sliding mode control · x r u r f ,b,g,e p n n p ∈ ∞ × × ⊂ ∈ ∈ piecewise continuous in...

1

Sliding Mode Control

2

Motivation ExampleConsider the problem of stabilizing the second order system

at the origin

Choose a surface Sliding Manifold (Surface)

How can we bring the trajectory to the manifold s = 0?

How can we maintain it there?

where h(x) and g(x) are unknown nonlinear function with

=0

3

Motivation ExampleTo bring s = 0, take the derivative of s as

Suppose2Rx∈∀,

Take the Lyapunov function , we have

Let

When

4

Motivation Examplewhen

000

101

<=>

⎪⎩

⎪⎨

⎧=

sss

-s

,,,

)sgn(

( )

5

Motivation Example

s(t) reaches zero in finite time

Once on the surface s = 0, the trajectory cannot leave it.

Why? because

6

SummarySome definitions:

The motion consists of a reaching phase and sliding phase.The manifold s = 0 is called sliding manifold.The control law u = -β(x)sgn(s) is called sliding mode control

Advantage:Sliding mode is robustness with respect to h and g . We only need to know their bounds. During the sliding phase, the motion is completely independent of h and g.

What is the region of validity?

7

Relay form of sling modelThe sliding model controller can be further simplified if in some domain of interest, h and g satisfy the inequality

We have

0t , |)(| )( 1 ≥∀≤⇒≤11

1 0actx

ac|x|

8

Relay form of sling model

Ω is positively invariant if

and the set sketched as Noted that 21 xx =&

Estimation of the region of attraction.

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Procedure for designing a sliding mode controller

The procedure for designing a sliding mode controller can be summarized by the following steps:

.

Design the sliding manifold s=0 to control the motion of the reduced order system.

Estimate the upper bound .

Take the control u = -β(x)sgn(s) is the switching (discontinuous) control.

This procedure exhibits model order reduction because the main design task is performed on the reduced-order system.

10

Chattering

Chattering results in low control accuracy,high heat losses in electrical power circuits, and high wear of moving mechanical parts.

Sliding model control leads chattering due to imperfections in switching devices anddelays.

The "zig-zag" motion (oscillation) shown inthe sketch, which is known as chattering.

It may also excite unmodeled high-frequency dynamics, which degrades the performance of the system and may even lead to instability

How can we reduce or eliminate chattering?

11

Two possible methods to reduce chattering

Two possible methods to reduce chattering:

• Use knowledge of h and g, i.e. not only their bounds.

• Replace the signum function sgn(s) by some high-slope saturation functions such as sat(s /ε ). But only ultimate stability can be achieved.

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Reduce the amplitude

Reduce the amplitude of the signum function

Because ρ is an upper bound on the perturbation term, it is likely to be smaller than that an upper bound on the whole function. Consequently, the amplitude of the switching component would be smaller.

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“Continuous " Sliding Mode ControllerReplace the signum function by a high-slope saturation function

The signum nonlinearity and its approximation are shown in Figure. The slope of the linear portion of sat(s/ε) is 1/ε. A good approximation requires the use of small ε .

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Performance of the "continuous " sliding mode controllerHow can we analyze the system?

0t , |)(| )( 1 ≥∀≤⇒≤11

1 0actx

ac|x|Thus

..and the set

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Performance of the "continuous " sliding mode controller

What happen inside the boundary layer?

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Performance of the "continuous " sliding mode controller

Inside the boundary layer:

||xa

|x|xa

|x|xaxa-θxx

1121

121

2111 1

)-()-(1

)(

11

11

θεθ

εθ

−−=

+−−≤&

The trajectories reach the positively invariant set

in finite time.

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Performance of the "continuous " sliding mode controller

011 2== x|xas

What happens inside Ωε?

Find the equilibrium points

00 =)(φ

Let

=0

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Performance of the "continuous " sliding mode controller

19

Performance of the "continuous " sliding mode controller

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Performance of the "continuous " sliding mode controller

011 2== x|xas

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Stabilization Sliding Mode Control

.RDt,x,ux,ut

RD

f ,B,G,ERu Rx

p

n

pn

××∞∈

⊂

∈∈

)[0, )( for )( in smooth lysufficient and in continuous piecewise

is function The origin. the contains that domain a in functions smooth

lysufficient are and input control the is state, the is where

δ

Consider the system

We assume that: • f ,B,E are known while G,δ can be uncertain.

• E(x) is nonsingular matrix, G(x) is a diagonal matrix whoseelements are positive and bounded away from zero, i.e.

.Dx∈all for

Goal: Find a feedback to stabilize (*) at the origin.

(*)

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Regular Form

Let T : be a diffeomorphism such thatnRD →

where I is the p X p identity matrix.

• Change of variables

transforms the system

This form is referred to as the regular form.

(*)

to

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Stabilization Sliding Mode Control•Design a sliding manifold such that when the motionis restricted to the manifold, the reduced-order model

has an asymptotically stable equilibrium point at the origin. The design of amounts to solving a stabilization problem for the system

with ξ viewed as the control input.

•Taking the derivative of s gives

Suppose that we can find the .We need to design u to bring s to zero.

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Stabilization Sliding Mode ControlAssume that the nominal value of G(x) is Ĝ(x) , the control is chosen as

Now we have

Assume that

(Known)

25

Stabilization Sliding Mode Control

obtain wecandidate, function Lyapunov a as )( Utilizing 221 ii s/V =

=iV&

Take

where

Then,

=iV&

which guarantees that si=0 reaches in a finite time.

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Design Procedure

)).(( : system order-reduced the stablize to 0 ) ( - manifold sliding a Design

a ηφηηηφξ

,fs

===

&

1)

2) Take the control u as

3) Estimate

4) Choose

Remark•This procedure exhibits model-order reduction because the main design task is preformed on the reduced-order system

•The key future of sliding model control is its robustness to matched uncertainties

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Stabilization Sliding Mode Control

Now use

||s-kg||sxx-kxgV,||s

||sxk||sxεssxxgV

s/V

iiii

i

iii

iii

ii

)()]( )())[-(( have we region the In

])()( )( sat )()[-(

inequality the satisfies derivative its ,)( function Lyapunov the Using

00 11

21

00

0

2

βρβε

βρβ

−≤+≤

≥

++≤

=

&

&

The trajectory reaches the boundary layer |s| ≤ ε in finitetime and remains inside thereafter.

Study the behavior of η

What do we know about this system and what do we need?

) )( ,( a sf += ηφηη&

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Stabilization Sliding Mode ControlTo study the behavior of η, we assume that, together with the sliding manifold design ξ= Φ(η) (continuously differentiable) Lyapunov functionV(η) that satisfies the inequalities

functions.ofclass areandwhere),( ),( all for 31 K,,,DT γαααξη 2∈

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Stabilization Sliding Mode Control

Define a class of K function:

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Stabilization Sliding Mode Control

is positively invariant and all trajectories starting in Ω reach

in finite time

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Stabilization Sliding Mode Control

state. initial any for holds conclusion foregoing the unbounded, radially is )( and globally hold sassumption the If time. finite in set invariant

positively the reaches and 0 all for bounded is ))( ),(( trajectory the , (0)) (0),( all for Then, . over hold sassumption the all Suppose

ηΩξη

ΩξηΩ

ε Vttt ≥

∈

Theorem 1: Consider the systems:

with

ξ= Φ(η) and V(η)

Assume that :

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Stabilization Sliding Mode Control

• The theorem shows that the "continuous " sliding mode controller achieves ultimate boundedness with an ultimate bound that can be controlled by the design parameter ε. It also gives conditions for global ultimate boundedness.

• Since the uncertainty δ could be nonvanishing at x=0, ultimate boundedness is the best we can expect, in general.

• If however, δ vanishing at x=0, then as shown in the next theorem, the origin is asymptotically stable.

Remark:

33

Stabilization Sliding Mode Control

stable allyasymptoticuniformly globally be willorigin the globally, hold

sassumption the If .attraction of region its of subset a is and stable llyexponentia is system loop-closed the of origin

the , 0 all for that such 0 exits there Then

stale llyexponentia is ))( ,( of origin The

, )(

. over hold sassumption the all Suppose :

0

Ω

εεε

ηφηη

ρ

Ω

∗∗ <<>

=•

==•

af

k

&

000

2 Theorem

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Unmatched Uncertainties

1δδδ

ξηδδξηξ

ξηδξηη

xT

,u,x,tuxExG,f,,f

b

a

bb

aa

∂∂

=⎥⎦

⎤⎢⎣

⎡

+++=

+=

where

)()()()()( )()(

&

&

Remark:-The sliding mode control usually applies to the system with matched uncertainties, i.e. the uncertainties enter the system at the same level (point) with the control.

-The sliding mode control cannot usually handle arbitrary unmatched uncertainties,

Suppose: The system (*) is modified as:

The system is transformed as

)()]()()()[()( xu,x,tuxExGxBxfx 1δδ +++=&

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Unmatched Uncertainties

unmatched. is term The uncertain matched the to added is term The)()()()()(

)()(

ab

bb

aa

.,u,x,tuxExG,f

,,f

δδδξηδδξηξ

ξηδξηη

+++=

+=&

&

.

,,f

a

aa

δηφ

ηφηδξηη

yuncertaint the of presence the in 0 origin the of stability asymptotic guarantee to have will of design The

))(()( to manifold sliding the on model order-reduced the changes It

=

+=&

Remark: • The difference between matched and unmatched uncertainties is that sliding model control guarantees robustness for any matcheduncertainty provided that an upper bound is known and the neededcontrol effort can be provided.

• There is no such guarantee for unmatched uncertainties.

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Example of Unmatched UncertaintiesExample Consider the second order system

The uncertainties are bounded by

With sliding model

This system is already in a regular form, there is no need to do atransformation.

.

Let

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Example of Unmatched Uncertainties

or

will stabilize the origin with sufficient small ε.

In this example we are able to design a controller to stabilize the systemwithout restrict the magnitude of the unmatched uncertainty. In general, this mat not be possible.

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Example of Unmatched Uncertainties

uxxx

xxx

++=

−+=

12222

2111 1

θ

θ

&

& )(

Example Consider the second order system

2111 1 xxx )( θ−+=&

We try to design x2 to robustly stabilize the origin x1=0.

Note that the system is not stabilizable at Hence, we must limit a to be less than one.

.θ 11=

).( taking by stabilized be can origin the Hence,]-)([)(

obtainwe Using

-a/kxxakxθkxxx

,kxx

110111

1

21

211

2111

12

>=−≤−−=

−=

&

.|x|akbxx

ssgnxkxxku

0

1

00222

21

>++=

−−+=

βββ

β

with)( where

)()()-( iscotnrollermodel sliding final The

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ExampleConsider the planar system:

Assume “nominal” system parameters:

and the bounds

cuxxbxaxx

=+=

2

21311

&

&

1.=== c, b, a 11

..c., .b., a 5150515020 <<<<<<

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Transforming to Standard FormRe-write the dynamics

)),((

)(

ux,xucx

x,xxxbxax

x

x

212

2121311

2

1

δ

δ

+=

++=

&

&

where

.)(),(

)()()(

uc

cc-ux,x

xxbbxaa-x,x

x

x

=

−+=

21

213121

2

1

δ

δ

Comparing to the general derivation,

)( )( 2 2121311 1

x,xδ,xxbxa,f,xξ,x xa δξηη =+===

)( ,)( ,)( ba 2120 x,xc,G,f xa δδξηξη ===

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Stabilizing the η Subsystem.xV xxφ 2

1211 2

1== define and -)( Choose α

.xabxbxxxxV

,xxx41

21

21

4111

2112

)()(

)( With

−−=−+==

−==

ααα

αφ

&&

Choose

desired. as 0)( then min max

1 ≤

≥

xV

,ba

&

α

Given the bounds on a and b, choose α=4.

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The Complete Control Law

)Δ( that so )(- Define 12

,v,xxvsxxs

21+==

&

φ

.xxbxaxc

ffGu

,vc

uu

ab-aeq

eq

)))((()(-

where Define

21311

1 21

1

+−=∂∂

+=

+=

αηφ

where

vccc-xxaaα

ccc-aα

xaaαccc-aα

δG,v,xx ba

+⎟⎠⎞

⎜⎝⎛ −++

⎟⎠⎞

⎜⎝⎛ −+=

∂∂

=

221

41

21

22

22

)()(-

)()(-

-)Δ( a δηφ

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Bounding Δ

|v|k|x|xkxk|| ++≤ 2212

411Δ

Note that

where

ccc-k

|b|b|c|c-cbαbbα

ccc-bαk

|a|a|c|c-caαaaα

ccc-aαk

≥

−+≥−+≥

−+≥−+≥

and ),()()(-

),()()(-

222

222

2

1

and where k>0 can be chosen less than one, as required. We have

.|x|xkxkx,xk|v|x,x|| 2212

4112121 +=+≤ )( where)(Δ ρρ

44

Simulation with the Signum Function

).αxsign(x)(

Let with)()( Define

212 +−=

>+=

-kx,xv

.bbx,xx,x

1

0

21

002121

β

ρβ

Define the model and control parameters

a=1.2, b=1.3, c=0.8

α = 50, k=0.5 k1=150

k2 = 100, b0=1.

45

Simulation with the Signum Functions

Note that: • s converges to zero in less than 0.1 second and |x2| grows relatively large

during this transient. • After s=0, the state converges asymptotically to zero.

46

Simulation with the Signum Function

Replace the signum function with the saturation function with slope ε.

Try ε values of 10, 0.1, and 0.001.

Things to note:

• The system behaves identically for s ≥ ε.

• After s < ε the system converges quickly to a small constant values.

47

Simulation with the Signum Functions

ε =10

ε =0.1

ε =0.001

48

TrackingConsider a single-input and single-output relative-degree ρ system:

)()]()[()()(

xhyu,x,tuxgxxfx

=+++= δδ1&

that is

We want to design a state feedback control so that the output y asymptotically tracks a reference signal r(t), where

Goal:

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TrackingThere exists a diffeomorphism T(x) over D transfer the system into the normal form by change the variables

Normal form:

Let

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TrackingThe change of variables yields

Our objective now is to design state feedback control to ensure thate(t) is bounded and converges to zero as t tends to infinity. Boundednessof e(t) will ensure boundedness of ξ, since R(t), and its derivatives are

bounded. We need also to ensure boundednessof η. This will follow from a minimum phase assumption.

is input-to-state stable.

In particular, we assume that the system

51

Tracking

We view as the control input. A controller is proposed so that

Then, the sliding manifold is

=0

52

TrackingWith

We can proceed by design u=v as a pure switching component

to cancel the known terms on the right-hand side.

We have

53

Tracking

Suppose

with a “continuous” controller

where 0.

It can be shown that with a “continuous” sliding model controller, thereexists a finite time T1, possibly dependent on ε and the initial states, andpositive constant k, independent of ε and the initial state, such that

.Ttkε|trty| 1≥≤− allfor)()(

What properties can we prove for this control?