Backstepping control of novel arc-shaped SMA actuator

Abstract

In this paper, we present modeling and control of a novel arc-shaped Shape Memory Alloy (SMA) actuator. The arc-shaped SMA actuator is developed to provide rotational motion with compliance for biologically inspired robots. We modeled the dynamics of proposed SMA actuator. Based on the dynamics structure, we have developed proportional integral derivative (PID), backstepping and integral backstepping controllers. We have tested experimentally these controllers with input, output and input-output combined disturbances. Based on tracking error, peak error, settling time and control effort, integral backstepping controller is the most suited controller for the actuator.

Appendices

Appendix A Backstepping control—proof

System dynamics of the arc-shaped SMA actuator is defined in Eqs. (17), (18) and (20). Based on this system dynamics, we can define

$$\begin{aligned}&e_1 = x_1 - x_{1_d} \end{aligned}$$

(36)

$$\begin{aligned}&\dot{e}_1 = x_2 - \dot{x}_{1_d} \end{aligned}$$

(37)

where $x_{1_d}$ is the desired trajectory. To achieve convergence for $e_1$, we can define Lyapunov function as follows

$$\begin{aligned}&V_1 = \frac{1}{2}e_1^2 \end{aligned}$$

(38)

$$\begin{aligned}&\dot{V}_1 =e_1\dot{e}_1 \end{aligned}$$

(39)

$$\begin{aligned}&= e_1[x_2- \dot{x}_{1_d}] \end{aligned}$$

(40)

To realize $\dot{V}_1<0$, we choose virtual control as

$$\begin{aligned} x_{2_d} = \dot{x}_{1_d}-k_1e_1. \end{aligned}$$

(41)

To achieve $x_2\rightarrow x_{2_d}$, we define new error

$$\begin{aligned}&e_2 = x_2 - x_{2_d} \end{aligned}$$

(42)

so,

$$\begin{aligned}&x_2 = e_2+x_{2_d}. \end{aligned}$$

(43)

Solving Eqs. (40), (41) and (43) simultaneously, we get

$$\begin{aligned}&\dot{V}_1 = e_1[e_2+x_{2_d}- \dot{x}_{1_d}] \end{aligned}$$

(44)

$$\begin{aligned}&\dot{V}_1 = e_1[e_2+\dot{x}_{1_d}-k_1e_1- \dot{x}_{1_d}] \end{aligned}$$

(45)

$$\begin{aligned}&\dot{V}_1 = -k_1e_1^2+e_1e_2. \end{aligned}$$

(46)

Next, we obtain

$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 x_3 - \dot{x}_{2_d} \end{aligned}$$

(47)

Now, to achieve convergence of $e_2$, we define Lyapunov function as

$$\begin{aligned}&{V}_2 = V_1 + \frac{1}{2}e_2^2 \end{aligned}$$

(48)

$$\begin{aligned}&\dot{V}_2 = \dot{V}_1+e_2\dot{e}_2 \end{aligned}$$

(49)

Simplifying Eq. (49) using (46) and (47), we get

$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2+e_1e_2+e_2[f_1(x)+g_1 x_3 - \dot{x}_{2_d} ] \end{aligned}$$

(50)

To realize $\dot{V}_2 < 0$, we define virtual law as

$$\begin{aligned}&x_{3d} =\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-e_1-k_2e_2] \end{aligned}$$

(51)

where $\vert f_1(x)-\hat{f}_1(x) \vert = \vert \tilde{f}_1(x) \vert \le F_1$ and $g_1 = [1+\Delta _1]\hat{g}_1$, $\vert \Delta _1\vert \le D_1$, $0<D_1<1$. Next, to achieve $x_3\rightarrow x_{3_d}$, we define new error

$$\begin{aligned}&e_3 = x_3 - x_{3d} \end{aligned}$$

(52)

so,

$$\begin{aligned}&x_3 = e_3+x_{3d}. \end{aligned}$$

(53)

Using Eqs. (53) and (51), we can simplify (47)

$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 [e_3+x_{3d}] - \dot{x}_{2_d} \nonumber \\&= f_1(x)+g_1 \Big [e_3+\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-e_1-k_2e_2]\Big ] - \dot{x}_{2_d} \end{aligned}$$

(54)

as $g_1 = [1+\Delta _1]\hat{g}_1$, $\vert \Delta _1\vert \le D_1$, $0<D_1<1$ and $\vert f_1(x)-\hat{f}_1(x) \vert = \vert \tilde{f}_1(x) \vert \le F_1$, therefore, we can write Eq. (54) as

$$\begin{aligned}&\dot{e}_2= f_1(x)+g_1 e_3+\Big [[1+\Delta _1][-\hat{f}_1(x)+\dot{x}_{2d}-e_1-k_2e_2]\Big ] - \dot{x}_{2_d} \nonumber \\&\dot{e}_2 \le f_1(x)+g_1 e_3+\Big [[1+D_1][-\hat{f}_1(x)+\dot{x}_{2d} \nonumber \\ & \quad-e_1-k_2e_2]\Big ] - \dot{x}_{2_d} \dot{e}_2 \le \tilde{f}_1(x)-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}+\Big [[1+D_1][-e_1-k_2e_2]\Big ] \end{aligned}$$

(55)

$$\begin{aligned}&\dot{e}_2 \le F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}+\Big [[1+D_1] \nonumber \\ & \quad [-e_1-k_2e_2]\Big ] \end{aligned}$$

(56)

Using Eq. (56), we can simplify (50)

$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2+e_1e_2+e_2\bigg [F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d} \nonumber \\ &\quad +\Big [[1+D_1][-e_1-k_2e_2]\Big ]\bigg ]\nonumber \\&\quad \dot{V}_2 = -k_1e_1^2 - [1+D_1]k_2e_2^2-D_1e_1e_2 \nonumber \\ &\quad +e_2\bigg [F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}\bigg ] \nonumber \\&\quad \dot{V}_2 = -k_1e_1^2 - [1+D_1]k_2e_2^2-D_1e_1e_2 \nonumber \\ &\quad +e_2\big [g_1 e_3 + V_{2_e}\big ] \end{aligned}$$

(57)

where

$$\begin{aligned} V_{2_e} = \bigg [F_1-D_1\hat{f}_1+D_1\dot{x}_{2d}\bigg ]. \end{aligned}$$

(58)

Next, to achieve convergence of $e_3$, we define Lyapunov function as

$$\begin{aligned}&V_3 = V_2+\frac{1}{2}e_3^2 \end{aligned}$$

(59)

$$\begin{aligned}&\dot{V}_3 = \dot{V}_2 + e_3\dot{e}_3 \end{aligned}$$

(60)

where

$$\begin{aligned}&\dot{e}_3 = \dot{x}_3 - \dot{x}_{3d} \end{aligned}$$

(61)

$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)u-\dot{x}_{3d} \end{aligned}$$

(62)

To achieve $\dot{V}_3 < 0$, we define control law as

$$\begin{aligned}&u = \frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3-\hat{g}_1e_2] \end{aligned}$$

(63)

Using Eq. (63), we can simplify (62) as

$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)\Big [\frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3 \nonumber \\ &\quad -\hat{g}_1e_2] \Big ]-\dot{x}_{3d} \end{aligned}$$

(64)

As $g_2 = [1+\Delta _2]\hat{g}_2$, $\vert \Delta _2\vert \le D_2$, $0<D_2<1$ and $\vert f_2(x)-\hat{f}_2(x) \vert = \vert \tilde{f}_2(x) \vert \le F_2$, therefore, we can write Eq. (64)

$$\begin{aligned} \dot{e}_3&= f_2(x)+[1+\Delta _2][-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3-\hat{g}_1e_2] -\dot{x}_{3d} \nonumber \\ \dot{e}_3&\le f_2(x)+[1+D_2][-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3-\hat{g}_1e_2] -\dot{x}_{3d} \nonumber \\ \dot{e}_3&\le \tilde{f}_2(x)-D_2\hat{f}_2+D_2 \dot{x}_{3d} +[1+D_2][-k_3e_3-\hat{g}_1e_2] \end{aligned}$$

(65)

$$\begin{aligned} \dot{e}_3&\le F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} +[1+D_2][-k_3e_3-\hat{g}_1e_2] \end{aligned}$$

(66)

Now, using Eq. (66), we can simplify (60) as

$$\begin{aligned} \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2+e_2\big [g_1e_3+V_{2_e}\big ] \nonumber \\&+ e_3\Big [ F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} +[1+D_2][-k_3e_3-\hat{g}_1e_2]\Big ] \nonumber \\ \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2\nonumber \\&+e_2\big [g_1e_3+V_{2_e}\big ] + e_3\Big [-\hat{g}_1e_2 +V_{3_e}\Big ] \end{aligned}$$

(67)

where

$$\begin{aligned} V_{3_e} = F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} -D_2\hat{g}_1e_2 \end{aligned}$$

(68)

As, $g_1 \le (1+D_1)\hat{g}_1$, we can write Eq. (67) as

$$\begin{aligned} \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ & \quad +(1+D_1)\hat{g}_1e_2e_3 - \hat{g}_1e_2e_3 + V_{t_e} \end{aligned}$$

(69)

where

$$\begin{aligned} V_{t_e} = e_2 V_{2_e} + e_3V_{3_e}. \end{aligned}$$

(70)

Therefore, we can write Eq. (69) as

$$\begin{aligned} \dot{V}_3 \le -k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ \quad +D_1\hat{g}_1e_2e_3+ V_{t_e} \end{aligned}$$

(71)

So, to achieve $\dot{V}_3 < 0$, we have to

$$\begin{aligned} \big [D_1\hat{g}_1e_2e_3+ V_{t_e}\big ] < \big [k_1e_1^2+[1+D_2]k_2e_2^2 \nonumber \\ &\quad +[1+D_2]k_3e_3^2 \big ] \end{aligned}$$

(72)

Therefore, if we shall choose $k_1, k_2$ and $k_3$ greater than zero and large enough to satisfy Eq. (72), we shall always have $\dot{V}_3 < 0$.

Appendix B Integral backstepping control—proof

Just like backstepping controller, procedure to derive integral backstepping controller is also same. It starts with system dynamics of the arc-shaped SMA actuator as defined in Eqs. (17), (18) and (20). Based on the dynamics, we define

$$\begin{aligned}&e_1 = x_1 - x_{1_d} \end{aligned}$$

(73)

$$\begin{aligned}&\dot{e}_1 = x_2 - \dot{x}_{1_d} \end{aligned}$$

(74)

where $x_{1_d}$ is the desired trajectory. To achieve convergence for $e_1$, we can define Lyapunov function as follows

$$\begin{aligned}&V_1 = \frac{1}{2}e_1^2 \end{aligned}$$

(75)

$$\begin{aligned}&\dot{V}_1 =e_1\dot{e}_1 \end{aligned}$$

(76)

$$\begin{aligned}&= e_1[x_2- \dot{x}_{1_d}] \end{aligned}$$

(77)

To realize $\dot{V}_1<0$, we choose virtual control as

$$\begin{aligned} x_{2_d} = \dot{x}_{1_d}-k_1e_1. \end{aligned}$$

(78)

To achieve $x_2\rightarrow x_{2_d}$, we define new error

$$\begin{aligned}&e_2 = x_2 - x_{2_d} \end{aligned}$$

(79)

so,

$$\begin{aligned}&x_2 = e_2+x_{2_d}. \end{aligned}$$

(80)

Solving Eqs. (77), (78) and (80) simultaneously, we get

$$\begin{aligned}&\dot{V}_1 = e_1[e_2+x_{2_d}- \dot{x}_{1_d}]\end{aligned}$$

(81)

$$\begin{aligned}&\dot{V}_1 = e_1[e_2+\dot{x}_{1_d}-k_1e_1- \dot{x}_{1_d}]\end{aligned}$$

(82)

$$\begin{aligned}&\dot{V}_1 = -k_1e_1^2+e_1e_2. \end{aligned}$$

(83)

Next, we obtain

$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 x_3 - \dot{x}_{2_d} \end{aligned}$$

(84)

Now, to achieve convergence of $e_2$, we define Lyapunov function as

$$\begin{aligned}&{V}_2 = V_1 + \frac{1}{2}e_2^2 \end{aligned}$$

(85)

$$\begin{aligned}&\dot{V}_2 = \dot{V}_1+e_2\dot{e}_2 \end{aligned}$$

(86)

Simplifying Eq. (86) using (83) and (84), we get

$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2+e_1e_2+e_2[f_1(x)+g_1 x_3 - \dot{x}_{2_d} ] \end{aligned}$$

(87)

To realize $\dot{V}_2 < 0$, we define virtual law as

$$\begin{aligned}&x_{3d} =\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-k_i\int e_1\;\text {d}t-e_1-k_2e_2]. \end{aligned}$$

(88)

This Eq. (88) is the brings major difference between backstepping controller and integral backstepping controller. We can compare Eq. (51) for backstepping controller with Eq. (88). Next, we define error

$$\begin{aligned}&e_3 = x_3 - x_{3d} - k_i \int e_1\;\text {d}t \end{aligned}$$

(89)

$$\begin{aligned}&x_3 = e_3+x_{3d}+k_i \int e_1\;\text {d}t \end{aligned}$$

(90)

$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 [e_3+x_{3d}+k_i \int e_1\;\text {d}t] - \dot{x}_{2_d} \end{aligned}$$

(91)

$$\begin{aligned}&\dot{e}_2 = f_1(x)+g_1 \Big [e_3+\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-k_i\int e_1\;\text {d}t \nonumber \\ & \quad -e_1-k_2e_2] +k_i \int e_1\;\text {d}t\Big ] - \dot{x}_{2_d} \end{aligned}$$

(92)

as $g_1 = [1+\Delta _1]\hat{g}_1$, $\vert \Delta _1\vert \le D_1$, $0<D_1<1$ and $\vert f_1(x)-\hat{f}_1(x) \vert = \vert \tilde{f}_1(x) \vert \le F_1$, therefore, we can write Eq. (92) as

$$\begin{aligned}&\dot{e}_2 = f_1(x)+g_1 e_3+\Big [[1+\Delta _1][-\hat{f}_1(x)+\dot{x}_{2d}-k_i \nonumber \\ & \quad \int e_1\;\text {d}t-e_1-k_2e_2] +k_i \int e_1\;\text {d}t\Big ] - \dot{x}_{2_d} \end{aligned}$$

(93)

$$\begin{aligned}&\dot{e}_2 \le f_1(x)+g_1 e_3+\Big [[1+D_1][-\hat{f}_1(x)+\dot{x}_{2d}-k_i \nonumber \\ & \quad \int e_1\;\text {d}t-e_1-k_2e_2] +k_i \int e_1\;\text {d}t\Big ] - \dot{x}_{2_d} \end{aligned}$$

(94)

$$\begin{aligned}&\dot{e}_2 \le \tilde{f}_1(x)-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t \nonumber \\ & \quad +\Big [[1+D_1][-e_1-k_2e_2]\Big ] \dot{e}_2 \le F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t +\Big [[1+D_1][-e_1-k_2e_2]\Big ] \end{aligned}$$

(95)

Using Eq. (95), we can simplify (87) as

$$\begin{aligned}&\dot{V}_2 \le -k_1e_1^2+e_1e_2\nonumber \\&\quad +e_2\bigg [F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t \nonumber \\ & \quad +\Big [[1+D_1][-e_1-k_2e_2]\Big ]\bigg ] \end{aligned}$$

(96)

$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2 - [1+D_1]k_2e_2^2-D_1e_1e_2 +e_2\bigg [g_1 e_3+V_{2_e}\bigg ] \end{aligned}$$

(97)

where

$$\begin{aligned} V_{2_e} = F_1-D_1\hat{f}_1+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t. \end{aligned}$$

(98)

Next, to achieve convergence of $e_3$, we define Lyapunov function as

$$\begin{aligned}&V_3 = V_2+\frac{1}{2}e_3^2 \end{aligned}$$

(99)

$$\begin{aligned}&\dot{V}_3 = \dot{V}_2 + e_3\dot{e}_3. \end{aligned}$$

(100)

Here,

$$\begin{aligned}&\dot{e}_3 = \dot{x}_3 - \dot{x}_{3d} - k_i e_1 \end{aligned}$$

(101)

$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)u-\dot{x}_{3d} - k_i e_1 \end{aligned}$$

(102)

Then, we define

$$\begin{aligned}&u = \frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}+k_ie_1-k_3e_3-\hat{g}_1e_2] \end{aligned}$$

(103)

Using Eq. (103), we can simplify (102) as

$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)\bigg [\frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}+k_ie_1 \nonumber \\ & \quad -k_3e_3-\hat{g}_1e_2] \bigg ]-\dot{x}_{3d} - k_i e_1 \end{aligned}$$

(104)

as $g_2 = [1+\Delta _2]\hat{g}_2$, $\vert \Delta _2\vert \le D_2$, $0<D_2<1$ and $\vert f_2(x)-\hat{f}_2(x) \vert = \vert \tilde{f}_2(x) \vert \le F_2$, therefore, we can write Eq. (104) as

$$\begin{aligned}&\dot{e}_3 \le \bigg [F_2-D_2\hat{f}_2(x)+[1+D_2]\Big [\dot{x}_{3d}+k_1e_1 \nonumber \\ & \quad -k_3e_3-\hat{g}_1e_2\Big ]-\dot{x}_{3d} -k_ie_1\bigg ] \end{aligned}$$

(105)

$$\begin{aligned}&\dot{e}_3 \le \bigg [F_2-D_2\hat{f}_2(x)+D_2\dot{x}_{3d}+D_2k_1e_1+ [1+D_2] \nonumber \\ & \quad \Big [-k_3e_3-\hat{g}_1e_2\Big ]\bigg ] \end{aligned}$$

(106)

Using Eq. (105), we can simplify (100) as

$$\begin{aligned}&\dot{V}_3 \le -k_1e_1^2-[1+D_1]k_2e_2^2+e_2\Big [g_1e_3+V_{2_e}\Big ] \nonumber \\&\quad + e_3\bigg [F_2-D_2\hat{f}_2(x)+[1+D_2]\Big [\dot{x}_{3d}+k_1e_1-k_3e_3-\hat{g}_1e_2\Big ] \nonumber \\ & \quad -\dot{x}_{3d} -k_ie_1\bigg ] \end{aligned}$$

(107)

$$\begin{aligned}&\dot{V}_3 \le -k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 +e_2\big [g_1e_3+V_{2_e}\big ] \nonumber \\ & \quad + e_3\Big [-\hat{g}_1e_2 +V_{3_e}\Big ] \end{aligned}$$

(108)

where

$$\begin{aligned} V_{3_e} = F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} -D_2\hat{g}_1e_2 \end{aligned}$$

(109)

SAs, $g_1 \le (1+D_1)\hat{g}_1$, we can write Eq. (108) as

$$\begin{aligned} \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ & \quad +(1+D_1)\hat{g}_1e_2e_3 - \hat{g}_1e_2e_3 + V_{t_e} \end{aligned}$$

(110)

where

$$\begin{aligned} V_{t_e} = e_2 V_{2_e} + e_3V_{3_e}. \end{aligned}$$

(111)

Therefore, we can write Eq. (110) as

$$\begin{aligned} \dot{V}_3 \le -k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ \quad +D_1\hat{g}_1e_2e_3+ V_{t_e} \end{aligned}$$

(112)

So, to achieve $\dot{V}_3 < 0$, we have to

$$\begin{aligned} \big [D_1\hat{g}_1e_2e_3+ V_{t_e}\big ] < \big [k_1e_1^2+[1+D_2]k_2e_2^2 \nonumber \\ \quad +[1+D_2]k_3e_3^2 \big ] \end{aligned}$$

(113)

Therefore, if we shall choose $k_1, k_2$ and $k_3$ greater than zero and large enough to satisfy Eq. (113), we shall always have $\dot{V}_3 < 0$.