# Sensor Fault Diagnostics of a Flexible Link Robot using Linear Matrix Inequalities

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#### Sensor Fault Diagnostics of a Flexible Link Robot using Linear Matrix Inequalities

Pradeep S. B

Department of EEE, Barton Hill Engineering College, Thiruvananthapuram

Lineesh A. S

Department of EEE, Trinity College of Engineering, Thiruvananthapuram

AbstractA stable observer design for Lipschitz non-linear systems like flexible link Robot is considering using Linear Matrix Inequalities. The methodology is applied to the system with three or more sensors in which state is observable through any one of the sensor measurements. By using this observer gain, we can find the residues, which are used for fault identification. Distance to unobservabilty and its relation to observer design is also considering here. A stable Observer is also designed for the flexible link robot by this method.

1. INTRODUCTION

Observer design for nonlinear systems has been a very

for design of the observer gain. This paper also presents the observer design for flexible link robot using distance to un observability technique.

2. FLEXIBLE LINK ROBOT

Consider a onelink manipulator with revolute joints actu- ated by a motor as shown in The elasticity of the joint can be well-modelled by a linear torsional spring. The elastic coupling of the motor shaft to the link introduces an additional degree of freedom. The states of this system are motor position and velocity, and the link position and velocity.

active field of research during the last decade. The introduction of geometric techniques led to great success in the development of controllers for nonlinear systems. Many attempts have

m

= m

(2)

been made to achieve results of equally wide applicability for state estimation. But the observer problem found to be more difficult than the controller problem. Krener and Respondek

1 = 1 (3)

(1985), Krener and Isidori (1983) and Xiao-Hua and Gao (1985) attempted a coordinate transformation so that the state estimation error dynamics were linear in the new coordinates.

m

K

= (1

Jm

m

B

m

m

) J m

K

+ u (4)

Jm

Necessary and sufficient conditions for the existence of such a coordinate transformation have been established but in practice

1

K

1

1

= J (1

m

mgh

1

1

) J sin 1

(5)

are extremely difficult to satisfy. Even if these conditions are satisfied, the construction of the observer still remains a difficult task due to the need to solve a set of simultaneous partial differential equations to obtain the actual transformation function.

This present paper assumes partial state measurement and concentrates on developing a fault diagnostic methodology for systems with Lipschitz non-linear dynamics. A continuous system is said to be globally Lipschitz if the derivative at all points of the function has absolute value, which is always less than a definite real number. That definite real number is called Lipschitz constant . The solution is unique and existing if the noninear function (x, u) satisfies the Lipschitz condition given by,

where, 1 is the link angle, m is the motor angular position, 1 is the link angular velocity and m is the motor angular velocity.

The system dynamics are nonlinear and of the form

x = Ax + (x, u) + g(y)u (6)

y = Cx (7)

(x, u) (x, u) (1)

x x

l A fault diagnostic technique using Linear Matrix Inequal- ity(LMI)for Robotic system with Lipchitz nonlinear dynamics is presented here. The system use three sensors, in which the state is observable through any one of the sensor mea- surements. The observer gain matrix is calculated for finding the residues in the system. Fault is determined using these residues. The paper not only provides sufficient conditions on the observer gain matrix for enabling isolation of faults but

Fig. 1. Robotic system with flexible joint

also develops an explicit and systematic numerical procedure

m

0 1 0 0

Also,

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where x = , A =

where x = , A =

m 48.6 1.25 48.6 0

C 0 1 0 0

1 0 0 0 1

m 19.5 0 19.5 0

Here

x

is the state vector and

A

is the system matrix.

Here

x

is the state vector and

A

is the system matrix.

1

M1 = =

M1 = =

C1A2 C1A3

C1A2 C1A3

60.75 1.5625 60.75 0

76 59 76 60.75

60.75 1.5625 60.75 0

76 59 76 60.75

C1A 48.6 1.25 0 0

Here

x

is the state vector and

A

is the system matrix.

Here

x

is the state vector and

A

is the system matrix.

0 0

21.6

0

21.6

0

(12)

g(y) = 0 , (x, u) = 0

0

0

C2

0 0 1 0

= =

= =

1 0 0 0

0 1 0 0

1 0 0 0

0 1 0 0

2

2

C2A2 C2A3

C2A2 C2A3

60.75 50.2 60.75 48.6

3387.42 2 3387.42 60.75

60.75 50.2 60.75 48.6

3387.42 2 3387.42 60.75

3.33 sin(x3) M C2A 48.6 1.25 48.6 0

and

C =

and

C =

and

C =

and

C =

The three sensors are

Two vectors orthogonal to null space of both M1

and

(13)

M2 are

y1 = C1x + v1 (8)

1

0

0

0

1

1

y2 = C2x + v2

(9)

v1 = 0 , v2 = 0

0 0

z = Ex + vz (10)

If there is no fault in the system, then both residues are zero.

The other vectors can be used in the LMI formulation are,

0 0

0

0

0

0

In the presence of a fault in the sensor z, both residues R1

v3 = 1 , v4 = 0

and R2 are nonzero.

Assumptions

1. The nonlinearity (x, u) is Lipschitz with a Lips-

0

The basis matrices for P are

1 0 0 0

0 0 0 0

P1 =

P1 =

1

0 0 0 0

, P2 =

, P2 =

0 1 0 0

chitz constant .

2. Only one of the sensors fail at a given time.

0 0 0 0

0 0 0 0

3. The signals v1(t), v2(t) and v3(t) are assumed to be zero, when the corresponding sensors are healthy.

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 1 0

If the observer gain matrix L is chosen such that,

If the observer gain matrix L is chosen such that,

Theorem 1

P3 = 0 0 1 0 , P4 = 0 1 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

1) LMI of equation (11) is satisfied

0 0 0 0

0 0 0 0

2) M1L = 0, M2L /= 0

P5 = 0 0 0 0

1. The nonlinear function (x, u) doesnt span the entire subspace that is spanned by the vector L orthogonal to null space of M1 and M2. Then the residues R1 and R2 grow in the unique subspace cor- responding to each fault in any of the three sensors.

Thus

1

0

0

0 0 0 1

0 0

0

0

0

0

0

0

0

P1v1 = 0 , P1v2 = 0 , P1v3 = 0

3. LINEAR MATRIX INEQUALITY (LMI) FORMULATION

0

0

= 0

= 0

l=1

Q 1

l=1

Q 1

0

0

0

0

The basic Linear Matrix Inequality (LMI) form is

0 0 0 0

q+s

T

i=1

i=1

r+p

T T

0

2

1

2

1

0

1

1

0

0

0

0

0

0

xi(A Pi + PiA) + 1(E ul + ulE) 1 Q > 0

(11)

2

2

2

2

2

2

4

4

3

3

1

1

(11)

P v , P v

= , P v

= , P v

=

0

0

0

0

0

0

0

0

Where Q = q+s xiPi, L = r+p ivj , = H1b,

i=1

j=1

i=1

j=1

i=1

i=1

H = (pv1 pv2 Pv3) and P = q+s xipi

0 0 0 0

0

1

0

0

0

0

0

0

0

1

0

0

In the equation 11 the matrices P and L are written in

P3v2 = 0 , P3v3 = 0 , P3v4 = 0 , P4v1 = 0

terms of their basis vectors.

The variables r, q, s, p are such that

dim(pivj )i=1,2,Â·Â·Â·q,j=1,2,Â·Â·Â·r = r

0 0 0 0

0

0

0 0 0 0

0

1

0

0

0

1

0

0

dim(pivj )i=1,2,Â·Â·Â·q+s,j=1,2,Â·Â·Â·r+p = r + p

P4v2 = 1 , P4v3 = 0 , P4v4 = 0

0

0

0

0

0

0

, P5v1 =

0

0

= r = 3, q = 3 and r + p = 4 = p = 1, s = 2

0

0

0

0

= r = 3, q = 3 and r + p = 4 = p = 1, s = 2

0

0

0

0

0

0

0

0

0

0

2 = 16 36.253

74.5 > 0 3

> 38.20

2

2

P5v2 = , P5v3 = 3

0 0

0 0

The basis vectors for this space are

Select 2 = 16.50

or 3 < 1.95 (17)

1

1

1

1

1 0 0 0

0

1

0

0

0

1

0

0

0.30

3

3

4

4

=

3

3

4

4

=

u1 = 0 , u2 = 0 , u3 = 1 , u4 = 0

0

0

0

1

0

0

0

1

Thus, u

+ u

2

2

3

4

3

4

+ u

+ u

16.5

(18)

0

0

0

1

0

0

0

1

5

5

x1 0 0 0

0

0

i=1

0 x4 x3 0

0 0 0 x5

i=1

0 x4 x3 0

0 0 0 x5

2

2

Select the observer gain matrix L such that the matrix ALE

ie, 4 + (b1 + c1 + d1 + 6.13)3 + (2b1 + 27.02c1 +

ie, 4 + (b1 + c1 + d1 + 6.13)3 + (2b1 + 27.02c1 +

P = xipi =

i=1

i=1

0 x2 x4

(14)

is stable

ie, 4 + (b1 + c1 + d1 + 6.13)3 + (2b1 + 27.02c1 +

ie, 4 + (b1 + c1 + d1 + 6.13)3 + (2b1 + 27.02c1 +

i=1

i=1

Thus the matrix P is positive definite if

x1 > 0 , x2 > 0 , x3 > 0 , x5 > 0 , x2x3 > 0 , x2 > 0

2.25d1 + 6.526)2 + (117.262c1 + 98.45d1 + 42.33b1 +

247) + 65.16b1 161.755c1 + 145.8d1 34.325 = 0

1

1

4 This is possible if d

> 22.60 or d1

< 162.48

1. To find the new Lipchitz constant

The Lipschitz condition is

(x, u) (x, u) =

x x

0.333| sin(x3 sin(x3))|

d1 > 142.86 or 3.35 < d1 < 5.723

d1 > 60.15 or d1 < 5.1311

d1 > 14.205 or d1 > 10.85

Select d1 = 143 , c1 = 1 , b1 = 16.5 0.25c1 = 16.75

Also the fourth order leading principle minor is positive

j(x1

x1)2) + (x2

x2)2) + (x3

<

)

)

x3)2)

2

0.333 sin(

(x3 x3) 2

2

2

) cos(

(x3 + x3) 2

< 0.333 =

x5 2.483×5 + 0.3608 < 0 0.1549 < x5 < 2.329

Select x5 = 0.42 4 = x44 = 143 Ã— 0.42 = 60.06

1

1

x3 x3

2

2

2 16.5

2 16.5

x1

1.123

0.30

x4

x5

x4

x5

0.25

0.42

0.25

0.42

x2 1

4.11

60.06

4.11

60.06

Equation 11 is reduced into

P =

P =

Hence, x = x3 = 0.07

, = =

P =

P =

2 2

2 2

4 1 x1 1 + 32 x1 + 48.6×2 D G

1

2.44

3

4

3

4

1 + 23 + 48.6×4 B E H

1 + 24 19.5×5 C F I

1 + 23 + 48.6×4 B E H

1 + 24 19.5×5 C F I

1 + 22 x1 + 48.6×2 A B C

> 0 (15)

Where A = 2.5×2 + 22 1 2(x2 + x2),

and L = =

For simulation, set frequency f = 1 Hz, so that u =

2 16.72

2 16.72

3

4

3

4

1

143

1

143

0.5 sin(2t). The actual and observed link angle, link velocity,

B = 3

+ 2

+ 1.25×4

48.6×2

2

2×4(x2

4

+ x3),

motor angular position and angular velocity are given below.

C = 4 + 2 x4, 2D 2 = 2 1 + 23 + 48.6×4 + 2,

E = 23 197.2×4 (x3 +x4), F = 3 +4 x3 +19.5×5,

5

5

G = 1 + 24 19.5×5 + 2, H = 3 + 4 + 19.5×5 x3, I = 24 2×2 1

2 2

2 2

Equation 15 implies that

41

1 x1 > 0 1

1.1678

> = 0.2919

4

Let 1 = 0.3, x1 = 0.1230, x2 = 1, x4 = 0.25

4

4

Then x2x3 x2 > 0 x3 = 0.07

The second and third order principal minors are positive implies

Fig. 2. Actual and Observed link axis Vs time

24.341 < 2 < 16.3 (16)

Fig. 3. Actual and Observed link axis Vs time

The observed motor angular rotation is obtained as

m(t) =

0.1727 cos(6.28t 1.4743) + 1.2113e0.193t cos(1.5015t

0.4634) 0.3653e0.4322t cos(8.276t 0.4281)

0.00098e0.10138t sin(41.668t 1.506)

0.00613e117.828t 0.002145e15.526t

Fig. 4. Actual and Observed motor angular position VS time

Fig. 6. Actual and Observed motor angular position VS time under fault

condition

Fig. 7. Actual and Observed motor angular velocity VS time under fault

condition

F+C) unobservable. It is defined mathematically as

C

C

(A, C) = minRmin (jI A\ (19)

1. Distance to Unobservability of Original System

The minimum singular value,

C

C

min (jI A\ = min((jI A)(jI A) + CT C)

Fig. 5. Actual and Observed motor angular velocity VS time

When fault occurs, the signal v (t) cannot be zero. Set v (t) =

(20)

We can find the minimum singular value for different values of . Solve the equation,

|I ((jI A)(jI A) + CT C)| = 0 (21)

That implies (a) when = 0, min = 0.47, (b) when

= 0.5, min = 0.5031, (c) when = 1.2, min = 0.7921,

(d) when = 3, min = 1.1347, (e) when = 5, min = 1.06654, (f) when = 8, min = 0.9035, (g) when = 12, min = 1.3486, (h) when = 20, min = 4.2866

2. Distance to Unobservability of the New system obtained by Similarity Transformation

z z

1.5 sin(500t). Thus under fault condition, the motor angular position and velocity are estimated are estimated and plotted as shown below. In this case, the residues are nonzero.

4. DISTANCE TO UNOBSERVABILITY

The distance to unobservability ofa pair (A,C) is defined as the smallest perturbation (E,F) that makes the pair (E+A,

The transformation X = T0z is used to obtain a new system with new Lipschitz constant. From the original state equation.

z = T01AT0z + T01(T0Z, u) + T01g(y)u (22)

is obtained, where

T0 = diag(1, 1, 1, )

Let = 20, then

2) I2 + 0 0

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I =

1

20 Ã— 3.33 = 0.1665

3) 0 > max(0, 0)

1. AIT P + PAI + PP + (I 2 (I 2 + 0)CIT CI

+ 0)I

Let A1 = T01AT0 Solve

1

1

1

1

|I ((jI A )(jI A ) + CT C)| = 0 (23)

= 0.7314, (c) when = 1.2,

= 0.7314, (c) when = 1.2,

= 0.8216, (d)

= 0.8216, (d)

That implies (a) when = 0, min = 0.7104, (b) when =

0.5, min

min

0.5, min

min

= 0

CI 2

of the above algebraic ricatti equation

of the above algebraic ricatti equation

Where P is the symmetric positive definite solution

0

0

0

0

CI 2

CI 2

2. I 2 + < 2 (AI, j(I 2 + ) CI \

min

min

when = 3, min = 0.9874, (e) when = 5, min = 1.0531,

(f) when = 8, min = 1.0717, (g) when = 12, min = 1.3739, (h) when = 20, min = 4.2793 The distance to unobservability for orginal and new system VS is shown in

To find the new Lipchitz constant

The new Lipschitz condition is

=

=

T01(T0x, u) T01(T0x, u)

x x

0.333| sin(x3 sin(x3))|

j(x1 x1)2) + (x2 x2)2) + (x3 x )2)

<

3

0.333 sin((x3 x3) ) cos( (x3 + x3) )

2 2 < 0.333

x3 x3

2

Thus the new Lipschitz constant is 0.333. Let,

a1 b1

b

b

Fig. 8. Distance to unobservability VS

L1 = a2 2 (29)

5. OBSERVER DESIGN BY SIMILARITY TRANSFORMATION

and

a3 b3

a4 b4

P =

P =

4.21 0.02 0.02 0.001

In this section a method for designing a stable observer for robot by new method is presented. The method is based on distance to observability. Similarity transformation is used

0.02 x 2.3 1

0.02 2.3 y h

0.001 1 h z

(30)

to reduce the Lipschitz constant and increase the distance to unobservability.

Let x = T0 xI so that Then the original system is transformed

into

The matrix P is positive definite. The elements of P is chosen such that 4.21 is an eigen value of P.

Let T0 = diag(1, 1, 1, 10)

Thus from the Algebraic Ricatti Equation (ARE),

xI = bT01(A L1C)T0xI + T01(T0xI, u)

+ T 1g(y)u + T 1L y (24)

2

2

2

a33 = a43 = a44 = a34 = 0 h + 20h + z + 1.11 + 0 = 0

0 0 1

p 3.90h + y2 218.159 + 0 = 0

y = CT0xI = CIxI (25)

The new Lipschitz constant is determined from

Where

(h + 10)y + z(h 1.95) = 46.3

T01(T0xI, u) T01(T0xI, u)

< (26)

a1 1 b1 0 0

x x

The observer for the above system is given by

I 48.6 a2 1.25 b2 48.6 0 (31)

A =

A =

a3 b3 0 10

1.95 0.1a4 0.1b4 1.95 0

xI =AIxI + T01(T0xI, u) + T01g(y)u + T01L1y

+ (I 2 + )LI(y CIxI)/ CI 2 (27)

Solve the algebraic Ricatti equation for AI with 0 = 16, 0 =

0

x = CIxI (28)

0.0123, z and y are obtained as z = 8.63587, y = 11.1242. Select x such that the matrix P is positive definite. From this x = 10.5 is obtained.

The observer gain LI is chosen to be LI =

P 1CIT

2

under the

Solve for the remaining variables, a1 = 17.1349, a2 =

10214.9583, a3 = 5449.3247, a4 = 53324.6597, b1 =

0

0

following conditions,

23646.6245, b2 = 52.2998 and b3 = 9.21073, are found.

1. AL1C is stable.

Also 2 (AI, jI2 +

cI

0

0

cI

\ I2

= 10.11192

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Fig. 9. Actual and Observed motor angular position

10214.9529 52.2999

10214.9529 52.2999

0.3332 16 = 86.139 > 0 is satisfied. Thus,

Fig. 10. Actual and estimated motor angular velocity

REFERENCES

a2 b2

a2 b2

a1 b1

L

=

L

=

17.1349 23646.624

=

=

1. Rajesh Rajamani and Ankur Ganguli, Sensor fault diagnostics for a class

(32)

of non-linear systems using linear matrix inequalities, International journal of control, vol. 77:10, pp. 920-930

(32)

of non-linear systems using linear matrix inequalities, International journal of control, vol. 77:10, pp. 920-930

1 a3 b3 a4 b4

5449.3247 9.2107

53324.6597 1.0000

2. Rajamani, R, 1998,Observer design for Lipschitz non-linear systems,

IEEE transactions on Automatic Control,43, 397-401.

Define x = T0xI as the estimate of x, then

x = Ax + (x, u) + g(y)u + L(y Cx) (33)

Where,

3. Vemuri, A.T., 2001, Sensor bias fault diagnosis in a class of non-linear systems, IEEE Transactions on Automatic Control,46,949-954.

4. Boyd,S., Ghaoui,L.E.,Feron.E.,and Balakrishnan,V.,1994, Linear matrix inequalities in System and control Theorey, vol.15.

5. Simani, S., Fantuzzi, C., and Beghelli, S., 2000, Diagnosis techniques

(I2 + )T LI

0.11904 0.00017

0.000167 0.050267

for sensor faults of industrial processes. IEEE Transactions on Control Systems Technology, 8, 848855.

L = L1 +

0 0

CI 2

= 0.0004236 0.0141

0.00035547 0.005805

(34)

6. N. Kazantzis and C. Kravaris, Nonlinear observer design using Lya- punovs auxiliary theorem, Systems & Control Letters, vol. 34, pp. 241247, 1998.

7. Gauthier, J. P., and Kupka, I., 1994, Observability and observers for

By using this observer gain matrix, actual and observed motor angular rotation and velocity are found.

m(t) =

0.1727 cos(6.28t 1.4743) + 1.2113e0.193t cos(1.5015t

0.4634) 0.3653e0.4322t cos(8.276t 0.4281)

0.00484e5.1858t sin(15.8409t 1.4193)

0.00478e13.036t cos(15504.4744)

nonlinear systems. SIAM Journal of Control Optimization, 32, 975994.

8. Golub, G. H., and Van Loan, C. F., 1993, Matrix Computations (Johns Hopkins University Press).

9. Green, M., and Limebeer, D. J. N., 1995, Linear Robust Control (New Jersey: Prentice Hall).

10. Hirsch, M., and Smale, S., 1974, Differential Equations, Dynamical Systems and Linear Algebra (Academic Press).

m(t) =

1.10685 sin(6.28t 1.47425) 1.83369e0.193t cos(1.5015t

1.9056) + 3.0276e0.4322t cos(8.276t 1.946)

0.00031346e13.0364t sin(15504.4743t)

m(t) = 1.17265 cos(6.28t 1.47425) + 1.211128e0.193t cos(1.5015t 0.46341)

0.36533e0.4322t cos(8.276t 0.4281)

m(t) =

1.10685 sin(6.28t 1.47425) 1.83369e0.193t cos(1.5015t

1.9056) + 3.0276e0.4322t cos(8.276t 1.946)

6. CONCLUSION

A stable observer is designed for flexible link robot using LMI. By using this observer gain matrix, we can identify the fault in the system. Distance to unobservability is studied and estimated for orginal and transformed system. A stable observer is designed for robot by using this method. Two

theorems based on Algebraic riccati equation are found.