}\Delta {W}_{1}\left(t,k\right)=\left(\frac{\partial e\left(t,k\right)}{\partial t}+{W}_{1}\left(t,k\right)\left(\sigma \left(x\left(t,k\right)\right)-\sigma \left(x\left(t,k+1\right)\right)\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left({\left[\sigma \left(x\left(t,k+1\right)\right)\right]}^{\text{T}}\sigma \left(x\left(t,k+1\right)\right)\right)}^{-1}×{\left[\sigma \left(x\left(t,k+1\right)\right)\right]}^{\text{T}}\end{array}$ (1-12)

2.3. 实时建模算法RAM的实现

3. RMA建模算法处理数据及仿真程序实现

3.1. 利用MATLAB实现RMA建模算法

3.1.1. 程序构建思路

3.1.2. 数据形态

3.2. 木材含水率数据与RMA建模平台的融合

Figure 1. Temperature and time at the experimental point

Figure 2. Humidity and time at the experimental point

Figure 3. Simulation of experimental data (temperature) by neural network model

Figure 4. Simulation of experimental data (humidity) by neural network model

4. 仿真实验结果及分析

$\left[\begin{array}{c}\text{d}{x}_{1}\\ \text{d}{x}_{2}\end{array}\right]=-A×\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]+\left[\begin{array}{cc}{w}_{11}& {w}_{22}\\ {w}_{33}& {w}_{44}\end{array}\right]×\left[\begin{array}{c}\mathrm{tanh}\left({t}_{1}\right)\\ \mathrm{tanh}\left({t}_{2}\right)\end{array}\right]$

$\begin{array}{l}{w}_{11}=-0.0008365{x}^{2}+0.0135x+0.4354\\ {w}_{22}=-0.0001135{x}^{2}+0.01734x+0.7403\\ {w}_{33}=0.0001015{x}^{2}-0.0003172x-0.01734\\ {w}_{44}=-0.0013{x}^{2}+0.0197x+0.8454\end{array}$

5. 结语

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https://doi.org/10.1109/81.678496

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