# how to implement super-twisting controller based on ... ?· how to implement super-twisting...

Post on 04-Jul-2018

214 views

Embed Size (px)

TRANSCRIPT

How to Implement Super-Twisting Controller based onSliding Mode Observer?

Asif Chalanga1

Shyam Kamal2 Prof.L.Fridman3 Prof.B.Bandyopadhyay 4 and Prof.J.A.Moreno5

124Indian Institute of Technology Bombay, Mumbai-India

3Facultad de Ingeniera Universidad Nacional Autonoma de Mexico (UNAM)

5Instituto de Ingeniera Universidad Nacional Autonoma de Mexico (UNAM)

VSS14, Nantes, June 29 July 2 2014

Standard Sliding Mode STC Based On STO For Perturbed Double Integrator STC based on Super-Twisting Output Feedback (STOF) HOSMO based Continuous

Outline

1 Standard Sliding Mode STC Based On STO For Perturbed Double Integrator

2 STC based on Super-Twisting Output Feedback (STOF)

3 HOSMO based Continuous Control of Perturbed Double Integrator

4 Numerical Simulation

5 Conclusion

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 2

Standard Sliding Mode STC Based On STO For Perturbed Double Integrator STC based on Super-Twisting Output Feedback (STOF) HOSMO based Continuous

Motivation

Consider the dynamical system of the following form

Second order system

x1 = x2x2 = u + 1y = x1 (2.1)

where y is the output variable, 1 is a non vanishing Lipschitz disturbanceand |1| < 0 .

Our aim is to reconstruct the states of the system and then designsuper-twisting controller based on the estimated information.

Although this is already reported in the literature,

We are going to show that existing methodology is not stand on thesound mathematical background.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 3

Standard Sliding Mode STC Based On STO For Perturbed Double Integrator STC based on Super-Twisting Output Feedback (STOF) HOSMO based Continuous

Motivation

The super-twisting observer dynamics

x1 = x2 + k1|e1|12 sign(e1)

x2 = u + k2sign(e1) (2.2)

where the error e1 = x1 x1.

The error dynamics is

e1 = e2 k1|e1|12 sign(e1)

e2 = k2sign(e1) + 1 (2.3)

So e1 and e2 will converge to zero in finite time t > T0, by selecting theappropriate gains k1 and k2.

For this, one can say that x1 = x1 and x2 = x2 after finite time t > T0.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 4

Motivation

Consider the sliding manifold of the form

s = c1x1 + x2. (2.4)

The time derivative of (2.4) (for designing the super-twisting control)

s = c1x1 + x2. (2.5)

After finite time t > T0, when observer start extracting the exact informationof the states, then one can substitute x1 = x2.

Also using (2.2) and (2.5), one can further write

s = c1x2 + u + k2sign(e1). (2.6)

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 5

Motivation

System (5.1) in the co-ordinate of x1 and s by using (2.4) and (2.5)

x1 = s c1x1s = c1x2 + u + k2sign(e1). (2.7)

Super-twisting control design (which is existing in the literature) as

u = c1x2 1|s|12 sign(s)

t02sign(s)d. (2.8)

where 1 and 2 are the designed parameters for the control.

The closed loop system after applying the control (2.8) to (2.7)

x1 = s c1x1

s = 1|s|12 sign(s)

t02sign(s)d + k2sign(e1) (2.9)

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 6

Motivation

Claim

Second order sliding motion is never start in the (2.9)

Mathematical discussion

Because of s contains the non-differentiable term k2sign(e1).

Which exclude the possibility of lower two subsystem of (2.9) to act asthe super-twisting algorithm.

So the second order sliding motion (so that s = s = 0 in finite time)cannot be establish.

In the next, we are going to propose the possible methodology of thecontrol design such that non-differentiable term k2sign(e1) is cancel out.

The lower two subsystem of (2.9) act as the super-twisting and finallysecond order sliding is achieved.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 7

Proposed method 1

The main aim here, is to design u, such that sliding motion occurs in finitetime.

Proposition 1

The control input u which is defined as

u = c1x2 k2sign(e1) 1|s|12 sign(s)

t02sign(s)d (2.10)

where, 1 > 0 and 2 > 0 are selecting according to (Levant), (Moreno),leads to the establishment second order sliding in finite time, which furtherimplies asymptotic stability of x1 and x2.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 8

Proposed method 1

Proof

The closed loop system after substituting (2.10) into (2.7)

x1 = s c1x1

s = 1|s|12 sign(s) +

= 2sign(s) (2.11)

Last two equation of (2.11) has same structure as super-twisting. Therefore,one can easily observe that after finite time t > T0, s = s = 0.

The closed loop system is given as

x1 = c1x1x2 = c1x1 (2.12)

Therefore, both states x1 and x2 are asymptotic stability by choosing c1 > 0.Also, when observer estimating the exact state x2 = x2 after finite time, thenx2 also going to zero simultaneously as x2.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 9

Existing result:STC based on Super-Twisting Output Feedback (STOF)

Step-1

Consider the following sliding sliding surface

s = c1x1 + x2 (3.1)

assuming that all states information are available.

Step-2

To realizing the control expression based on super-twisting , take the firsttime derivative of sliding surface s using (3.1)

s = c1x1 + x2 (3.2)

Step-3

Now substitute x1 and x2 from (5.1) into (3.2),

s = c1x2 + u + 1 (3.3)

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 10

Existing result:STC based on Super-Twisting Output Feedback (STOF)

Step-4

Now design control as

u = c1x2 1|s|12 sign(s)

t02sign(s)d (3.4)

After substituting the control (3.4) into (3.3),

s = 1|s|12 sign(s) +

= 2sign(s) + 1. (3.5)

Now select 1 > 0 and 2 > 0 according to (moreno2012), which leads tosecond order sliding in finite time provided 1 is Lipschitz and |1| < 0.When s = 0, then x1 = x2 = 0 asymptotically same as discussed above byselecting c1 > 0.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 11

Existing result:STC based on Super-Twisting Output Feedback (STOF)

The control (3.4) is not implementable because we do not haveinformation of x2, so replace x2 by x2.

It is argued that after finite time x1 = x1 and x2 = x2, therefore controlsignal applied to original system (5.1) is

u = c1x2 1|s|12 sign(s)

t02sign(s)d (3.6)

where s = c1x1 + x2,

Without considering the the dynamics of x2 for which control derivation isexplicitly dependent and it contains the discontinuous term k2sign(e1).

One can easily see that average value of this discontinuous term isequal to negative of the disturbance.

So control (3.6) we are applying for the real system is only approximatenot the exact. However, the exact controller is always discontinuouswhich already discussed and mathematically proved in the abovesection.

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 12

Higher Order Sliding Mode Observer based Continuous Control ofPerturbed Double Integrator

This methods gives the correct way to implement continuous STC, when onlyoutput information of the perturbed double integrator (5.1) is available.

The HOSMO dynamics to estimate the states for the system (5.1) is given as

x1 = x2 + k1|e1|23 sign(e1)

x2 = x3 + u + k2|e1|13 sign(e1)

x3 = k3sign(e1) (4.1)

Let us define the error e1 = x1 x1 and e2 = x2 x2 and the error dynamics is

e1 = e2 k1|e1|23 sign(e1)

e2 = x3 k2|e1|13 sign(e1) + 1

x3 = k3sign(e1) (4.2)

Prof.L.Fridman How to Implement Super-Twisting Controller based on Sliding Mode Observer? 13

Standard Sliding Mode STC Based On STO For Perturbed Double Int

Recommended