Adaptive functional-based neuro-fuzzy PID incremental controller structure

Abstract

This paper presents an adaptive functional-based neuro-fuzzy PID incremental controller structure that can be tuned either offline or online according to required controller performance. First, differential membership functions are used to represent the fuzzy membership functions of the input–output space of the three-term controller. Second, controller rules are generated based on the discrete proportional, derivative, and integral functions for the fuzzy space. Finally, a fully differentiable fuzzy neural network is constructed to represent the developed controller for either offline or online controller parameter adaptation. Two different adaptation methods are used for controller tuning, offline method based on controller transient performance cost function optimization using bees algorithm and online method based on tracking error minimization using back-propagation with momentum algorithm. The proposed control system was tested to show the validity of the controller structure over a fixed PID controller gains to control SCARA^® type robot arm.

Appendix: SCARA® robot dynamics

The equations of motion can be described by a set of differential or difference equations. The equation set consists of two parts, the kinematics equations and the dynamic equation. Robot arm kinematics deals with the geometry of robot arm motion as a function of time (position, velocity, and acceleration) without reference to the forces and moments that cause this motion, while the dynamics of robot is the study of motion with regard to forces and torques.

In robotics manipulators, there are two methodologies used for dynamic modeling.

• (a) Newton–Euler formulation.

• (b) Lagrangian formulation.

An analytical approach based on the Lagrange’s energy function, known as Lagrange–Euler method, results in a dynamic solution that is simple and systematic. In this method, the kinetic energy (K) and the potential energy (P) are expressed in terms of joint motion trajectories. The resulting differential equations then provide the forces (torques) which drive the robot. Closed-form equations result in a structure that is very useful for robot control design and also guarantee a solution. The dynamics of n-link manipulators are conveniently described by Lagrangian dynamics. In the Lagrangian approach, the joint variables, q = (q _{1, …,} q _n )^T, serve as a suitable set of generalized coordinates. The kinetic energy is a quadratic function of the vector $\dot{q}$ of the form:

$$K = \frac{1}{2}\sum\limits_{i,j = 1}^{n} {d_{ij} (q)\dot{q}_{i} \dot{q}_{j} } = \frac{1}{2}\dot{q}^{T} D(q)\dot{q}$$

(27)

where the n × n “inertia matrix” D(q) is symmetric and positive definite for each q. The gravitational potential energy P = P(q) is independent of $\mathop q\limits^{.}$. The Euler–Lagrange equations for such a system can be derived as follows. Since

$$L = K-P = \frac{1}{2}\sum\limits_{i,j = 1}^{n} {d_{ij} (q)\dot{q}_{i} \dot{q}_{j} - P(q)}$$

(28)

where L is the Lagrangian, then the dynamic equations of an n-joint robotic manipulator described by Lagrange’s equations can be expressed as:

$$D(q)\textit{\"{q}} + C(q,\dot{q})\dot{q} + G(q) + B(\dot{q}) = \tau$$

(29)

where q is the generalized coordinates of the robot arm, $\dot{q}_{i}$ is the first derivative of q _i ; D(q) is the symmetric, bounded, positive-definite inertia matrix; vector C(q, $\dot{q}$) $\dot{q}$ presents the centrifugal and Coriolis torque; G(q), B($\dot{q}$) and τ represent the gravitational torque, friction and applied joint torque, respectively.

D(q) (n × n matrix) expressed as:

$$D(q)\mathop = \limits^{\varDelta } \sum\limits_{k = 1}^{n} {\left\{ {\left[ {A^{k} (q)} \right]^{T} m_{k} A^{k} (q) + \left[ {B^{k} (q)} \right]^{T} D_{k} (q)B^{k} (q)} \right\}}$$

(30)

where A ^k and B ^k represent 3 × n Jacobian submatrices, m _k is the mass of link k, and D _k is the n × n link inertia tensor which depends on q. The equation of velocity coupling vector C(q, $\dot{q}$) is:

$$C(q,\dot{q}) = \sum\limits_{k = 1}^{n} {\mathop C\nolimits_{kk}^{i} } (q)\dot{q}_{k}^{2} + \sum\limits_{k = 1}^{n} {\sum\limits_{j = k} {\mathop C\nolimits_{kj}^{i} (q)} \dot{q}_{k} \dot{q}_{j} }$$

(31)

The equation of gravity loading vector (n × 1) is:

$$G_{i} (q)\mathop = \limits^{\varDelta } - \sum\limits_{k = 1}^{3} {\sum\limits_{j = i}^{n} {g_{k} m_{j} A_{ki}^{j} (q)} }$$

(32)

and the frictional force model for joint k is expressed as:

$$B_{k} (\dot{q}) = b_{k}^{v} \dot{q}_{k} + \text{sgn} \,\,(\dot{q}_{k} )\left[ {\mathop b\nolimits_{k}^{d} + (\mathop b\nolimits_{k}^{s} - \mathop b\nolimits_{k}^{d} )e^{f} } \right]$$

(33)

where f = $\frac{{ - \left| {\dot{q}_{k} } \right|}}{\varepsilon }$, $b_{k}^{v}$ represent the coefficient of viscous friction, $b_{k}^{d}$ is the coefficient of dynamic friction, and $b_{k}^{s}$ is the coefficient of static friction for joint k, and ε is a small positive parameter. The dynamic equation derived by using the Euler–Lagrangian method for the first two arms of the SCARA configuration will be as follows [6]:

First joint:

$$\begin{aligned} \tau_{1} & = \left[ {\left( {\frac{{m_{1} }}{3} + m_{2} } \right)a_{1}^{2} + m_{2} a_{1} a_{2} C_{2} + \frac{{m_{2} }}{3}a_{2}^{2} \textit{\"{q}}_{1} } \right] \\ &\quad + \frac{{m_{2} }}{2}a_{1} a_{2} C_{2} + \frac{{m_{2} }}{3}a_{2}^{2} \textit{\"{q}}_{2} - m_{2} a_{1} a_{2} S_{2} \left( {\dot{q}_{1} \dot{q}_{2} + \frac{{\dot{q}_{2}^{2} }}{2}}\right) + b_{1} \left(\dot{q}_{1} \right) \\ \end{aligned}$$

(34)

Second joint:

$$\tau_{2} = \left[ {\frac{{m_{2} }}{2}a_{1} a_{2} C_{2} + \frac{{m_{2} }}{3}a_{2}^{2} } \right]\textit{\"{q}}_{1} + \frac{{m_{2} }}{3}a_{2}^{2} \textit{\"{q}}_{2} + \frac{{m_{2} }}{2}a_{1} a_{2} S_{2} \dot{q}_{1}^{2} + b_{2} (\dot{q}_{2} )$$

(35)

where m _i is the mass and a _i is the length of link i, C and S represent the cos(q) and sin(q), respectively. Let the state defines as x ^T = [q ^T, v ^T], where v = q. Since n = 2, the manipulator inertia tensor is a symmetric 2 × 2 matrix. From the coefficients of the joint accelerations, the distinct components of D(q) are:

$$D_{11} (q) = \left( {\frac{{m_{1} }}{3} + m_{2} } \right)a_{1}^{2} + m_{2} a_{1} a_{2} C_{2} + \frac{{m_{2} }}{3}a_{2}^{2}$$

(36)

$$D_{12} (q) = \frac{{m_{2} }}{2}a_{1} a_{2} C_{2} + \frac{{m_{2} }}{3}a_{2}^{2}$$

(37)

Because D(q) is symmetric, then

$$D_{12} (q) = D_{21} (q)$$

(38)

$$D_{22} (q) = \frac{{m_{2} }}{3}a_{2}^{2}$$

(39)

$$\varDelta (q) = D_{11} D_{22} - D_{12} D_{21}$$

(40)

$$\dot{q}_{1} = v_{1} ,\,\,\dot{q}_{2} = v_{2}$$

(41)

$$\begin{aligned} \dot{v}_{1} & = \frac{1}{\varDelta (q)}\left[ {D_{22} (q)\left\{ {\tau_{1} + a_{1} a_{2} S_{2} m_{2} \left( {v_{1} v_{2} + \frac{1}{2}v_{2}^{2} } \right) - b_{1} (\dot{q}_{1} )} \right\}} \right. \\ &\quad \left. { - D_{12} (q)\left\{ {\tau_{2} - \frac{{m_{2} }}{2}a_{1} a_{2} S_{2} v_{1}^{2} - b_{2} (\dot{q}_{2} )} \right\}} \right] \\ \end{aligned}$$

(42)

$$\begin{aligned} \dot{v}_{2} & = \frac{1}{\varDelta (q)}\left[ { - D_{21} (q)\left\{ {\tau_{1} + a_{1} a_{2} S_{2} m_{2} \left( {v_{1} v_{2} + \frac{1}{2}v_{2}^{2} } \right) - b_{1} (\dot{q}_{1} )} \right\}} \right. \\ &\left. {\quad + D_{11} (q)\left\{ {\tau_{2} - \frac{{m_{2} }}{2}a_{1} a_{2} S_{2} v_{1}^{2} - b_{2} (\dot{q}_{2} )} \right\}} \right] \\ \end{aligned}$$

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