Trajectory tracking control based on non-singular fractional derivatives for the PUMA 560 robot arm

Abstract

In this paper, a novel hybrid fractional-order control strategy for the PUMA-560 robot manipulator is developed and presented, which combines the derivative of Caputo–Fabrizio and the integral of Atangana–Baleanu, both in the Caputo sense. The fractional-order dynamic model of the system (FODM) is also considered which consists of two models, the robot manipulator model, and the model of the induction motors which are the actuators that drive their joints. The fractional model of the manipulator is obtained using the Euler–Lagrange formulation. On the other hand, for controlling each one of the induction motors, fractional-order controllers ${{\mathop{\mathrm{PI}}\nolimits } ^{\vartheta }}$ based on Atangana–Baleanu in the Caputo sense integral were developed. And for the trajectory tracking control, fractional-order controllers ${{\mathop{\mathrm{PD}}\nolimits } ^{\xi }}$ were developed based on the fractional derivative of Caputo–Fabrizio in the Caputo sense. Also, ordinary PI and PD controllers were developed for the PUMA robot control to compare their performance with the fractional-order controllers. The results obtained demonstrated that the fractional-order controllers have a better capability for tracking trajectory tasks than the integer-order controllers, even when changes of the desired trajectory and external disturbances are considered. Additionally, an end-effector trajectory tracking task for manufacturing applications is also considered. All numerical simulations were performed by using the same orders and gains, demonstrating that the proposed fractional-order ${{\mathop{\mathrm{PI}}\nolimits } ^{\vartheta }}$ and ${{\mathop{\mathrm{PD}}\nolimits } ^{\xi }}$ controllers are robust, under different operating conditions, for tracking trajectory tasks. The fractional-order controllers and the integer-order controllers were tuned applying the cuckoo search optimization algorithm where the root-mean-square error (RMSE) was chosen as the cost function to minimize.

Appendices

Appendix A: Elements of the dynamic model

By choosing the values for the parameters of the PUMA 560 arm proposed in [9], see Table 10, the inertial, Coriolis and centrifugal matrices and gravitational forces vector defined in (3.8), (3.9), (3.10), and (3.11), respectively, are given by

$$ \textstyle\begin{array}{l@{\quad }l@{\quad }l} m_{11}=2\text{.}57+1\text{.}38C_{2}^{2}+0\text{.}30S_{23}^{2}+0\text{.}744C_{2}S_{23} & & m_{33}=1\text{.}16 \\ m_{12}=0\text{.}690S_{2}-0\text{.}134C_{23}+0\text{.}0238C_{2} & & m_{34}=-0\text{.}00125S_{4}S_{5} \\ m_{13}=-0\text{.}134C_{23}-0\text{.}00397S_{23} & & m_{35}=0\text{.}00125C_{4}C_{5} \\ m_{14}=0 & & m_{36}=0 \\ m_{15}=0 & & m_{44}=0\text{.}20 \end{array} $$

$$ \textstyle\begin{array}{l@{\quad }l@{\quad }l} m_{16}=0 & & m_{45}=0 \\ m_{22}=6\text{.}79+0\text{.}744S_{3} & & m_{46}=0 \\ m_{23}=0\text{.}333+0\text{.}372S_{3}-0\text{.}0110C_{3} & & m_{55}=0 \text{.}18 \\ m_{24}=0 & & m_{56}=0 \\ m_{25}=0 & & m_{66}=0\text{.}19 \\ m_{26}=0 & & \end{array}$$

$$\begin{aligned} b_{112} =&-2\text{.}76S_{2}C_{2}+0\text{.}744C_{223}+0\text{.}60S_{23}C_{23}-0\text{.}0213\left ( 1-2S_{23}^{2}\right ) \\ b_{113} =&0\text{.}744C_{2}C_{23}+0\text{.}6S_{23}C_{23}+0\text{.}022C_{2}S_{23}-0\text{.}0213\left ( 1-2S_{23}^{2}\right ) \\ b_{114} =&-0\text{.}0025S_{23}C_{23}S_{4}S_{5}+0\text{.}00086C_{4}S_{5}-0\text{.}00248C_{2}C_{23}S_{4}S_{5} \\ b_{115} =&-0\text{.}0025\left ( S_{23}^{2}S_{5}-S_{23}C_{23}C_{4}C_{5} \right ) -0\text{.}00248C_{2}\left ( S_{23}S_{5}-C_{23}C_{4}C_{5} \right ) +0\text{.}00086S_{4}C_{5} \\ b_{123} =&0\text{.}267S_{23}-0\text{.}00758C_{23} \\ b_{116} =&b_{124}=b_{125}=b_{126}=b_{134}=b_{135}=b_{136}=b_{145}=b_{146}=b_{156}=0 \\ b_{212} =&b_{213}=b_{216}=0 \\ b_{214} =&0\text{.}00164S_{23}-0\text{.}0025C_{23}C_{4}S_{5}+0 \text{.}00248S_{2}C_{4}S_{5}+0\text{.}0003S_{23}\left ( 1-2S_{4}^{2}\right ) \\ b_{215} =&-0\text{.}0025C_{23}S_{4}C_{5}+0\text{.}00248S_{2}S_{4}C_{5}-0\text{.}000642C_{23}S_{4} \\ b_{223} =&0\text{.}022S_{3}+0\text{.}744C_{3} \\ b_{224} =&-0\text{.}00248C_{3}S_{4}S_{5} \\ b_{225} =&-0\text{.}0025S_{5}+0\text{.}00248\left ( C_{3}C_{4}C_{5}-S_{3}S_{5} \right ) \\ b_{234} =&-0\text{.}00248C_{3}S_{4}S_{5} \\ b_{235} =&-0\text{.}0025S_{5}+0\text{.}00248\left ( C_{3}C_{4}C_{5}-S_{3}S_{5} \right ) \\ b_{226} =&b_{236}=b_{245}=b_{246}=b_{256}=0 \\ b_{314} =&-0\text{.}0025C_{23}C_{4}S_{5}+0\text{.}00164S_{23}+0 \text{.}0003S_{23}\left ( 1-2S_{4}^{2}\right ) \\ b_{315} =&-0\text{.}0025C_{23}S_{4}C_{5}-0\text{.}000642C_{23}S_{4} \\ b_{325} =&-0\text{.}0025S_{5} \\ b_{335} =&-0\text{.}0025S_{5} \\ b_{345} =&-0\text{.}0025S_{4}C_{5} \\ b_{312} =&b_{313}=b_{316}=b_{323}=b_{324}=b_{326}=b_{334}=b_{336}=b_{346}=b_{356}=0 \\ b_{412} =&-0\text{.}00164S_{23}+0\text{.}0025C_{23}C_{4}S_{5}-0 \text{.} 00248S_{2}C_{4}S_{5}-0\text{.}0003S_{23}\left ( 1-2S_{4}^{2} \right ) \\ b_{413} =&0\text{.}0025C_{23}C_{4}S_{5}-0\text{.}00164S_{23}-0\text{.} 0003S_{23} \left ( 1-2S_{4}^{2}\right ) \\ b_{415} =&-0\text{.}00064S_{23}C_{4} \\ b_{425} =&0\text{.}000642S_{4} \\ b_{435} =&0\text{.}000642S_{4} \\ b_{414} =&b_{416}=b_{423}=b_{424}=b_{426}=b_{434}=b_{436}=b_{445}=b_{446}=b_{456}=0 \\ b_{512} =&0\text{.}0025C_{23}S_{4}C_{5}-0\text{.}00248S_{2}S_{4}C_{5}+0 \text{.}000642C_{23}S_{4} \\ b_{513} =&0\text{.}0025C_{23}S_{4}C_{5}+0\text{.}000642C_{23}S_{4} \end{aligned}$$

$$\begin{aligned} b_{514} =&0\text{.}000642S_{23}C_{4} \\ b_{523} =&0\text{.}0025S_{5} \\ b_{524} =&-0\text{.}000642S_{4} \\ b_{534} =&-0\text{.}000642S_{4} \\ b_{515} =&b_{516}=b_{525}=b_{526}=b_{535}=b_{536}=b_{545}=b_{546}=b_{556}=0 \\ b_{612} =&b_{613}=b_{614}=b_{615}=b_{616}=b_{623}=b_{624}=b_{625} \\ =&b_{626}=b_{634}=b_{635}=b_{636}=b_{645}=b_{646}=b_{656}=0 \end{aligned}$$

$$\begin{aligned} c_{12} =&0\text{.}69C_{2}+0\text{.}134S_{23}-0\text{.}0238S_{2} \\ c_{13} =&0\text{.}5\left ( 0\text{.}267S_{23}-0\text{.}00758C_{23} \right ) \\ c_{11} =&c_{14}=c_{15}=c_{16}=0 \\ c_{21} =&-0\text{.}5\left [ -2\text{.}76S_{2}C_{2}+0\text{.}744C_{223}+0\text{.}60S_{23}C_{23}-0\text{.}0213\left ( 1-2S_{23}^{2}\right ) \right ] \\ c_{23} =&0\text{.}5\left ( 0\text{.}022S_{3}+0\text{.}744C_{3}\right ) \\ c_{22} =&c_{24}=c_{25}=c_{26}=0 \\ c_{31} =&-0\text{.}5\left [ 0\text{.}744C_{2}C_{23}+0\text{.}60S_{23}C_{23}+0\text{.}022C_{2}S_{23}-0\text{.}0213\left ( 1-2S_{23}^{2} \right ) \right ] \\ c_{32} =&-0\text{.}5\left ( 0\text{.}022S_{3}+0\text{.}744C_{3}\right ) \\ c_{34} =&-0\text{.}00125C_{4}S_{5} \\ c_{35} =&-0\text{.}00125C_{4}S_{5} \\ c_{33} =&c_{36}=0 \\ c_{41} =&-0\text{.}5\left ( -0\text{.}0025S_{23}C_{23}S_{4}S_{5}+0 \text{.}00086C_{4}S_{5}-0\text{.}00248C_{2}C_{23}S_{4}S_{5}\right ) \\ c_{42} =&0\text{.}5\left ( 0\text{.}00248C_{3}S_{4}S_{5}\right ) \\ c_{43} =&c_{44}=c_{45}=c_{46}=0 \\ c_{51} =&-0\text{.}5\left [ (-0\text{.}0025\left ( S_{23}^{2}S_{5}-S_{23}C_{23}C_{4}C_{5} \right ) \right ] -0\text{.}00248C_{2}\left ( S_{23}S_{5}-C_{23}C_{4}C_{5}\right ) \\ &{}+0\text{.}00086S_{4}C_{5} \\ c_{52} =&-0\text{.}5\left [ -0\text{.}0025S_{5}+0\text{.}00248\left ( C_{3}C_{4}C_{5}-S_{3}S_{5} \right ) \right ] \\ c_{53} =&0\text{.}5\left ( 0\text{.}0025S_{5}\right ) \\ c_{54} =&c_{55}=c_{56}=0 \\ c_{61} =&c_{62}=c_{63}=c_{64}=c_{65}=c_{66}=0 \end{aligned}$$

$$\begin{aligned} g_{2} =&-37\text{.}2C_{2}-8\text{.}4S_{23}+1\text{.}02S_{2} \\ g_{3} =&-8\text{.}4S_{23}+0\text{.}25C_{23} \\ g_{4} =&0\text{.}028S_{23}S_{4}S_{5} \\ g_{5} =&-0\text{.}028\left ( C_{23}S_{5}+S_{23}C_{4}C_{5}\right ) \\ g_{1} =&g_{6}=0 \end{aligned}$$

The trigonometric functions have been abbreviated by writing $S_{i}$ and $C_{i}$ for $\sin \left ( {{q_{i}}} \right )$ and $\cos \left ( {{q_{i}}} \right )$, respectively. Similarly, when a trigonometric operation is applied to the sum of several angles we write $S_{ij}$ for $\sin \left ( {{q_{i}} + {q_{j}}} \right )$ and $C_{ij}$ for $\cos \left ( {{q_{i}} + {q_{j}}} \right )$.

Appendix B: Jacobian matrix

The Jacobian matrix for the end-effector has the form [7]

$$ {\mathbf{{J}}}_{0}^{6} = \left [ { \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {{J_{11}}}&{{J_{12}}}& \cdots &{{J_{16}}} \\ {{J_{21}}}&{{J_{22}}}& \cdots &{{J_{26}}} \\ \vdots & \vdots & \ddots & \vdots \\ {{J_{61}}}&{{J_{62}}}& \cdots &{{J_{66}}} \end{array}\displaystyle } \right ], $$

(B.1)

where

$$ \textstyle\begin{array}{l@{\quad }l@{\quad }l} J_{11} = - {S_{1}}\left ( {{a_{3}}{C_{23}} + {d_{4}}{S_{23}} + {a_{2}}{C_{2}}} \right ) - {d_{23}}{C_{1}} & & J_{44} = {C_{1}}{S_{23}} \\ J_{12} = {C_{1}}\left ( { - {a_{3}}{S_{23}} + {d_{4}}{C_{23}} - {a_{2}}{S_{2}}} \right ) & & J_{45} = - {C_{1}}{C_{23}}{S_{4}} - {S_{1}}{C_{4}} \\ J_{13} = {C_{1}}\left ( { - {a_{3}}{S_{23}} + {d_{4}}{C_{23}}} \right ) & & J_{46} = \left ( {{C_{1}}{C_{23}}{C_{4}} - {S_{1}}{S_{4}}} \right ){S_{5}} + {C_{1}}{S_{23}}{C_{5}} \\ J_{14} = {J_{15}} = {J_{16}} = 0 & & J_{51} = 0 \\ J_{21} = {C_{1}}\left ( {{a_{3}}{C_{23}} + {d_{4}}{S_{23}} + {a_{2}}{C_{2}}} \right ) - {d_{23}}{S_{1}} & & J_{52} = {J_{53}} = {C_{1}} \\ J_{22} = {S_{1}}\left ( { - {a_{3}}{S_{23}} + {d_{4}}{C_{23}} - {a_{2}}{S_{2}}} \right ) & & J_{54} = {S_{1}}{S_{23}} \\ J_{23} = {S_{1}}\left ( { - {a_{3}}{S_{23}} + {d_{4}}{C_{23}}} \right ) & & J_{55} = - {S_{1}}{C_{23}}{S_{4}} + {C_{1}}{C_{4}} \\ J_{24} = {J_{25}} = {J_{26}} = 0 & & J_{56} = \left ( {{S_{1}}{C_{23}}{C_{4}} + {C_{1}}{S_{4}}} \right ){S_{5}} + {S_{1}}{S_{23}}{C_{5}} \\ J_{31} = {J_{34}} = {J_{35}} = {J_{36}} = 0 & & J_{61} = 1 \\ J_{32} = - \left ( {{a_{2}}{C_{23}} + {d_{4}}{S_{23}} + {a_{2}}{C_{2}}} \right ) & & J_{62} = {J_{63}} = 0 \\ J_{33} = - \left ( {{a_{3}}{C_{23}} + {d_{4}}{S_{23}}} \right ) & & J_{64} = {C_{23}} \\ J_{41} = 0 & & J_{65} = {S_{23}}{S_{4}} \\ J_{42} = {J_{43}} = - {S_{1}} & & J_{66} = - {S_{23}}{C_{4}}{S_{5}} + {C_{23}}{C_{5}} \end{array} $$

The modified Denavit–Hartenberg parameters shown in Table 7 were taken from [9] and used for numerical simulations.

Table 7 Modified Denavit–Hartenberg parameters

Table 8 Parameters of the second and third desired trajectories

Table 9 RMSE for end-effector tracking trajectory

Table 10 Mass and length of the links

Appendix C: Inverse kinematics

Formally, the inverse kinematics problem can be defined as the problem of finding the required joint variables to place the tool frame in a position and orientation desired, i.e. given

$$ {\mathbf{{T}}}_{0}^{6} = \left [ { \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {{r_{11}}}&{{r_{12}}}&{{r_{13}}}&{{p_{x}}} \\ {{r_{21}}}&{{r_{22}}}&{{r_{23}}}&{{p_{y}}} \\ {{r_{31}}}&{{r_{32}}}&{{r_{33}}}&{{p_{z}}} \\ 0&0&0&1 \end{array}\displaystyle } \right ], $$

(C.1)

determine ${q_{1}},{q_{2}}, \ldots ,{q_{6}}$. On the other hand, when the last three axes intersect (spherical wrist), the origins of the frames 4, 5, and 6 are located at the point of intersection:

$$ \tilde{{\mathbf{{p}}}} = {\mathbf{{T}}}_{0}^{1}{\mathbf{{T}}}_{1}^{2}{\mathbf{{T}}}_{2}^{3}{{{ \tilde{\mathbf{{p}}}}}_{3}}, $$

(C.2)

where ${\tilde{{\mathbf{{p}}}}}$ and ${{\tilde{{\mathbf{{p}}}}}_{3}}$ are the homogeneous vectors of the center of the spherical wrist expressed in the frames 0 and 3 respectively. According to Denavit–Hartenberg modified convention [18], the homogeneous transformation between frame $i$ and frame $i-1$ can be expressed as

$$ {\mathbf{{T}}}_{i - 1}^{i} = \left [ { \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} {{C_{{\theta _{i}}}}}&{ - {S_{{\theta _{i}}}}}&0&{{a_{i - 1}}} \\ {{S_{{\theta _{i}}}}{C_{{\alpha _{i - 1}}}}}&{{C_{{\theta _{i}}}}{C_{{ \alpha _{i - 1}}}}}&{ - {S_{{\alpha _{i - 1}}}}}&{ - {S_{{\alpha _{i - 1}}}}{d_{i}}} \\ {{S_{{\theta _{i}}}}{S_{{\alpha _{i - 1}}}}}&{{C_{{\theta _{i}}}}{S_{{ \alpha _{i - 1}}}}}&{{C_{{\alpha _{i - 1}}}}}&{{C_{{\alpha _{i - 1}}}}{d_{i}}} \\ 0&0&0&1 \end{array}\displaystyle } \right ]. $$

(C.3)

Substituting $i = 4$ in Eq. (C.3) and from Table 7, ${{{{\tilde{\mathbf{p}}}}}_{3}}$ can be obtained of the fourth column of the resulting matrix. Similarly, the homogeneous transformation ${\mathbf{{T}}}_{2}^{3}$ is obtained by using Eq. (C.3),

$$ {{\tilde{\mathbf{p}}}} = {\mathbf{{T}}}_{0}^{1}{\mathbf{{T}}}_{1}^{2}\left [ { \textstyle\begin{array}{c} {{f_{1}}\left ( {{q_{3}}} \right )} \\ {{f_{2}}\left ( {{q_{3}}} \right )} \\ {{f_{3}}\left ( {{q_{3}}} \right )} \\ 1 \end{array}\displaystyle } \right ], $$

(C.4)

where

$$ \textstyle\begin{array}{l} {{f_{1}}\left ( {{q_{3}}} \right ) = {a_{3}}{C_{3}} + {d_{4}}{S_{3}} + {a_{2}}}, \\ {{f_{2}}\left ( {{q_{3}}} \right ) = {a_{3}}{S_{3}} - {d_{4}}{C_{3}}}, \\ {{f_{3}}\left ( {{q_{3}}} \right ) = {d_{3}}}. \end{array} $$

(C.5)

Applying Eq. (C.3) for ${\mathbf{{T}}}_{0}^{1}$ and ${\mathbf{{T}}}_{1}^{2}$ in (C.4), we obtain

$$ \tilde{{\mathbf{{p}}}} = \left [ { \textstyle\begin{array}{c} {{C_{1}}{g_{1}} - {S_{1}}{g_{2}}} \\ {{S_{1}}{g_{1}} + {C_{1}}{g_{2}}} \\ {{g_{3}}} \\ 1 \end{array}\displaystyle } \right ], $$

(C.6)

where

$$ \textstyle\begin{array}{l} {{g_{1}} = {C_{2}}{f_{1}}\left ( {{q_{3}}} \right ) - {S_{2}}{f_{2}} \left ( {{q_{3}}} \right ),} \\ {{g_{2}} = {f_{3}}\left ( {{q_{3}}} \right ) + {d_{2}}}, \\ {{g_{3}} = - {S_{2}}{f_{1}}\left ( {{q_{3}}} \right ) - {C_{2}}{f_{2}} \left ( {{q_{3}}} \right )}. \end{array} $$

(C.7)

Let $r$ be the magnitude squared of ${\mathbf{{p}}}$

$$ r = g_{1}^{2} + g_{2}^{2} + g_{3}^{2}. $$

(C.8)

Substituting (C.7) into (C.8), it is obtained

$$ r = f_{1}^{2} + f_{2}^{2} + f_{3}^{2} + d_{2}^{2} + 2{d_{2}}{f_{3}}. $$

(C.9)

Equation (C.9) and the $Z$-component from (C.6) can be formulated as a system of two equations in the form

$$\begin{aligned} &r = {k_{3}} , \end{aligned}$$

(C.10a)

$$\begin{aligned} &{p_{z}} = - {k_{1}}{S_{2}} + {k_{2}}{C_{2}}, \end{aligned}$$

(C.10b)

where

$$ \textstyle\begin{array}{l} {k_{1}} = {f_{1}}, \\ {k_{2}} = - {f_{2}}, \\ {k_{3}} = f_{1}^{2} + f_{2}^{2} + f_{3}^{2} + d_{2}^{2} + 2{d_{2}}{f_{3}}. \end{array} $$

(C.11)

Then, after making the following substitutions in the (C.10a), a quadratic equation in $u$ results which may be solved for $q_{3}$:

$$ u = \tan \left ( {\frac{{{q_{3}}}}{2}} \right ),\quad {C_{3}} = \frac{{1 - {u^{2}}}}{{1 + {u^{2}}}},\quad {S_{3}} = \frac{{2u}}{{1 + {u^{2}}}}. $$

(C.12)

Once $q_{3}$ is calculated, (C.10b) and (C.6) can be solved for $q_{2}$ and $q_{1}$ respectively. Having obtained $q_{1}$, $q_{2}$ and $q_{3}$, ${\mathbf{{R}}}_{0}^{3}$ can be calculated and the orientation ${\mathbf{{R}}}_{3}^{6}$ of frame 3 relative to the frame 0 is given by

$$ {\mathbf{{R}}}_{3}^{6} = {\left ( {{\mathbf{{R}}}_{0}^{3}} \right )^{ - 1}}{\mathbf{{R}}}_{0}^{6}. $$

(C.13)

The last three joint angles $q_{4}$, $q_{5}$ and $q_{6}$ can be solved by using the $Z$-$Y$-$Z$ Euler angle solution given in [18]. The complete inverse kinematics solution for the PUMA-560 is

$$ \begin{aligned} &{q_{1}} = {\tan ^{ - 1}}\left ( {\frac{{{p_{y}}}}{{{p_{x}}}}} \right ) - {\tan ^{ - 1}}\left ( {\frac{{{g_{2}}}}{{{g_{1}}}}} \right ), \\ &{q_{2}} = {\tan ^{ - 1}}\left ( {\frac{{ - {k_{1}}}}{{{k_{2}}}}} \right ) \pm {\tan ^{ - 1}}\left ( { \frac{{\sqrt{k_{1}^{2} + k_{2}^{2} + p_{z}^{2}} }}{{{p_{z}}}}} \right ), \\ &{q_{3}} = {\tan ^{ - 1}}\left ( {\frac{{{d_{4}}}}{{{a_{3}}}}} \right ) \pm {\tan ^{ - 1}}\left ( { \frac{{\sqrt{a_{3}^{2} + d_{4}^{2} - K} }}{K}} \right ), \\ &{q_{4}} = {\tan ^{ - 1}}\left ( {\frac{{{{r'}_{33}}}}{{{{r'}_{13}}}}} \right ), \\ &{q_{5}} = {\tan ^{ - 1}}\left ( { \frac{{\sqrt{{{\left ( {{{r'}_{21}}} \right )}^{2}} + {{\left ( {{{r'}_{22}}} \right )}^{2}}} }}{{ - {{r'}_{23}}}}} \right ), \\ &{q_{6}} = {\tan ^{ - 1}}\left ( { \frac{{ - {{r'}_{22}}}}{{{{r'}_{21}}}}} \right ), \end{aligned} $$

(C.14)

where

$$ K = \frac{{r - a_{2}^{2} - a_{3}^{2} - {{\left ( {{d_{2}} + {d_{3}}} \right )}^{2}} - d_{4}^{2}}}{{2{a_{2}}}}, $$

(C.15)

and the elements ${r'}_{ij}$ correspond to the rotation matrix obtained in the right side of (C.13) and they are given by

$$ \textstyle\begin{array}{l} {{r'}_{13}} = {r_{13}}{C_{1}}{C_{23}} + {r_{23}}{S_{1}}{C_{23}} - {r_{33}}{S_{23}}, \\ {{r'}_{21}} = - {r_{11}}{C_{1}}{S_{23}} - {r_{21}}{S_{1}}{S_{23}} - {r_{31}}{C_{23}}, \\ {{r'}_{22}} = - {r_{12}}{C_{1}}{S_{23}} - {r_{22}}{S_{1}}{S_{23}} - {r_{32}}{C_{23}}, \\ {{r'}_{23}} = - {r_{13}}{C_{1}}{S_{23}} - {r_{23}}{S_{1}}{S_{23}} - {r_{33}}{C_{23}}, \\ {{r'}_{33}} = - {r_{13}}{S_{1}} + {r_{23}}{C_{1}}. \end{array} $$

(C.16)

Appendix D: Skew-symmetry of matrix $\dot{\mathbf{D}}\left ( {\mathbf{{q}}} \right ) - 2{\mathbf{{C}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right )$

The skew-symmetry property is an important relationship between the inertia matrix ${\mathbf{{D}}}\left ( {\mathbf{{q}}} \right )$ and the matrix ${\mathbf{{C}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right )$ [44, 48].

Proposition D.1

If ${\mathbf{{N}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right ) = \dot{\mathbf{D}} \left ( {\mathbf{{q}}} \right ) - 2{\mathbf{{C}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right )$, then ${\mathbf{{N}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right )$ is skew-symmetric, i.e., their components $n_{jk}$ satisfy $n_{jk} = - n_{kj}$.

Proof

According to the chain rule, the $kj$th component of $\dot{\mathbf{D}}\left ( {\mathbf{{q}}} \right )$ is given by

$$ {{\dot{d}}_{kj}} = \sum _{i = 1}^{n} { \frac{{\partial {d_{kj}}}}{{\partial {q_{i}}}}{{\dot{q}}_{i}}} .$$

(D.1)

Thus, the $kj$th component of ${\mathbf{{N}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right ) = \dot{\mathbf{D}} \left ( {\mathbf{{q}}} \right ) - 2{\mathbf{{C}}}\left ( {{\mathbf{{q}}},{\dot{\mathbf{q}}}} \right )$ can be obtained as

$$ \begin{aligned} {n_{kj}} & = {{\dot{d}}_{kj}} - 2{c_{kj}} = \sum _{i = 1}^{n} {\left [ {\frac{{\partial {d_{kj}}}}{{\partial {q_{i}}}} - \left ( {\frac{{\partial {d_{kj}}}}{{\partial {q_{i}}}} + \frac{{\partial {d_{ki}}}}{{\partial {q_{j}}}} - \frac{{\partial {d_{ij}}}}{{\partial {q_{k}}}}} \right )} \right ]} {{ \dot{q}}_{i}} \\ & = \sum _{i = 1}^{n} {\left [ { \frac{{\partial {d_{ij}}}}{{\partial {q_{k}}}} - \frac{{\partial {d_{ki}}}}{{\partial {q_{j}}}}} \right ]} {{\dot{q}}_{i}}. \end{aligned} $$

(D.2)

Since ${\mathbf{{D}}}\left ( {\mathbf{{q}}} \right )$ is symmetric, ${d_{ij}} = {d_{ji}}$, so that by interchanging the indices k and j in (D.2) it follows that

$$ {n_{jk}} = - {n_{kj}} ,$$

(D.3)

which completes the proof. □

Appendix E: Stability analysis of the numerical scheme

The stability analysis of the fractional scheme (8.8) is examined below. Recall that, given the fractional-order differential equation

$$ {}_{0}^{\mathit{CF}}D_{t}^{\alpha }x\left ( t \right ) = f\left ( {t,x\left ( t \right )} \right ), $$

(E.1)

the solution can be obtained by

$$ {x_{n + 1}} = {x_{n}} + \left [ { \frac{{1 - \alpha }}{{B\left ( \alpha \right )}} + \frac{{3\alpha h}}{{2B\left ( \alpha \right )}}} \right ]f\left ( {{t_{n}},{x_{n}}} \right ) - \left [ {\frac{{1 - \alpha }}{{B\left ( \alpha \right )}} + \frac{{\alpha h}}{{2B\left ( \alpha \right )}}} \right ]f\left ( {{t_{n - 1}},{x_{n - 1}}} \right ). $$

The above equation can be expressed as

$$ {x_{n + 1}} = {x_{n}} + Rf\left ( {{t_{n}},{x_{n}}} \right ) - Sf \left ( {{t_{n - 1}},{x_{n - 1}}} \right ), $$

(E.2)

where

$$ R = \left [ {\frac{{1 - \alpha }}{{B\left ( \alpha \right )}} + \frac{{3\alpha h}}{{2B\left ( \alpha \right )}}} \right ], \qquad S = \left [ {\frac{{1 - \alpha }}{{B\left ( \alpha \right )}} + \frac{{\alpha h}}{{2B\left ( \alpha \right )}}} \right ]. $$

By using (E.1), (E.2) becomes

$$ {x_{n + 1}} = \left ( {1 + R} \right ){x_{n}} - S{x_{n - 1}}. $$

(E.3)

According to the Von-Neumann stability analysis [11], the terms in the above equation can be written as

$$ \textstyle\begin{array}{l} {x_{n + 1}} = {{\hat{x}}_{n + 1}}{e^{\left ( {n + 1} \right )i\Delta t}}, \\ {x_{n}} = {{\hat{x}}_{n}}{e^{\left ( n \right )i\Delta t}}, \\ {x_{n - 1}} = {{\hat{x}}_{n - 1}}{e^{\left ( {n - 1} \right )i\Delta t}}. \end{array} $$

(E.4)

By substituting (E.4) into (E.3)

$$ {\hat{x}_{n + 1}}{e^{\left ( {n + 1} \right )i\Delta t}} = \left ( {1 + R} \right ){\hat{x}_{n}}{e^{\left ( n \right )i\Delta t}} - S{\hat{x}_{n - 1}}{e^{\left ( {n - 1} \right )i\Delta t}}, $$

(E.5)

which reduces to

$$ {\hat{x}_{n + 1}}{e^{i\Delta t}} = \left ( {1 + R} \right ){\hat{x}_{n}} - S{\hat{x}_{n - 1}}{e^{ - i\Delta t}}. $$

(E.6)

If a recursive formula is applied, when $n = 0$, we get

$$ {{\hat{x}}_{1}}{e^{i\Delta t}} = \left ( {1 + R} \right ){{\hat{x}}_{0}}. $$

(E.7)

By taking the norms of both sides of (E.6), we have

$$ \left \| {{{\hat{x}}_{1}}{e^{i\Delta t}}} \right \| = \left \| {\left ( {1 + R} \right ){{\hat{x}}_{0}}} \right \| , $$

(E.8)

so that

$$ \left \| {{{\hat{x}}_{1}}} \right \| = \left ( {1 + R} \right )\left \| {{{\hat{x}}_{0}}} \right \| . $$

(E.9)

If

$$ h \le 0.55, \qquad \alpha > \frac{1}{{1 - 1.9167h}}, $$

(E.10)

and we assume that

$$ \forall n \ge \left | {{{\hat{x}}_{n}}} \right | < \left | {{{\hat{x}}_{0}}} \right | ,$$

(E.11)

then

$$ \begin{aligned} \left | {{{\hat{x}}_{n + 1}}} \right | & \le \left ( {1 + R} \right )\left | {{{\hat{x}}_{n}}} \right |\left | {{e^{ - i\Delta t}}} \right | - S\left | {{{\hat{x}}_{n - 1}}} \right |\left | {{e^{ - 2i \Delta t}}} \right | \\ & \le \left ( {1 + R} \right )\left | {{{\hat{x}}_{n}}} \right | - S \left | {{{\hat{x}}_{n - 1}}} \right |. \end{aligned} $$

(E.12)

By hypothesis, we have [41]

$$ \begin{aligned} \left | {{{\hat{x}}_{n + 1}}} \right | & < \left ( {1 + R} \right )\left | {{{\hat{x}}_{0}}} \right | - S\left | {{{\hat{x}}_{0}}} \right | \\ & < \left ( {1 + R - S} \right )\left | {{{\hat{x}}_{0}}} \right | \\ & < \left ( {1 + R - S} \right ) < 1. \end{aligned} $$

(E.13)

Thus, the fractional scheme (8.8) is conditionally stable.

Appendix F: Nomenclature

${}_{a}^{\mathit{CFC}}D_{t}^{\varsigma }$	Caputo–Fabrizio–Caputo fractional-order derivative.
${}_{a}^{\mathit{ABC}}I_{t}^{\vartheta }$	Atangana–Baleanu–Caputo fractional-order integral.
${}_{a}^{\mathit{RL}}I_{t}^{\vartheta }$	Riemann–Liouville fractional-order integral.
h	integration step size.
$\left (i_{d},i_{q}\right )$	stator currents on d − q axis (A).
$\left (u_{d},u_{q}\right )$	stator voltage inputs on d − q axis (V).
$\psi _{d}$	rotor flux linkage (Wb).
$\psi _{d_{\mathit{ref}}}$	reference rotor flux linkage (Wb).
$n _{p}$	number of pair pole pairs.
ω	rotor angular speed (rad/seg).
$\omega _{d}$	desired angular speed (rad/seg).
J	moment of inertia of the motor (kg m2).
$\tau _{L}$	load torque (M m).
$\tau _{\mathit{em}}$	electromagnetic torque (M m).
$\tau _{d}$	desired load torque (M m).
$\tau _{f}$	friction torque (M m).
M	mutual inductance (H).
$L_{s}$	stator inductance (H).
$L_{r}$	rotor inductance (H).
$R_{s}$	stator resistance (Ω).
$R_{r}$	rotor resistance (Ω).
$v_{d}$, $v_{q}$	input voltages of the stator (V).
θ	position angular of the induction motor (rad).
$f_{c}$	Coulomb friction coefficient (N m).
$f_{v}$	viscous friction coefficient (N m s).