Slide Mode Control (S.M.C.)

Slide Mode Control(SMC)

SOLO HERMELIN

Updated: 6.03.12

1

Table of Content

SOLO Slide Mode Control (SMC)

2

Sliding Mode Control - Introduction

Control Statement of Sliding Mode

Existence of a Sliding Mode

Reachability: Attaining Sliding Manifold in Finite TimePicard-Lindelöf Existence and Uniqueness of a Differential Equations Solutions

Uniqueness of Sliding Mode SolutionsSliding Motion Surface Keeping

Controller DesignDiagonalization MethodOther Methods – Relays with Constant GainsOther Methods – Linear Feedback with Switched Gains

Other Methods – Linear Continuous Feedback

Other Methods – Univector Nonlinearity with Scale FactorChattering

Table of Content (continue - 1)


3

Higher Order Sliding Mode Control

Sliding Order and Sliding Set

Second Order Sliding Modes

The Twisting Controller (Levantosky, 1985)

The Problem Statement

The Super-Twisting Controller

The Super-Twisting Controller (Shtessel version)Sliding Mode Control for Linear Time Invariant (LTI) Systems

Regular Form of a LTISliding Surface of a Regular Form of a LTIUnit Vector Approach for a Controller of a Regular Form of a LTI

Output Feedback Variable Structure Controllers and State Estimators for Uncertain Dynamic Systems

System Sliding Surface

Sliding Modes and System ZerosProperties of the Sliding ModesDesign of a Sliding Surface (Hyperplane)



Sliding Mode Control for Nonlinear Systems

Continuous Sliding Mode ControlSecond Order Sliding Mode Control

Sliding Mode Observers

Sliding Mode Observers of Target Acceleration



Slide Mode Control Examples

Control System of a Kill Vehicle

Equations of Motion of a KV (Attitude)

Fu, L-C, et all, Control System of a Kill Vehicle

Fu, L-C, et al, Solution for Attitude Control of KV

Fu, L-C, et al, Zero SM Guidance of a KV

Crassidis , et al -Attitude Control of the Kill Vehicle

Midcourse Intercept of a Ballistic in a Head On Scenario Above a Minimal Altitude

HTK Guidance Using 2nd Order Sliding Mode

Slide Mode Control (SMC) is a type of Variable Structure Control (VSC) that possessRobust Characteristics to System Disturbances and Parameter Uncertainties. A VSC System is a special type of Nonlinear System characterized by a discontinuous control which change the System Structure when the States reach the Intersection of Sets of Sliding Surfaces. The System behaves independently of its general dynamical characteristics and system disturbances once the controller has driven the System into a Sliding Mode.


Such a High-Level of Performance requires High-Quality Actuators requiring a Very Fast Responding and Fast Switching Action. This translates to a Very Wide Bandwidth Actuators.

6

Sliding Mode Control - Introduction

A Few Examples are presented:• Midcourse Intercept of a Ballistic in a Head On Scenario Above a Minimal Altitude, a solution using Sliding Mode given in 1975 (before the development of the Sliding Mode Method)• Control of a Kill Vehicle (because of the beauty of quaternion mathematics)• Hit-to-Kill (HTK) Guidance Law using a Second Order Sliding Mode Control.

SOLO

In control theory, Sliding Mode Control, or SMC, is a form of Variable Structure Control (VSC). It is a nonlinear control method that alters the dynamics of a nonlinear system by application of a high-frequency switching control. The state-feedback control law is not a continuous function of time. Instead, it switches from one continuous structure to another based on the current position in the state space. Hence, sliding mode control is a variable structure control method. The multiple control structures are designed so that trajectories always move toward a switching condition, and so the ultimate trajectory will not exist entirely within one control structure. Instead, the ultimate trajectory will slide along the boundaries of the control structures. The motion of the system as it slides along these boundaries is called a Sliding Mode and the geometrical locus consisting of the boundaries is called the sliding (hyper)surface. The sliding surface is described by σ = 0, and the sliding mode along the surface commences after the finite time when system trajectories have reached the surface. In the context of modern control theory, any variable structure system, like a system under SMC, may be viewed as a special case of a hybrid dynamical system.

Sliding Mode Control (SMC)

7

http://en.wikipedia.org/wiki/Sliding_mode_control

SOLO


Consider a Nonlinear Dynamical System Affine in Control:

nxmnmn xBtxftutx

tuxBtxftx

RRRR

,,,,

,

The components of the discontinuous feedback are given by:

mixiftxu

xiftxutu

ii

iii ,,2,1

0,

0,

where σi (x) = 0 is the i-th component of the Sliding Surface, and

0,,, 21 Tm xxxx is the (n-m) dimensional Sliding Manifold

The sliding-mode control scheme involves:

1.Selection of a Hypersurface or a Manifold (i.e., the Sliding Surface) such that the system trajectory exhibits Desirable Behavior when confined to this Manifold. 2.Finding discontinuous feedback gains so that the System Trajectory intersects and stays on the Manifold.


0x

0x

0x

Sliding Manifold

1x

2x

8

http://en.wikipedia.org/wiki/Nonlinear_system


SOLO


Consider a Nonlinear Dynamical System Affine in Control:

nxmnmn xBtxftutx

tuxBtxftx

RRRR

,,,,

,

mi

xiftxu

xiftxutu

ii

iii ,,2,1

0,

0,

where σi (x) = 0 is the i-th component of the Sliding Surface, and

0,,, 21 Tm xxxx is the (n-m) dimensional Sliding Manifold

A Sliding-Mode exists, if in the Vicinity of the Switching Surface, σ (x) = 0, the Velocity Vector of the State Trajectory, , is always Directed Toward the Switching Surface.

tx

Because sliding mode control laws are not continuous, it has the ability to drive trajectories to the sliding mode in finite time (i.e., stability of the sliding surface is better than asymptotic). However, once the trajectories reach the sliding surface, the system takes on the character of the sliding mode (e.g., the origin x=0 may only have asymptotic stability on this surface).


0x

0x

0x

Sliding Manifold

1x

2x

9

http://en.wikipedia.org/wiki/Nonlinear_system


SOLO


The Existence of the Sliding Mode requires Stability of the State Trajectory to theSliding Surface , σ (x) = 0, at least in a Neighborhood of the Sliding Surface, i.e., the System State must approach the surface at least asymptotically.

From a Geometrical point of view, in the Vicinity of the Switching Surface, σ (x) = 0, the Velocity Vector of the State Trajectory, , is always Directed Toward the Switching Surface.

tx


The Existence Problem can be seen as a Generalized Stability Problem hence the Second Method of Lyapunov provides a natural setting for Analysis.

Aleksandr Mikhailovich Lyapunov

1857 - 1918

0x

0x

0x

Sliding Manifold

1x

2x

10


http://en.wikipedia.org/wiki/Aleksandr_Lyapunov


http://en.wikipedia.org/wiki/File:Alexander_Ljapunow_jung.jpg

SOLO




1857 - 1918

DefinitionA Domain D in a Manifold σ (x) = 0 is a Sliding Mode Domain if for each ε > 0, there is a δ >0, such that any trajectory starting within a n-dimensional δ-vicinity of D may leave the n-dimensional δ-vicinity of D only through the n-dimensional ε-vicinity of the boundary of D.

Second Method of Lyapunov

11

Sliding Manifold

D

Trajectory 0x





SOLO




1857 - 1918

For the (n-m) dimensional domain D to be the Domain of a Sliding Mode, it is sufficient that in some n-dimensional domain Ω ϵ D, there exists a function V (x,t,σ) continuously differentiable with respect to all of its arguments, satisfying the following conditions:

1.V (x,t,σ) is positive definite with respect to σ, i.e., V (x,t,σ) > 0, with σ ≠ 0 and arbitrary x,t, and V (x,t,σ=0) = 0; and on

the sphere ||σ|| = ρ, for all x ϵ Ω and any t the relations

holds, hρ and Hρ, depend on ρ (hρ ≠ 0 if ρ ≠0)

2.The Total Time Derivative of V (x,t,σ) for the System Dynamicshas a negative supremum for all x ϵ Ω except for x on the

Switching Surface where the control input are undefined, and hence the derivative of V (x,t,σ) does not exist.


,,txV

H

h

0,,,sup

0,,,inf

HHtxV

hhtxV

12





SOLO




Unfortunately, there are no standard methods to find Lyapunov Functions for Arbitrary Nonlinear Systems.

Existence of Sliding Mode

Consider a Lyapunov Function candidate:

0002

1

2

12

xxVxxxxV T

where ||*|| is the Euclidean norm (i.e. ||σ (x)||2 is the Distance away from the SlidingManifold where σ (x)=0 ). V (σ (x)) is Globally Positive Definite.

tuxBtxftx ,

A Sufficient Condition for the Existence of the Sliding Mode is:

0td

d

td

d

d

Vd

td

Vd T

in a neighborhood of the surface σ (x)=0.

utxBtxfxd

d

td

xd

xd

d

td

d,,

The feedback control law u (x) has a direct impact on . td

d 13


SOLO




Existence of Sliding Mode (continue – 1)

0002

1

2

12

xxVxxxxV T

tuxBtxftx ,

0td

d

td

d

d

Vd

td

Vd T

Roughly speaking (i.e., for the scalar control case when m = 1), to achieve , the feedback control law u (x) is picked so that σ and have opposite sign, that is

utxBtxfxd

d

td

xd

xd

d

td

d,,

• u (x) makes negative when σ (x) is positive. xd

d

0td

dT

td

d

• u (x) makes positive when σ (x) is negative. xd

d

14


SOLO




Reachability: Attaining Sliding Manifold in Finite Time

tuxBtxftx ,

To ensure that the Sliding mode σ (x) = 0 in a Finite Time, dV/dt must be Strongly Bounded Away From Zero. That is, if it vanished to quickly, the Attraction to the Sliding Mode will only be Asymptotic. To ensure that the Sliding Mode is entered in Finite Time

Vtd

Vd

where μ > 0 and 0 < α < 1 are constant

01

0

00

1

111

0

ttVVdVtd

Vd

V S

V

V

This shows that the time necessary to reach the Sliding Manifold σ [x(ts)] = 0 is bounded by:

1001

10

0

VttS 15


SOLO




Note that for all single input functions a suitable Lyapunov function is:

xtxV 2

2

1,,

which is Globally Positive Definite.

tutxBtxfxd

xdx

td

xd

xd

xdx

td

xdxtxV

td

d,,,,

Suppose that we can find u (t) such that in the neighborhood of V (x,t,σ=0)=0 we have: 0,, x

td

xdxtxV

td

d

0sgn

td

xd

td

xdx

td

xd

x

x 00 tttxtx integration

This shows that the time necessary to reach the Sliding Manifold σ [x(ts)] = 0 is bounded by:

0

0

txttS

Reachability: Attaining Sliding Manifold in Finite Time (continue – 1)

16


SOLO




Region of Attraction

Reachability: Attaining Sliding Manifold in Finite Time (continue – 2)

For the Dynamic System given by and for the Sliding Surface σ (x) = 0, the subspace for which the Sliding Surface is Reachable is given by 0: xxRx Tn

tuxBtxftx ,

When Initial Conditions come from this Region, the Lyapunov Function Candidate is a Lyapunov Function and the Space Trajectories are sure to move toward Sliding Mode Surface σ (x) = 0. Moreover, if the Reachable Condition is satisfied, the Sliding Mode will reach σ (x) = 0 in Finite Time.

2/xxxV T

10 VV

17


SOLO

Picard-Lindelöf Existence and Uniqueness of a Differential Equations Solutions


Lipschitz Continuity Condition

Charles Émile Picard

1856 - 1941

Ernst Leonard Lindelöf1870 - 194618

Rudolf Otto Sigismund Lipschitz

1832 – 7 1903

In mathematical analysis, Lipschitz continuity, named after Rudolf Lipschitz, is a strong form of uniform continuity for functions.

A Function f (x) is called Lipschitz continuous if there exists a real constant K ≥ 0 such that, for all x1 and x2 in X:

1212 xxKxfxf

Picard–Lindelöf Theorem

Consider the initial value problem

0000 ,,,, tttxtxtxftd

xd

Suppose f is Lipschitz continuous in x and continuous in t. Then, for some value ε > 0, there exists a Unique Solution x(t) to the initial value problem within the range [t0-ε,t0+ε].


http://upload.wikimedia.org/wikipedia/en/1/1a/Charles_%C3%89mile_Picard.jpg

SOLO





1857 - 1918

Uniqueness of Sliding Mode Solutions

The Nonlinear Dynamical System Affine in Control:

nxmnmn xBtxftutx

tuxBtxftx

RRRR

,,,,

,

mixiftxu

xiftxutu

ii

iii ,,2,1

0,

0,

with the switching control, do not formally satisfy the classical Picard-Lindelöf Existence and Uniqueness Solutions, since they have discontinuou right-hand sides. Moreover the right-hand sides usually are not defined on the discontinuous surfaces.

Charles Émile Picard 1856 - 1941

Ernst Leonard Lindelöf1870 - 1946

Existence and Uniqueness of Differential Equations with Discontinuous Right-hand Sides is was addressed by different researchers. One of the straightforward approaches is the Method of Filippov (Filippov Aleksei Fedorovich, “Differential Equations with Discontinuous Right Hand Sides”, Kluwer, Dordrecht, the Nederlands) 19





http://upload.wikimedia.org/wikipedia/en/1/1a/Charles_%C3%89mile_Picard.jpg

SOLO




Uniqueness of Sliding Mode Solutions (continue – 1)

utxftx ,,

0,

0,

xiftxu

xiftxutu

Method of Filippov (Filippov Aleksei Fedorovich)

Consider the n-order Single Input System:

with the following Control Strategy:

The System Dynamics is not defined on σ (x) = 0. Filippov has shown that the solution on the Surface σ (x) = 0 is given by the equation:

10,,1,, 0111 nxnxnx futxfutxftx

utxfnx ,,1

H

0x

utxfnx ,,1

utxf

utxff

nx

nxnx

,,1

,,

1

10

1

The term α is a function of the System States and

can be specified in such a way that the “average” dynamics f0 is tangent to the Surface σ (x) = 0 .

20


SOLO



Sliding Motion Surface Keeping

0,

equxBx

txfx

txx

x

0x The Dynamic System will stay on the Sliding Surface σ (x) = 0, if the equivalent control ueq will keep

xBx nxm

mxn

If is nonsingular, i.e., the System has a kind of Controllability

that assures that we can find a controller to move a trajectory closer to σ (x) = 0,

then txfx

xBx

ueq ,1

txfx

xBx

Itx ,1

The Dynamic System Equation is

Note that using ueq any trajectory that starts at σ (x) = 0, remains on it, Since . As a consequence the Sliding Manifold σ (x) = 0 is an Invariant Set.

0x

Return to Chattering

21


SOLOSliding Mode Control (SMC)

Controller Design

tutxBx

txQtu ,,* 1

We must choose Switched Feedback capable of forcing the Plant State Trajectories to the Switching Surface and maintaining a Sliding Mode Condition. We assume that the Sliding Surface has already been designed.

Diagonalization Method

The Diagonalization Method converts the multi – input design problem into m single-input design problems.

The Method is based on the construction of a new control vector u* through a nonsingular transformation of ueq:

where Q (x,t) is an arbitrary mxm Diagonal Matrix with elements qi (x,t), i=1,…,m, such that inf |qi (x,t)| > 0 for all t ≥ 0 and all x.

22



Controller Design (continue – 1)

tutxQtutxBx

tutxBx

txQtu *,,,,* 1

Diagonalization Method (continue – 1)

where Q (x,t) is an arbitrary mxm Diagonal Matrix with elements qi (x,t), i=1,…,m, such that inf |qi (x,t)| > 0 for all t ≥ 0 and all x.

For existence and reachability of a Sliding Mode is enough to satisfy .0td

dT

mitutxqtxfxd

d

td

dortutxQtxf

xd

d

td

xd

xd

d

td

dii

i

i ,,1,,,, **

To satisfy the existence and reachability we choose each control u*i as

tutxQtxB

xtu

xwhentxfx

utxq

xwhentxfx

utxq

i

i

n

jj

jii

i

i

n

jj

jii

*,,

0,,

0,, 1

1

*

1

*

23



Controller Design (continue – 2) Other Methods

A possible structure for the control is: miuuu iNii eq,,2,1

Where is continuous and

uiN is the discontinuous part

i

i txfx

xBx

ueq

,1

NNeq

Neq

uxBx

uxBx

uxBx

txfx

uuxBx

txfxtd

d

0

,

,

Several Design Methods are applicable

24





Other Methods – Relays with Constant Gains


where is continuous and


i

i txfx

xBx

ueq

,1

NuxBxtd

d

00

01

01

sgn0sgn1

xif

xif

xif

x

xxxxB

xu ii

i

iiN

This controller satisfies the reaching condition since:

00sgn xifxxxtd

xdx iiiiii

ii 25





Other Methods – Linear Feedback with Switched Gains




i

i txfx

xBx

ueq

,1

NuxBxtd

d

00

001

jjij

iiij

ijij

i

iN xif

xifxxB

xu


011 niniii

i xxxtd

xdx 26




Control Methods (continue – 5)

Other Methods – Linear Continuous Feedback




i

i txfx

xBx

ueq

,1

NuxBxtd

d

xLu nxnN


00 xifxLxtd

xdx TT

where Lnxn is a Positive Definite Constant Matrix

27





Other Methods – Univector Nonlinearity with Scale Factor




i

i txfx

xBx

ueq

,1

NuxBxtd

d

xxxx

xxu T

N

22

&0


002

xifxtd

xdxT

28



Chattering

0x

0x

0x

Sliding Manifold

1x

2x

The ChatteringEffect

The Chattering Effect

Due to the presence of external disturbance, noise and inertia of the sensors and actuators the switching around the Sliding Surface occurs at a very high (but finite) frequency. The main consequence is that the Sliding Mode take place in a small neighbor of the Sliding Manifold , which is called Boundary Layer, and whose dimension is inversely proportional with the Control Switching Frequency.

The effect of High Frequency Switching is known as Chattering.

The High Frequency Switching propagate through the System exciting the fast dynamics and undesired oscillations that affect the System Output To prevent the Chattering Effect different techniques are used. One of the techniques is the use of

continuous approximations of sign (.) (such as sat (.) function, the tanh (.) function,..) in the implementation of the Control Law. A consequence of this method is that theInvariance Property is Lost.

Invariance Definition29




Sliding Order and Sliding Set

The Sliding Order r is the number of continuous total derivative, including the zero one, of the function σ = σ (t,x) whose vanishing defines the equations of the Sliding Manifold.

The Sliding Set of r – th order associated in the Manifold σ (t,x) = 0 is defined by the equalities 01 r

which forms an r – dimensional condition on the State of the Dynamic System.The corresponding motion satisfying the equalities iscalled an r – order Sliding Mode with respect to theManifold σ (t,x) = 0 .

0x

0 xx

0x

Second Order Sliding Mode Trajectory

30





The Problem Statement

0x

0 xx

0x


Consider a Dynamic Single Output System of the form:

11 ,,,,

,nxnn xbtxftutx

tuxbtxftx

RRRR

Let σ (t,x) = 0 be the chosen Sliding Manifold, then the Control Objective is to enforce a Second Order Sliding Mode on the Sliding Manifold σ (t,x) = 0 , i.e.,

in Finite Time.

Let analyze the following two cases:

Case A: relative degree

01

u

r

Case B: relative degree 0,02

uu

r 31

0,, xtxt





The Problem Statement (continue – 1)

0x

0 xx

0x


tuxbtxftx ,

Case A: relative degree 01

u

r

tuuxtuxt

tuxtbxtfxtx

xtt

AA

,,,,

,,,,

xtbxtx

xt

tuxtbxtfxtx

xtt

uxt

A

A

,,:,

,,,,:,,

The control u is understand as an internal disturbance affecting the drift term φA.The control derivative is used as an auxiliary control used to steer σ and to 0.Note that affect the dynamics.

uu

32

0,,: xtxtSurfaceSliding





The Problem Statement (continue – 2)

0x

0 xx

0x


tuxbtxftx ,


uu

r

tuuxtuxt

tuxbxtfxtx

xtt

BB ,,,,

,,,0

xtbuxtx

xt

xtfuxtx

uxtt

uxt

B

B

,,,:,

,,,,,:,,

It is assumed that , which means that the sliding variable has relative degree two. In this case the actual actuator u is discontinuous.

0, xtB 33






The Problem Statement (continue – 3) 0x

0 xx

0x


tuxbtxftx ,


uu

r

Case A: relative degree 01

u

r

Both Cases A and B can be dealt with an uniform treatment, because the structure of the System to be stabilized is the same, i.e. a 2nd Order System with Affine relevant control signal (the control derivative in Case A, the actual control u in Case B). u

xttvxtty

xttyty

xtty

,,

,

,

2

21

1

Case A: relative degree r = 1

tutv

uxtA

,,

Case B: relative degree r = 2

tutv

uxtB ,,

Assume

21 ,0 GxtG

34


tutv

uxtA

,,






1yx

2yx

021 yyxx

021 yyxx

021 yyxx

021 yyxx

Twisting Algorithm Trajectory in the Phase Plane

O


This Algorithm provides twisting around the origin of the phase plane . This means that trajectories perform rotations around the origin while converging in finite time to the origin of the phase plane.

O

xttvxtty

xttyty

xtty

,,

,

,

2

21

1

1;0

1;0

1

211

211

uyyifysignV

uyyifysignV

uifu

tu

M

m

21 ,0 GxtG

35


tutv

uxtA

,,

Levant, Arie ( formerly Levantosky, Lev )

tuxbtxftx ,







This Algorithm provides twisting around the origin of the phase plane . This means that trajectories perform rotations around the origin while converging in finite time to the origin of the phase plane.

O

xttvxtty

xttyty

xtty

,,

,

,

2

21

1

0

0

211

211

yyifysignV

yyifysignVtutv

M

m

21 ,0 GxtG

The following conditions must be fulfilled for the finite time convergence

mM

m

m

mM

VGVG

GV

GV

VV

21

1

2

04

36

Case B: relative degree r = 2

tutv

uxtB ,,

1yx

2yx

021 yyxx

021 yyxx

021 yyxx

021 yyxx

Twisting Algorithm Trajectory in the Phase Plane

O

tuxbtxftx ,





Second Order Sliding Modes 1yx

2yx

Super-Twisting Algorithm Trajectory in the Phase Plane

The Super-Twisting Controller

121

1111

ysigntu

tuysigntytu

The control is given by:

xttvxtty

xttyty

xtty

,,

,

,

2

21

1

21 ,0 GxtG

5.00

4

12

21

222

1

21

G

G

G

G

t

dtysignysigntytu0

12111

37


tuxbtxftx ,


tutv

uxtB ,,




Second Order Sliding Modes 1yx

2yx

Super-Twisting Algorithm Trajectory in the Phase PlaneThe Super-Twisting Controller (continue)

01111

01101

2

12

1

21

1

1

yifysigny

yifysigntu

uifysign

uifutu

tututu

The control is given by:

xttvxtty

xttyty

xtty

,,

,

,

2

21

1

21 ,0 GxtG

5.00

4

12

21

222

1

21

G

G

G

G

38


utxftx ,,General Nonlinear System

tuxbtxftx ,





Sliding Order Sliding Modes

1yx

2y

The State-Flow

1

2/1

112 ysignyy

1

2/1

112 ysignyy

21

2/1

11

1

3/1

12

1

2

yysigny

ysigny

y

ym

Isoclines

Trajectory

The Super-Twisting Controller (Shtessel version)

0

0,

,

21

3/1

122

121

2/1

111

1

ysigntyty

xttyysigntyty

xtty

The 2nd Order Sliding Mode is Given by

is Finite Time Stable, i.e., is Asymptotically Stable with a Finite Settling Time for any solution and any initial conditions.

Proof Let choose the following Lyapunov Function candidate:

000

0004

3

2,

21

213

4

12

22

21

yandyifonly

yandyifyy

yyV

6/5

1211

3/1

12221

2/1

111

3/1

12

2211

3/1

1222

11

21

21

,

yysigntyyyysignyysigny

yyyysignyyy

Vy

y

VyyV

td

d

yy

00, 1

6/5

12121 yifyyyVtd

d 39




alpha1=3, alpha2 =3

alpha1=1, alpha2 =9

-5000 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000-2000

-1000

0

1000

X1

X1d

ot

alpha1=1, alpha2 =1

-4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000-400

-200

0

200

400

X1

X1do

t

alpha1=3, alpha2 =1

-2000 -1000 0 1000 2000 3000 4000 5000-400

-200

0

200

X1X

1dot

-3000 -2000 -1000 0 1000 2000 3000 4000 5000-1000

-500

0

500

X1

X1d

ot

alpha1=3, alpha2 =9

-4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000-1000

-500

0

500

1000

X1

X1d

ot

alpha1=9, alpha2 =1

-1000 0 1000 2000 3000 4000 5000-1000

-500

0

500

X1

X1d

ot

-1000 0 1000 2000 3000 4000 5000-1000

-500

0

500

X1

X1d

ot

alpha1=9, alpha2 =3

alpha1=9, alpha2 =9

-2000 -1000 0 1000 2000 3000 4000 5000-1000

-500

0

500

X1

X1d

ot

Time = 100 sec

-4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000-1000

-500

0

500

X1

X1d

ot

alpha1=1, alpha2 =3

We can see that to speed-up theConvergence to Origin we mustIncrease alpha1 and keep alpha2 Small relative to alpha1.

40





000

0002,

21

21

0

3/1

2

22

21

1

yandyifonly

yandyifdzzsignzy

yyV

y

>> [X,Y]=meshgrid(-0.5:0.05:0.5);>> alpha2=1;>> Z=0.5*Y.^2+0.75*alpha2*abs(X).^1.3334+eps;>> mesh(X,Y,Z,'EdgeColor','black')>> contour(X,Y,Z)

-0.5

0

0.5

-0.5

0

0.50

0.1

0.2

0.3

0.4

0.5

1y2y

21, yyV

MATLAB:

1y

2y

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

41





Theorem: Assume a Lyapunov Function that satisfies

00,&000

000, 121

21

2121

yifyyVtd

d

yandyifonly

yandyifyyV

Assume that in addition exists a Domain , that includes the origin(y1 = 0, y2 = 0), and in this domain

Dyy 21,

10&00,, 2121 kyyVkyyVtd

d

then V (y1, y2) → 0 in a Finite Time ts.

Proof:

0,, 2121 yyVtd

dyyVk

0

,

,1

21

21 yyV

yyVd

ktd

0td

q.e.d.

1001

,,,

1

11

21121

1

0

2100

0

k

yyVyyVyyV

ktt t

ttS S

We can see that ts at which V (y1, y2) → 0 is Finite.

Let integrate between an initial time t0 to a time ts at which V (y1, y2) → 0.

42





Let find, if in our case, exists a Domain D, that includes the origin, and satisfies:


d

We have:

000

0004

3

2,

21

213

4

12

22

21

yandyifonly

yandyifyy

yyV

00, 1

6/5

12121 yifyyyVtd

d

6

5

121213

4

12

22

21

,1

4

3

2, y

ktd

yyVd

ky

yyyV

6

5

1213

4

12

10

2

33

4

12

22

21 2

3

4

3

2,

2

1

122

2

yk

yyy

yyVyy

Define the Domain D that includes the origin by: 12

31

3

4

12

2

2 yandyy

1

6

5

1

1

213

4

122

3y

ky

or:

If since16

51

1

6

5 11

3

4,1

1

6

51 yand

from the Figure, and choosing some k>0, we can see that exists some small |y1s| such that for

we have3

4

102

2

2101 2

3yyandyy 100,, 2121 yyVkyyV

td

d

1

21

2

1

6

5

3

4

10 3

2

ky

equality 1yx

2y

11 y

11 y11

y

1y1y

11

111 y

3

4

102

2

20

1

21

2

1

6

5

3

4

10

23

32

yy

ky

10y10y

20y

20y

1

1

1

1111

1111

yyyy

yyyy

1y

2y

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

43





Therefore for:

Return to Table of Content

3

4

10221

2

1

3

4

102202101 2

3,

2

3,

0yyyVandyyyyy t

V (y1, y2) → 0 in Finite Time ts .

1001

, 121

00

k

yyVtt t

S

-0.5

0

0.5

-0.5

0

0.50

0.1

0.2

0.3

0.4

0.5

1y2y

21, yyV

1y

2y

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

44




Sliding Mode Control for Linear Time Invariant (LTI) Systems

Regular Form of a LTI

Consider the following LTI:

mnRBRARuRxuBxAx nxmnxnmn

Assume rank (B) = m (i.e., matrix B is full rank) and the pair (A,B) is controllable.

Perform the Singular Value Decomposition (SVD) of B:

H

Bxmmn

B

Bnxm mxm

mxm

nxnVUB 1

1

1 0

where H means Transpose of a matrix and complex conjugate of it’s elements, and:

mBH

BH

BBnBH

BH

BB IVVVVIUUUU 11111111 ;

mxmmnxnn

mmB

diagIdiagI

diagmxm

1,,1,1,1,,1,1

0,,,, 21211

45




Regular Form of a LTI (continue - 1)

Consider the following LTI:nxmnxnmn RBRARuRxuBxAx

Define the Orthogonal Transformation Matrix:

H

nxnxmmnmn

mmnmx

BnxnH

Bmnmxm

mnxmmn

nxn TI

IUTU

I

IT

nxnnxn

0

0

0

0: 1

11

midiagVUB

immxmmxm

mxm

nxn BBBBH

Bxmmn

B

Bnxm ,,10,,0 111

1

1

S.V.D. of B:

0det000

2211

11

mxm

mxmmxmmxm

mxm

mxm

BBV

VBTxmmn

HBB

xmmnHB

B

xmmn

nxmnxn

Perform the following Transformation of Variables: xTx nxnr : 46




Regular Form of a LTI (continue - 2)

Consider the following LTI:nxmnxnmn RBRARuRxuBxAx

Perform the following Transformation of Variables: xTx nxnr :We obtain:

0det

021

22

1

2221

1211

2

1

1

1

1

1

mxm

mxmmx

xmn

mxmmnmx

xmmnmnxmn

mx

xmn BuBx

x

AA

AA

x

xx mx

xmmn

r

r

r

r

r

This is called the Regular Form of the LTI

47




Sliding Surface of a Regular Form of a LTI

Consider the following Regular Form of a LTI:

1

22

1

2221

1211

2

1 0

1

1

1

1

mx

xmmn

r

r

r

r

r uBx

x

AA

AA

x

xx

mxmmx

xmn

mxmmnmx

xmmnmnxmn

mx

xmn

Define a Switching Function s (t)

0det0 222112

1

211

1

1

mxm

mx

xmn

mxmmnmxSxSxS

x

xSSts rr

r

r

S

mx

Therefore, on the Sliding Surface: txMtxSStx rr

M

r 1111

22

112112121111 rrrr xMAAxAxAx Then we have:

By analogy to the “Classical” State-Feedback Theory, it can be seen that this is the same problem of finding the State-Feedback matrix M for the Regulator Form where xr2(t) plays the role of the “control” signal.

48




Sliding Surface of a Regular Form of a LTI (continue - 1)

111

212112121111 r

M

rrr xSSAAxAxAx

Then we have:

The Stability and Performance of the System depends on the Controllability of the Regular Form Pair (A11, A12). It can be shown that (A11, A12) is Controllable, if and only if the pair (A , B) is Controllable.

Controllability of the Regular Form is

222221221

221212112

22

122

1

0B

AAAA

AAAAB

A

AB

I

nrankfullBABAB

m

xmmn

rn

rrrr

49





The design of the Sliding Mode Controller must achieve:

• The design of the Matrix S=[S1,S2] to obtain the required performance and Stable Dynamics for the Closed-Loop Sliding Mode System.

• The design of the Control Law to ensure that the Sliding Surface is Reached and Maintained.

For the Sliding Surface Reachability Condition let define the Lyapunov Function:

00

00

2

1:

tststsif

tststsiftststV

T

T

T

The Reachability Condition is:

0& someforttstststd

tVd T 50






• The design of the Matrix S = [S1,S2] to obtain the required performance and Stable Dynamics for the Closed-Loop Sliding Mode System.


0 tsts


nxmnxnmn RBRARxtDuRxxtDBuBxAx ,,,

xtD , are the uncertainties in the input

On the Sliding Surface

0, xtDBuBxAStxSts eq

The Equivalent Control that Maintains the System on the Sliding Surface is

xtDBSxASBStueq ,1 51






• The design of the Matrix S = [S1,S2] to obtain the required performance and Stable Dynamics for the Closed-Loop Sliding Mode System.



nxmnxnmn RBRARxtDuRxxtDBuBxAx ,,,

0,1 tsxtDBSxASBStueq

xtDBSBSBIxASBSBIx n

P

n

S

,0

11

xAPx S52




Unit Vector Approach for a Controller of a Regular Form of a LTI

Consider the System:

where is an unknown but bounded (matched) uncertainty that satisfies:

mnm

nxmnxnmnm RRRfRBRARuRxuxtfuBxAx xx,, 1

uxtfm ,,

xttukuxtfm ,,, By using the Orthogonal Transformation T we obtain:

uxtfB

uBx

x

AA

AA

x

xx rm

xmmn

mx

xmmn

r

r

r

r

r

mxmmxmmx

xmn

mxmmnmx

xmmnmnxmn

mx

xmn ,,00

21

22

1

2221

1211

2

1

1

1

1

1

The Switching Function can be written:

0det 22

11

122

2

121

S

x

xISSS

x

xSSts

r

rm

Mr

r 53




Unit Vector Approach for a Controller of a Regular Form of a LTI (continue – 1)

Define the Sub-System:

Let Differentiate this:

uxtfBS

uBSs

x

SASSASMASASMASAS

SAMAA

uxtfB

uBs

x

SSS

I

AA

AA

SS

I

s

x

rmr

rmrr

,,00

,,0000

2222

1

12222

12121222212121111

12121211

22

1

121

122221

1211

21

1

or:

ts

tx

SSS

I

tx

tx

tx

tx

SS

I

ts

tx r

r

r

r

r

T

r

S

1

121

122

1

2

1

21

10

&0

uxtfB

uBsS

x

AAMMAMMAAMA

AMAA

sS

xrm

rr ,,00

221

2

1

221212221121

121211

12

1

54





The Sub-System:

In order to force s to zero, Φ must satisfy a Lyapunov Equation of the type:

uxtfB

uBsS

x

AA

AA

sS

xrm

rr ,,00

221

2

1

2221

1211

12

1

Choose:NonlinearLinear uuu

suBSsSASxASs Linearr 22

122221212

sSASxASBSu rLinear 122221212

122

mT IPP 22

sP

sPBSxtu rNonlinear

2

2122, Choose:

and:

The Linear Part must keep the System on the Sliding Manifold 0ss

The Nonlinear Part must force the System to Reach the Sliding Manifold.

55





The Sub-System:

uxtfB

uBsS

x

AA

AA

sS

xrm

rr ,,00

221

2

1

2221

1211

12

1

Choose:NonlinearLinear uuu

sSASxASBSu rLinear 122221212

122

sP

sPBSxtu rNonlinear

2

2122,

The Linear Part must keep the System on the Sliding Manifold 0ss

The Nonlinear Part must force the System to Reach the Sliding Manifold.

uxtfBsP

sPBSxtBsSASxASBSBsSAxAsS rmrrr ,,, 2

2

21222

122221212

1222

1222121

12

uxtfBSsP

sPxtss rmr ,,, 22

2

2 56





For the Sub-System:

002 sifsPsV T

Choose a potential Lyapunov Function that shows the Reachability of the Sliding Surface:

uxtfBSsP

sPxtss rmr ,,, 22

2

2

m

T

T

mTT fBS

sP

sPsPssPfBS

sP

sPssPssPsV

td

Vd22

2

22222

2

222

mTT

I

TT fBSPssPPssP

sPPs 222222

22 21

2

mTT fBSPssPss 2222 22

57





For the Sub-System:

002 sifsPsV T

Choose a potential Lyapunov Function that shows the Reachability of the Sliding Surface:

uxtfBSsP

sPxtss rmr ,,, 22

2

2

mTT fBSPssPssV 2222 22

mmT fBSsPfBSPs 222222 Using Cauchy-Schwarz Inequality:

mfBSsPsV 2222

We want to choose ρ (t,x) such that and xtukBSfBS m ,2222

02 222 mfBSsPsV 58





mfBSsPsV 2222

We want to choose ρ (t,x) such that and xtukBSfBS m ,2222

02 222 mfBSsPsV

002 sifsPsV T

1

221

22

BSuBSuuuuuu LinearLinearNonlinearLinearNonlinearLinear

k

xtukBS Linear

1

,22

122

1

22221

22221

22 1 BSBSBSBSIBSBSI

automatically fulfilled for k > 1

Define η (t,x) > 0 such that1

22

BSuu Linear

xtBSkukBSxtukBSkfBS Linearm ,,1

22222222

59



where u1 and u2 are the known and unknown inputs, respectively.

System Description and Notation

212211

21212211

21

2121

mmppCrankmBrankmBrank

RCRyxCy

RBRBRARuRuRxuBuBxAx

pxn

pxnp

nxmnxmnxnmmn

nxmnxm

Assumptions:

• There exists a known nonnegative scalar function such that yt, tytu ,2

• The pairs (A,B1), (A,B2) are controllable and (A,C) is observable with the matrices B1, B2 and C being of full rank

• p ≥ m1+m2, that means that bthe number of output channels is greater or equal then the number of inputs, and rank (CB1) =m1, rank (CB2) – m2


60




where u1 and u2 are the known and unknown inputs, respectively.

System Sliding Surface

xCy

uBuBxAx

2211

Assume that the Sliding Surface is of the type: xSx 1On the Sliding Surface we must have: 0&0 xx

0221111 uBuBxASxSx If (S B1) is nonsingular:

2211

111 uBxASBSueq

On the Sliding Surface the dynamics of the System is given by:

01

11

111

xS

xASBSBIx

Note that in the Sliding Mode the System is governed by a reduced order of differential equations with the eigenvalues of , and are not affected by the unknown inputs/disturbances.

ASBSBI 11

111


61




The Transmission Zeros of the System are defined as the solutions for λ of:

Sliding Modes and System Zeros


11

1

1

1

00

0det mn

S

BAIrankwhichforor

S

BAIz nn

The solutions are not affected if we multiply the Square Matrix by Non-Singular Square Matrices that are not functions of λ:

00

00

00

0

1

11

11

1

11

111TSM

NBTTFBAITrank

N

T

IF

I

S

BAI

M

Trank

xmm

nxn

m

nn

mxm

nxn

00

0

000

0

1

11

11

1

11

111TSM

NBTTSHAITrank

N

T

S

BAI

I

HI

M

Trank

xmm

nxnn

m

n

mxm

nxn

xnmm

nxmnxnmn

RSRxS

RBRARuRxuBxAx11

11

11

1111


62




Sliding Modes and System Zeros (continue – 1)


00

det1

1

S

BAIz n

The System Zeros are not affected under the following set of transformations:

• Nonsingular State Transformation xTx ~

• State Feedback ( A+B F)

• Output Injection (A + H S)

• Nonsingular Input Control Transformations uNu ~

• Nonsingular Output Signal Transformations M~

xnmm

nxmnxnmn

RSRxS

RBRARuRxuBxAx11

11

11

1111


63






11

1

11

1

11

11

1

detdet

0

0det

0det

BAISAI

BAIS

BAII

IS

AI

S

BAIz

nn

n

nn

m

nn

Hence:

AI

zBAIS

nn

detdet 1

11

xnmm

nxmnxnmn

RSRxS

RBRARuRxuBxAx11

11

11

1111


64





xnmm

nxmnxnmn

RSRxS

RBRARuRxuBxAx11

11

11

1111


1

111111

1111

1

11

11

11

111

11

detdet

0det

0

0det

0det

BASBSBIISASBSBII

S

BASBSBII

IASBS

I

S

BAI

S

BAIz

nn

n

m

nnn


65



using:


Let compute:

11

111111

11111

1

111111

1111

detdetdetdet

detdet

1 BSASBSBIIBSASBSBII

BASBSBIISASBSBIIzm

nn

nn

1

111111 BASBSBIIS n

111111

mmmnnmmmmnnnnmmmmmmnnmmm CBDCBIDCCBDC

0since 1

1111111

11

11111

11

111111

SBSBISBS

BSBSBIIISBASBSBIIS nnn

Therefore: 1

111

111 detdet 1 BSzASBSBII mn

We found that the Poles of the System on the Sliding Surface S1x = 0 are defined by the Zeros of the triple (A, B1, S1).


66




Properties of the Sliding Modes

The following are a Summary of the Properties of the Sliding Modes:

• The System behaves as a Reduced Order motion which (apparently) does not depend on the control signal u (t).

• There are (n-m) States that determines the dynamics of the Closed Loop System.

• The Closed-Loop Sliding Motion depends only on the choise of the Sliding Surface.

1111 112112121111 xmnxmmnmnxmnmxxmmnxmnmnxmnxmn rmnmxrrr xMAAxAxAx

• The Poles of the Sliding Motion are given by the Invariant Zeros of the System Triple (A, B, S)

67



Design of a Sliding Surface (Hyperplane)

The following “Classical” Methods can be used to obtain the Matrix 11

2 SSM • Quadratic Minimization

• Robust Eigen-structure Assignment

• Direct Eigen-structure Assignment

68



Design of a Sliding Surface (Hyperplane) by Quadratic Minimization


1

22

1

2221

1211

2

1 0

1

1

1

1

mx

xmmn

r

r

r

r

r uBx

x

AA

AA

x

xx

mxmmx

xmn

mxmmnmx

xmmnmnxmn

mx

xmn

And the following Optimization (Minimization) Problem:

SS t

rT

rrT

rrT

r

t

rT

r tdxQxxQxxQxtdxQxJ 222221211111 22

1

2

1

Let rewrite:

112

122121112

122222112

1222

1121

221211121

22221121

2222221121

222

1121

22221

221211121

22221

221211121

222222221

221212222

222211222121

rTT

rrT

r

T

rT

r

rTT

rrTT

rT

rr

T

rT

r

rT

I

Trr

T

I

Trr

T

I

Trr

I

Trr

Tr

rT

rrT

rrTT

r

xQQQxxQQxQxQQx

xQQQxxQQQxQQxxQxQQx

xQQQQQxxQQQQQxxQQQxxQQQxxQx

xQxxQxxQx

69



Design of a Sliding Surface (Hyperplane) by Quadratic Minimization (continue -1)


1

22

1

2221

1211

2

1 0

1

1

1

1

mx

xmmn

r

r

r

r

r uBx

x

AA

AA

x

xx

mxmmx

xmn

mxmmnmx

xmmnmnxmn

mx

xmn

And the following Optimization (Minimization) Problem:

St

rT

r

T

rT

rrTT

r tdxQQxQxQQxxQQQQxJ 1121

222221121

2221121

22121112

1

Define: 1121

222121

221211 :&:ˆr

Tr

T xQQxvQQQQQ

Therefore:

St

Tr

Tr tdvQvxQxJ 221111

ˆ2

1

vAxQQAAxAxAx rT

rrr 121121

2212112121111

70



Design of a Sliding Surface (Hyperplane) by Quadratic Minimization (continue - 2)

We ended up with the following Optimization Problem:

The “Optimal Control” is:

1121

222121

22121111121

221211 :&:ˆ,:ˆr

Tr

TT xQQxvQQAAAQQQQQ

where P1 is given by the Riccati Equation:

St

Tr

Tr tdvQvxQxJ 221111

ˆ2

1

vAxAx rr 121111ˆ

11121

22* rxPAQv

0ˆˆˆ112

12212111 QPAQAPAPPA TT

We obtained: 11121121

221121

222 * rrTT

rT

r xMxQPAQxQQvx

Therefore: TT QPAQSSM 121121

2211

2

where S2 is arbitrary.

71



Consider a Sliding Manifold σ (x,t). We want to design the Control ueq that keepsthe trajectory on the manifold:

tutftxtd

deq ,,

72

where f (σ,t) is a known or unknown but bounded function.



tutxftxtd

deq ,,

73

When f (σ,t) is a known function a solution for ueq is:

0,2/1 signtxftueq

Continuous Sliding Mode Control

We obtain: 0,2/1 signtx

td

d

Let choose the following Lyapunov Function:

00

00

2

1 2

V

02/32/1 sign

td

d

td

Vd

kforkkVktd

Vd0

2/31

4

30

22/3

Therefore σ→0 in a Finite Time

When f (σ,t) is a unknown we can use an Sliding Mode Observer to estimate it.

2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

12

14

16

18

20



tutxftxtd

deq ,,

74

Continuous Sliding Mode Control (continue – 1)

Lyapunov Function:

00

00

2

1 2

V

>> [X,Y]=meshgrid(-0.5:0.05:0.5);>> Z=0.5*Y.^2+0.5*X.^2+eps;>> mesh(X,Y,Z,'EdgeColor','black')>> contour(X,Y,Z)

MATLAB:

-0.5

0

0.5

-0.5

0

0.50

0.1

0.2

0.3

0.4

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5



tutxftxtd

deq ,,

75

When f (σ,t) is a known function a Second Order SM Control for ueq is:

0,, 100

2/1

1

0

t

t

eq dtsignsigntxftu

Second Order Sliding Mode Control

We obtain: 0,0

0

2/1

1 t

t

dtsignsigntxtd

d

Let choose the following Lyapunov Function:

0,00

0,00

22, 0

2

0

0

2

z

zzdsign

zzV

Rewrite:

signz

zsign

0

2/1

1



tutxftxtd

deq ,,

76

Second Order Sliding Mode Control (continue -1)

0,00

0,00

22, 0

2

0

0

2

z

zzdsign

zzV

Let check:

2/1

10

2/1

1000,

zsignsignsignzsignzzzVtd

d

z

For: 02

,0

0

2

zatC

Cz

zVMax

kforkCkCz

kVktd

Vd

12/1

012/1

0

12/102/1

0100

22/1

10 02

Therefore σ→0 in a Finite Time

When f (σ,t) is a unknown we can use an Sliding Mode Observer to estimate it.



tutxftxtd

deq ,,

77

Second Order Sliding Mode Control (continue -2)

Lyapunov Function:

>> [X,Y]=meshgrid(-0.5:0.05:0.5);>> rho0=2;>> Z=0.5*Y.^2+rho0*abs(X).^1+eps;>> mesh(X,Y,Z,'EdgeColor','black')>> contour(X,Y,Z)

MATLAB:

0,00

0,00

2, 0

2

z

zzzV

-0.5

0

0.5

-0.5

0

0.50

0.5

1

1.5

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5


Sliding Mode Observers

In most of the Linear and Nonlinear Unknown Input Observers proposed so far, the necessary and sufficient conditions for the construction of Observers is that the Invariant Zeros of the System must lie in the open Left Half Complex Plane, and the transfer Function Matrix between Unknown Inputs and Measurable Outputs satisfies the Observer Matching Condition.

Observers are Dynamical Systems that are used to Estimate the State of a Plant using its Input-Output Measurements; they were first proposed by Luenberger.

David G. LuenbergerProfessor

Management Science and Engineering

Stanford University

In some cases, the inputs of the System are unknown or partially unknown, which led to the development of the so-called Unknown Input Observers (UIO), first for Linear Systems. Motivated by the development of Sliding-Mode Controllers, Sliding Mode UIOs have been developed.

The main advantage of using Sliding-Mode Observer over their Linear counterparts is that while in Sliding, they are Insensitive to the Unknown Inputs and, moreover, they can be used to Reconstruct Unknown Inputs which can be a combination of System Disturbances, Faults or Nonlinearities.

78

79

Guidance of Intercept

Sliding Mode Observers of Target Acceleration

Kinematics:

tataRRtd

dMT 11

We want to Observe (Estimate) the Unknown Target Acceleration Component :

taT 1

Define:0:1_ vtaestAt

EstT

mEstEst AvRztd

d 00

The Differential Equation of the Observer will be a copy of the kinematics:

mM Ata

:1

Define the Observer Error: EstEstO Rz 0:

Define the Sliding Mode Observers that must drive σO→0:

22

11

1232

201

2/1

0121

1

3/2

10

1.1

5.1

2

vz

vz

vzsigntv

zvzsignvztv

zsigntv OO

Missile Command Acceleration

estAtdRsignRRsignRRNa

EstEstRSM

EstEstEstEstEstEstEstEstEstEstC _'

2

3/1

2

2/1

1

t1, t2, t3 are Design Parameters

Observer 4: Variation of 1

22

11

122

201

2/1

012/1

1

1

3/23/10

1.1

5.1

2

vz

vz

vzsignLv

zvzsignvzLv

zsignLv OO

Observer 1:

02

11

3/1

21

1

2/1

10

z

vz

signv

zsignv

OOO

OOO

02

11

21

1

2/1

10

z

vz

signv

zsignv

OO

OOO

L is a Design Parameter are Design Parameters21, OO are Design Parameters21, OO

Observer 2: Observer 3:

80


Sliding Mode Observer of Target Acceleration: MATLAB Listing

% Nonlinear Sliding Mode Target Acceleration ObserversAt_est=0;v0=0;z0=x1;z1=0;z2=0;Observer=1;%First Observer ParameterL=10;%Second Observer ParametersalphaO1=30;alphaO2=1;%Third Observer Parametersrho1=20;rho2=3;%Fourth Observer Parameterst1=10;t2=3;t3=1;

%Second Order Sliding Mode SigmaSM=Range_est*Lamdadot_est; y2 = alpha1*sign(SigmaSM)*abs(SigmaSM)^0.5+x2; x2_dot =alpha2*sign(SigmaSM)*abs(SigmaSM)^(1/3); %Nonlinear Sliding Mode Target Acceleration Observers z0_dot=v0-Rdot_est*Lamdadot_est-Am; SigmaO=z0-SigmaSM; if(Observer==1) v0=-2*L^(1/3)*abs(SigmaO)^(2/3)*sign(SigmaO)+z1; v1=-1.5*L^(1/2)*abs(z1-v0)^(1/2)*sign(z1-v0)+z2; v2=1.1*L*sign(z2-v1); z1_dot=v1; z2_dot=v2; v0_dot=0; At_est=v0; end if(Observer==2) v0=-alphaO1*abs(SigmaO)^(1/2)*sign(SigmaO)+z1; v1=-alphaO2*abs(SigmaO)^(1/3)*sign(SigmaO); z1_dot=v1; z2_dot=0; v0_dot=0; At_est=v0; end if (Observer==3) v0=-rho1*abs(SigmaO)^(1/2)*sign(SigmaO)+z1; v1=-rho2*sign(SigmaO); z1_dot=v1; z2_dot=0; v0_dot=0; At_est=v0; end if(Observer==4) v0=-2*t1*abs(SigmaO)^(2/3)*sign(SigmaO)+z1; v1=-1.5*t2*abs(z1-v0)^0.5*sign(z1-v0)+z2; v2=1.1*t3*sign(z2-v1); z1_dot=v1; z2_dot=v2; v0_dot=0; At_est=v0; end %Missile Acceleration Command and Autopilot Ac=-N*Rdot_est*Lamdadot_est+y2+At_est;

N = 3;alpha1 =10;alpha2 = 1;

%Nonlinear Sliding Mode Target Acceleration % Observer State Integration z0=z0+z0_dot* delta_time; z1=z1+z1_dot* delta_time; z2=z2+z2_dot* delta_time; v0=v0+v0_dot* delta_time;

22

11

1232

201

2/1

0121

1

3/2

10

1.1

5.1

2

vz

vz

vzsigntv

zvzsignvztv

zsigntv OO

22

11

122

201

2/1

012/1

1

1

3/23/10

1.1

5.1

2

vz

vz

vzsignLv

zvzsignvzLv

zsignLv OO

02

11

3/1

21

1

2/1

10

z

vz

signv

zsignv

OOO

OOO

02

11

21

1

2/1

10

z

vz

signv

zsignv

OO

OOO

81


22

11

1232

201

2/1

0121

1

3/2

10

1.1

5.1

2

vz

vz

vzsigntv

zvzsignvztv

zsigntv OO



22

11

122

201

2/1

012/1

1

1

3/23/10

1.1

5.1

2

vz

vz

vzsignLv

zvzsignvzLv

zsignLv OO

Observer 1:

02

11

3/1

21

1

2/1

10

z

vz

signv

zsignv

OOO

OOO

02

11

21

1

2/1

10

z

vz

signv

zsignv

OO

OOO



Scenario: R0=10000 m, Rdot=-1000 m/s, alpha1=10, alpha2=1, N=3, Ldot0=0.05 rad/sA step pulse Target acceleration At=100 m/s2 starting at t=3 s and finishing at t=7sWith Not Noise

10L 1,30 21 OO 3,20 21 OO 1,3,10 321 ttt

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-100

0

100

z1

0 1 2 3 4 5 6 7 8 9 10-10

0

10

Sigm

aO

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-20

0

20

z1

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Sigm

aO

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-10

0

10

z1

0 1 2 3 4 5 6 7 8 9 10-20

0

20

Sigm

aO

0 1 2 3 4 5 6 7 8 9 10-100

0

100

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-200

0

200

z1

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Sigm

aO

Sliding Mode Observer of Target acceleration - MATLAB Results

82


22

11

1232

201

2/1

0121

1

3/2

10

1.1

5.1

2

vz

vz

vzsigntv

zvzsignvztv

zsigntv OO



22

11

122

201

2/1

012/1

1

1

3/23/10

1.1

5.1

2

vz

vz

vzsignLv

zvzsignvzLv

zsignLv OO

Observer 1:

02

11

3/1

21

1

2/1

10

z

vz

signv

zsignv

OOO

OOO

02

11

21

1

2/1

10

z

vz

signv

zsignv

OO

OOO



0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-100

0

100

z1

0 1 2 3 4 5 6 7 8 9 10-100

0

100

Sigm

aO

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-10

0

10

z1

0 1 2 3 4 5 6 7 8 9 10-20

0

20

Sigm

aO

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-20

0

20

z1

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Sigm

aO

0 1 2 3 4 5 6 7 8 9 10-200

0

200

Atest

0 1 2 3 4 5 6 7 8 9 10-1000

0

1000

z0

0 1 2 3 4 5 6 7 8 9 10-100

0

100

z1

0 1 2 3 4 5 6 7 8 9 10-20

0

20

Sigm

aO

Scenario: R0=10000 m, Rdot=-1000 m/s, alpha1=10, alpha2=1, N=3, Ldot0=0.05 rad/sA step pulse Target acceleration At=100 m/s2 starting at t=3 s and finishing at t=7sWith Lamda_dot Noise Filtered with Time Constant of 200msec

10L 1,30 21 OO 3,20 21 OO 1,3,10 321 ttt


83



Scenario: R0=1000 m, Rdot=-1000 m/s, alpha1=30, alpha2=1, N=3, Ldot0=0.05 rad/sNo Target acceleration , No Measurement Noises

Observer Output0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-505

Atest

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

0100

z0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-101

z1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

00.1

Sigm

aO

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

100

X1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

X1 do

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

X2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-500

0

500

Am

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

0

0.1

Lam

dadot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.5

0

0.5

Lam

dadot

2

84


Sliding Mode Observer of Target acceleration - MATLAB ResultsScenario: R0=1000 m, Rdot=-1000 m/s, alpha1=30, alpha2=1, N=3, Ldot0=0.05 rad/sA step pulse Target acceleration At=100 m/s2 starting at t=0.3 s and finishing at t=0.6sWithout Noise

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

100

X1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

X1 dot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

X2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-500

0

500

Am

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

0

0.1

Lam

dadot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5

Lam

dadot

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

0100

Atest

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

0100

z0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20

020

z1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10

010

Sigm

aO

Observer Output

85




Scenario: R0=1000 m, Rdot=-1000 m/s, alpha1=30, alpha2=1, N=3, Ldot0=0.05 rad/sA step pulse Target acceleration At=100 m/s2 starting at t=0.3 s and finishing at t=0.6sWith Lamda_dot Noise Filtered with Time Constant of 20msec

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

100

X1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

X1 do

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

X2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-500

0

500

Am

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

0

0.1

Lam

da do

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5

Lam

da do

t2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

0200

Atest

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

0100

z0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20

020

z1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10

010

Sigm

aO

ObserverOutput 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.1

0

0.1

Lam

dadot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5

Lam

dadot

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.05

0

0.05N

oise

Lamda

dot

Lamda_dotNoise


86

Slide Mode Control Examples



Fu, L-C, et all, Control System of a Kill Vehicle

Fu, L-C, et al, Solution for Attitude Control of KV

Fu, L-C, et al, Zero SM Guidance of a KV

Crassidis , et al -Attitude Control of the Kill Vehicle



87



Problem: Develop the Midcourse Guidance Law to intercept in a Head On Scenario a Ballistic Missile, by an Vertical Launched Interceptor at altitude above a given value hmin.

Solution Method:

• Start with a Planar Approximation where the Ballistic Trajectory is a known Straight Line and the Launch Point define the Plane.• Assume a Constant Velocity Bounded Maneuver Interceptor and don’t consider gravitation.• Define the Straight Line Ballistic Trajectory as the Sliding Curve and define a Linear Guidance Law and a Lyapunov Volume around the Sliding Curve such that the Entering Interceptor Trajectory will not escape and will Converge to the Ballistic Trajectory in a Head On situation.• Using Optimal Control Theory define the Interceptor Trajectory from the Launching Point that will enter the Sliding Lyapunov Volume, above the minimal height hmin, in Minimum Time.• Define the Interceptor Midcourse Guidance Law as the combination of the Optimal Trajectory that reaches the Sliding Lyapunov Surface followed by the Linear Guidance Law to reach the Ballistic Trajectory in Head On.• Transfer the Planar Guidance Law to a Three Dimensional Law with the same structure.• Simulate the result using a real 6 DOF Interceptor Model.

R

R

R

R

minh

sw

x

zh

BallisticTrajectory

88



R

R

R

R

minh

sw

x

yh

BallisticTrajectory

• Start with a Planar Approximation where the Ballistic Trajectory is a known Straight Line and the Launch Point define the Plane.

In this plane choose a Coordinate System xOy, having the x axis on the Ballistic Straight Line approximation pointing toward the Ballistic Target, the origin at the intersection of the straight line with the ground and y axis pointing above the ground.

• Assume a Constant Velocity Bounded Maneuver Interceptor and don’t consider gravitation.

Under those conditions the Interceptor equation of motion are given by:

MAXMM

M

M

M

aaV

a

td

xd

Vtd

yd

Vtd

xd

sin

cos MV

x

yh

Interceptor

BallisticTrajectory

yMa

where γ is the angle between Interceptor Velocity Vector and the Ballistic Trajectory. To drive the Interceptor to fly on the Ballistic Trajectory we want to bring γ, y and dy/dt to zero, therefore let choose a Interceptor Guidance Law as:

MAXMMMM aaVktd

ydkyk

td

dVa

321

89



• Define the Straight Line Ballistic Trajectory as the Sliding Curve and define a Linear Guidance Law and a Lyapunov Volume around the Sliding Curve such the Entering Interceptor Trajectory will not escape and will Converge to the Ballistic Trajectory in a Head On situation.

Define:

cos1sin0

MM VdV

Choose a Lyapunov Function candidate:

0cos12

1

2

1:, 1

1

22

1

2 kk

Vy

k

VyyV MM

We can see that:

,...2,1,0200

000,

jjandy

andyyV

MAXMM

M

M

M

aaV

a

td

xd

Vtd

yd

Vtd

xd

sin

cos

MAXMMMM aaVktd

ydkykVa 321Guidance Law:

MV

x

yh

Interceptor

BallisticTrajectory

yMa

90




Choose a Lyapunov Function candidate:

0,...2,1,0200

00cos1

2

1:, 1

1

22

kjjandy

y

k

VyyV M

&0,,0sinsin

sinsinsinsin,

3212

1

322

1

2

321

1

2

1

2

kkkforVk

kV

k

k

kkyV

k

k

VVy

td

d

k

V

td

ydy

td

yVd

MM

M

MM

M

We can see that the Lyapunov Function confirms the convergence of the system to y=o, γ =0.

MAXMMMMM

M

M

aakkyV

kk

td

yd

V

ky

V

k

V

a

td

d

Vtd

yd

321

321 sin

sin

Guidance Law



The contour of the Lyapunov Function in the y, γ plane is defined by

constk

VyyV M 2

1

22

2

1cos1

2

1:,

,,, yVyVyV

MV

ky

y

2sin2,0

0

11maxmin,

maxmin,

2

2

y

1

22,

k

VyV M

1

2

,k

VyV M

2

,2 yV

1

1

2

k

VM

1

2

2

k

VM

31t

2t

3t

321

331

ttt

Trajectory

The Maximum Contour Ω where the trajectoriesConverge to y=0, γ = 0 is given for

1

maxmin,

1

maxmin,

22

k

Vy

k

V MM

Note: when

the Guidance Law will assureconvergence

1

2

1

22 2

cos12

1:,

k

V

k

VyyV MM




To define the values k1, k2, k3 we use Isoclines method to define the Trajectories behavior:

MAXMMM

M

aaVkkyV

k

td

d

Vtd

yd

321 sin

sinThe Trajectories in y, γ plane are:

The slope (inclination) of the Trajectories in y, γ plane is given by: NVkky

V

kV

d

yd

MM

M

321 sin

sin

The curves in y, γ plane for which the slope N is constant are called Isoclines and are given by:

31

21

sin kk

V

N

Vk

k

Vy MMM

N

312

31

21

0

2

sin

,2,1,0,0

kk

Vy

k

VN

kk

Vk

k

VyN

kkN

M

k

VN

M

MMN

N

M




O Singular Stable Points

* Singular Non-Stable Points

31

21

sin kk

V

N

Vk

k

Vy MMM

N

312

31

21

0

2

sin

,2,1,0,0

kk

Vy

k

VN

kk

Vk

k

VyN

kkN

M

k

VN

M

MMN

N

M

Isoclines



To define the values k1, k2, k3 we use Isoclines. We found that if

econvergencyoscillatornonkkk

econvergencyoscillatorkkk

321

321

MAXMMMM aaVktd

ydkykVa 321Guidance Law:




Un-Saturated AccelerationRegion M

MAX

MM

MAX

V

akky

V

k

V

a 32

1 sin


From which: 21

3211

321

1 :sinsin: yk

akk

k

Vy

k

akk

k

Vy MAXMMAXM

2

2

1

2

3

2

1

2

32

1k

aMAX

1k

aMAX

1k

aMAX

1k

aMAX

0

0

31

kk

Vy M

UnsaturatedRegion

SaturatedRegion

SaturatedRegion

y



In can be shown that all the trajectories that enter the non-saturated region for will stay in the un-saturated region, and will finally reach the origin (y=0,γ=0).

11


where:

21

223

21

2232

12

2

31

21

22

21

1

/

//

sin/

sin:

MAXM

MAXM

MAXMMAXM

aVkkk

aVkkkaVkk

k

aVkk

k

The unsaturated region around the origin bounded by –γ1<γ<γ1 , defined as Ω1 , is a Capture Zone for the trajectories.

2

1

2

1

1k

aMAX

1k

aMAX

0

0

UnsaturatedRegion Saturated

Region

SaturatedRegion

11

1

y

97



0

sin MVyy

0y

sinSystem Equations

MaMAXa

MAXaMV

1

2k 1kMVk3

Ca

MAXM aa

&2

Controller

0


Block Diagram of the Linear Guidance Law that assure convergence of the Interceptor Trajectory to the Ballistic Target Trajectory in a Head On situation.

98



• Using Optimal Control Theory define the Interceptor Trajectory from the Launching Point that will enter the Sliding Lyapunov Volume, above the minimal height hmin, in Minimum Time.

Choose the Coordinate System xOy, having the x axis on the Ballistic Straight Line approximation pointing toward the Ballistic Target, the origin at the intersection of the straight line with the ground and y axis pointing above the ground.

• Assume a Constant Velocity Bounded Maneuver Interceptor and don’t consider gravitation.

freeTTttt

Tzuuz

TygivenyyzVy

MAX

MAXM

,001

0,/01

0,0sin

0

0

We want to reach the Ox line, in minimum time T, ( y (T)=0 ), and with angle γ (T)=0. Start without considering the constraint of minimal height hmin. The system equations are:

The Hamiltonian of the Optimal Problem is : 321 sin uzVH MAXM

0

cos

0

3

12

1

t

H

zVy

H

y

H

MAXMMAX

133

11

Ttt

JTt

constTt

TtJuu

minmin

99




Since H is not an explicit function of time, we have H =constant,

Therefore:

1sin 21 uzVH MAXM

0coscos 211

2

uuzVuzVtd

HdMAXMMAX

zMAXMAXM

constutttu

or

uzconstzttt

utd

HdMAX

21

2212

2

0

01

200

0

The optimum is obtained using: 1sinminargminargminarg 21* uzVHJu MAXM

uuu

212

22*

00

0

ttt

signu

100




The trajectories that ends at the origin are given by integrating the equations of state assuming optimal control, and are given by:

212

22*

00

0

ttt

signu

freeTTttt

Tzuuz

TygivenyyzVy

MAX

MAXM

,001

0,/01

0,0sin

0

0

1&00cos1

1&00cos1

*

*

uTV

ty

uTV

ty

MAX

MII

MAX

MI

signusignV

tyMAX

MIII *

, cos1

or

Extremal (non necessarlly Optimal) Trajectories

101




Singular Trajectories:

21

2

2

2

0&0

0&0

0&0

ttt

ttI

ttI

ttI

nn

uzVtHtI

MAXM 21 sin1

For Singular Trajectories to occur for t1<t<t2, the following conditions must be satisfied:

0cos

0cos

0

0sin1

12

1

2

1

MAXMMAX

zMAXMAXM

MAXM

zVt

uzVtI

t

zVtI

2

MAXz

MV/11 0u

102



The Optimal Trajectories are function of the Initial Conditions y0, γ0. If the constraint of minimal convergence altitude hmin is disregarded, the Optimal Trajectories are given by:• Turn toward the x axis in Ox direction with maximum turn rate.• If γ = +π/2 or - π/2 and the trajectory is toward Ox and the distance is y > R (Turning Radius), than the trajectory will be the straight line normal to Ox. When y = R turn, with maximum turn rate toward Ox direction, on yI,II (t), to reach y (T) = 0 and γ (T) = 0.• If during the first turn we get close to Ox line before reaching γ = +π/2 or - π/2 , we reverse the maximum turn, on yI,II (t), to reach y (T) = 0 and γ (T) = 0.

2/2/2/

cos1,*

signRysignyysignu IIII

cos1,

*

signRy

signu

III

Converging toy=0. γ=0

II

cos1

2/

0*

signRy

u

On Singular ArcsIII

There are three classes of optimal paths defined by:

103



MAX

0

u sin MVyy

0y

sinSystem Equations

cos12

MAX

M

a

Vsign

0

&2

OptimalController

0u

1

1

MAX


Block Diagram of the Optimal Law that assure convergence in minimum time of the Interceptor Trajectory to the Ballistic Target Trajectory in a Head On situation.

104



• Define the Interceptor Midcourse Guidance Law as the combination of the Optimal Trajectory that reaches the Sliding Lyapunov Surface followed by the Linear Guidance Law to reach the Ballistic Trajectory in Head On.

• Start with the Optimal Law toward the Sliding Surface until reaching the Captive Volume Ω1.

• Switch to Linear Guidance Law that keeps the trajectory inside the Captive Zone Ω1 and converges to the origin y=0 and γ = 0, to the Head On with the Ballistic Target.

105



MAX

0

u sin MVy

y

0y

sinSystem Equations

cos12

MAX

M

a

Vsign

0

&2

OptimalController

0u

1

1

MAX

Ca

Controller

1k2kMVk3

MaMAXa

MAXa

,&

y

aa MAXC

MAXMV 1

Block Diagram of the Guidance Law that starts with the Optimal Law that assure convergence in minimum time to the Capture Zone, followed by the Linear Law that assre convergence to the Ballistic Target Trajectory in a Head On situation.

106



Minimal Altitude hmin of the Trajectory

From the Figure we can see that for y0>0 and γ0>0and for y0<0 and γ0<0There may be situation when the minimum time trajectory is not feasible and we must change it.

107




00 coscos RRhh

00min cos1 Rhhh

From the Figure

Therefore

We want to find hmin as function ofy0, γ0

RRyRRdd

dRl

lh

0001

01

0

cos/cos/

sin

cos

000000 sincossincossin Ryh

1sincossincossin 000000min Ryh

108



MV

Ma

minh

Minimum Time TrajectoryAbove Minimum Height hmin

M

Ballistic TrajectoryApproximation

d

h

h

Ix1

Ix1

Convergence Point to the Trajectory

n

P


A solution to the altitude problem is to choose on the Ballistic Trajectory the pointof the minimum altitude hmin as the point below which convergence of the Intercept Trajectory is not acceptable. By doing this if exists a Singular Arc it will be at |γ| < π/2

109



MV

Ma

minh


M


d

h

h

Ix1

Ix1

Convergence Point to the Trajectory

n

P

• Transfer the Planar Guidance Law to a Three Dimensional Law with the same structure.• Simulate the result using a real 6 DOF Interceptor Model.

Define:

M – the Interception PositionP - the point at the Ballistic Missile Trajectory at altitude hmin.

Ix1 – unit vector in the Straight Line Ballistic Trajectory Approximation, pointing up.

h

– the vector pointing from P to M.

d

– the vector distance from Straight Line Ballistic Trajectory to M, pointing to M.

dhxf I

:1

MV

– Interceptor Velocity at M.

– angle between to .MV

Ix1

– angle between to .hMV

h

hV

hV

M

Mh

1sin

M

IM

V

xV 1sin 1

110



Minimum Time Trajectory

d

MVh

Ma

minh


M


• Transfer the Planar Guidance Law to a Three Dimensional Law with the same structure.• Simulate the result using a real 6 DOF Interceptor Model.

MAXC

MIM

MIMM

MM

MMC aa

VxV

VxVVk

dVV

dVVdkdka

1

1321

MAXCMM

h

h

MMM

MMhMC aahVV

hV

k

hVV

hVVVka

sin

321

32 /sin

cos11:

kkk

dVkk

Rd

SWSWSWM

SWSW

When the denominators are close to zero, they will be bounded above zero, to prevent numerical problems.

Guidance Law B: Converges to the Ballistic Trajectory in H.O. Scenario

Guidance Law A: Reaches the Ballistic Trajectory above Altitude hmin

SW

Guidance Law BMinimum Time TrajectoryAbove Minimum Height hmin

Ballistic TrajectoryApproximation SW

SWSW Rd cos1

RR

Guidance Law A

When the Interceptor distance to the Ballistic trajectory is less than dSW (defined bellow) we switch to Guidance Law B.



x

yI

z

xB yB

zB

Assume a Divert Attitude Control System (DACS).The Divert Control Thrusts are located near the Mass Center of the Vehicle, and are aligned with the two axes perpendicular to the longitudinal axis of the Kill Vehicle, so as to generate the Pure but Arbitrary Divert Motion (axB, ayB, azB), where Attitude Control Thrusts are located and aligned such that only Three Pure Rotational Moments about the principal axes are produced (TxB,TyB,TzB). All those Thrusts are Pulse Type, i.e., they only have ON/OFF states with fixed amplitude.

Thrust Command

Thrust Output

111

See “Kill Vehicle Guidance & Control Using Sliding Mode” Presentation for more details

The goal of the Control System is to track a design quaternion and corresponding angular velocity .

dqd

112

SOLO

Desired Attitude and Angular Rate of the Kill Vehicle (KV)

x

y

z

I

xB

yB

zB

DesiredDirection

ICGV

ICGa

d1

dddd

dddd

dddd

dddddddddddd

dddddddddddd

dddddddddddd

Bd

Bd

TBdI

Bd

BdI

Bd

I

zI

yI

xI

qqqq

qqqq

qqqq

qqqqqqqqqqqq

qqqqqqqqqqqq

qqqqqqqqqqqq

xCxCd

d

d

d

3120

3021

23

22

21

20

23

22

21

2010323120

32102

32

22

12

03021

203121302

32

22

12

0

2

2

0

0

1

22

22

22

111

1

1

1

Kill Vehicle (KV)Control

Since the KV roll is free, let choose q1d = 0

We obtain:0

30

20 2

1,

2

1,

2

11

q

dq

q

dq

dq yI

dzI

dxI

d

The Main KV Engine is aligned to xB direction of the KV.The KV Divert Thrusters act normal to to the xB direction. Suppose that we want that the KV xB direction shall follow a given direction and it’s inertial derivative . The rotation position of the KV is free.The Desired KV Attitude is defined as Bd. The following relation must be satisfied:

d1I

dtd

d1

113

SOLO



We found the quaternion from inertia (I) to Desired Body (Bd) Attitude:

03

0210 2

1,

2

1,0,

2

11

q

dq

q

dqq

dq yI

dzI

ddxI

d

Taking the derivatives:

td

qd

q

dd

td

d

qtd

qdq

td

qd

q

dd

td

d

qtd

qdq

td

qdqd

td

d

dtd

qdq d

d

yIyI

d

dd

d

d

zIzI

d

dd

ddxI

xI

dd

02

00

33

02

00

22

11

00

2

11

2

1,

2

11

2

1,0,1

112

1

d

d

d

d

BdI

q

q

q

q

q

3

2

1

0

where:

Using those results we can find:

d

d

d

d

dddd

dddd

ddddI

IBd

q

q

q

q

qqqq

qqqq

qqqq

3

2

1

0

0123

1032

2301

2

td

d

td

qd

Iqqdt

dq

d

d

dxddBdI

BdI

TIIBd

0

33022

SOLO Kill Vehicle (KV)ControlControl System of a Kill Vehicle

x

yI

z

xB yB

zB

114



115


SOLO

120

23

22

21

20

04210

T

BIIB

qqqqq

qqkqjqiqqq

Coordinate Systems

x

yI

z

xB

yB

zB

xBTyBT

zBT

Inertial Coordinates (x,y,z), and Body Coordinates at the Body Center of Gravity (xB, yB, zB)

The Rotation Matrix from Inertial Coordinates (x,y,z), to Body Coordinates at the Body Center of Gravity (xB, yB, zB), CI

B, is defined via the Quaternions:

1 kjikkjjii

kijji

ijkkj

jkiik

i

j

k

IB

BI

BI

BI

BI

BIIB

qqqqqqqqq

qqkqjqiqqq

*123

22

21

20

*

04210

**

1

Quaternion Complex Conjugate: kkjjii

***

1ˆˆ2/sinˆ2/cos0 nnnq T

116


117

SOLO Kill Vehicle (KV) Control

Product of Quaternions

1 kjikkjjii

kijji

ijkkj

jkiik

BAABBA

BABA

B

B

A

ABA qq

qqqqqq

00

0000

A

A

BxBB

TBB

B

B

AxAA

TAA

B

B

A

ABA

q

Iq

qq

Iq

qqqqq

0

330

00

330

000

3210321000

BBBBAAAAB

B

A

ABA qkqjqiqqkqjqiq

qqqq

Using this definition the Product of two Quaternions is given by:BA qq

Let define:

or in Matrix Product Form:


118

SOLO

Product of Quaternions

Let compute:

1

0

1

0

00

1

20

001

AAAAAA

AT

AA

A

A

A

AAA

qq

q

qqqq


Kill Vehicle (KV) Control

SOLO

Coordinate Transformations

AA

BA

ABAB

B

vqI

q

v

qqvq

vv 22

000

0

00*

Given the vector described in (A) coordinates by , and in (B) coordinates byv Av

Bv

2

32

22

12

010323120

32102

32

22

12

03021

203121302

32

22

12

0

0

22

22

22

22

qqqqqqqqqqqq

qqqqqqqqqqqq

qqqqqqqqqqqq

qIC BA

q

x

T

q

xB

A

Iq

Iq

qqq

qqq

qqq

qqq

qqqq

qqqq

qqqq

CT

330

330

012

103

230

321

0123

1032

2301

119

The Rotation Matrix from A to B, CAB, is defined as:

Equations of Motion of a KV (Attitude) Kill Vehicle (KV) Control

SOLO

Differential Equation of the Quaternions

tqtttqtqtd

d BA

AAB

BAB

BA

BA

2

1

2

1

AAB

AAB

BAB

BAB

AAB

BAB

qqdt

ddt

qd

00

0

2

1

2

12

1

2

1

BAB

- Angular Rotation Rate Vector from (A) to (B) in (B) Coordinates

120

BAq - Quaternion defining rotation

from (A) to (B)

AAB

- Angular Rotation Rate Vector from (A) to (B) in (A) Coordinates

tqttqt BA

BAB

BA

AAB



SOLO

Differential Equation of the Quaternions

AAB

q

x

T

BAB

q

x

T

BA

BA

BA

IqIq

qdt

d

330330

2

1

2

1

0

0

0

:

xByB

xBzB

yBzB

BAB

zB

yB

xB

BAB

This can be rewritten as:

121

AAB

AAB

BAB

BAB qq

dt

d

00 2

1

2

1

AAB

BABdt

qd

2

1

2

10

tqtttqtqtd

d BA

AAB

BAB

BA

BA

2

1

2

1



SOLO

Differential Equation of the Quaternions B

AB

- Angular Rotation Rate Vector from (A) to (B) in (B) Coordinates

BAB

BA

BA

BAB

BA qqq

dt

d

2

1

2

1

We have:

330330

330

&: xT

xBA

T

x

T

BA IqIqq

Iq

q

3333

1

xxTT IIqqqq

1

2 Tx

Tx

TT qqIqqIqqqq 4444

130 xT qq 3

ABT

BAT qqqq 4

122

See Development



123

Equations of Motion

SOLO

zz

yy

xx

z

y

x

aV

aV

aV

Vz

Vy

Vx

Given a Rigid Body with Center of Gravity Position , Velocity ,

Acceleration Commands

z

y

x

R ICG

z

y

x

ICG

V

V

V

V

z

y

x

ICG

a

a

a

a

Translational Motion

Rotational Motion

BCG

BIB

BCG

BIB

BIB

B

CGB

IBB

CG TJJJ

Inertial Coordinates (x,y,z), and Body Coordinates at the Body Center of Gravity (xB, yB, zB)

0

0

0

:

xByB

xBzB

yBzB

BIB

zB

yB

xB

BIB

zB

yB

xB

BCG

T

T

T

T

Torque Commands

x

yI

z

xB

yB

zB

xBTyBT

zBT



Attitude Control of the Kill Vehicle (KV)


The Rotational Errors are defined by:

1.The Error Quaternion is defined as the desired rotation from the Body (B)Present Position to the Body Desired (Bd) Position

0qq

0q

q BI

d

dBdI

qq

0

2. The Angular Rate Error is defined as the difference between Desired Rotation Rate and the Actual KV Rotation Rate

IBIBd

IBIBdBBd

:

The goal of the Control System is to track a design quaternion and corresponding angular velocity , that are related by .

dq

d

ddd qq 2

1

x

yI

z

xB

yB

zB

xBTyBT

zBT

124

Fu, L-C et al, Control System of a Kill Vehicle

Equations of Motion


The Error Quaternion The Error Quaternion is defined as the desired rotation from the Present Position to the Desired Position

,0qq

,0qq B

I ddBdI qq

,0

ddd

dd

d

dBdI

BI qq

qqqqqq

qq

00

0000*0

BdI

BI

q

BdB

BI

BdI

BdB

BI

IBI

BdB

BdI

IBdI

Bd qqqqqqqqvqqqvqv ****

BdBdIB

BdIBd

BdIBd

BdI

BI

BdI

BIB

BdI

BdI

BI

BdIBd

BdI

BI

BdI

BIB

BI

BdI

BI

BdI

BI

qqqqqqqq

qqqqqtd

dqqq

td

dq

td

d

BIB

2

1

2

1

2

1

2

1

2

1

2

1

**

1

**

**

**

BI

BdI

TBdI

BI

T

d

dx

BdI

TBI qqqq

qIqqqq

0

3300 &


dq

d

ddd qq 2

1

125

Equations of Motion


The Error Quaternion The Error Quaternion is defined as the desired rotation from the Present Position to the Desired Position

,0qq

,0qq B

I ddBdI qq

,0

BdBd qqqtd

d

2

1

2

1


dq

d

ddd qq 2

1

126

BBBdB

BBdB

BdB

qq

qqqqqqtd

d

Bd

BdB

2

1

2

1

2

1 *

tqttqt BA

AAB

BA

BAB

since

BB

B

qdt

ddt

qd

02

12

1

330

:

x

T

BA

Iq

q

330

:

x

T

BA

Iq

q



We want to Rotate the KV in the prescribed direction

0,,: 321 ia ppppdiagPPS Define the Auxiliary Error

The Rotational Errors are defined by:

1.The Error Quaternion is defined as the desired rotation from the

present position to the desired position

0qq

0q

q BI

d

dBdI

qq

0

2. The Angular Rate Error is defined as the difference between Desired Rotation Rate and the Actual KV Rotation Rate

IB

IBd

IBIBdBBd

:

The Control Task is to bring Sa to zero in order to obtain the Desired Attitude with the Desired Angular Rate.

Define the Lyapunov Function:

aT

aaa SSSV

2

1: x

yI

z

xB

yB

zB

xBTyBT

zBT

127

BdI

BI qqq

*



We want to Rotate the KV in the prescribed direction

0,, 321 ia ppppdiagPPS

Define the Auxiliary Error

Define the Lyapunov Function: 02

1: a

Taaa SSSV

PSSSSV

td

d Taa

Taaa

BBqdt

d

02

1

BCG

B

CGB

IBB

CGB

IB

B

CGB

IB

B

CG

B

CGB

IB TJJJJJ 111

Define the Angular Rate Error IBIBdBBd

:

We have:

Therefore:

x

yI

z

xB

yB

zB

xBTyBT

zBT

128

BCG

B

CGB

IBB

CGB

IB

B

CGB

IB

B

CG

B

CGB

IBdBBT

a

Taaa

TJJJJJqPS

PSSVtd

d

11102

1



BCG

B

CGT

a

BCG

B

CGB

IBB

CGB

IB

B

CGB

IB

B

CG

B

CGB

IBdBBT

a

Taaa

TJDS

TJJJJJqPS

PSSVtd

d

1

11102

1

To obtain System Stability let choose aB

CGaB

CG SJWT

*sgn

where Wa > 0 is a design parameter and

3* ,,:sgn RSSSS

Ssign

Ssign

Ssign

S Tazayaxa

az

ay

ax

a

x

yI

z

xB

yB

zB

xBTyBT

zBT

129

BIBd

BIB

BCG

BIB

B

CGB

IB

B

CG

B

CGBB JJJJqPD

1102

1:



aaaaa

aT

aaT

aB

CG

B

CGT

aT

aaa

WDSSWDS

SSWDSTJDSPSSVtd

d

11111

*1 sgn

To obtain System Stability let choose aB

CGaB

CG SJWT

*sgn

If we choose Wa > ||D||1+ς where ς > 0 (the Thrust Pulse Amplitude is high enough) we have:

01 aaa SSV

td

d

This result implies that Sa is bounded and will converge to zero in finite time.

PPSa 0

x

yI

z

xB

yB

zB

xBTyBT

zBT

130

azayaxaTaa SSSSSS

sgn

1where

131



Provided that Sa =0 the System Dynamics is constrained by the following differential equation:

PPSa 0

PPqq

dt

d BB 00 2

1

2

1

Pdt

qd TBT

2

1

2

10

Define Another Lyapunov Function Candidate: T

eV :

PPqPPqtd

dV

td

d TTTTe

0

002

Since: 11 02

0 qCq T

PPqVtd

dPC TT

eT 0

Hence will converge to zero exponentially, and so will because 0 PSa

x

yI

z

xB

yB

zB

xBTyBT

zBT


Zero-Sliding Guidance Law of the Kill Vehicle (KV)


x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

Let develop the Guidance Law to intercept a Target with Velocity and Acceleration TzTyTxT

IT

TzTyTxT

IT

TTTT

IT aaaaVVVVzyxR ,,,,,,,,

IT

IT

IT

IT

aV

VR

Define the Relative Quantities

ICG

IT

ICG

IT

RI

CGI

TI

CGI

T

ICG

IT

aaRRR

VRRtRRRRRRRVVRRR

RRRRR

:

11111:

1:

2

111:R

RRRR

R

R

R

RRRRRRtRRRVR

2

322

R

RR

R RR

RRR

R

RRRRRRRRRRV

td

d

RR

RRIRRRRRR

RR

RRRR

R

RRR

T

VR

22

2

422

1

132



Let develop the Guidance Law to intercept a Target with Velocity and Acceleration

2

111:R

RRRR

R

R

R

RRRRRRtRRRVR

uRaRVRaRaRVR

RRVRRR

RRIRRRRRR

RV

R

RRRV

td

d

ITR

auI

CGI

TR

R

R

T

VR

RR

ICG

R

BBABBA

BA

BA

,,

,1

:

2

,

22

42

133

tRtRtd

dRRRtRtRRRRR

R

RRRRRRtRtRtRRRRR

RRRRRRR

VR

RRRVR

RR

R

RR

11121112

111121

1,

22

32222222

22

42

A

22

22

22

22222

11

100

010

001

yxzyzx

zyzxyx

zxyxzy

zyx

z

y

x

zyx

T

RRRRRR

RRRRRR

RRRRRR

RRRR

R

R

R

RRRR

RRIR

B



Let define the input as

uVa

Vif

VifV

V

RaVaRuR RMAXM

R

R

R

R

MAXRMAX

sgn*

00

0sgn* BBB

01since2

nRRVRandV

V

R

RRIuRuuR R

R

RT

BB

x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

134



uRaRVRVtd

d ITRR

BBA ,

with the input

00

0sgn*

R

R

R

R

MAXRMAXM

zB

yB

xB

Vif

VifV

V

aVa

a

a

a

u

Define the Lyapunov Function

002

1

2

1:

2

RRR

T

RR VVVVVV

RMAXMI

TR

T

RR

T

RR VaaRVRVVtd

dVVV

td

d sgn*, BA

RR

T

R

T

R

RVT

R

T

R

R

T

R

RV

R

T

RR

T

R

VVVVR

RRIVRV

VVR

RRRRRRRR

RV

R

RRVVRV

TR

TR

sgn*

1,

0

2

2

02

2

42

B

A

RMAXI

T

T

RRR

T

RR VaaVVR

RRV

td

dVVV

td

d 2

2

x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

135



uRaRVRVtd

d ITRR

BBA ,

with the input

00

0sgn*

R

R

R

R

MAXMRMAXM

Vif

VifV

V

aVau

The Lyapunov Function

002

1

2

1:

2

RRR

T

RR VVVVVV

10

1

1

0

2

2

2

someandtVR

RRa

V

Vaif

VaVaaVVR

RRV

td

dVVV

td

d

RI

T

R

T

RMAXM

RMAXMRMAXMI

T

T

RRR

T

RR

x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

136

10

1

10

2

12

22

someandtV

R

RRa

V

VaifVaV

td

dR

IT

R

T

RMAXMRMAXMR

Therefore 0 RVR



x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

137

We found

dtaV

Vd

MAXM

R

R

2

2

2

0

0

0

ttaVV sMAXMt

Rt

Rs

MAXMt

RMAXM

s a

RV

att

00

00

10

10

1

10

2

12

22

someandtV

R

RRa

V

VaifVaV

td

dR

IT

R

T

RMAXRMAXR

u

udud

2

Therefore in a Finite Time.0 RVR



138


R1During the Accelerated Phase the Main KV Engine (aligned to xB direction) must point in the direction. In the Coast Phase (Main KV Engine off) we want the KV Divert Thrusters to be normal to direction, therefore again the xB must be aligned to direction.

R1R1

x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

the input of the Zero-Sliding Guidance Law is

00

0sgn*

R

R

R

R

MAXRMAXM

zB

yB

xB

Vif

VifV

V

aVa

a

a

a

u

Let compute:

R

RR

R

R

RR

R

R

R

R

R

R

td

dd

td

d

I

1

12

Using the defined and we can compute the desired Attitude quaternionand the Desired Body Angular Rate by the following procedure:

I

dtd

d1d1

BdIq

IIBd

R

RRd

11

139

SOLO

Desired Attitude and Angular Rate of the Kill Vehicle (KV)We found the quaternion from inertia (I) to Desired Body (Bd) Attitude:

03

0210 2

1,

2

1,0,

2

11

q

dq

q

dqq

dq yI

dzI

ddxI

d


d

d

d

d

BdI

q

q

q

q

q

3

2

1

0

where:


Slide Mode Control (SMC)

td

qd

q

dd

td

d

qtd

qdq

td

qd

q

dd

td

d

qtd

qdq

td

qdqd

td

d

dtd

qdq d

d

yIyI

d

dd

d

d

zIzI

d

dd

ddxI

xI

dd

02

00

33

02

00

22

11

00

2

11

2

1,

2

11

2

1,0,1

112

1


d

d

d

d

dddd

dddd

ddddI

IBd

q

q

q

q

qqqq

qqqq

qqqq

3

2

1

0

0123

1032

2301

2

td

d

td

qd

Iqqdt

dq

d

d

dxddBdI

BdI

TIIBd

0

33022

Integration of Attitude Control and Guidance Law of the Kill Vehicle (KV)


gT

ggT

aaT

aga SSSSSSSSV

2

1

2

1:,

where

To check the Stabilization of the System Performing Attitude Control and the Guidance Law let choose the following Lyapunov Function

x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

Rga VSPS

:&:

The Torque Command of the Attitude Control is Modified as

gaB

CGaB

CG SSJWT

*sgn

140


Crassidis , et all -Attitude Control of the Kill Vehicle (KV)


Optimal Control Analysis

In order to determine the Optimal Switching Surface we want to optimize the following:

02

1:

pdtqqqqpS TTt

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

Constrained by:

Hamiltonian of the Optimization Problem is given by:

Htd

dBIq

by choosing . tS is the time of arrival at the Sliding Surface.B

IB

where is the co-state 4x1 vector that must satisfy the Euler-Lagrange Equations:


dq

d

ddd qq 2

1

BIB

BI

BI q

dt

qd

2

1

BIB

BI

TBIB

BIBd

TBIB

BIBd

BI

BdI

BdI

TTBI qqqqqpH

2

1

2

1:

141

BI

BdI

TBdI

BI

T

d

dx

BdI

TBI qqqq

qIqqqq

0

3300 &



BIB

TBI

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

BI

TBIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

qqqqqp

qqqqqpH

T

T

2

1

2

1

2

1

2

1:

Euler-Lagrange Equations:


dq

d

ddd qq 2

1

02

1:

pdtqqqqpS TTt

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

Hamiltonian of the Optimization Problem is given by (using ): TB

IBI

T qq

BI

BI

BdI

BdI

T

qqqqpH

td

dBI

2

1

Optimal Control Analysis) (continue – 1)

142



BIB

BI

TBIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI qqqqqpH

T

BIB

BIB

2

1

2

1minmin

02

1*

BI

TT

BIB

BIBd qHB

IB


dq

d

ddd qq 2

1

02

1min:min

pdtqqqqpS TT

BIB

BIB t

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

The minimum is given by the Minimum of the Hamiltonian when changingB

IB

The condition for optimality is:

from which:


143


BI

TBIBd

BIB q 2

1*



BI

TBIBd

q

xB

IBdB

IB qIq

2

1

2

1 0

330

*

Let choose the following Sliding Vector 0:

BI

BdI

TBIBd

BIBa qqkkS

This Sliding Vector is optimal if it minimizes the П functional, that has the optimal

BdI

BI

Tqqqq

BI

BdI

TS

BI

TBIBd

BIB qqkqqkq

BdI

BI

TBI

BdI

Ta

0*

2

1

BdIqk2Since this is true for all we must have Bd

IBI

T qq

From we obtain BIBd

BdI

TBdI qtd

qd

2

1 BIBd

BdI

T qktd

d


dq

d

ddd qq 2

1

02

1min:min

pdtqqqqpS TT

BIB

BIB t

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

144



Optimal Control Analysis (continue – 4)

By choosing the Sliding Vector 0:

BI

BdI

TBIBd

BIBa qqkkS

dBdI

T qktd

d


dq

d

ddd qq 2

1

02

1min:min

pdtqqqqpS TT

BIB

BIB t

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

we obtained:

Solving the Optimization Problem by using Euler-Lagrange we obtained:

B

ITB

IBdBI

BdI

BdI

TBI

BI

BdI

BdI

T

qqqqqpqqqpH

td

dBI 2

1

2

1

2

1 *

BI

TBIBd

BIB q 2

1*

we obtain the following:

B

ITB

IBdBI

BdI

BdI

TBIBd

BdI

T qqqqpqk2

1

2

1 145





dq

d

ddd qq 2

1

02

1min:min

pdtqqqqpS TT

BIB

BIB t

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

we obtain the following relation:

BdI

BI

TBdI

BIBd

BI

BI

BdI

BdI

TBIBd

BdI

T qqqkqkqqqPqk 2

BI

TBIBd

BI

BdI

BdI

TBIBd

BdI

T qqqqpqk 4

1

2

1

BdIqk2Using the relation we obtain:

BI

BdI

TBdI

BIBd

BI

BI

BdI

BdI

Tqqqq

BIBd

BdI

T qqqkqkqqqpqkBI

BdI

TBdI

BI

T

2

or:

we have a identity if: pk


146




dq

d

ddd qq 2

1

02

1min:min

pdtqqqqpS TT

BIB

BIB t

BIB

BIBd

TBIB

BIBd

BI

BdI

TBI

BdI

BIB

we obtained: BI

TBIBd

BIB q 2

1*

BdIqk2Using the relation we obtain:

0*

kSkqqkqqk aBI

BdI

TBdI

BI

TBIBd

BIB

2

2

2*

22

1kpdtkp

St

T

k

Using

Tk

T k

td

qd

22

10

S

tt

T tqqkdtdt

qdkdtk

SS

0

1

002* 22


147




dq

d

ddd qq 2

1

S

tt

T tqkdtdt

qdkdtk

SS

002* 122

Both represent the same orientation. But represents the shortest rotation and smallest П*, and represents the longest rotation and a larger П*.

,, 00 qandq

,0q

,0q

In order to obtain the shortest rotation the following Sliding Vector is chosen:

0sgn 0

BI

BdI

TBdIB

BIBa qqqkS

0sgn qk

It is assumed that is non-zero for a finite time.0q


148




dq

d

ddd qq 2

1

0sgn 0

BI

BdI

TBdIB

BIBa qqqkS

0sgn qk

Tqk

T qk

td

qd0

sgn0 sgn

22

1 0

S

ttt

T tqkdtdt

qdkdtq

dt

qdkdtk

SSS

00

002* 122sgn2

Using the fact that

0

0

0

00

20

00

200 sgn

2

2q

dt

qd

q

q

dt

qd

q

dtqd

q

dt

qd

dt

qd

we obtain:


149




dq

d

ddd qq 2

1

2

00

1

0

sgn0 1sgn

2sgn

22

12

00

qqk

qk

td

qdTq

Tqk

T

S

ttt

T tqkdtdt

qdkdtq

dt

qdkdtk

SSS

00

002* 122sgn2

We can see that

To prove this let use the following Lyapunov Function:


0110 20

20

0

TTqthereforeqwhentd

qd

0002

1

ifonlyVV T

02sgn

2

1

2

1sgn

2

1

2

1

0000

3300

sgn

330

0

Vqkqkqqk

IqqkIqtd

Vd

TT

xT

qk

xTT

Hence V is a Lyapunov Function, for k > 0, and in a Finite Time.0 T

150



151

Missile-Target Kinematics

SOLO

MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

2211 111 ttRS

2

1

1

2

12

2

1

1

1

1

0

0

0

1

1

1

t

t

R

t

t

R

td

d

S

S

Choose a Cartesian System related to the Line of Sight .

21 1,1,1 ttR

(Example: A Real Seeker)

S - Rotation Rate of around

R1

21 1,1 tt

0

0

0

:

1

2

12

S

S



See “HTK Guidance Implementation”Presentation for more details

152

Missile-Target Kinematics

SOLO

MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

2211 111 ttRS

2

1

1

2

12

2

1

1

1

1

0

0

0

1

1

1

t

t

R

t

t

R

td

d

S

S

2121112122 111111 tRtRRtRtRRRRRRR

11121122212212 1111111 tRRtRRtRRtRRRRRRR SS

MTSS aatRRRtRRRRRR

221111222

22

1 12121

MTS

S

MT

MT

aatRRR

tRRRRRRR

VVtRtRRRRRRRR

RRRRR

2211

11222

22

1

2211

12

121

11111

1

(Example: A Real Seeker)

0

0

0

:

1

2

12

S

S

Choose a Cartesian System related to the Line of Sight .

21 1,1,1 ttR

Slide Mode Control (SMC)HTK Guidance Using 2nd Order Sliding Mode

153

True Proportional Navigation (TPN) Guidance

SOLO

MV

MV

TV

RRR 1

n1

t1

R1

RR 1

R

TaMa

MR TR

R

Missile

Target

I

taRNta TM 1'1

RNRR

RaRR M

'2

122'

00

N

R

R

Differentiating we obtain 3'

000 2'

N

R

R

R

RN

2'NFor we have when0 0R

For we have when3'N 0 0R

For the missile is on collision course even for an accelerating missile0

TPN is sensitive to the knowledge of . We want to design a Robust Guidance LawBased on TPN improved by a Second Order Sliding Mode

taT 1

R

RdN

d2'

Slide Mode Control (SMC)HTK Guidance Using 2nd Order Sliding Mode

SOLO Sliding Mode Control (SMC)

Shtessel, Y., Tournis, C., Shkolnikov, I., “Guidance and Autopilot for Missile Steered by Aerodynamic Lift and Divert Thrusters Using Second Order Sliding Modes”, AIAA 2006-6784, AIAA Guidance, Navigation and Control Conference and Exhibit, 21-24 Aug. 2006, Keystone, Colorado, AIAA 2006-6784

Start from MTR aanRtRRRRRR

11212 For the tangential component we have

tataRR MT 112

tataRRRRdt

dMT 11

The Target acceleration component is unknown and is Estimated

They proposed the Sliding Surface

tataRtd

dR MT 11

The proposed Guidance Law is a Smooth Asymptotic Second Order Sliding Mode Control Algorithm with a Finite Reaching Time (an Improvement of TPN):

2

3/1

2

2/1

11'1

SMEst

TEstEstCommand

M dsignsigntaRNta

The 2nd Order Sliding Mode can be defined by:

The simulations show Robustness to Hit-to-Kill in presence of Uncertainties and Measurement Noises.

0

0,

,

21

3/1

122

121

2/1

111

1

ysigntyty

xttyysigntyty

xtty


155

SOLO Guidance of Intercept

12221

22112

1'

1'

2

1

RSMRtaRNa

RSMRtaRNa

EstSEst

TC

EstSEst

TC

t

t

21 11121

tataatt CCCC

22

21 tt CCC aaACIa - Magnitude of Acceleration Command

tta

at

a

a

C

C

C

CC

tt 1111 2121

- Direction of Acceleration Command

MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

22211

11122

112

112

tataRRR

tataRRR

MTS

MTS

Target-Missile Kinematics

Second Order Sliding Mode HTK Guidance Command in Seeker Coordinates

0&001'

0&001'

111211

222222

Rtd

dRRSMR

R

RNR

td

d

Rtd

dRRSMR

R

RNR

td

d

Ideally this Guidance Law will give.

The Robust 2nd Order SM will suppress disturbances.

dtRsignRRsignRRSM 3/1

2

2/1

12


156

3. HTK Guidance Implementation (continue - 1)SOLO Guidance of Intercept

MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

The 2nd Order Sliding Mode HTK Guidance Law will give:

01'3/1

2

2/1

1 dRsignRRsignRRR

RNR

td

d

Define: ::1Ry

RsignRy

yRsignRRR

RNR

td

dy

3/1

22

2

2/1

11 1'

Let choose the following Lyapunov Function candidate:

000

0004

3

2,

21

21

3/4

12

22

21

yandyifonly

yandyifyy

yyV

1'&0&001', 1

6/5

121

3/4

1221 NRyifyyR

RNyyV

td

d

6/5

121

3/4

121

3/1

12221

2/1

1111

3/1

12

2211

3/1

1222

11

21

1'1'

,

2

1

yyR

RNysigntyyyysignyy

R

RNysigny

yyyysignyyy

Vy

y

VyyV

td

d

yy

Therefore V (y1,y2) → 0 (i.e. ) in a Finite Time 0:,0: 21 Rtd

dyRy

157


To prove that V (y1,y2) → 0 (i.e. ) in a Finite Time tS wemust show that exists a Domain D, that includes the origin, un which

0:,0: 21 ii R

td

dyRy


d

We have:

000

0004

3

2,

21

213

4

12

22

21

yandyifonly

yandyifyy

yyV

001', 1

0sin

3/4

12

6/5

12121

yifyR

RNyyyV

td

d

Rcenegative

td

yyVd

ky

ky

yyyV 216

5

1213

4

12

22

21

,1

4

3

2,

6

5

1213

4

12

10

2

33

4

12

22

21 2

3

4

3

2,

2

1

122

2

yk

yyy

yyVyy

Define the Domain D that includes the origin by: 12

31

3

4

12

2

2 yandyy

1

6

5

1

1

213

4

122

3y

ky

or:

1

21

2

1

6

5

3

4

10 3

2

ky

equality

If since16

51

1

6

5 11

3

4,1

1

6

51 yand

from the Figure, and choosing some k>0, we can see that exists some small |y1s| such that for

we have3

4

102

2

2101 2

3yyandyy 100,, 2121 yyVkyyV

td

d

1yx

2y

11 y

11 y11

y

1y1y

11

111 y

3

4

102

2

20

1

21

2

1

6

5

3

4

10

23

32

yy

ky

10y10y

20y

20y

1

1

1

1111

1111

yyyy

yyyy

1y

2y

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5



Therefore for:


3

4

10221

2

1

3

4

102202101 2

3,

2

3,

0yyyVandyyyyy t

V (y1, y2) → 0 in Finite Time ts .

1001

, 121

00

k

yyVtt t

S

-0.5

0

0.5

-0.5

0

0.50

0.1

0.2

0.3

0.4

0.5

1y2y

21, yyV

1y

2y

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Therefore V (y1,y2) → 0 (i.e. ) in a Finite Time 0:,0: 21 Rtd

dyRy

159


MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

Robustness against Disturbances


011'3/1

2

2/1

1 dRsignRRsignRtaRR

RNR

td

d

eDisturbanc

T

Assume an Uncompensated Disturbance (Difficult toobtain a good Estimation). Then we have:

RsignRy

tayRsignRRR

RNR

td

dy T

3/1

22

2

2/1

11 11'

Let choose the same Lyapunov Function candidate:

6/5

121

3/4

121

3/1

12221

2/1

1111

3/1

12

2211

3/1

1222

11

21

1'11'

,

2

1

yyR

RNysigntyytayysignyy

R

RNysigny

yyyysignyyy

Vy

y

VyyV

td

d

yy

T

tayyR

RNyyV

td

dT 11',

6/5

121

3/4

1221

taT 1

000

0004

3

2,

21

21

3/4

12

22

21

yandyifonly

yandyifyy

yyV

We can see that as long as 0,11' 21

6/5

121

3/4

12 yyVtd

dtayy

R

RN T

but close to y1 → 0 , the sign of depends on . 21, yyVtd

dtaT 1


160


R1During the Accelerated Phase the Main KV Engine (aligned to xB direction) must point in the direction. In the Coast Phase (Main KV Engine off) we want the KV Divert Thrusters to be normal to direction, therefore again the xB must be aligned to direction.

R1R1

x

y

z

I

xB

yB

zB

Target

ITa

ITV

ICGV

ICGa

R

the input of the 2nd Order-Sliding Guidance Law is

Let compute: RRRR

RRRR

Rtd

dd

td

d

I

111

11

11

Using the defined and we can compute the desired Attitude quaternionand the Desired Body Angular Rate by the following procedure:

I

dtd

d1d1

BdIq

IIBd

R

RRd

11

3. HTK Guidance Implementation (continue - 3)

12221

22112

1'

1'

2

1

RSMRtaRNa

RSMRtaRNa

EstSEst

TC

EstSEst

TC

t

t

I

Rtd

dRRRRR

td

dR

td

d111

MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

161

SOLO

Desired Attitude and Angular Rate of the Kill Vehicle (KV)We found the quaternion from inertia (I) to Desired Body (Bd) Attitude:

03

0210 2

1,

2

1,0,

2

11

q

dq

q

dqq

dq yI

dzI

ddxI

d


d

d

d

d

BdI

q

q

q

q

q

3

2

1

0

where:

Slide Mode Control (SMC)3. HTK Guidance Implementation (continue - 4)

td

qd

q

dd

td

d

qtd

qdq

td

qd

q

dd

td

d

qtd

qdq

td

qdqd

td

d

dtd

qdq d

d

yIyI

d

dd

d

d

zIzI

d

dd

ddxI

xI

dd

02

00

33

02

00

22

11

00

2

11

2

1,

2

11

2

1,0,1

112

1


d

d

d

d

dddd

dddd

ddddI

IBd

q

q

q

q

qqqq

qqqq

qqqq

3

2

1

0

0123

1032

2301

2

td

d

td

qd

Iqqdt

dq

d

d

dxddBdI

BdI

TIIBd

0

33022

162


MV

MV

TV

RRR 1

R1

RR 1

R

TaMa

MR TR

21t

11t

2

S1

Missile

Target

MATLAB Simulation

Kinematics:

tataRR MT 112

tataRRtd

dMT 11

EstEstRSM

EstEstEstEstEstEstEstEstEstEstC dRsignRRsignRRNa

2

3/1

2

2/1

1'

MissileMCMCMissile

M taatatd

da

sta

/11

1

11

Missile Guidance & Dynamics:

Range and Range_Rate: Given R0 and Rdot(t)

Estimation:

t

t

dRRtR0

0

Noise

NoiseRRRdot

NoiseRRR

Est

RBiasEST

RBiasEst

163

SOLO %Plotfigureh(1) = subplot(311);plot(t_out,output1);gridylabel('X1') h(2) = subplot(312);plot(t_out,output2);gridylabel('X1_dot') h(3) = subplot(313);plot(t_out,output3);gridylabel('X2')linkaxes(h,'x') figureh(1) = subplot(511);plot(t_o,output11);gridylabel('Lamda_dot') h(2) = subplot(512);plot(t_o,output12);gridylabel('Lamda_dot2') h(3) = subplot(513);plot(t_out,output13);gridylabel('Noise_Lamdadot')linkaxes(h,'x') figureh(1) = subplot(411);plot(t_out,output7);gridylabel('Am') h(2) = subplot(412);plot(t_out,output9);gridylabel('Lamda_dot') h(3) = subplot(413);plot(t_out,output10);gridylabel('Lamda_dot2')linkaxes(h,'x') figureh(1)=subplot(211);plot(output1,output2);gridxlabel('X1')ylabel('X1dot')

%Solution of Sliding Mode HTK% Solo - 7/2011%-------------------------------------------------------% Three State Variables% X1 = R*Lamdadot% X2 = Internal State of Sliding Mode% X3 = Range%---------------------------------------------------------- %Initial Conditionsclear;integ_steps = 1000;Range = 10000;Rdot = -1000;x3 =Range;x3_dot = Rdot;Lamdadot=0.05;N = 3;alpha1 =9;alpha2 = 1;x1 = Range*Lamdadot;x2 = 0;time = 0;time_end=0;tfin = -Range/Rdot;delta_time = tfin/integ_steps;Am=0;Tau_Missile=0.001;x4=0;% Estimated Noises and BaiasesNoise_Range=0;Noise_Rdot=0;Noise_Lamdadot=0;Bias_Range=0;Bias_Rdot=0;Sigma_Ldot=0.002;j = 1;k = 1; for i=1:(integ_steps) %Define Derivates %Disturbance At if time>0&time<0 At=100; else At=0; end

%Estimated Range, Rdot and Lamdadot Range_est=abs(x3+Bias_Range+Noise_Range)+eps; Rdot_est=Rdot+Bias_Rdot+Noise_Rdot; Lamda_dot=x1/(abs(x3)+1); Noise_Lamdadot=normrnd(0.,Sigma_Ldot); Noise_Lamdadot=0; Lamdadot_est=Lamda_dot+Noise_Lamdadot; %Second Order Sliding Mode SigmaSM=Range_est*Lamdadot_est; y2 = alpha1*sign(SigmaSM)*abs(SigmaSM)^0.5+x2; x2_dot =alpha2*sign(SigmaSM)*abs(SigmaSM)^(1/3); %Missile Acceleration Command and Autopilot Ac=-N*Rdot_est*Lamdadot_est+y2; if Tau_Missile<0.005 Am=Ac; x4_dot=0; else Am=x4; x4_dot=(Ac-Am)/Tau_Missile; end %Kinematics Computations x1_dot = -Rdot*x1/(abs(x3)+eps)+At-Am; x3_dot = Rdot; Lamda_dot2=(x1_dot-Lamda_dot*Rdot)/(abs(x3)+1); %Integration (Euler) time = time + delta_time; x1 = x1+ x1_dot*delta_time; x2 = x2+ x2_dot*delta_time; x3 = x3+ x3_dot*delta_time; x4 = x4+ x4_dot*delta_time; % For Plot t_out(j) = time; output1(j) = x1; output2(j) = x1_dot; output3(j) = x2; output4(j) = x2_dot; output5(j) = x3; output6(j) = y2; output7(j) = Am; output8(j) = At; output9(j) = Lamda_dot; output10(j) = Lamda_dot2; output13(j) = Noise_Lamdadot; j = j+1; tgo=tfin - time; if (tgo<1) k=k+1; time_end=time_end+delta_time; t_o(k)=time_end; output11(k) = Lamda_dot; output12(k) = Lamda_dot2; endend

MATLAB Simulation

164


-100 0 100 200 300 400 500-200

-100

0

100

X1

X1d

ot

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

da do

t

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

da do

t2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-200

-100

0

100

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-30

-20

-10

0

X2

alpha1=3, alpha2=1N=3, At=0


0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-30

-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500A

m

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-300

-200

-100

0

100

X1

X1do

t


0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2-100 0 100 200 300 400 500

-400

-200

0

200

X1

X1do

t

MATLAB Simulation Results - 1Increasing alpha1 IncreasesConvergence Time

165


-100 0 100 200 300 400 500-200

-100

0

100

X1

X1do

t

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-200

-100

0

100

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-30

-20

-10

0

X2



-100 0 100 200 300 400 500-200

-100

0

100

X1

X1do

t

0 1 2 3 4 5 6 7 8 9 10-500

0

500A

m

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-200

-100

0

100

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-100

-50

0

X2


-200 -100 0 100 200 300 400 500-200

-100

0

100

X1

X1d

ot

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-200

-100

0

100

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-200

-100

0

100

X2

MATLAB Simulation Results - 2Increasing alpha2 DecreasesConvergence Time

166

SOLO


0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 dot

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-400

-200

0

200

X1

X1do

t

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-2000

-1000

0

1000

X1 dot

0 1 2 3 4 5 6 7 8 9 10-30

-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-2

0

2

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-2000

0

2000

Lam

dadot

2

-100 0 100 200 300 400 500-1500

-1000

-500

0

500

X1

X1do

t




0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-300

-200

-100

0

100

X1

X1do

t

MATLAB Simulation Results - 3Increasing N DecreasesConvergence Time

167

SOLO


0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 dot

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-400

-200

0

200

X1

X1do

t


alpha1=9, alpha2=1N=3, At=50 at 6<t<7

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-400

-200

0

200

X1

X1do

t


0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-400

-200

0

200

X1

X1d

ot

MATLAB Simulation Results - 4Robustness to UncompensatedTarget Acceleration.

168

SOLO


0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

da

dot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

da

dot2

-100 0 100 200 300 400 500-400

-200

0

200

X1

X1dot


0 1 2 3 4 5 6 7 8 9 100

200

400

600

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

X1 do

t

0 1 2 3 4 5 6 7 8 9 10-40

-20

0

X2


0 1 2 3 4 5 6 7 8 9 10-400

-200

0

Am

0 1 2 3 4 5 6 7 8 9 100

0.05

Lam

da

dot

0 1 2 3 4 5 6 7 8 9 100

50

Lam

da

dot2

0 50 100 150 200 250 300 350 400 450 500-400

-300

-200

-100

0

X1

X1dot

MATLAB Simulation Results - 5

169

SOLO

alpha1=9, alpha2=1R0=10000, N=3, At=0

0 1 2 3 4 5 6 7 8 9 10-500

0

500

X1

0 1 2 3 4 5 6 7 8 9 10-400

-200

0

200

X1 dot

0 1 2 3 4 5 6 7 8 9 10-20

-10

0

X2

0 1 2 3 4 5 6 7 8 9 10-500

0

500

Am

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

Lam

dadot

0 1 2 3 4 5 6 7 8 9 10-50

0

50

Lam

dadot

2

-100 0 100 200 300 400 500-400

-200

0

200

X1

X1do

t


alpha1=20, alpha2=1R0=2000, N=3, At=0

0 0.5 1 1.5 2 2.5-50

0

50

100

X1

0 0.5 1 1.5 2 2.5-200

-100

0

100

X1 do

t

0 0.5 1 1.5 2 2.5-3

-2

-1

0

X2

0 0.5 1 1.5 2 2.5-500

0

500

Am

0 0.5 1 1.5 2 2.5-0.05

0

0.05

Lam

dadot

0 0.5 1 1.5 2 2.5-50

0

50

Lam

dadot

2

-20 0 20 40 60 80 100-200

-100

0

100

X1

X1d

ot

alpha1=30, alpha2=1R0=1000, N=3, At=0

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-400

-200

0

200

X1 do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-1.5

-1

-0.5

0

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.05

0

0.05

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

Lam

dadot

2

-10 0 10 20 30 40 50-300

-200

-100

0

100

X1

X1d

ot


EstEstRSM

EstEstEstEstEstEstEstEstEstEstC dRsignRRsignRRNa

2

3/1

2

2/1

1'

Alpha1 must be increased when R0 decreases

170


alpha1=30, alpha2=1R0=1000, N=3, At=0

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-400

-200

0

200

X1 do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-1.5

-1

-0.5

0

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.05

0

0.05

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

Lam

dadot

2

-10 0 10 20 30 40 50-300

-200

-100

0

100

X1

X1d

ot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-400

-200

0

200

X1 do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-2

-1

0

X2

alpha1=30, alpha2=1R0=1000, N=3, At=50,0<t<0.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500A

m

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.05

0

0.05

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

Lam

dadot

2

-10 0 10 20 30 40 50-300

-200

-100

0

100

X1

X1d

ot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-400

-200

0

200

X1 dot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-2

-1

0

X2

alpha1=30, alpha2=1R0=1000, N=3, At=50,0.5<t<1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-5000

0

5000

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-5

0

5

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-5000

0

5000

Lam

dadot

2-10 0 10 20 30 40 50

-300

-200

-100

0

100

X1X

1dot


171


alpha1=30, alpha2=1R0=1000, N=3,

Lamdadot_Noise = nrmrand(0.,0.002)

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

X1 dot

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.5

1

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-1000

0

1000

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.2

0

0.2

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lam

dadot

2

-10 0 10 20 30 40 50-500

0

500

X1

X1do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.2

0

0.2

Lamda

dot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lamda

dot2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.01

0

0.01

Noise

Lamdad

ot

alpha1=30, alpha2=1R0=1000, N=3,

Lamdadot_Noise = 0

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-400

-200

0

200

X1 do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.5

1

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.1

0

0.1

Lam

da do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lam

da do

t2

-10 0 10 20 30 40 50-400

-200

0

200

X1

X1dot


172


alpha1=30, alpha2=1R0=1000, N=3,

Lamdadot_Noise = nrmrand(0.,0.002)

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

X1 dot

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.5

1

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-1000

0

1000

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.2

0

0.2

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lam

dadot

2

-10 0 10 20 30 40 50-500

0

500

X1

X1do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.2

0

0.2

Lamda

dot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lamda

dot2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.01

0

0.01

Noise

Lamdad

ot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

X1 do

t

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.5

1

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.2

0

0.2

Lamd

adot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lamd

adot2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.05

0

0.05

Noise

Lamda

dot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-1000

0

1000

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.2

0

0.2

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

Lam

dadot

2

-10 0 10 20 30 40 50-500

0

500

X1

X1d

ot

Alpha1=30,alpha2=1R0=1000,n=3

Lamdadot_Noise=nrmrand(0.,0.010)


173

Guidance of InterceptAlpha1=30,alpha2=1

R0=1000,n=3Lamdadot_Noise=nrmrand(0.,0.010)

At=100, 0< t< 7

0 0.2 0.4 0.6 0.8 1 1.2 1.4-50

0

50

X1

0 0.2 0.4 0.6 0.8 1 1.2 1.4-500

0

500

X1 dot

0 0.2 0.4 0.6 0.8 1 1.2 1.40

1

2

X2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.1

0

0.1

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-200

0

200

Lam

dadot

2

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.05

0

0.05

Nois

eLam

dado

t

0 0.2 0.4 0.6 0.8 1 1.2 1.4-1000

0

1000

Am

0 0.2 0.4 0.6 0.8 1 1.2 1.4-0.1

0

0.1

Lam

dadot

0 0.2 0.4 0.6 0.8 1 1.2 1.4-200

0

200

Lam

dadot

2

-10 0 10 20 30 40 50-500

0

500

X1

X1do

t


174

Guidance of InterceptAlpha1=30,alpha2=1

R0=1000,n=3Lamdadot_Noise=nrmrand(0.,0.002)

At=100, 0< t< 7


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

100

X1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

X1 do

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

X2

Tau_Missile = 0.05

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-500

0

500

Am

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

0

0.1

Lam

dadot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

0

100

Lam

dadot

2

-10 0 10 20 30 40 50 60-400

-200

0

200

X1

X1do

t

Tau_Missile = 0

-10 0 10 20 30 40 50 60-300

-200

-100

0

100

X1

X1do

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

100

X1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

X1 dot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

X2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-500

0

500

Am

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

0

0.1

Lam

dadot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

Lam

dadot

2

Tau_Missile = 0.20

-50 0 50 100 150 200-2

0

2

4x 10

4

X1

X1d

ot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

0

200

X1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5x 10

4

X1 do

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

X2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5000

0

5000

Am

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

Lam

dadot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5x 10

4

Lam

dadot

2


References


Fu, L-C, Tsai, C-W, Yeh, F-K, “A Nonlinear Missile Guidance Controller with Pulse Type Input Device”, Proc. Of the American Control Conference, San Diego, CA, June 1999, pp. 3753-3757

Yeh, F-K , Chien, H-H, Fu, L-C, “Design of Optimal Midcourse Guidance Sliding Mode Control for Missile with TVC”, IEEE Trans. On Aerospace & Electronic Systems, Vol. 39, No. 3, July 2003, pp. 824-837

Crassidis, J.L., Vadali, S.,R., Markley, F.L., “Optimal Variable-Structure Control Tracking of Spacecraft Maneuvers”, Proceedings of the Flight Mechanics Symposium, NASA Goddard Space Flight Center, Greenbel, MD, May 1999


Vecchio Claudio, “Sliding Mode Control: Theoretical Development and Applications to Uncertain Mrchanical Systems”, PhD Thesis, University of Pavia, 2008

Kolemen, E, Kasdin, N.,J., “Advanced Guidance, Navigation, and Control (GNC) Algorithm Development to Enhance Lethality of Interceptors Against Maneuvering Targets”, Radiance Technologies, Inc., MDA Phase SBIR Final Report, 5 Jan. 2005 175

References (continue)


Shtessel, Y., Tournis, C., Shkolnikov, I., “Guidance and Autopilot for Missile Steered by Aerodynamic Lift and Divert Thrusters Using Second Order Sliding Modes”, AIAA 2006-6784, AIAA Guidance, Navigation and Control Conference and Exhibit, 21-24 Aug. 2006, Keystone, Colorado

Shkolnikov, I., Shtessel, Y., Lianos, D., “Integrated Guidance-Control System of a Homing Interceptor: Sliding Mode Approach”, AIAA 2001-4218, AIAA Guidance, Navigation and Control Conference and Exhibit, 6-9 Aug. 2001, Montreal, Canada

Tournis, C., Shtessel, Y., Shkolnikov, I., “Autopilot for Missile Steered by Aerodynamic Lift and Divert Thrusters Using Nonlinear Dynamic Sliding Modes”, AIAA 2005-6382, AIAA Guidance, Navigation and Control Conference and Exhibit, 15-18 Aug. 2005, San Francisco, California

Shima, T., Idan, M., Golan, O., “Sliding Mode Control for Integrated Missile Autopilot-Guidance”, AIAA 2004-4884, AIAA Guidance, Navigation and Control Conference and Exhibit, 16-19 Aug. 2004, Providence, Rhode Island

176

References (continue)


Levant, A., Pridor, A., Gitizadeh, R., Yaesh, I., Ben-Asher, J., Z., “Aircraft Pitch Control Via Second Orfer Sliding Techniques”, AIAA 2001-4218, AIAA Guidance, Navigation and Control Conference and Exhibit, 6-9 Aug. 2001, Montreal, Canada

177

Hermelin, S., “HTK Guidance Implementation”,

Hermelin, S., “Kill Vehicle Guidance & Control”,

Hermelin, S., “Notes on Rotation”,

178

SOLO

Given:

00

00

RtRaRRtd

d

RtRRRtd

d

Integrating those equation:

t

T

t

T

dttaTRdttRTRtR 1111

t

T

t

T

dtdttatTTRTRtR2

211

where t = T is given and must be known TRTR

,

Define the Zero-Effort- Miss:

t

T

t

T

t

T

dtdttadttatTTRtTtRtRTtZ2

21111:,

The time t = T is not necessarily equal to tmin (the time to reach the Rmin)

Note

MISSILE-TARGET KINEMATICS AND ZERO EFFORT MISS

GUIDANCE OF INTERCEPT

179

2. If the missile is in Collision-Course with the target for that t. 0,

TtZ

1. If no accelerations are applied than gives the miss-distance (Zero-Effort Miss)

0 tata MT

TRTZ

SOLO

We have:

tRtTtatTdttatatTdttaTtZt

T

t

T

1111,

TRTZ

1

2

Integrating equation we obtain:1

t

T

dttatTTZTtZ 11,

It is easy to show, integrating by parts, that:

t

T

dttatT 111

11

1

dttadu

tTv

t

T

t

T

t

T

dtdttadttatT2

21111

We can see that:

t

T

t

T

t

T

dtdttadttatTTRtTtRtRTtZ2

21111:,

MISSILE-TARGET KINEMATICS AND ZERO EFFORT MISS (Continue)GUIDANCE OF INTERCEPT

180

UNBOUNDED MISSILE & TARGET ACCELERATION DIRECTIONSBOUNDED MAGNITUDES

SOLO

Optimal Control Approach (following Anderson G.M. – see References)

Given tTaaTtZtd

TtZdMT

,,

Minimize the Cost Function 2/min2/minmin22

TZTRJMMM aaa

Assumptions MAXMMMAXTT aaaa

&

tTaaH MTZ

:The Hamiltonian of the Optimal Problem is

The Euler-Lagrange Equations and the Transversality Conditions are:

TttHZZ 00 TZTJT TZZ

TttTZtZ 0

tTaaTZtH MT

from which

Solution

181

SOLO

Optimal Control Approach (Continue)

Pontryagin Minimum Principle HJMM aa

minmin

tTaaTZH MT

0

0min

TZuniquenon

TZaTZ

TZ

HaMAXM

aM

M

Therefore

From which, for we have 0

TZ

tTaTZ

TZatZ MAXMT

We can see that for the direction of is constant and for aM MAX =const:

constaT

TtZ ,

gogo

t

gogoMAXMT

tTtt

T

MAXMT dttaTZ

TZaTZdttTa

TZ

TZaTZTtZ

0

,

2/min2/minmin22

TZTRJMMM aaa

22

2

1

2

1, tTa

TZ

TZtTaTZTtZ

MAXMT


182


SOLO

2/min2/minmin22

TZTRJMMM aaa


For the cases when 002

1

2

1, 2

02

00 TZtTatTaTtZ MAXMTt

we can use an infinite number of control strategies to reach Z (T) =0.Gerald M. Anderson (see References) suggested the following strategies:

1. Control Law (Optimal Control P1) OCP1 . 0TZ

The simplest way is to apply the maximum missile acceleration aM MAX in direction, to reduce it to zero (reaching the collision course), and then to use to maintain for .

TtZ ,

TM aa

0,&0, TtZTtZ Ttt 1

TttRtTRTtZa

tttRtTRTtZaTtZ

TtZ

a

T

MAXM

M

1

10

,

,,

,

from which we obtain:

constafor

Ttt

ttttTaTtZ

TtZa

TtZ T

MAXMT

1

10

0

,

,

,

constaconstta MAXMT &

Tildocs # 7592302 v1183


SOLO

2/min2/minmin22

TZTRJMMM aaa


TttRRtTTtZ

RtTRTtZ

1

00,

0,

tTtRtRtTtRtRTtZ 11111 0,

11 00, tRtRtRtTtRTtZTttfor

TttRtttRtTtR

tttRtRdttRtRtRTttfort

t

11111

11111

1

We can see that so for t1 ≤ t ≤ T the Interceptor is on a Collision Course. 0

TR

The conditions that the Interceptor is on a Collision Course for t1 ≤ t ≤ T are:

Note

184

SOLO


To find out what are the conditions to reach the Sliding Manifold (Collision Course) let define the Lyapunov Function

Ttif

TtifTtZTtZTtV

0

0,,

2

1,

tTaaTtZ

TtZtT

TtZ

TtZaaTtZ

td

TtZdTtZ

td

TtVdMAXMTMAXMT

,

,

,

,,

,,

,

2/min2/minmin22

TZTRJMMM aaa


Assume that

100,

,

someandTtforaaa

TtZ

TtZMAXMMAXMT

or

10,

,

1

1

1

someandTtfora

TtZ

TtZaa T

MAXTMAXM

then

Ttif

TtiftTa

td

TtVdMAXM 0

0,

This means that we will reach the Sliding Manifold in a finite time t1 ≤ Twhere t1 is the first time at which , and we don’t need to know the Target acceleration.

0,

TtZ 0, 1111

tRtTtRTtZ

185

SOLO


Maintaining the Sliding Surface

2/min2/minmin22

TZTRJMMM aaa


10,,

,tttRtTRTtZa

TtZ

TtZa MAXMM

To reach the Sliding Surface we use

To maintain the Sliding Surface we must choose TttRtTRTtZ 1,

TttRtTRTtZaa TM 1,

Since present target acceleration is unknown we will use target acceleration estimation

taT

errortata TT

TttRtTRTtZaa TM 1,ˆ

In this case and will result after some time to leaving the Sliding Surface. Then we will use again

0ˆ, TTMT aatTaatTTtZ

TttRtTRTtZ 1,

TttRtTRTtZaTtZ

TtZa MAXMM 1,

,

,

The Trajectory will chatter around the Sliding Surface.

References Sliding Mode in Guidance and Autopilot (continue)


Kolemen, E., Kasdin, N., J.,“MDA Phase I SBIR Final Report – Advanced Guidance , Navigation and Control (GNC) Algorithm Development to Enhance Lethality of Interceptors Against Maneuvering Targets”, Radiance Technologies, Inc., 5 January 2005

186

References


Arie Levant Homepage & Publications http://www.tau.ac.il/~levant/

http://www.ece.uah.edu/Individual_Staff_Pages/ShtesselY.html

John L. Crassidis Publications http://www.acsu.buffalo.edu/~johnc/pub.htmJohn L. Crassidis Homepage http://www.acsu.buffalo.edu/~johnc/index.html

Vadim Utkin Homepage http://ece.osu.edu/people/utkin

Leonid G. Fridman Homepage and Papers http://www.depi.itch.edu.mx/lfridman/ http://verona.fi-p.unam.mx/~lfridman/papers-ya.php?

orden=0&tipo=num&sentido=DESC&year=ALL&topic=ALL

Giorgio Bartolini Homepage http://www.diee.unica.it/~giob/infoit.html

187

188

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

189

ROTATIONS

Mathematical Computation of a Rotation

SOLO

AB

C

O

n

v

1v

We saw that every rotation is defined by three parameters:

• Direction of the rotation axis , defined by by two parameters.n

• The angle of rotation , defines the third parameter. Let rotate the vector around by a large angle , toobtain the new vector

OAv n

OBv1

From the drawing we have:

CBACOAOBv1

vOA

cos1ˆˆ

vnnAC Since direction of is: sinˆˆ&ˆˆ vnnvnn

and it’s length is:

AC

cos1sin v

sinˆ vnCB

Since has the direction and the

absolute valueCB

vn

ˆsinsinv

sinˆcos1ˆˆ1 vnvnnvv

190

ROTATIONS

Computation of the Rotation Matrix

SOLO

We have two frames of coordinates A and B defined by the orthogonal unit vectors and AAA zyx ˆ,ˆ,ˆ BBB zyx ˆ,ˆ,ˆ

The frame B can be reached by rotating the A framearound some direction by an angle . n

We want to find the Rotation Matrixthat describes this rotation from A to B.

,ˆ33 nRC xBA

sinˆˆcos1ˆˆˆˆˆ



AAAB

AAAB

AAAB

znznnxz

ynynnxy

xnxnnxx

Let write those equations in matrix form.

0

0

1

sinˆ

0

0

1

cos1ˆˆ

0

0

1

ˆ AAAAB nnnx

0

0

0

ˆ

xy

xz

yz

A

nn

nn

nn

n 0ˆ ntrace

AxAz

Ay

Bz

By

BxO

n

Rotation Matrix

Equations of Motion



IB

- Angular Rotation Rate Vector from (I) to (B) in (B) Coordinates

BIB

BI

BI

BIB

BI qqq

dt

d

2

1

2

1

100

010

001

0

0

0

0

0

0222

zyx

xy

xz

yz

xy

xz

yz

zyx

z

y

x

T

Let compute:

330330

330

&: xT

xBI

T

x

T

BI IqIqq

Iq

q

332

0

330

330 xT

x

T

xT Iq

Iq

Iqqq

33xTT Iqqqq

1

191

Equations of Motion



IB


BIB

BI

BI

BIB

BI qqq

dt

d

2

1

2

1

Let compute:

330330

330

&: xT

xBI

T

x

T

BI IqIqq

Iq

q

332

0

0

0

0

0

330

330x

TTT

x

x

T

T

Iqq

q

Iq

Iq

qq

Tx

TT qqIqqqq 44

Tx

T

T

qq

T

T

Iqq

Txx

qq

T

q

qq

Iq

q

0

02

0

332

013

20

33

0

0

2

192

Equations of Motion



IB


BIB

BI

BI

BIB

BI qqq

dt

d

2

1

2

1

Let compute:

330330

330

&: xT

xBI

T

x

T

BI IqIqq

Iq

q

13

0

000

330 0 xxT qq

qIqqq

ABT

BABAABB

BAxAABA

T qqqqq

Iqqq

00

0330

3

4

193

Equations of Motion



IB


BIB

BI

BI

BIB

BI qqq

dt

d

2

1

2

1

Let compute:

330330

330

&: xT

xBI

T

x

T

BI IqIqq

Iq

q

33

003300

330

330

xBT

A

BABAABxBAT

BA

BxB

TB

AxAABAT

Iqq

qqIqq

Iq

Iqqq

5

Return to Differential of the Quaternion

194

Lyapunov Stability Analysis of Linear Time Invariant (LTI) Systems

SOLO Stability Analysis of a Linear Systems

Consider the following LTI1

1nxnxn

nx xAdt

xd

where Anxn is a constant non-singular square matrix

Theorem: The equilibrium state x = 0 of the LTI System is Asymptotically Stable if and only if given any Positive Definite Hermitian Matrix Q (Q*T = Q) (or Positive Definite Symmetric Matrix Q (QT = Q) ), there exists a Positive Definite Hermitian Matrix P (or Positive Definite Real Symmetric Matrix P) such that

A*P+P A = - QThe scalar function V (x) = x*P x is a Lyapunov Function for the LTI System.


1857 - 1918

Proof of Theorem (if)

Let prove first that if exists a Positive Definite Hermitian Matrix P such that A*P+P A =- Q,then the equilibrium state x = 0 is Asymptotically Stable. Define

00* xxPxxV

0********DefinitePositive

Q

xQxxAPPAxxAPxxPAxxPxxPxxV

Hence V (x) is a Lyapunov Function for the LTI System, and x = 0 is Asymptotically Stable. 195




1nxnxn

nx xAdt

xd


A*P+P A =- QThe scalar function V (x) = x*P x is a Lyapunov Function for the LTI System.

Proof of Theorem (only if)

Let prove first that only if the equilibrium state x = 0 is Asymptotically Stable, there exists a Positive Definite Hermitian Matrix P such that A*P+P A =- Q.

DefinitePositiveQXAXXAX 00* Start with the Differential Equation

That has the Solution (proof by substitution) tAtA QeetX *

AtdXtdXAXXtdX

000

*0

Since A is a Stable Matrix, X (∞)=0

0:*0

0

*

0

Q

tAtA tdQeetdXPwhereAPPAQ 196




1nxnxn

nx xAdt

xd


A*P+P A =- QThe scalar function V (x) = x*P x is a Liapunov Function for the LTI System.

Corollary: A Necessary and Sufficient condition for x = 0 to be an Asymptotically Stable Solution for the LTI System is that exists a Positive Definite Hermitian Matrix P (or Positive Definite Real Symmetric Matrix P) such that

A*P+P A = - I

197

198

Slide Mode Control (SMC)Higher Order Sliding Modes and Arbitrary-Order Exact Robust DifferentiationA. Levant, Proceedings of the European Conference 2001, pp.996 - 1001

1

21

2/1

211111

11

1/

1

201

/1

011111

10

1/

00000

,

,

,

,

nnnn

nnnnnnnnn

iii

inin

iiiiii

nn

nn

vzsignz

zvzsignvzvvz

zvzsignvzvvz

zvzsignvzvvz

ztfzsigntfzvvz

1

0

1/1

011

10

1/

0

20

1/1

011

10

1/

000

nnnn

n

n

nn

i

nin

ii

nn

nn

vzsignz

ztfzsigntfzz

ztfzsigntfzz

ztfzsigntfzz

ztfzsigntfzz

This can be rewritten as

Levant proposed the following Arbitrary-Order Exact Robust Differentiator of a function f (t).The assumptions are that the functions have Lipschitz constant C>0, . The following (not unique) differentiation algorithm is proposed:

000

2

0

1

0 ,,,,,,ni

fffff

Ctf

i

0

λ0.λ1,…,λn > 0 are differentiator (Sliding Mode) parameters to be defined.

The positive parameters κ0.κ1,…,κn > 0 are calculated on basis the λi parameters.

SOLO

199

Slide Mode Control (SMC)Higher Order Sliding Modes and Arbitrary-Order Exact Robust Differentiation

% SLDING MODE% First Order Differetiation%Solo 7/2011 %Initial Conditionsclear;integ_steps=1000;time=0;tfin=10;delta_time=tfin/integ_steps;%Function Parametersf0=1;phase0=0;sigmanoise=0.01;noise=1;%Differentiation ParametersLamda0=12;Lamda1=1;z0=0;z1=0;f=0;j=1; for i=1:integ_steps % Define function to Differentiate if time>5&time<8 f=sin(2*pi*f0*time+phase0); fdot=2*pi*f0*cos(2*pi*f0*time+phase0); else f=0; fdot=0; end if noise==1 noisef=normrnd(0,sigmanoise); else noisef=0; end

f_output=f+noisef; %Sliding Mode First Order Differentiator v0=-Lamda0*abs(z0-f_output)^0.5*sign(z0-f_output)+z1; v1=-Lamda1*sign(z1-v0); z0_dot=v0; z1_dot=v1; df_est=v0; errd=fdot-df_est; %Integration (Euler) time=time+delta_time; z0=z0+z0_dot*delta_time; z1=z1+z1_dot*delta_time; %Plot Data t_out(j)=time; output1(j)=f; output2(j)=fdot; output3(j)=noisef; output4(j)=z0; output5(j)=v0; output6(j)=z1; output7(j)=v1; output8(j)=errd; j=j+1; endlinkaxes(h,'x')

%Plot figure h(1)=subplot(311); plot(t_out,output1);grid ylabel('f') h(2)=subplot(312); plot(t_out,output2);grid ylabel('fdot') h(3)=subplot(313); plot(t_out,output3);grid ylabel('noisef') linkaxes(h,'x') figure h(1)=subplot(311); plot(t_out,output5);grid ylabel('fdot_est') h(2)=subplot(312); plot(t_out,output2);grid ylabel('fdot') h(3)=subplot(313); plot(t_out,output8);grid ylabel('errd') linkaxes(h,'x') figure h(1)=subplot(411); plot(t_out,output5);grid ylabel('fdot_est') h(2)=subplot(412); plot(t_out,output4);grid ylabel('z0') h(3)=subplot(413); plot(t_out,output6);grid ylabel('z1') h(3)=subplot(414); plot(t_out,output7);grid ylabel('v1') linkaxes(h,'x')

MATLAB Program for First-Order Differentiation

SOLO

200


0 1 2 3 4 5 6 7 8 9 10-1

0

1

f

0 1 2 3 4 5 6 7 8 9 10-10

0

10

fdot

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

nois

ef

0,1852sin

8&500

0000 fttf

tttf

01.0,0, meantgausstnnoise

Input to Differentiator: tntftfi

0,1852cos2

8&500

00000 fttff

tttf

Figure 1

SOLO

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0 1 2 3 4 5 6 7 8 9 10-10

0

10

fdot

est

0 1 2 3 4 5 6 7 8 9 10-10

0

10

fdot

0 1 2 3 4 5 6 7 8 9 10-5

0

5

errd

,0vfest

0,1852cos2

8&500

00000 fttff

tttf

estfferr

010111

11

10

2/1

000

00

,

,

vzsignvzv

vz

ztfzsigntfzv

vz

ii

First-Order Differentiator1

12

1

0

Figure 2

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0 1 2 3 4 5 6 7 8 9 10-10

0

10fd

otest

0 1 2 3 4 5 6 7 8 9 10-2

0

2

z0

0 1 2 3 4 5 6 7 8 9 10-0.5

0

0.5

z1

0 1 2 3 4 5 6 7 8 9 10-1

0

1

v1

010111

11

10

2/1

000

00

,

,

vzsignvzv

vz

ztfzsigntfzv

vz

ii

First-Order Differentiator1

12

1

0

Figure 3

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% SLDING MODE% Third Order Differentiation%Solo 7/2011

%Initial Conditionsclear;integ_steps=1000;time=0;tfin=10;delta_time=tfin/integ_steps;%Function Parametersf0=1;phase0=0;sigmanoise=0.01;noise=0;%Differentiation ParametersLamda0=20;Lamda1=40;Lamda2=1.;Lamda3=0.2;z0=0;z1=0;z2=0;z3=0;f=0;j=1;

for i=1:integ_steps % Define function to be differentiate if time>5&time<8 f=sin(2*pi*f0*time+phase0); fdot=2*pi*f0*cos(2*pi*f0*time+phase0); fdot2=-(2*pi*f0)^2*sin(2*pi*f0*time+phase0); fdot3=-(2*pi*f0)^3*cos(2*pi*f0*time+phase0); else f=0; fdot=0; fdot2=0; fdot3=0; end if noise==1 noisef=normrnd(0,sigmanoise); else noisef=0; end f_output=f+noisef;

%Sliding Mode Third-Order Differentiator v0=-Lamda0*abs(z0-f_output)^(3/4)*sign(z0-f_output)+z1; v1=-Lamda1*abs(z1-v0)^(2/3)*sign(z1-v0)+z2; v2=-Lamda2*abs(z2-v1)^(1/2)*sign(z2-v1)+z3; v3=-Lamda3*sign(z3-v2); z0_dot=v0; z1_dot=v1; z2_dot=v2; z3_dot=v3; df_est=v0; df2_est=v1; df3_est=v2; errd1=fdot-df_est; errd2=fdot2-df2_est; errd3=fdot3-df3_est; %Integration (Euler) time=time+delta_time; z0=z0+z0_dot*delta_time; z1=z1+z1_dot*delta_time; z2=z2+z2_dot*delta_time; z3=z3+z3_dot*delta_time; %Plot Data t_out(j)=time; output1(j)=f; output2(j)=fdot; output3(j)=fdot2; output4(j)=fdot3; output5(j)=noisef; output6(j)=z0; output7(j)=v0; output8(j)=z1; output9(j)=v1; output10(j)=z2; output11(j)=v2; output12(j)=z3; output13(j)=v3; output14(j)=errd1; output15(j)=errd2; output16(j)=errd3; output17(j)=z0-f_output; output18(j)=z1-v0; output19(j)=z2-v1; output20(j)=z3-v2; j=j+1; end

%Plot figure h(1)=subplot(511); plot(t_out,output1);grid ylabel('f') h(2)=subplot(512); plot(t_out,output2);grid ylabel('fdot') h(3)=subplot(513); plot(t_out,output3);grid ylabel('fdot2') h(4)=subplot(514); plot(t_out,output4);grid ylabel('fdot3') h(5)=subplot(515); plot(t_out,output5);grid ylabel('noisef') linkaxes(h,'x') figure h(1)=subplot(311); plot(t_out,output7);grid ylabel('fdot_est') h(2)=subplot(312); plot(t_out,output2);grid ylabel('fdot') h(3)=subplot(313); plot(t_out,output14);grid ylabel('errd1') linkaxes(h,'x') figure h(1)=subplot(311); plot(t_out,output9);grid ylabel('fdot2_est') h(2)=subplot(312); plot(t_out,output3);grid ylabel('fdot2') h(3)=subplot(313); plot(t_out,output15);grid ylabel('errd2') linkaxes(h,'x')

figure h(1)=subplot(311); plot(t_out,output11);grid ylabel('fdot3_est') h(2)=subplot(312); plot(t_out,output4);grid ylabel('fdot3') h(3)=subplot(313); plot(t_out,output16);grid ylabel('errd3') linkaxes(h,'x') figure h(1)=subplot(511); plot(t_out,output7);grid ylabel('v0') h(2)=subplot(512); plot(t_out,output6);grid ylabel('z0') h(3)=subplot(513); plot(t_out,output8);grid ylabel('z1') h(4)=subplot(514); plot(t_out,output9);grid ylabel('v1') h(5)=subplot(515); plot(t_out,output10);grid ylabel('z2') linkaxes(h,'x') figure h(1)=subplot(411); plot(t_out,output17);grid ylabel('z0-f') h(2)=subplot(412); plot(t_out,output18);grid ylabel('z1-v0') h(3)=subplot(413); plot(t_out,output19);grid ylabel('z2-v1') h(4)=subplot(414); plot(t_out,output20);grid ylabel('z3-v2') linkaxes(h,'x')

MATLAB Program for Third-Order DifferentiationSOLO

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Figure 1

0 1 2 3 4 5 6 7 8 9 10-101

f

0 1 2 3 4 5 6 7 8 9 10-10

010

fdot

0 1 2 3 4 5 6 7 8 9 10-50

050

fdot

2

0 1 2 3 4 5 6 7 8 9 10-500

0500

fdot

3

0 1 2 3 4 5 6 7 8 9 10-505

x 10-3

nois

ef

Input to Differentiator: tntftfi

0,1852sin

8&500

0000 fttf

tttf

0,1852cos2

8&500

0000

1

fttff

tttf

0,1852sin2

8&500

00002

2

fttff

tttf

0,1852cos2

8&500

00003

3

fttff

tttf

001.0,0, meantgausstnnoise

Higher Order Sliding Modes and Arbitrary-Order Exact Robust DifferentiationA. Levant, Proceedings of the European Conference 2001, pp.996 - 1001

SOLO

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Figure 2

0 1 2 3 4 5 6 7 8 9 10-10

0

10

fdot

est

0 1 2 3 4 5 6 7 8 9 10-10

0

10fd

ot

0 1 2 3 4 5 6 7 8 9 10-5

0

5

errd

1

0,1852cos2

8&500

0000

1

fttff

tttf

tf est

1

estfferrd11

1


2.0

.,1,

.,40,

.,20,

32333

2

3

2312

2/1

122222

1

2

1201

3/2

011111

0

1

010

4/3

00000

vzsignz

vfzvzsignvzvvz

vfzvzsignvzvvz

vfztfzsigntfzvvz

est

est

estii

Third OrderDifferentiator

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Figure 3

0 1 2 3 4 5 6 7 8 9 10-200

0

200

fdot

2 est

0 1 2 3 4 5 6 7 8 9 10-50

0

50fd

ot2

0 1 2 3 4 5 6 7 8 9 10-200

0

200

errd

2

0,1852sin2

8&500

00002

2

fttff

tttf

tf est

2

estfferrd22

2


2.0

.,1,

.,40,

.,20,

32333

2

3

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2/1

122222

1

2

1201

3/2

011111

0

1

010

4/3

00000

vzsignz

vfzvzsignvzvvz

vfzvzsignvzvvz

vfztfzsigntfzvvz

est

est

estii


SOLO

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Figure 4

0 1 2 3 4 5 6 7 8 9 10-20

0

20

fdot

3 est

0 1 2 3 4 5 6 7 8 9 10-500

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500

fdot

3

0 1 2 3 4 5 6 7 8 9 10-500

0

500

errd

3

0,1852cos2

8&500

00003

3

fttff

tttf

tf est

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estfferrd33

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2.0

.,1,

.,40,

.,20,

32333

2

3

2312

2/1

122222

1

2

1201

3/2

011111

0

1

010

4/3

00000

vzsignz

vfzvzsignvzvvz

vfzvzsignvzvvz

vfztfzsigntfzvvz

est

est

estii


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Figure 5

0 1 2 3 4 5 6 7 8 9 10-10

0

10

v0

0 1 2 3 4 5 6 7 8 9 10-2

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2

z0

0 1 2 3 4 5 6 7 8 9 10-10

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z1

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v1

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2

z2


2.0

.,1,

.,40,

.,20,

32333

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3/2

011111

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1

010

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vfzvzsignvzvvz

vfzvzsignvzvvz

vfztfzsigntfzvvz

est

est

estii


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0 1 2 3 4 5 6 7 8 9 10-0.2

0

0.2

z0-f

0 1 2 3 4 5 6 7 8 9 10-5

0

5z1

-v0

0 1 2 3 4 5 6 7 8 9 10-200

0

200

z2-v

1

0 1 2 3 4 5 6 7 8 9 10-20

0

20

z3-v

2

Figure 6


2.0

.,1,

.,40,

.,20,

32333

2

3

2312

2/1

122222

1

2

1201

3/2

011111

0

1

010

4/3

00000

vzsignz

vfzvzsignvzvvz

vfzvzsignvzvvz

vfztfzsigntfzvvz

est

est

estii


SOLO

Arie LevantAssociate professor

Applied Mathematics Dept.,School of Mathematical

Sciences,Tel-Aviv University,

Ramat-Aviv.Tel-Aviv 69978

Israel

John L. Crassidis Professor

Mechanical and Aerospace EngineeringUniversity of Buffalo

Yuri B. Shtessel University

of AlabamaDepartment of Electrical and

Computer Engineering

Vadim Utkin

Department of Electrical & Computer Engineering

OHIO STATE UNIVERSITY

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Slide Mode Control (SMC)SOLO

Leonid FridmanUniversidad Nacional Autonoma de Mexico

Christopher EdwardsDepartment of Engineering

Leicester University

Sarah K. SurgeonDepartment of Engineering

Leicester University

211


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Filippov Aleksei Fedorovich – Professor of Differential Equations of Mechanics and Mathematics Faculty (1978). Born in Moscow. He graduated from the Mechanics and Mathematics Faculty of Moscow State University in 1950, Doctor of Physical and Mathematical Sciences (1976), Professor (1980). The participant of Great Patriotic War, he was awarded medals "For the victory over Germany in the Great Patriotic War of 1941 – 1945.", "Veteran of Labour", "For Valiant Labor. To mark the 100 anniversary of Lenin's birth, commemorative medals. Laureate. MV Lomonosov Moscow State University for his excellent lecturer's skill and create a unique textbook on the mathematics Problems in Differential Equations "(1993). He was awarded the honorary title "Honorary Professor of Moscow State University" (1996). Research interests: differential equations, theory of diffraction, differential equations with discontinuous right-hand side, the differential inclusion. Total teaching experience AF Filippov is 40 years. Each year, AF Filippov, special courses on differential equations and their research, which are intended both for students and for graduate students and young scientists. Produced 12 PhDs. Published 65 scientific papers.


Filippov Aleksei Fedorovich(1923 – 2006)

SOLO

Modern Sliding Mode Control Theory : New Perspectives and Applications (Lecture Notes in Control and Information Sciences) 〈 Vol. 375), G. Bartolini, A. Pisano, L. Friedman, E. Usai, Ed., Springer Verlag 2008

“Sliding mode control: theory and applications”Christopher Edwards,

Sarah K. SpurgeonTaylor & Francis, 1998

213


Sabanovic, A., Fridman, L., and Spurgeon, S.K., Variable Structure Systems: from principles to implementation, IEE Book Series, 2004 214


215


Slide Mode Control (S.M.C.)

Science

Transcript of Slide Mode Control (S.M.C.)