Optimal synchronization of teleoperation systems via cuckoo optimization algorithm

Abstract

The main goal of controller design in bilateral teleoperation systems is to achieve stability and transparency in presence of different factors such as time delay in communication channel and modeling uncertainties. The contribution of this paper is designing an optimal controller for synchronization of bilateral teleoperation systems with the objectives of reducing the factors. This requires optimizing a set of parameters of the so-called synchronization control law. To this reason, a novel meta-heuristic algorithm namely cuckoo optimization (CO) algorithm was employed. Comparative simulations are performed to demonstrate the feasibility of the proposed control technique. The results indicate that CO is a powerful search and optimization technique that may yield better solutions to the problem in hand than those obtained using other algorithms including biogeography-based optimization, imperialist competitive and artificial bee colony.

Appendix

Proof

According to (28), we have

$$\begin{aligned} F_m&= K_D\dot{e}_m (t)+K_P e_m (t)\nonumber \\&= K_D (\dot{e}_m +K_D^{-1}K_P e_m) \end{aligned}$$

(42)

Choosing $\lambda =K_D^{-1} K_P$ and using Eqs. (26) and (27), Eq. (42) can be rewritten as

$$\begin{aligned} F_m&= K_D (\dot{e}_m +\lambda e_m)\nonumber \\&= K_D e_{rm} \nonumber \\&= K_D [r_s (t-T)-r_m (t)] \end{aligned}$$

(43)

Similarly, from Eq. (29), we get

$$\begin{aligned} F_s&= K_D (\dot{e}_s +\lambda e_s)\nonumber \\&= K_D e_{rs} \nonumber \\&= K_D [r_m (t-T)-r_s (t)] \end{aligned}$$

(44)

Fig. 12

Parameter estimation profile using ABC in scenario 2

Fig. 13

Parameter estimation profile using BBO in scenario 2

Fig. 14

Parameter estimation profile using IC in scenario 2

Fig. 15

Parameter estimation profile using CO in scenario 2

Fig. 16

Position tracking using CO in scenario 2

Fig. 17

Force tracking using CO in scenario 2

In addition, under Assumption A1., the following equation can be derived [47].

$$\begin{aligned} \int _0^t {(F_e^T r_s -F_h^T r_m)\hbox {d}t} \ge 0 \end{aligned}$$

(45)

In order to analyze the stability, consider the following Lyapunov function candidate

$$\begin{aligned} V=V_1+V_2+V_3 \end{aligned}$$

(46)

where

$$\begin{aligned} V_1&= r_m^T M_m r_m +r_s^T M_s r_s\nonumber \\&+2\int _0^t (F_e^T r_s - F_h^Tr_m)\hbox {d}t\end{aligned}$$

(47)

$$\begin{aligned} V_2&= K_D \int _{t-T}^t (r_m^T r_m +r_s^T r_s)\hbox {d}t +e_m^T \lambda K_D e_m\nonumber \\&+\,e_s^T\lambda K_D e_s \end{aligned}$$

(48)

$$\begin{aligned} V_3&= \tilde{\theta }_m^T \Gamma ^{-1}\tilde{\theta }_m +\,\tilde{\theta }_s^T\Lambda ^{-1}\tilde{\theta }_s \end{aligned}$$

(49)

The time derivative of the Lyapunov function along the trajectory of the system is given as follows: For derivative of $V_1$, we have

$$\begin{aligned} \dot{V}_1&= 2r_m^T M_m \dot{r}_m +2r_s^T M_s \dot{r}_s +r_m^T \dot{M}_m r_m +r_s^T \dot{M}_s r_s\nonumber \\&+\,2F_e^T r_s -2F_h^T r_m \nonumber \\&= 2r_m^T (-C_m r_m +F_h +Y_m \tilde{\theta }_m +F_m)\nonumber \\&+\,2r_s^T (-C_s r_s -F_e +Y_s \tilde{\theta }_s +F_s)\nonumber \\&+\,r_m^T \dot{M}_m r_m +r_s^T \dot{M}_s r_s +2F_e^T r_s -2F_h^T r_m \nonumber \\&= 2r_m^T F_m +2r_s^T F_s +2r_m^T Y_m \tilde{\theta }_m +2r_s^T Y_s \tilde{\theta }_s \nonumber \\&+\,r_m^T (\dot{M}_m -2C_m)r_m +r_s^T (\dot{M}_s -2C_s)r_s \end{aligned}$$

(50)

From P2. $(\dot{M}_i (q)-2C_i (q,\dot{q}), i=m,s)$ is a skew symmetric matrix. Thus

$$\begin{aligned} \dot{V}_1 =2r_m^T F_m +2r_s^T F_s +2r_m^T Y_m \tilde{\theta }_m +2r_s^T Y_s \tilde{\theta }_s \end{aligned}$$

(51)

For derivative of $V_2$, we have

$$\begin{aligned} \dot{V}_2&= K_D [r_m^T r_m +r_s^T r_s -r_m^T (t-T)r_m (t-T)\nonumber \\&-r_s^T (t-T)r_s (t-T)]\nonumber \\&+\,2e_m^T \lambda K_D \dot{e}_m +2e_s^T \lambda K_D \dot{e}_s\nonumber \\&= K_D [r_m -r_s (t-T)]^{T}[r_m +r_s (t-T)]\nonumber \\&+\,K_D [r_s -r_m (t-T)]^{T}[r_s +r_m (t-T)] \nonumber \\&+\,2e_m^T \lambda K_D \dot{e}_m +2e_s^T \lambda K_D \dot{e}_s\nonumber \\&= K_D [r_m -r_s (t-T)]^{T}[2r_m -r_m +r_s (t-T)]\nonumber \\&+\,K_D [r_s -r_m (t-T)]^{T}[2r_s -r_s +r_m (t-T)]\nonumber \\&+\,2e_m^T \lambda K_D \dot{e}_m +2e_s^T \lambda K_D \dot{e}_s\nonumber \\&= 2r_m^T K_D [r_m -r_s (t-T)]\nonumber \\&+\,2r_s^T K_D [r_s -r_m (t-T)]\nonumber \\&-\,K_D [r_m -r_s (t-T)]^{T}[r_m -r_s (t-T)]\nonumber \\&-\,K_D [r_s -r_m (t-T)]^{T}[r_s -r_m (t-T)] \nonumber \\&+\,2e_m^T \lambda K_D \dot{e}_m +2e_s^T \lambda K_D \dot{e}_s \nonumber \\&= -2r_m^T K_D [r_s (t-T)-r_m]\nonumber \\&-\,2r_s^T K_D [r_m (t-T)-r_s] \nonumber \\&-\,K_D [r_m -r_s (t-T)]^{T}[r_m -r_s (t-T)]\nonumber \\&-\,K_D [r_s -r_m (t-T)]^{T}[r_s -r_m (t-T)] \nonumber \\&+\,2e_m^T \lambda K_D \dot{e}_m +2e_s^T \lambda K_D \dot{e}_s \end{aligned}$$

(52)

From Eqs. (26) and (27), we have

$$\begin{aligned} \dot{V}_2&= -2r_m^T K_D [r_s (t-T)-r_m]\nonumber \\&-\,\,2r_s^T K_D [r_m (t-T)-r_s] \nonumber \\&-\,\,(\dot{e}_m +\lambda e_m)^{T}K_D (\dot{e}_m +\lambda e_m)\nonumber \\&-\,\,(\dot{e}_s+\lambda e_s)^{T}K_D (\dot{e}_s +\lambda e_s)\nonumber \\&+\,\,2e_m^T \lambda K_D \dot{e}_m +2e_s^T \lambda K_D \dot{e}_s \end{aligned}$$

(53)

Substituting (43) and (44) into Eq. (53), it yields

$$\begin{aligned} \dot{V}_2&= -2r_m^T F_m -2r_s^T F_s-\dot{e}_m^T K_D \dot{e}_m\nonumber \\&-\,\,\dot{e}_s^T K_D \dot{e}_s -e_m^T \lambda ^{T}K_D \lambda e_m-e_s^T \lambda ^{T}K_D \lambda e_s\nonumber \\ \end{aligned}$$

(54)

The derivative of $V_3$ is given by

$$\begin{aligned} \dot{V}_3&= 2\tilde{\theta }_m^T \Gamma ^{-1}\dot{\tilde{\theta }}_m+ 2\tilde{\theta }_s^T \Lambda ^{-1}\dot{\tilde{\theta }}_s\nonumber \\&+\,\,\tilde{\theta }_m^T \dot{\Gamma }^{-1}\tilde{\theta }_m+ \tilde{\theta }_s^T \dot{\Lambda }^{-1}\tilde{\theta }_s \nonumber \\&= -2\tilde{\theta }_m^T \Gamma ^{-1}\dot{\hat{\theta }}_m-2 \tilde{\theta }_s^T \Lambda ^{-1}\dot{\hat{\theta }}_s \end{aligned}$$

(55)

Using Eqs. (24) and (25), it is straightforward to see that

$$\begin{aligned} \dot{V}_3&= -2\tilde{\theta }_m^T \Gamma ^{-1}\Gamma Y_{^{m}}^T r_m -2\tilde{\theta }_s^T \Lambda ^{-1}\Lambda Y_{^{s}}^T r_s \nonumber \\&= -2\tilde{\theta }_m^T Y_{^{m}}^T r_m -2\tilde{\theta }_s^T Y_{^{s}}^T r_s \end{aligned}$$

(56)

From there

$$\begin{aligned} \dot{V}&= \dot{V}_1 +\dot{V}_2 +\dot{V}_3\nonumber \\&= (2r_m^T F_m +2r_s^T F_s +2r_m^T Y_m \tilde{\theta }_m \nonumber \\&+\,2r_s^T Y_s \tilde{\theta }_s) +\,(-2r_m^T F_m -2r_s^T F_s -\dot{e}_m^T K_D \dot{e}_m \nonumber \\&-\,\dot{e}_s^T K_D \dot{e}_s -\,e_m^T \lambda ^{T}K_D \lambda e_m -e_s^T \lambda ^{T}K_D \lambda e_s)\nonumber \\&+\,(-2\tilde{\theta }_m^T Y_{^{m}}^T r_m -2\tilde{\theta }_s^T Y_{^{s}}^T r_s)\nonumber \\&= -\,\dot{e}_m^T K_D \dot{e}_m -\dot{e}_s^T K_D \dot{e}_s \nonumber \\&-\,e_m^T \lambda ^{T}K_D \lambda e_m -e_s^T \lambda ^{T}K_D \lambda e_s \end{aligned}$$

(57)

As is evident above, the time derivative of the Lyapunov function $\dot{V}$ is negative semi definite. To illustrate that $\dot{V}$ is the uniformly continuous in time, it is necessary to show that $\ddot{V}$ is finite. Taking the time derivative from (57) leads to

$$\begin{aligned} \ddot{V}&= -2\dot{e}_m^T K_D \ddot{e}_m -2\dot{e}_s^T K_D \ddot{e}_s\nonumber \\&-2e_m^T \lambda ^{T}K_D \lambda \dot{e}_m -2e_s^T \lambda ^{T}K_D \lambda \dot{e}_s \end{aligned}$$

(58)

It is obvious that $\ddot{V}$ satisfies the bounded condition when $\dot{e}_m, \dot{e}_s, \ddot{e}_m$ and $\ddot{e}_s$ are bounded. Because of $\dot{V}$ is negative semi definite, $r_m,r_s, \tilde{\theta }_m, \tilde{\theta }_s$ are bounded from (57). Since $r_m$ and $r_s$ are bounded, it follows that $q_s,q_m, \dot{q}_s$ and $\dot{q}_m$ are bounded and thus $e_s,e_m,\dot{e}_s$ and $\dot{e}_m$ are also bounded. According to Eq. (4), $\ddot{q}_s$ and $\ddot{q}_m$ and thus $\ddot{e}_s$ and $\ddot{e}_m$ are also bounded. It concludes that $\ddot{V}$ satisfies the bounded condition. Now, from Barbalat’s lemma [52], $\dot{V}$ is uniformly continuous and then $\dot{V}$ converge to zero. This proves that the signals $\dot{e}_m, \dot{e}_s ,e_m ,e_s$ converge to zero as well as the system is asymptotically stable. This completes the proof. $\square$