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 智能系统学报  2020, Vol. 15 Issue (2): 193-203  DOI: 10.11992/tis.202001013 0

### 引用本文

JIANG Feng, YIN Xunfeng, YI Chunzhi, et al. A review of the research and application of calculating joint torque by electromyography signals[J]. CAAI Transactions on Intelligent Systems, 2020, 15(2): 193-203. DOI: 10.11992/tis.202001013.

### 文章历史

1. 哈尔滨工业大学 计算机科学与技术学院，黑龙江 哈尔滨 150001;
2. 哈尔滨工业大学 机电工程学院，黑龙江 哈尔滨 150001

A review of the research and application of calculating joint torque by electromyography signals
JIANG Feng 1, YIN Xunfeng 2, YI Chunzhi 2, YANG Chifu 2
1. College of Computer Science and Technology, Harbin Institute of Technology, Harbin 150001, China;
2. School of Mechatronics Engineering, Harbin Institute of Technology, Harbin 150001, China
Abstract: Surface electromyography is a neuronal signal of the human body that is easily detected. It extensively provides information about human motion. For a deeper understanding of human body dynamics, the use of electromyographic signals and biological models to examine the relationship between myoelectric signals and muscle forces or corresponding joint torques, is of great importance. This paper summarizes:the research results of solving human joint torque using electromyography signals; introduces the process of measuring and optimizing the neuromusculoskeletal model;gives some model physiological parameters to provide reference for future research; and provides application of the method in solving human joint torque in the current stage. Some of the problems encountered in the solution process are then evaluated, and the development prospect of the method is summarized, providing a reference for future research.
Key words: surface electromyography    neural activation    signal processing    muscle activation    musculoskeletal model    joint moment    joint dynamics    parameter identification

1 基本原理

2 神经激活求解

 $u(t) = \alpha \times e\Bigg(t - \frac{d}{{{T_E}}}\Bigg) - {\beta _1} \times u(t - 1) - {\beta _2} \times u(t - 2)$ (1)

 ${\beta _1} = {\gamma _1} + {\gamma _2}(\left| {{\gamma _1}} \right| < 1;\left| {{\gamma _2}} \right| < 1)$ (2)
 ${\beta _2} = {\gamma _1} \cdot {\gamma _2}\left( {\left| {{\gamma _1}} \right| < 1;\left| {{\gamma _2}} \right| < 1} \right)$ (3)
 $\alpha - {\beta _1} - {\beta _2} = 1$ (4)
3 肌肉激活求解

 $\left\{ \begin{array}{l} a(t) = d\ln (cu(t) + 1) ,\;\;\;\;\;0 \leqslant u(t) < 0.3\\ a(t) = mu(t) + b,\;\;\;\;\;\;\;\;\;\;\;\;0.3 \leqslant u(t) < 1 \end{array} \right.$ (5)

 $a(t) = \frac{{{{\rm{e}}^{Au(t) - 1}}}}{{{{\rm{e}}^A} - 1}}$ (6)

4 神经肌肉骨骼模型

 $\begin{array}{c} \;\;{F^{mt}}(\theta,t) = {F^t} = [F_A^m + F_P^m]\cos\; \phi= \\ \;[{f_A}(l)f(v)a(t)F_o^m + {f_P}(l)F_o^m]\cos\; \phi \end{array}$ (7)

4.1 肌肉力与肌肉纤维长度关系

 $\begin{array}{c} {f_A}(l) = \sin ({b_1} {l^2} + {b_2}l + {b_3})\\ ({b_1} = - 1.317,{b_2} = - 0.403,{b_3} = 2.454) \end{array}$ (8)

 ${f_A}(l) = \left\{ {\begin{array}{*{20}{l}} {{\delta _0} + {\delta _1}l + {\delta _2}{l^2}}&{0.5 \leqslant l \leqslant 1.5}\\ 0&{\text{其他}} \end{array}} \right.$ (9)

 $\begin{array}{l} {f_P}(l) = {A_P} \cdot [{{\rm{e}}^{{K_{pe}}\frac{{{l^m} - l_o^m}}{{l_o^m}}}} - 1] \\ {A_P} = 0.129,{K_{pe}} = 4.525 \\ \end{array}$ (10)

 ${f_P}(l) = \frac{{{{\rm{e}}^{10(l - 1)}}}}{{{{\rm{e}}^5}}} = {{\rm{e}}^{10l - 15}}$ (11)

 $l_o^m(t) = {l_o}(\lambda (1 - a(t)) + 1)$ (12)

4.2 肌肉力与肌肉纤维收缩速度关系

 $H = ax$ (13)

 $E = {F^m}x + H = ({F^m} + a)x$ (14)

 $({F^m} + a)\frac{{{\rm{d}}x}}{{{\rm{d}}t}} = ({F^m} + a){v^m}$ (15)

 $({F^m} + a){v^m} = b(F_o^m - {F^m})$

 ${F^m} = \frac{{F_o^mb - a{v^m}}}{{b + {v^m}}}$ (16)

 ${F^m} = \Bigg(\frac{{F_o^mb - a{v^m}}}{{b + {v^m}}}\Bigg){f_A}(l)$ (17)

 ${F^m} = \Bigg(F_{{\rm{Ecc}}}^mF_o^m - (F_{{\rm{Ecc}}}^m - 1)\dfrac{{F_o^mb' + a'{v^m}}}{{b' - {v^m}}}\Bigg){f_A}(l)$ (18)

 $\left\{ {\begin{array}{*{20}{l}} {f(v) = \dfrac{{0.3\Bigg[\dfrac{{{v^m}}}{{v_{\max }^m}} + 1\Bigg]}}{{ - \dfrac{{{v^m}}}{{v_{\max }^m}} + 0.3}}},&{\dfrac{{{v^m}}}{{v_{\max }^m}} < 0}\\ {f(v) = \dfrac{{2.34\dfrac{{{v^m}}}{{v_{\max }^m}} + 0.039}}{{1.3\dfrac{{{v^m}}}{{v_{\max }^m}} + 0.039}}},&{\dfrac{{{v^m}}}{{v_{\max }^m}} \geqslant 0} \end{array}} \right.$ (19)

4.3 肌腱力求解

 ${\varepsilon ^t} = \frac{{{l^t} - l_s^t}}{{l_s^t}}$ (20)

 ${F^t} = \left\{ {\begin{array}{*{20}{l}} 0,\qquad {\varepsilon \leqslant 0}\\ {1\;480.3{\varepsilon ^2}},&{0 < \varepsilon < 0.012\;7}\\ {37.5\varepsilon - 0.237\;5},&{\varepsilon \geqslant 0.012\;7} \end{array}} \right.$ (21)
 Download: 图 3 肌腱力与应变关系 Fig. 3 Relationship between tendon stress and strain
4.4 羽状角求解

 ${F^t} = {F^m}\cos \;\phi$ (22)

 $\phi (t) = {\sin ^{ - 1}}\Bigg(\frac{{l_o^m\sin {\phi _0}}}{{{l^m}(t)}}\Bigg)$ (23)

4.5 肌肉肌腱长度

 ${l^{mt}} = {l^t} + {l^m}\cos \;\phi$ (24)

 ${l^t} = {l^{mt}} - {l^m}\cos \;\phi$ (25)

 $\begin{array}{c} L({Q_1},{Q_2},{Q_3},{Q_4}) = {a_1} + {a_2}{f_1}({Q_1},{Q_2},{Q_3},{Q_4}) +\\ {a_3}{f_2}({Q_1},{Q_2},{Q_3},{Q_4}) + \cdots + {a_n}{f_{n - 1}}({Q_1},{Q_2},{Q_3},{Q_4}) \end{array}$ (26)

 ${l^{mt}} = {b_0} + {b_1} \theta$ (27)

4.6 肌肉力整合求解

 $\begin{gathered} {F^{mt}}(\theta ,t) = {F^t} = [F_A^m + F_P^m]\cos \phi = \hfill \\ [{f_A}(l)f(v)a(t)F_o^m + {f_P}(l)F_o^m]\cos \;\phi \hfill \\ \end{gathered}$

 $f(v) = \frac{{{F^t} - f_p^lF_o^m\cos \;\phi }}{{{f_A}(l)a(t)F_o^m\cos \;\phi }}$ (28)

4.7 关节力矩求解
 ${M^j}(\theta,t) = \sum\limits_{i = 1}^m {({r_i}(\theta ) \cdot F_i^{mt}(\theta,t))}$ (29)

 $r(\theta ) = \frac{{\partial {l^{mt}}(\theta )}}{{\partial \theta }}$ (30)
5 参数辨识

Ai等[46]进行了两组优化，第1组根据受试者特定的肌肉骨骼几何模型进行估计；第2组采用通用的肌肉骨骼模型。之后将两组原始肌电信号和运动学数据都输入到优化模型(非线性最小二乘和levenberg-marquardt算法)中，使得估计力矩和参考力矩差值最小。Menegaldo等[47]采用式(31)对参数进行优化评价指标：

 ${\rm{RMSE}} = \dfrac{1}{{{\rm{T}}{{\rm{M}}_{{\rm{MAX}}}}}}\sqrt {\frac{{\displaystyle\sum\limits_{t = 1}^N {{{({\rm{TM}}(t) - {\rm{TS}}(t))}^2}} }}{N}} \times 100{{\% }}$ (31)

Li等[49]采用遗传算法对未知生理参数进行优化，优化目标函数为

 $\min \sum\limits_{i = 1}^n {{{({\theta _{ci}} - {\theta _{mi}})}^2}}$

 $\min \sum\limits_{t = 1}^n {\Bigg({M_t} - \sum\limits_{j = 1}^m {{r_{tj}} \cdot {F_{kj}}} \Bigg)}$