Thesis - Analysis and Control of Simple Nonlinear Limb Model

8/12/2019 Thesis - Analysis and Control of Simple Nonlinear Limb Model

1/82

Analysis and Control of a Simple NonlinearLimb Model

by

David Csercsik

Submitted to the department of Control Engineering andInformation Technology in partial fulllment of the requirements

for the degree of

Master of Science in Electrical Engineering

at the

Budapest University of Technology and Economics

Supervisor: Department Consultant:

Prof. Katalin Hangos Dr. Istvan Harmati

Systems and Control Laboratory Department of Control EngineeringComputer and Automation and Information Technology

Research Institute Budapest Faculty of Control Engineeringand Informatics

Budapest University of Technologyand Economics

BME2005


2/82

Contents1 Introduction 5

1.1 Literature review on human locomotion analysis and control . 51.2 Notation list . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Structure of the diploma work . . . . . . . . . . . . . . . . . . 7

2 Construction of a simple limb model in state-space 82.1 Limb and muscle modelling . . . . . . . . . . . . . . . . . . . 82.2 System description . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Modelling assumptions . . . . . . . . . . . . . . . . . . 92.3 Muscle properties . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3.1 Force-length characteristics ( F L (lCE )) . . . . . . . . . 102.3.2 Force-contraction velocity characteristics ( F v(vCE )) . . 112.3.3 Passive force (F P E ) . . . . . . . . . . . . . . . . . . . . 122.3.4 Muscle activation ( q ) . . . . . . . . . . . . . . . . . . 132.3.5 Muscle force (F CE ) . . . . . . . . . . . . . . . . . . . . 13

2.4 Tendon properties . . . . . . . . . . . . . . . . . . . . . . . . . 142.4.1 The tendons dynamical equations . . . . . . . . . . . . 14

2.5 Dynamic equations at the limb level . . . . . . . . . . . . . . . 142.5.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5.2 Joint torques . . . . . . . . . . . . . . . . . . . . . . . 152.5.3 Movement . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.6 State-space model equations and its variables . . . . . . . . . 172.6.1 Variables in the state-space model . . . . . . . . . . . . 172.6.2 State-space equations . . . . . . . . . . . . . . . . . . . 172.6.3 The domain of the model and the model parameters . . 19

2.7 Model verication . . . . . . . . . . . . . . . . . . . . . . . . . 212.8 A biological extension: A simple Gamma-loop model . . . . . 24

2.8.1 The simplied model of the Gamma-loop . . . . . . . . 242.8.2 Simulation results . . . . . . . . . . . . . . . . . . . . . 26

3 Model analysis 283.1 Linear analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 303.1.1 Linearization around a steady-state point . . . . . . . . 303.1.2 Analysis of the linearized model at the rst steady-

state point . . . . . . . . . . . . . . . . . . . . . . . . . 313.1.3 Analysis of the linearized model at the second steady-

state point . . . . . . . . . . . . . . . . . . . . . . . . . 333.2 Relative degree . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2.1 The structure graph . . . . . . . . . . . . . . . . . . . 35

1


3/82

3.3 Controllability . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.3.1 Algorithm for constructing the controllability distribu-

tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.3.2 Structural controllability and observability . . . . . . . 373.3.3 Structural controllability analysis of the model . . . . . 37

3.4 Observability . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.4.1 Algorithm for constructing the observability

co-distribution . . . . . . . . . . . . . . . . . . . . . . . 373.4.2 Structural observability analysis of the model . . . . . 38

3.5 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.5.1 Lyapunov-stability . . . . . . . . . . . . . . . . . . . . 39

3.5.2 Strong asymptotic -stability . . . . . . . . . . . . . . . 393.5.3 Lyapunovs rst theorem . . . . . . . . . . . . . . . . . 393.5.4 Estimating the stability region in the neighborhood

of the steady-state point with quadratic Lyapunov-function candidate . . . . . . . . . . . . . . . . . . . . 40

3.6 Zero dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.6.1 Zero dynamics state-space equations . . . . . . . . . . 443.6.2 Computing the zeroing input . . . . . . . . . . . . . . 453.6.3 Zero-dynamics analysis results . . . . . . . . . . . . . . 45

4 Controller design 464.1 Performance specication . . . . . . . . . . . . . . . . . . . . . 46

4.1.1 Control aims . . . . . . . . . . . . . . . . . . . . . . . 464.1.2 Control domain . . . . . . . . . . . . . . . . . . . . . . 464.1.3 Disturbances . . . . . . . . . . . . . . . . . . . . . . . 46

4.2 Linear SISO LQ regulator . . . . . . . . . . . . . . . . . . . . 484.2.1 Simulation results of the SISO LQ conroller . . . . . . 50

4.3 Linear MISO LQ regulator . . . . . . . . . . . . . . . . . . . . 524.3.1 Simulation results of the MISO LQ controller without

disturbance . . . . . . . . . . . . . . . . . . . . . . . . 524.3.2 Trajectory following results of the MISO LQ controller

with disturbance . . . . . . . . . . . . . . . . . . . . . 544.3.3 Regulation results of the MISO LQ controller with dis-

turbance . . . . . . . . . . . . . . . . . . . . . . . . . . 554.4 Input-Output linearization . . . . . . . . . . . . . . . . . . . . 56

4.4.1 Simulation results of the IOL-PP controller withoutdisturbance . . . . . . . . . . . . . . . . . . . . . . . . 58

4.4.2 Trajectory following results of the IOL-PP controllerwith disturbance . . . . . . . . . . . . . . . . . . . . . 59

2


4/82


5/82

Abstract

Motivation and Aim:Even the simplest limb model exhibits strongly nonlinear dynamic behavior that

calls for applying the results of nonlinear systems and control theory. The analysisand control of limb models are important in the elds of designing and control-ling articial limbs, muscle prosthesis and in neuro-physiological investigations.The aim of this study is to investigate the possibility to applying input-outputlinearization [11] for nonlinear control of a simple limb model.

Material and Methods:A nonlinear input-affine state-space model has been developed for a simple one-

joint system with a exor and an extensor muscle (see gure 1) which is suitable fornonlinear systems analysis and control. The model takes the nonlinear propertiesof the force-length relation and the force-contraction velocity relation into account.

Exerted forces depend linearly on the activation state of muscles, and a vis-coelestic tendon is considered following the principles in [28], [24] and [22]. Thismodel has been extended with a simple model of the gamma-loop mechanism, butonly the non-extended model is used for the control studies. The inputs of themodel are the normalized activation signal of muscles, the output is the joint angle,and the number of state variables is 8.

As preliminary model analysis we performed stability, controllability and ob-servability analysis of the linearized model around steady-state points. Moreover,the relative degree of the model and the stability of its zero-dynamics [3] werealso determined. Both regulating and servo controllers were designed for the sim-ple limb model and were compared to the standard reference case being an LQ-controller designed for the locally linearized system. A pole-placement control wasdesigned for the input-output linearized system, and also a fuzzy controller wasdesigned.

Result:The model was veried against engineering intuition and proved to be suitable

for controller design purposes. The model analysis showed that the nonlinear limb

model was controllable and was in the edge of stability because of the Hamlitonianproperties of the model. The relative degree of the model is 3 for both of the inputswith a stable zero dynamics. Therefore input-output linearization was applicableand a 3rd order linear system was obtained in the new coordinates. A simple pole-placement controller was designed for this input-output linearized model. Boththe pole-placement, the fuzzy, and the reference LQ-controller were suitable forcontrol purposes but the fuzzy and the LQ-controller were more sensitive to thedisturbances.

4


6/82

1 IntroductionHuman locomotion is a complex movement of a body. For a successful move-ment, interactions among muscular-skeletal system and central nervous sys-tem (CNS) are needed. The CNS controls the movement by sending activation signals to the correct skeletal muscle at correct moment. The effect of this signal is that the muscle initiates a movement by exerting force to a body segment. Even in the case of a simple movement the contribution of a large number of muscles of different size and shape is necessary. A common move-ment such as locomotion require more muscles. With a feedback system that uses different kind of receptors, the CNS controls the movement. In healthy

humans all these complex coordinated actions lead unconsciously to a smooth movement. [8]The study of human locomotion has gained more attention recently with

the developement of analytic and computational tools with which to examineit. A much researched subject within the eld today is the effort to modelhuman motor control systems using control theoretic methods. Analytic,computational, and experimental studies of locomotion can produce modelsthat provide further insight into the design and functioning of human motorcontrol systems, as well as provide directions for research into diagnostics andtherapies for muscle- and nerve-related disorders affecting these systems.

The aim of this work is to investigate the difficulties related to the con-trol of a such strongly nonlinear system, as a human limb, and to design acontroller for a simple nonlinear limb model. For this aim we need a modelsuitable for the mathematical tools of linear and nonlinear control theoryand system analysis - a nonlinear state space model.

1.1 Literature review on human locomotion analysisand control

Several control models exist that take different important aspects of humanlocomotion control into account, in various cases of control tasks. The mostrelevant results of recent investigations on locomotion analysis and controlare listed and summarized in the following list:

A multi level control model including timing and learning was devel-oped by Levine and Loeb [18].

A study proposed by Levine and Zajac [20] has shown, that in the caseof the pedaling problem, the control to achieve maximal accelerationfor a simple skeletal system is bang-bang.

5


7/82


8/82

Hungarian scientic background of locomotion analysis:

Modelling of limb movement patterns based on neuronal activity has beendeveloped in Hungary for some years. Differential neuro-muscular-skeletalstructures have been studied by Laczko et al. [15, 14, 13] using mathematicalmodels and computer simulation.

A neuro-mechanical model was developed by Fazekas in [6, 7]. It takesinto account the one and two-joint muscles. It handles the force-length andforce-frequency relationship, passive force, geometric and inertial propertiesof the limb, the maximal isometric forces and the gravitational effect.

1.3 Structure of the diploma workThe description of the construction method and the derived equations of the simple limb model can be found in chapter 2. Chapter 3 describes theanalysis results of the simple limb model. In chapter 4 the description, andresults of the applied control methods can be found. Conclusions of the workand possible future works are summarized in chapter 5.

7


9/82

2 Construction of a simple limb model in state-space

2.1 Limb and muscle modellingOur aim is to create a model of a simple one-joint system with a exor andan extensor muscle (see in gure 1) which is suitable for nonlinear systemsanalysis and control.

The models of dynamics of a one-joint system with two muscles containsthe equations of multi-rigid-limb system and the equations of the dynamicsof muscle contraction. The crucial component of this system is the model

that generates the exerting muscle forces.

Muscle modelling:Different muscle models exist that focus different important aspects of

muscle functioning. In some of the models individual muscle characteristicsare used, and some of the models deal with generalized muscle characteristicsthat are valid for all muscle. But most of the existing models deal only witha partial functioning of the muscle such as a force depending on the musclelength or muscle geometry or ring rates of nerves etc.

There are some models that deal with the movement pattern generation

according to the force exertion and neuronal input. For example Cheng [3]created this kind of model.

A simple nonlinear limb model:In this chapter the model equations are derived from rst engineering (med-

ical) principles based on the detailed model of Fazekas [8], Zajac [28] and VanSoest [21]. Therefore the equations are transformed into a state-space modelform.

2.2 System descriptionA simple two dimensional limb is considered that consists of two musclesand one joint. The schematic picture of the limb is shown in gure 1. Theinput of the model consists of the activation signal or stimuling signal of eachmuscles as a function of time. Activation signals are normalized, i.e. theirvalues are between zero and one.

In general case, the output of the model is the joint angle.

8


10/82

Figure 1: The system

2.2.1 Modelling assumptions

Dimension: The model is two dimensional.

Structure: We model 2 muscles, one joint.

Segments: The bones are totally rigid.

Gravity: Gravity appears in the direction -y only.

Geometry: We do not model the geometry of bones and the muscle.

Muscle Properties: The model takes into account only the followingproperties of the muscle: Force-length dependency, Force-contractionvelocity dependency, Passive Force. The characteristics are the sameby the exor, and the extensor muscle.

Moment arms: Moment arms are constants.

Aponeurosis: There is not aponeurosis.

Pennate effect: We do not deal with the pennate effect.

Tendon: We suppose viscoelastic tendon.

9


11/82

Fatigue: In general case, we do not model fatigue, potentialization andshort time histories in the muscles.

Activation dynamics: We use rst-order activation dynamics model.

Viscosity: We do not model viscosity in the muscle.

2.3 Muscle propertiesTo construct a state-space limb model, at rst we need to examine the proper-ties of its components (muscles, tendon, etc). We start with the mathematicalequations describing the behaviour of the muscles.

2.3.1 Force-length characteristics ( F L (lCE ))

The equation describes the quotient of the actual force, and the maximalforce of the muscle, at given length ( lCE ). We use a parabolic function:

F L (lCE ) = clCE loptCE

2

2clCE loptCE

+ c + 1 (1)

where lCE [m] is the actual length of the muscle, loptCE [m] is the optimal lengthof the muscle, at which it can produce its maximal force. The coefficient c

is dened by: c = 12 where is a constant, describing the range of length,where the muscle is able to work. As Soest [22], we dene = 0.56 .

The equation (1) is only valid in the range of 1 lCEloptCE 1 + outside this range FL is 0. We can see the characteristics in gure 2.

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.2

0.4

0.6

0.8

1

lCE

F / F

m a x

Figure 2: F L(lCE )

10


12/82

2.3.2 Force-contraction velocity characteristics ( F v(vCE ))

The force produced by a muscle depends on its actual contraction velocity.We use a similar function to Hillequation [10], extended by Van Soest [21]that is:

F v(vCE ) =

(1+ A ref )B ref v CEl optCE

+ B ref Aref if vCE > 0

c1

v CEl optCE

c3 c2 otherwise

(2)

where

c1 = B ref (1 F asy )2

2+2 A ref , c2 = F asy , c3 = B ref (1 F asy )

2+2 A ref

Furthermore, vCE is the actual contraction velocity of the muscle (posi-tive, if the muscle is contracting, and negative if the muscle is extracting),Aref and Bref are constants, F asy is a constant, which shows at excentric con-traction the quotient of the actual available force, and the maximal isometricforce.

This function is not continously differentiable, so to avoid the problems,we use an approximating smooth function that ts to the requirements of nonlinear analysis (the parameters of the function were found in experimentalways):

F v(vCE ) = 3/ 2 arctan(9 / 5 vCE 925 ) 1 + 167200 (3)

The original and the approximated functions are shown in gure 3.

3 2 1 0 1 2 30

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

vCE

[m/s]

F / F

m a x

equation of Hillaxpproximation

Figure 3: F v(vCE ) and its approximation

11


13/82

2.3.3 Passive force ( F P E )

For describing the passive force F P E as a function of lCE we use modiedsigmoid functions for both the exor and extensor muscles:

F P E = (1 sigm [7( lCE + 0.3333)]10)F maxP E (4)

where the function identier sigm stands for the function:

y = sigm[x] = 11+ e 100 x

with F maxP E being the maximal passive force. We denote the exor muscles

passive force by F P E 1 and the extensor muscles passive force by F P E 2.

1.04 1.06 1.08 1.1 1.12 1.14 1.160

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

lCE /lCEopt

F P E

/ F P E m a x

Figure 4: The function F P E

12


14/82

2.3.4 Muscle activation ( q )

The differential equation of the muscle shows the connection between q (t), theactivation state of the muscle and the activation signal u(t). With u(t) [0, 1]the equation taken from Zajac [28] is:

dq dt

= 1 act

( + [1 ]u(t) q + 1 act

u(t) (5)

where act [s] is the activation time, showing how quick the muscle reacts onthe external activation signal coming from the nerval system. is a constant,describing the correlation between the decrease of the activation state andthe external activation signal. If = 1 then the external activation signaldoes not affect the decrease of the activation state, if = 0 then it stronglyaffects it. q 1(t) denotes the activation state of the exor muscle, and q 2(t)denotes the activation state of the extensor muscle.

2.3.5 Muscle force ( F CE )

We can compute the force of the muscle with the characteristics above in thefollowing form:

F CE = F max F L (lCE )F v(vCE )q + F P E (6)

where F max is the maximal force of the muscle. We use F CE 1 for the exormuscles force and F CE 2 for the extensor muscles force.

F CE 1 = F max F L (lCE 1)F v (vCE 1)q 1 + F P E 1F CE 2 = F max F L (lCE 2)F v (vCE 2)q 2 + F P E 2 (7)

where q 1 is the exor muscles activation state q 2 is the extensor muscles acti-vation state, lCE 1 [m] is the length of the exor muscle, lCE 2 [m] is the lengthof the extensor muscle, vCE 1 [m/s] is the contraction velocity of the exor

muscle and vCE 2 [m/s] is the contraction velocity of the extensor muscle.

13


15/82

2.4 Tendon properties2.4.1 The tendons dynamical equations

We need tendons to transport the force from the muscles to the bones. Wedescribe the tendons behaviour with the following equations:

l T t = vT

v T t =

k t ( lT lslackt )+ s T vT F T zT

(8)

where the rst equation is valid only, if lT > l slackT . lT [m] is the actual lengthof the tendon, lslackT [m] is the tendons length in the case, when forces do

not appear. vT [m/s] is the extracting velocity of the tendon (it is positive,if the tendon extracts), kT [N/m] is the elasticity constant of the tendon,sT [Ns/m] and z T [Ns 2/m ] are constants referring to the dynamics of thetendon, F T [N] is the force acting on the tendon.

2.5 Dynamic equations at the limb level2.5.1 Forces

Besides the force of the muscle, we use a sigmoid function for modelling theforces of ligaments and bones in the end-positions of the joint. These func-

tions are sigmoid in both of the angle of the joint and in the angle velocityto take unelastic properties (the properties making the joint available to dis-sipate motion energy) of the joint into account. The equation for describingthe forces of the ligaments is as follows:

F lig = F maxlig sigm[0.3( )]sigm [0.5] (9)

where [rad] is the external joint angle, F maxlig [N ] is the maximum force of the ligaments and bones, [rad] is a constant, describing at which angle theforce appears and [rad/s] is the joint angle velocity. We use F lig 1 for theforce, which appears at 0 angle, when the limb can not be more extended (at

the maximum length of the exor muscle), so it wants to ex the limb, andF lig 2 for the extensor ligamental force.

The exing forces can be computed as:

F flexor = F CE 1 + F lig 1 (10)

The extending forces can be computed as:

F extensor = F CE 2 + F lig 2 (11)

14


16/82

We suppose, for the sake of simplicity, that the forces of the ligamentsand bones appear at the same point as the forces of the muscle.

Furthermore, apart from the most extended and most exed states, wecan neglect the forces of ligaments (and in some cases the passive force too),to further simplify the model for symbolic analysis.

In this model the force of the tendon is of the same size as the force of themuscle and ligaments, it just appears in the opposite direction. For example:F T 1 = F flexor where F T 1 [N] is the tendons force.

2.5.2 Joint torques

The movement is determined not directly by the forces conveyed by thetendon, but by their torques. For example the exor tendons torque can bewritten as:

M 1 = F T 1d = F flexor d (12)

where d [m] is the moment arm, the distance between the axis of the joint,and the point, where the forces appear.The resultant joint torque can be computed as:

M = M 1 M 2 (13)

2.5.3 Movement

Now we know how the forces are generated, and transported to the bones.Let us then examine how the limb moves. We use the form of Newtons IIlaw applied to rotation movement as follows:

t =

t =

1+ ml 2COM

(M + mlCOM cos( + )gy)(14)

where [rad] is the joint angle, [rad] is the angle between the globalcoordinate-systems x axis, and the not-moving upper segment of the limb(in our model is always equal to / 2), [rad/s] is the angle velocity, [kgm2] is the moment of inertia dened to the mass-centre point of the bone,m [kg] is the mass of the moving limb part, lCOM [m] is the distance betweenthe moving limb parts center of mass point and the joint axis, M [Nm] is theresulting joint torque, and g = [gx , gy] [m/s 2] is the vector of gravitationalacceleration.

15


17/82

Whats left, is to dene the connection between the joint angle, andthe muscle length. For the limb the actuator system does not matter, soat the level of the segments we talk about muscle-tendon complexes. So,what we can directly compute from the joint angle, is the length of thesemuscle-tendon complexes. We will denote these muscle-tendon complexeswith CE + T in the subscript. We can compute the length of the muscle-tendon complexes approximately with the following equations:

In the case of exor muscle:

lCE + T =

d2 + ( d prox )2 2dd prox sin

(t)

2 +

d2 + ( ddist )2 2dddist sin

(t)

2(15)

In the case of extensor muscle:

lCE + T = d2 + ( d prox )2 + 2 dd prox sin (t)2 + d2 + ( ddist )2 + 2 dddist sin (t)2(16)

where lCE + T [m] is the length of the muscle-tendon complex, d [m] is themoment arm, d prox [m] is the distance between the joint, and the origin of

the muscle, ddist

[m] is the distance between the joint, and the insertion of the muscle, [rad]is the joint angle. The extracting velocity of the muscle-tendon complex can be computed as:

vCE + T = dlCE + T

dt (17)

From these values we can compute the length, and contraction velocityof the muscles, if we know the length and extraction velocity of the tendons.Both for exor and extensor muscles:

lCE (t) = lCE + T (t) lT (t)

vCE (t) = vCE + T (t) + vT (t)(18)

16


18/82

2.6 State-space model equations and its variables2.6.1 Variables in the state-space model

We have the following state variables:

Joint angle: [rad]

Joint angle velocity [rad/s]

Muscle activation state (for exor and extensor muscle): q 1, q 2

Tendon length (of the exor and the extensor muscle): lT 1, lT 2 [m]

Tendon extracting velocity (for exor and extensor muscle): vT 1, vT 2[m/s]

So we get: x = [q 1 q 2 lT 1 lT 2 vT 1 vT 2]T

We have the following input variables:

Flexor muscles activation signal: u1(t)

Extensor muscles activation signal: u2(t)

The output of the model is the joint angle: y= = x3

2.6.2 State-space equations

dq1dt =

1 act ( + [1 ]u1(t)) q 1 +

1 act u1(t)

dq2dt =

1 act ( + [1 ]u2(t)) q 2 +

1 act u2(t)

ddt =

ddt =

1+ ml 2COM

(M (q 1, q 2, , , l T 1, lT 2, vT 1, vT 2) + mlCOM cos( + )gy)

dl T 1dt = vT 1

dl T 2dt = vT 2

dv T 1dt =

k t ( lT 1 lslackt )+ s T lT 1 F flexor (q1 ,,,l T 1 ,v T 1 )zT

dv T 2dt =

k t ( lT 2 lslackt )+ s T vT 2 F extensor (q2 ,,,l T 2 ,v T 2 )zT

(19)

17


19/82

With the notation of x1,x2...,x8 for state space variables the equationsare as follows:

dx 1dt =

1 act ( + [1 ]u1(t)) x1 +

1 act u1(t)

dx 2dt =

1 act ( + [1 ]u2(t)) x2 +

1 act u2(t)

dx 3dt = x4

dx 4dt =

1+ ml 2COM

(M (x1, x2, x3, x4, x5, x6, x7, x8) + mlCOM cos(x3 + )gy)

dx 5dt = x7

dx 6dt = x8

dx 7dt =

k t (x 5 lslackt )+ s T x 7 F flexor (x 1 ,x 3 ,x 4 ,x 5 ,x 7 )zT

dx 8dt =

k t (x 6 lslackt )+ s T x 8 F flexor (x 2 ,x 3 ,x 4 ,x 6 ,x 8 )zT

(20)

y = x3

18


20/82


21/82


22/82

2.7 Model vericationThe model verication has been performed by simulation where the modelresponse has been tested against engineering intuition. The following testcase was used: The activation signal of the exor muscle was equal to 1 fromt=0 to t=0.5, 0 from t=0.5 to t=1, 0.5 from t=1 to t=1.5 and 0.6 from t=1.5to t=2. The extensor muscles activation signal was equal to 0 from t=0 tot=0.5, 1 from t=0.5 to t=1, 0.5 from t=1 to t=1.5 and 0.1 from t=1.5 tot=2, as it can be seen in the following gure.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2u1u

2

Figure 5: The muscle activation signals

The muscle activation states are shown in gure 6, as functions of time(q 1 is the exor muscles activation, q 2 is the extensors):

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1q1,q2

t [s]

q 1

, q 2

q1q2

Figure 6: The muscle activation states

21


23/82

Figure 7 shows and as functions of time:

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 230

25

20

15

10

5

0

5

10

15alpha,omega

t [s]

a l p h a

[ r a

d ] , o m e g a

[ r a

d / s ]

alphaomega

Figure 7: and

The length of the tendons - lT 1 (the exor muscles tendon) and lT 2 (the

extensor muscles tendon) - are depicted in gure 8:

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20.1

0.101

0.102

0.103

0.104

0.105

0.106

0.107

0.108

0.109l1,l2

t [s]

l 1 [ m ] , l 2 [ m ]

l1l2

Figure 8: The length of the tendons

The length of the muscles are seen in gure 9 with lCE 1 being the exormuscles length and lCE 2 being the extensor muscles length:

22


24/82

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5lCEf,lCEe

t [s]

l C E f [ m ] , l C E e [ m ]

lCEf

lCEe

Figure 9: The length of the muscles

In gures 7 and 8 we can clearly see the strong forces of the ligamentsappearing at the maximal exed/extended states.

Taking into account the aim of the model construction, the simplica-tions used in the construction process and the simulation results, the modelbehaves as one expects, thus the model is veried and is acceptable for con-troller design purposes.

23


25/82

2.8 A biological extension: A simple Gamma-loop modelThe -loop is a well known solution in human and other species for someproblems emerging in muscle actuation. The short description of the mech-anism, and its componetents can be found in the appendix. (see more in[23])

Figure 10: The servomechanical -loop as depicted in [23]

2.8.1 The simplied model of the Gamma-loop

Simplifying assumptions The receptors of the muscle spindles can sense only the length differ-

ence between the muscle bers inside the muscle spindles ls , and thesurrounding muscle bers, of which length are commensurable to theworking muscle length lCE .

The activation signal of the spinal neurons is linear in the signal of themuscle spindless detector.

24


26/82

In this case, theres no other input to the system, only the musclespindles activation signals.

In this case, we can extend our model with two more state-space variables:With the lengths of the muscle bers of the muscle spindles ( ls 1 in the caseof the exor muscle, and ls 2 in the case of the extensor muscle).

If we suppose, that the normal length of the muscle spindles (and thesurrounding muscle bers) is always one tenth of the working muscle, we candescribe the behavior of the new variables with the following equations:

dls 1dt = cP (lCE 1/ 10 ls 1) cG us 1(t)

dls 2dt

= cP (lCE 1/ 10 ls 2) cG us 2(t) (21)

where cP [1/s] is a constant showing how quick the length of the muscle spin-dles (and the muscle bers inside the muscle spindles) follows the length of the working muscle (lCE 1, lCE 2), cG is a constant showing how sensitive themuscle spindle is to the external activation signal of the descending tracts,and us (t) is the activation signal of the descending tracts acting on the themuscle bers inside the muscle spindles. If we consider the normalized valuesof lCE 1/ 10 ls 1 and lCE 2/ 10 ls 2 as the signals of the muscle spindless recep-tors we can compute the activation states of the working muscles (denotedwith q 1 and q 2 in the equation (20)):

q 1 = cG (lCE 1/ 10 ls 1)q 2 = cG (lCE 2/ 10 ls 2) (22)

25


27/82

2.8.2 Simulation results

We act on the exor muscles spindle with the following activation signal:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.2

0.4

0.6

0.8

1

1.2

t [s]

u f

Figure 11: the activation signal of the exor muscles spindle

We can see the muscle lengths, and the length of the muscle spindles(multiplied with 10 for the better comparability) in the next gure:

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

t [s]

l e n g t h s [ m ]

lCE1 ,lCE2 ,10l s1 ,10l s2

lCE1 lCE2 10l s110l s2

Figure 12: the length of the muscles and the muscle spindles

26


28/82


29/82


30/82

g1(x) =

1tau act (1 )x1 + 1tau act

0

0

0

0

0

0

0

g2(x) =

0

1tau act (1 )x2 + 1

tau act

0

0

0

0

0

0

0

h(x) = x3

We can observe, that the state function f is nonlinear, but both the inputfunctions g1 and g2 and the output function h are linear.

29


31/82

3.1 Linear analysisThe well-known standard methods of linear analysis can only be applied forlinear time-invariant (LTI) systems with the following state-space model:

x = Ax + Buy = Cx (24)

3.1.1 Linearization around a steady-state point

We call x0(u0) a steady-state point of the system, if

x = f (x)|x 0 +m

i=1gi (x)u i (t)|u 0 = 0

From the input-affine state space model in equation (23) we can obtain a LTImodel by linearizing it locally around a steady-state point x0(u0), as in [2].

Let us take the centered variables:

x = x x0 u = u u0

To obtain the LTI model form:

x = Ax + B uy = C x

(25)

we can compute the matrixes A, B and C in the neighborhood of the steady-state point ( x0, u0) in the following way:

A = J (f,x ) |x (0) + J (g,x ) |x (0) u(0)

B = g(x0)

C = J (h,x ) |x (0)

(26)

where J (v,x ) is the Jacobian-matrix of the multivariate function v:

J (v,x ) j,i = v jx i

30


32/82

Two steady-state points were selected for linear analysis purposes:

The most extended steady-state of the limb (very low effect of gravity)- both of the muscles are totally inactive.

/ 2 joint angle (no effects of ligaments and passive force)

3.1.2 Analysis of the linearized model at the rst steady-statepoint

Equation (27) contains the state matrix A in the rst steady state point:

A = 10 4

0.0042 0 0 0 0 0 0 00 0.0042 0 0 0 0 0 00 0 0 0.0001 0 0 0 0

0.0112 0.0102 0.0047 0.0004 0.000 0.000 0 00 0 0 0 0 0 0.0001 00 0 0 0 0 0 0 0.0001

0.052 0 0.0004 0.0002 1.25 0 1.25 00 0.0048 0.000 0.000 0 1.25 0 1.25

(27)The rank of A is 8.We obtain the following results at the steady-space point in the neigh-

borhood of the most extended state of the limb:Steady-state conditions (in case of u1 = u2 = 0):

x1,0 = 0x2,0 = 0x3,0 0.004516x4,0 = 0x5,0 0.1x6,0 = 0.1x7,0 = 0x8,0 = 0

(28)

The A martixs eigenvalues are the following:

31


33/82

103

1.245 0.005 0.0022 + 0.0065i 0.0022 0.0065i 1.2450 0.005 0.0417 0.0417

(29)

If the eigenvalues of the state-matrix A have negative real part, then thelinear system is stable, and this is the case in the rst steady-state point.

Next we analyze the controllability and observability testmatrices, con-structed from the A, B and C matrices of the linearized system - see [17].

In the rst steady-state point:

B =

65.7255 00 83.3330 00 0

0 00 00 00 0

, C = [0 0 1 0 0 0 0 0]

Controllability testmatrix:

M c = [B AB......A n 1B] (30)

If the rank of the controllability testmatrix is equal to n (the number of

states) the system is state-controllable, which means also output controlla-bility in our case.

Observability testmatrix:

M 0 =

C CA...

CAn 1

(31)

32


34/82

If the rank of the observability testmatrix is equal to n (the number of states)the system is observable.

The results of the analysis in the rst steady-state point is as follows::

The linearized system is stable.

The linearized system is controllable.

The linearized system is observable.

3.1.3 Analysis of the linearized model at the second steady-statepoint

We can nd an another steady state-point at / 2 joint angle with the follow-ing steady-state conditions (of course, in this case the exor muscle is active- we take a case, when only the exor muscle is active - u1 = 0.0554 - andthe extensor muscle is totally inactive - u2 = 0):

x1(0) = 0 .1050x2(0) = 0x3(0) = / 2x4(0) = 0x5(0) = 0 .101439x6(0) = 0 .1x7(0) = 0x8(0) = 0

(32)

In this point:

A = 10 4

0.0052 0 0 0 0 0 0 00 0.0042 0 0 0 0 0 00 0 0 0.0001 0 0 0 0

0.0095 0.0083 0.0009 0.000 0.0206 0.000 0.0037 00 0 0 0 0 0 0.0001 00 0 0 0 0 0 0 0.0001

0.0045 0 0.0004 0.000 1.2597 0 0.1267 00 0.0039 0.000 0.000 0 1.2500 0 0.1250

(33)

The rank of the matrix A in this case is 8.

33


35/82

The eigenvalues are as follows:

103

1.25700.0000 + 0.0030i0.0000 0.0030i 0.0100 1.2399 0.0101 0.0520 0.0420

(34)

B,

C are the same as above, and M C , M O are computed the same way. Theresults of the analysis in the second steady-state point is as follows:

The linearized system is at the edge of stability(it has a pole-pair with 0 real-part)

The linearized system is controllable

The linearized system is observable

3.2 Relative degreeThe SISO (single input, single output) nonlinear system

x = f (x) + g(x)uy = h(x) (35)

has relative degree r at a point x0 if

1. LgLkf h(x) = 0 for all x in a neigborhood of x0 and all k > r 1

2. LgLr 1f h(x0) = 0

where Lgh(x) = dh (x )dx g(x) is the Lie-derivative of h(x) along g. Instead of performing the Lie-derivatives we can easily determine the relative degreeof a system, using graph-theoretic methods [2]. We have to determine thelength of the shortest directed path from the input to the output vertex inthe structure graph (see below) , and this length -1 is the relative degree of the system.

34


36/82

3.2.1 The structure graph

The structure graph of a system is constructed in the following way:

The vertices of the graph are the state-variables, the input variables,and the output variables.

A directed path leads form the vertex V 1 to the vertex V 2 if and only if the variable V 2 depends on V 1 (If V 2 is a state-variable, this means thatV 1 can be found in the state-equation describing the time derivative of V 2. An output variable depends on a state variable if and only if thestate variable can be found in its output equation).

Figure 15 shows the structure graph of our simple limb model.In our case the length of the shortest directed path from u1 (or form u2)

to y is 4, therefore the relative degree of the system is 3.

Figure 15: The structure graph

35


37/82

3.3 ControllabilityIn general case we study the controllability of the original nonlinear state-space model (see in equation (20)) with the tool of the controllability distri-bution ( c) (see [11]).

3.3.1 Algorithm for constructing the controllability distribution

1. Starting point

0 = span {g1, g2} (36)

2. Development of the controllability distribution

k = k 1m

i=0[gi , k 1] (37)

where:[gi , k 1] = span {[gi , 1], ..., [gi , l]}

where 1...l are the vectors spanning k 1, and [f, g ] denotes the Lie-product of f and g:

[f, g ](x) = g(x)x f (x) f (x)x g(x)

g (x )x is the Jacobian matrix of the function g with respect to its independent

variable x .

3. Stopping condition

If k = k 1, then c = k

If the dimension of the controllability distribution is equal to n (the numberof states), then the system is controllable.

In our case the symbolic expressions needed for computing of the con-trollability distribution are so complex, that it is impossible to determinethe Lie-brackets with the available methods, so we use another method forcontrollability analysis.

36


38/82

3.3.2 Structural controllability and observability

Structural controllability and observability [2] apply for a set of systems withthe same structure, i.e. with the same structure graph.

A set of linear or linearized systems with structure matrixes ([ A ],[ B ],[ C ])is structurally controllable or observable if

The matrix [ A ] is of full structural rank (maximal possible rank within the class specied by the structure matrix)

The system structure graph S is input connectable or output connectable.For controllability there should be at least one directed path from any of

the input variables to each of the state variables. For observability there should be at least one directed path from each of the state variables toone of the state variables.

Structural controllability implies that the points where the system is notcontrollable are forming a set with 0 measure.

3.3.3 Structural controllability analysis of the model

Because the rank of the matrix A of the linearized system is 8 in both in-vestigated steady-state points, its structural rank is 8. We can apply the

structural controllability method for the entire set of the locally linearizedsystems.We can see in gure 15 , that we can nd directed paths from the inputs

to the states, so the system is structurally controllable. This is in a goodagreement with the results of the linear controllability analysis (see section3.1).

3.4 Observability3.4.1 Algorithm for constructing the observability

co-distribution

Similarly to the controllability-analysis, in the general case we study the ob-servability of the model with the observability co-distribution o .

1. Starting point

0 = span {g1, g2} (38)

2. Development of the observability co-distribution

37


39/82

k = k 1m

i=0Lgi k 1 (39)

where: Lgi k 1 is the Lie-derivative of k 1 along gi , and the expansion tothe spanned sub-space is the same as above.

3. Stopping condition If k = k 1, then o = k

If the dimension of the controllability distribution is equal to n (the numberof states), the system is controllable.

3.4.2 Structural observability analysis of the model

Similarly to the controllability, we study structural observability. In gure15 we can nd directed paths from the states to the output, so the system isstructurally observable. This is in a good agreement with the results of thelinear observability analysis (see section 3.1)

38


40/82

3.5 StabilityWe can easily determine the stability of a linear system, by analyzing itseigenvalues as in section 2.1. In the nonlinear case, we use the Lyapunov-theory.

3.5.1 Lyapunov-stability

deniton:If (t) is a solution of the differential equation x = f (x, t ), we say that

the solution (t) is a Lyapunov stable solution, if for t0 [a, ) and for > 0 (, t 0) > 0, that if x(t) is a solution andx(t0) (t0) < (, t 0), then t [t0, ) x(t) (t) < .

3.5.2 Strong asymptotic -stability

deniton:If (t) is a solution of the differential equation x = f (x, t ), we say that the

solution (t) = 0 is a strong asymptotic-stable solution, if for t0 [a, )and for > 0 (, t 0) > 0, that if x(t) is a solution and

x(t0) (t0) < (, t 0), then t [t0, ) x(t) (t) 0.

3.5.3 Lyapunovs rst theorem

Let us suppose, that x0 is a steady-state point and V (x) is a positive denitescalar-valued function (Lyapunov-function). If

V (t, x ) = dV (t, x (t))

dt 0

then the steady-state point is Lyapunov-stable.

If V (t, x ) = dV (t, x (t))

dt < 0

then the steady-state point x0 is asymptotical stable in strong sense (seemore in [17]).

39


41/82

3.5.4 Estimating the stability region in the neighborhoodof the steady-state point with quadratic Lyapunov-functioncandidate

Let us take take the second steady-space point, and the following quadraticLyapunov function candidate:

V [x(t)] = ( x x )T Q(x x ) (40)

where x is equal to the vector (32), and Q is a diagonal unit weightingmatrix. This results in the Lyapunov-function candidate:

V [x(t)] = ( x 1 0.105)2 + x 22 + ( x 3 1/ 2 )2 + x 24

+ ( x 5 0.101)2 + ( x 6 0.1)2 + x 27 + x28 (41)

If we compute dV dt , we can give a conservative estimation of the stabilityregion. The points were dV dt is negative, belong to the strong asymptoticallystable region of the steady-state point.

dV dt

= V x

dxdt

= V x

f (x) (42)

where f (x) = f (x) + g(x)C (x) with C (x) beeing the static feedback law.In the case of the rst steady-state point C (x) = u1 = 0.0554.

At the steady-space point (32) the value of V t is of course zero, becausethe system is on the edge of stability.Along = x3, lT 1 = x5 and lT 2 = x6 the function dV dt does not change, becauseof the Hamiltonian properties of the system (the description of Hamiltoniansystems can be found in [9]).

We can study the behavior of dV dt in the neighborhood of the state-pointdescribed in (32), along the directions parallel to the axes x1, x2, x4, x7, x8.So, we can get cuts of the function (42). The following gures depict dV dt asa function of state-space variable pairs.

40


42/82


43/82

0.20.1

00.1

0.20.3

0.40.5

0.4

0.2

0

0.2

0.48000

7000

6000

5000

4000

3000

2000

1000

0

q 1vT1 [m/s]

d V / d t

Figure 19: The change of V t along vT 1 and q 1

3.6 Zero dynamicsWe call the systems behavior zero dynamics in the case, when its is outputforced to be identically zero (all time derivatives of the output are zero). Weneed to examine the zero dynamicss stability for controller design purposes.

Let us study the zero-dynamics for the input-output pair: u1 (the exormuscles activation signal), and h(x) = x3 / 2. In this case y 0 outputmeans 90 joint angle.

= x3 / 2 x3 = 0 = x4 0 x4 = 0

From equation (20):

x4 = 1

+ ml2COM (M + mlCOM cos(x3 + )gy) = 0 (43)

with = / 2 and = / 2 we get cos( + ) = 1.

Next we dene a constant denoted by H 1:

mlCOM gy .= H 1 < 0 where gy = 10m/s s .

Equation (43) can be equal to 0, if only if the joint torque M is equal toH 1, thus equation (43) implies

M = H 1

42


44/82

We know from equation (12) that:

M = M f M e = F flexor d F ext d = ( F CE 1 + F lig 1)d (F CE 2 + F lig 2)d

In the position = / 2 we can neglect the forces of ligaments (because theyonly appear at the position near to the maximal exed/extended states),and we suppose, that the extensor muscle is totally inactive, so we can alsoneglect F CE 2. This simplies the equation above to

M = F CE 1d = ( F max F L (lCE )F v (vCE )x1 + F P E )d

Passive force of the muscle does also not appear in this position, so we neglectit, and obtain the simplied equation for the joint torque:

M = F CE 1d = ( F max F L (lCE )F v(vCE )x1)d (44)

If / 2, d prox = ddist = 0 .2m and d = 0.01 we can compute the exormuscles length from (18) as:

lCE = 0.3368 x 5

So, we can determinate F L(lCE ), which will only depend on x5= lT 1 by usingequation (1).

Furthermore, if

1 lCEloptCE 1 +

then we can compute the exor muscles contracting velocity from the time

derivative of the muscle length:

vCE = dlCEdt

If we take = / 2, vCE will depend only on x7 = vT 1

Next we can compute F v(vCE ) by using the equation (3).

43


45/82

From equation (44) and M = H 1 we get:

H 1dF max F L (lCE (x5))F v(vCE (x7))

= x1.=

H 2F L (lCE (x5))F v(vCE (x7))

(45)

where H 2 is a newly dened constant.

In this case zero-dynamics means that the torques originating from the ten-dons dynamics have to be balanced by the change of muscle activation state.The states of tendon change the length (and so the contracting velocity) of muscles - in this way they change the muscles force. This means the changeof the joint torques. The muscle activation also changes the muscles force,balancing the equilibrium of the joint torques.

3.6.1 Zero dynamics state-space equations

With the simplications of the zero dynamics above, and from equation (20)we can write the zero-dynamics state-space model in the following form:

x1 = dq1dt = 1 act ( + [1 ]u1(t))

H 2F L ( lCE (x 5 )) F v (vCE (x 7 )) +

1 act u1(t)

x2 = dq2dt = 0

x3 = ddt = x4 = 0

x4 = ddt = 1

+ ml 2COM ((F max F L (lCE (x5))F v(vCE (x7))

H 2F L ( lCE (x 5 )) F v (vCE (x 7 )) )d + H 1) = 0

x5 = dlT 1dt = x7

x6 = dlT 2dt = x8

x7 = dvT 1dt = k t (x 5 lslackt )+ s T x 7 (F max F L ( lCE (x 5 )) F v (vCE (x 7 ))) H 2F L ( l CE ( x 5 )) F v ( v CE ( x 7 ))

zT

= k t (x 5 lslackt )+ s T x 7 + F max H 2

zT

x8 = dvT 2dt = k t (x 6 lslackt )+ s T x 8

zT (46)

44


46/82

3.6.2 Computing the zeroing input

If we know the form of q 1 = x1 as the function of time in equation (45), wecan compute its time-derivative:

dx1dt

= ddt

H 2F L (lCE )(x5)F v(vCE )(x7)

(47)

= H 21

(F L (lCE )(x5)F v(vCE )(x7))2

F v (vCE )(x7) ddt F L (lCE )(x5) + F L(lCE )(x5)

ddt F v (vCE )(x7)

By the derivation rule of composite functions we can compute ddt F L (lCE (x5))which will depend on x5 and x7.

Similarly, we can also compute ddt F v(vCE ).

If we susbstitue the results to (47) we can compute dx 1dt .

If we take the rs the rst equation of (46), and we know the form of x 1

t ,we can compute the zeroing input as follows:

dx1dt

= 1 act

x1 + 1 act

[1 ]x1 + 1 act

u(t)

and therefore

u(t) =dx 1dt +

1 act x1

1 act [1 ]x1 + 1 act

(48)

where act and are the same as in equation (5).

3.6.3 Zero-dynamics analysis results

If we apply the input described in equation (48) to the open-loop system, wecan notice that the output is indeed identically zero.

If we analyze the zero-dynamics in some points of the state-space (thiswas done mostly in steady-state points) we get the result that the linearizedzero-dynamics is stable in every investigated case. So we can suppose, thatthe zero-dynamics of the system is stable. This result is important for input-output type controller design.

45


47/82

4 Controller designThe aim of this chapter is to propose different controllers to the simple limbmodel and to investigate and compare their performance. For this purposewe return back to the original model ( -loop not included) described in equa-tion (20). In addition, we suppose that all of the state-space variables areavailable for the control system for a state feedback.

4.1 Performance specicationAny controller design starts with the explicite denition of the criteria forthe control systems performance.

4.1.1 Control aims

Based on the most simple tasks a human upper limb should perform, wedetermine three basic tasks for the control system:

Stability: We expect the closed loop system to be stable in the regionwe apply the selected control method.

Regulation and Trajectory following: We expect the applied controlsolving this two basic tasks in limb movement control.

Disturbance-rejection: We expect the controller being not seriouslysensitive for disturbances.

4.1.2 Control domain

Our aim is to control the output of the system in the neighborhood of = / 2angle. The values of other the state-space variables can change in the domain

described in section 2.6.3. The value of the actuation signal can changebetween 0 and 1 as described in 2.3.4 and 2.6.3.

4.1.3 Disturbances

We complete the model with three disturbances:

An external mass on the edge of the moving limb-part: This can be aweight lifted by the limb.

46


48/82

A simplied fatigue model - the drift of the act parameter: As with timethe ATP concentration decreases in the working muscle, the activationtime increases.

Spontaneous muscle activity: This phenomenon can be considered forexample, as a simplied effect of the Parkinsons disease.

The state-space equations, modied by the disturbances are in the fol-lowing form:

x1 = 1 act ( t ) ( + [1 ]u1(t)) x1 + 1

act ( t ) u1(t) + d1(t)

x2 = 1 act ( t ) ( + [1 ]u2(t)) x2 + 1

act ( t ) u2(t) + d2(t)

x3 = x4

x4 = 1+ ml 2COM (M (x1, x2, x3, x4, x5, x6, x7, x8)

+ mlCOM cos(x3 + )gy + mD lfa cos(x3 + )gy)

x5 = x7

x6 = x8

x7 = k t (x 5 lslackt )+ s T x 7 F flextor (x 1 ,x 3 ,x 4 ,x 5 ,x 7 )

zT

x8 = k t (x 6 lslackt )+ s T x 8 F extensor (x 2 ,x 3 ,x 4 ,x 6 ,x 8 )

zT

(49)

where mD is the external mass, lfa is the length of the moving limb-part (theforearm), d1(t) and d2(t) are the spontaneous muscle activities as functionsof time.

47


49/82

Figure 20: The system with the external mass

4.2 Linear SISO LQ regulatorIn order to design a linear SISO (single input, single output) LQ regulator[17] we use the linearized system model obtained around the steady-statepoint at / 2 rad and the centered variables x = x x0 u = u u0, as inequation (26). We use the exor muscles activation signal as the only inputto the system. The other input, u2 is set to be identically zero.

x1(0) = 0 .4031x2(0) = 0x3(0) = / 2x4(0) = 0x5(0) = 0 .101439

x6(0) = 0 .1x7(0) = 0x8(0) = 0

(50)

u0 = 0.2528

We can apply an LQ servo controller (see below) to minimize the cost func-tion:

(x Qx + u Ru) dt (51)48


50/82


51/82

4.2.1 Simulation results of the SISO LQ conroller

In this case (used as reference case) no disturbances were used. In the fol-lowing gure we can see the reference signal and the output of the system.On the gures the real (decentered) values of the variables are shown.

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a [ r a d ] , r [ r a d ]

alfar reference signal

Figure 21: The output and the reference signal

In this case the reference signal was r(t) = 2 + 16 sin (3t)

The muscle activation states ( q 1, q 2) and the muscle activation signals (oractuation signals - u1, u2) are depicted in the following gures:

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

q1,q2

t [s]

q 1 , q 2

q1

q2

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

u1,u2

t [s]

u 1 , u 2

u1

u2

Figure 22: Muscle activation states and muscle activation signals

If we use the same reference signal, and the following matrices for thepenalty function,

50


52/82

Q =

1 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 0.1 0 0 0 0 0 00 0 0 0.01 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 1000

(54)

R = 1we get the results depicted on the following gures:

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a [ r a

d ] , r

[ r a

d ]


0 0.5 1 1.5 2 2.50

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

q1,q2

t [s]

q 1 , q 2

q1q2

Figure 23: The output, the reference signal and the muscle activations

We can see, that the closed loop system follows the reference signal withmore delay. The appearance of the delay in the trajectory following can beexplained with the systems relative degree (3). In other words between theinput and the output 3 integrators can be found.

51


53/82

4.3 Linear MISO LQ regulatorWe are able to use both of the muscle activity signals as input, and extendthe LQ-servo controller to the MISO (multiple input, single output) system,for more efficiency and roboustness with the same linearization method as insection 4.1.

We use the same Q as in equation (54) for the penalty function, and:

R = 10 00 10 (55)

Solving the Ricatti-equation with MATLAB, we get the following value forK:

K = 0.554 0.498 3.924 0.137 0.702 0.011 0.003 0 62.934

0.632 0.730 4.885 0.174 0.942 0.008 0.003 0 77.713

4.3.1 Simulation results of the MISO LQ controller without dis-

turbanceIn this case the reference signal was the same as above -

r (t) = 2

+ 16

sin (3t)

In the following gure we can see the reference signal and the output of the system:

52


54/82


55/82


56/82

4.3.3 Regulation results of the MISO LQ controller with distur-bance

If the reference signal is constant, we apply d1(t) and d2(t) for generatingspontaneous muscle activation for disturbance. This method can be consid-ered, for example, as a simple model of the effect of the Parkinsons disease.

d1(t) = |1.5 sin (14 t) sin (16 t)|d2(t) = |1.5 sin (15 t) sin (17 t)|

The results are depicted in the following gures:

0 0.5 1 1.5 2 2.51.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a [ r a d ] , r [ r a d ]



0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

q1,q2

t [s]

q 1 , q 2

q1q2

0 0.5 1 1.5 2 2.50.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.22

0.24

u1,u2

t [s]

u 1 , u 2

u1u2

Figure 29: Muscle activation states, and activation signals

55


57/82

4.4 Input-Output linearizationIn this section we use only the exor muscles activation signal as input tothe system (the extensor muscles activation signal is identicaly zero), to geta SISO (single input single output) structure.

As described in [9] for a nonlinear n-dimensional SISO system with rela-tive degree r < n we need to apply the feedback

u = 1

Lg Lr 1

f h(x)( Lrf h(x) + v(t)) (56)

and a suitable nonlinear co-ordinate transformation to obtain:

a linear subsystem of order r which is inuenced by the chosen inputu - including the external input v(t) and

a nonlinear subsystem described by the zero dynamics

This means, that the state-space model of the input-output linearized closedloop system is the following in the new co-ordinates:

z 1 = z 2 z r +1 = q r +1 (z )z 2 = z 3 z r +2 = q r +2 (z )

... ...z r 1 = z r z n = q (z )z r = v y = z 1

(57)

where Lf h(x) denotes the Lie-derivative of h(x) along f .In our case r=3, so a three dimensional subsystem will be linear in the new

coordinates z 1, z 2, z 3. z 1, z 2, z 3 can be determined by using the co-ordinate-transformation z i = Li

1f h(x).

The remaining part is described by the zero dynamics, which has to bestable to apply the control. In section 3.6 we have investigated the stabilityof the zero dynamics and have found that is stable.

After input-output linearization (IOL) we can apply any kind of the linearcontrollers, for example a pole-placement control (PP) for v(t) as the newinput, with the correction of the reference signal r(t) as described in [17].The signal ow diagram of the pole-placement servo controller applied tothe input-output linearized system is depicted in gure 30 . If we choosethe poles to P 1,2,3 = 50, we get the the result K = [125000 7500 150],Nx = [1;0; 0], Nu = 0.

56


58/82

If we use poles with lower magnitude, the reference following will be lessaccurate (more delay appears), and if we use greater ones, the actuationsignal will be larger. This contradicts to the control requirements describedin section 2.3.4 (the value of the actuation signal can be only in the range of [0, 1]).

Figure 30: Input-output linearization and pole placement control withreference signal

57


59/82


60/82

4.4.2 Trajectory following results of the IOL-PP controller withdisturbance

If we use the same disturbances as in section 4.3.2 and the same referencesignal with the poles P 1,2,3 = 50 we get the following results:

Figure 33: and the reference signal by input-output linearization control

with pole-placement

0 0.5 1 1.5 2 2.50

0.02

0.04

0.06

0.08

0.1

0.12

0.14

q1,q2

t [s]

q 1 , q 2

q1q2

0 0.5 1 1.5 2 2.50.05

0.055

0.06

0.065

0.07

0.075

0.08

u1

t [s]

u 1


59


61/82

4.4.3 Regulation results of the IOL-PP controller with distur-bance

We use the same functions for d1(t) and d2(t) as described in subsection 4.3.3,and set the poles to P 1,2,3 = 50. The results can be seen on the followingtwo gures:

0 0.5 1 1.5 2 2.5

1.6

1.7

1.8

1.9alfa,reference signal

t [s]

a l f a

, r e f s .

alfareference signal

Figure 35: and the reference signal by input-output linearization control

and pole-placement

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

q1,q

2

t [s]

q 1 , q 2

q1

q2

0 0.5 1 1.5 2 2.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

u1

t [s]

u 1


The activation of the extensor muscle originates only from the used dis-turbance.

60


62/82

4.5 Fuzzy Inference SystemSoft-computing methods are also quite prevalent for controlling nonlinear sys-tems [16], because of the application of these methods do not need the explicitknowledge of mathematical model of the system. We study the acceptabil-ity of a simple Fuzzy-controller for the model, with rules and membershipfunctions designed in intuitive and experimental ways.

4.5.1 Basic concepts in fuzzy theory

At rst, we dene the basic concepts used in the theory of fuzzy systems,and fuzzy control.

Fuzzy set:The fuzzy set A denes on a set X the degree of belonging to A for the ele-

ments x X . A can be dened with the pair ( X, A ), where A : X [0, 1]is the membership function of A.

The main difference between the fuzzy and the conventional set theory isthe continual membership function. In conventional set theory an elementis either member of the set, or it is not. In fuzzy set theory for example avalue of a temperature-variable can be the member of the warm set with 0.5

degree.

Operations on fuzzy sets:In our case we use the following operations:

Union: A B (x) = max (A (x), B (x))

Intersection: A B (x) = min (A (x), B (x))

Complement: A = 1 A (x)

Direct product of fuzzy sets:Let A1 be a fuzzy set on the set X i with the membership function A i (xi ),

Ai = ( X i , A i ), i = 1, 2..., n . The membership function of the fuzzy directproduct A1 A2 ... An is dened by

A 1 A 2 ... A n (x1, x2,...,x n ) = A 1 (x1) A 2 (x2) ... A n (xn )

where is a suitable T-norm, for example min (see more in [16]).

61


63/82

Fuzzy relation:

R : X 1 X 2 ... X n [0, 1]R fuzzy relation (X 1 X 2 ... X n , R )

where R (x1, x2,...,x n ) denes the possibility of the elements x1, x2,...,x nbeing in relation R.

Cylindrical extension:

If the fuzzy relation R is dened on X i 1 ... X i r where (i1,...,i r ) 1,...,n ,then the cylindrical extension of R to the set X 1 ... X n is

ce (R )(x1,...,x n ) .= R (xi 1 ,...,x i r )

Projection:If R is a fuzzy relation dened on X 1 ... X n , the projection of the relation

R to the space X i1 ... X i r is a relation P = P roj X i 1 ... X i r , which relationsmembership function is with the notation { j1,...,j n r } = {1,...,n }/ {i1,...,i r }

P (xi 1 ,...,x i r ) .= sup

x j1

,...,x jn r

R (x1,...,x n )

Fuzzy composition:Let it be

X 1 ... X m 1 X m 1 ... X r the set of the fuzzy relation RX m ... X r X r +1 ... X n the set of the fuzzy relation SX 1 ... X m 1 X r +1 ... X n the set of the fuzzy relation R S

We get the fuzzy composition with cylindrical extending ( ce) both of therelations to the X 1 ... X n product-space, take the intersection of therelations, and project the result to the X 1 ... X m 1 X r +1 ... X nproduct-space.

R S = P roj X 1 ... X m 1 X r +1 ... X n [ce(R) ce(S )]

R S (x1,...,x m 1, x r ,...,x n ) = sup x m ,...,x r (R (x1,...,x r ) S (xm ,...,x n ))

Fuzzy composition is important in fuzzy controller design, becouse (as wewill later see) this method is the connection between the rules of a controller,and the input data.

62


64/82

Fuzzy rule base:We use Mamdani-implication: a b a b

Implication: R: If x is A, then y is B. Ri = A B is a relation overX Y .

We have to note that this method is not compatible with the Bool-algebra,but it can be easily computed, and it is used in the most case in the practice.

The projection of the relation R = A B to the Y fuzzy set gives thecomposition rule of the fuzzy implication :

Proj Y (R) = P roj Y (A B) = A B

where is the fuzzy composition.

The general form of the fuzzy implication can be described with the R1, R 2,...,R nrelations ( R = R i ) of the rule-base.

R i : if x 1 is X i1 and ... and x n is X in then y is Y

i

In the case of fuzzy control, the R = R i rule base (the summation of the

if...then... implications) from the linguistical statements of the professionalknowledge.

Matching data to a relation:Let x j the measured value of the variable x j . Let x j match the A j fuzzy

set. In the case of Ri , let the fuzzy set A j be dened on the set X i j .

In this case

The cylindrical extended values of the input data:

D = ce(A1) ... ce(An ), D (x1,...,x n , y) = A 1 (x1) ... A n (xn )

The fuzzy form of the Ri Mamdani implication (as a relation):

R i = ce(X i1) ... ce(X in ) ce(Y i ),R i (x1,...,x n , y) = X i1 (x1) ... X in (xn ) Y i (y)

63


65/82

In the general case of a fuzzy controller x could be, for example, the differ-ence between the reference signal, and the output of the system, and y couldbe the actuation signal.

The composition rule of the fuzzy implication (using that ce(A j ) ce(B j ) = ce(A j B j ) in the case of identical sets):

D R i = P roj Y (D R i )D R i (y) = sup

x 1 ,...x N D (x1, ...xN , y) R i (x1, ...xN , y) =

supx 1 ,...x N [A 1 (x1) ... A N (xN )] [X i1 (x1) ... X

iN (xN ) Y

i

(y)] =sup

x 1 ,...x N [A 1 (x1) X i1 (x1)] ... [A N (xN ) X iN (xN )] Y i (y) =

[supx 1

A 1 (x1) X i1 (x1)] [supx N A N (xN ) X iN (xN )] Y i (y)

we can dene the

ij = supx j

A j (x j ) X ij (x j )

i = i1 ... iN

ring rates, soD R i (y) = y Y i (y)

Figure 37: The illustration of the rule If x1 is xi1 and x2 is xi2 then y is yi

64


66/82

The Mamdani min-max implication algorithm:

1. Evaluate ij j = 1,...,N ring rates, and i for all Ri relation

ij = X ij (x j ) i = min j ( ij )

2. evaluate D R i for all y

D R i (y) = Y i (y) if Y i (y) < i i otherwise

3. Evaluate the output of the fuzzy system for all y

D R (y) = maxi

(D R i (y))

65


67/82

4.5.2 The applied fuzzy controller

We used the Mamdani min-max implication algorithm described above.We used the following variables as input for the fuzzy inference system(weighted with constants):

The difference between the reference signal and the output:r (t) x3(t) = r (t) (t)

The joint angle velocity: x4= (t)

The exor muscles activation state: x1(t) = q 1(t)

The extensor muscles activation state: x2(t) = q 2(t)

Where x1, x2, x3 and x4 denotes the state-space variables of the model (notthe input of the fuzzy inference system).

The output of the fuzzy inference system were the muscle activity signals(weighted with constants).

We used the following membership functions as input ( r (t) wasweighted with 10):

Figure 38: The membership functions of r(t)

66


68/82


69/82

Figure 42: The membership functions of u1

Figure 43: The membership functions of u2

We used the following rules:

If r is ps, then u1 is 5 and u2 is 1

If r is ns , then u1 is 1 and u2 is 3

If r is pb, then u1 is 6 and u2 is 1 If r is nb, then u1 is 1 and u2 is 4

If r is not nb and omega is pb, then u1 is 1 and u2 is 6

If r is not pb and omega is nb, then u1 is 7 and u2 is 1

If r is z and omega is ps, then u1 is 3 and u2 is 5

If r is z and omega is ns, then u1 is 3 and u2 is 1

68


70/82

If r is not pb and q 1 is big, then u1 is 2 and u2 is 4

If If r is not nb and q 2 is big, then u1 is 5 and u2 is 2

4.5.3 Simulation results of the fuzzy controller without distur-bance

If we use the same reference signal as in 4.3.1, and apply the fuzzy controllerdescribed above, we get the following results:

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a

[ r a

d ] , r

[ r a

d ]


Figure 44: The output and the reference signal by fuzzy control

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

q1,q

2

t [s]

q 1 , q 2

q1q2

0 0.5 1 1.5 2 2.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

u1,u2

t [s]

u 1 , u 2

u1u2


69


71/82

4.5.4 Trajectory following results of the Fuzzy controller with dis-turbance

In the next two gures, we can see the effect of disturbance to the fuzzycontrol system. The applied disturbance was the same as in subsections4.3.2 and 4.4.2 with the difference, that the external mass was only 0.333 kg.

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a [ r a d ] , r [ r a d ]



0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

q1,q2

t [s]

q 1 , q 2

q1q2

0 0.5 1 1.5 2 2.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

u1,u2

t [s]

u 1 , u 2

u1u2


We can see, that the fuzzy control system is more sensitive to distur-bances.

70


72/82

4.5.5 Regulation results of the Fuzzy controller with disturbance

The applied disturbance, the spontaneous muscle activity, was the same asin subsections 4.3.3 and 4.4.3.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11.65

1.7

1.75

1.8

1.85

1.9

1.95

2alfa,r

t [s]

a l f a [ r a d ] , r [ r a d ]



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

q1,q2

t [s]

q 1 , q 2

q1q

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

u1,u2

t [s]

u 1 , u 2

u1u

2


71


73/82

4.6 Comparison of the different control methodsWe compared the results of the MISO LQ regulator designed for the locallylinearized model, the IOL-PP control, and the fuzzy control. The main com-parison viewpoints were accuracy, output signal smoothness and disturbancesensitivity.

4.6.1 Comparison of the results without disturbance

If we compare the trajectory following of the different control methods inideal environment, we get the following results:

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8

alfa,r

t [s]

a l f a [ r a

d ] , r

[ r a

d ]


0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8

alfa,reference signal

t [s]

a l f a

, r e

f s .

alfareference signal

Figure 50: Trajectory following with lin. MISO LQ and IOL-PP control

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a [ r a d

] , r

[ r a

d ]


Figure 51: Trajectory following with fuzzy control

72


74/82

4.6.2 Comparison of the results with disturbance

Let us compare the results of the most difficult task (the trajectory followingwith disturbance) for the three investigated controllers, depicted in the guresbelow:

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8

1.85alfa,r

t [s]

a l f a

[ r a d

] , r

[ r a

d ]


Figure 52: Trajectory following with lin. MISO LQ and IOL-PP control

0 0.5 1 1.5 2 2.51.35

1.4

1.45

1.5

1.55

1.6

1.65

1.7

1.75

1.8alfa,r

t [s]

a l f a [ r a

d ] , r

[ r a

d ]


Figure 53: Trajectory following with fuzzy control

73


75/82

4.6.3 Comparison results: a discussion

In the gures 50 and 51 we can see, that in ideal circumstances (i.e. withoutdisturbances) all of the the LQ regulator, the IOP-PP and the fuzzy con-troller provides acceptable output, that is within the tolerance limit. TheLQ regulator and the IOP-PP provides more smooth output than the fuzzycontroller, but we have to note, that the fuzzy controller requires only 4state-space variables for computing the feedback (the other methods requireall 8 state-space variables for the feedback), and does not require the mathe-matical equations of the model. On the other hand, the required computingcapacity is the highest the case of the fuzzy controller.

In the gures 52 and 53 we can observe, that the input-output linearizedsystem is more stable in the case of the appearing disturbances, and providesa more smooth and more accurate output signal, than the other solutions.We have to note, that the fuzzy controller was tested mainly in ideal circum-stances, and designed mainly for trajectory following. Probably the tuningof the membership functions and rules would lead to more roboustness in anenvironment accused with disturbances in both cases of the control tasks.

Moreover, we can note, that in some cases of the fuzzy and the MISO-LQcontrol of the locally linearized system, the control uses both of the musclesin the same time. This method (using both muscles at the same time) is notregarded to be the optimal, if we consider, for example, the ATP consumptionin the muscles. Because of the extensor muscles activity the exor musclehas to work on a higher activity level. Still this method of the linear controlprovides more roboustness (compared to the SISO-LQ control methods).

74


76/82

5 Conclusions and future workWe have seen that the methods of nonlinear systems and control theory canbe successfully applied to the investigated simple limb model. The problemthat we had to solve in the case of the applying nonlinear control (input-output linearization) was the complexity of the functions needed for thelinearizing feedback. In fact, both of the control tasks (trajectory followingand regulation) can be better solved with nonlinear control methods, as itcan be clearly seen in the results of the simulation. On the other hand, if the expectations are not too high or the computing capacity is low, a linearcontrol also can be successfully applied. If the model equations or some

important parameters are not available, a fuzzy control can also be applied.The possible tasks for the research in the future could be for example:

Extend the state-space model for more muscles and/or joints.

Extend the model to 3D.

Extend the control to the system including the gamma-loop.

Utilize the Hamiltonian properties of the system for a nonlinear PDcontrol.

Utilize the properties of the linearized subsystem for an adaptive fuzzycontrol.

Explore the similarities and differences of the control of the model, andthe physiological control of a real human limb, search the control loopsin the central nervous system.

Analyize the muscle activity states generated by the model in the caseof various movement patterns, and compare them with EMG signalsrecorded on real patients doing the same movements, for the develop-ment of the model to a more realistic form.

Extend the control with a simple model of adaptation motor learning,for example as an adaptation to changes in the model parameters or inthe disturbance parameters.

75


77/82

6 AcknowledgementsI would like to say thanks for helping my work to the following people, andinstitutes: My supervisor Prof. Katalin Hangos, my department consultantDr. Istv an Harmati, Csaba Fazekas, the Systems and Control Laboratory atthe Computer and Automation Research Institute Budapest, Dr. Peter V an.

76


78/82

7 Appendix

7.1 The gamma-loop

Figure 54: The servomechanical -loop

7.1.1 Muscle spindles

Muscle spindles are found within the eshy portions of muscles, embeddedin so-called extrafusal muscle bers. They are composed of 3-10 intrafusalmuscle bers, of which there are two types, nuclear bag bers and nuclearchain bers and the axons of sensory neurons. Axons of motoneurons alsoterminate in muscle spindles; they make synapses at either or both of theends of the intrafusal muscle bers and regulate spindle sensitivity. Musclespindles are encapsulated by connective tissue, and are aligned parallel toextrafusal muscle bers, unlike Golgi tendon organs, which are oriented inseries.

77


79/82

The muscle spindle has both sensory and motor components. Primary andsecondary sensory bers spiral around and synapse on the central portionsof intrafusal ber

Documents

Thesis - Analysis and Control of Simple Nonlinear Limb Model