Control of Robot Manipulators in Joint Space - Part 5 pptx

As-sume that the partial derivatives of the elements of the matrix A are bounded functions, that is, that there exists a finite constant δ such that Define now the vectorial function [Ax −

Mathematical Support

In this appendix we present additional mathematical tools that are employed

in the textbook, mainly in the advanced topics of Part IV It is recommendedthat the graduate student following these chapters read first this appendix,specifically the material from Section A.3 which is widely used in the text Asfor other chapters and appendices, references are provided at the end

A.1 Some Lemmas on Linear Algebra

The following lemmas, whose proofs may be found in textbooks on linearalgebra, are used to prove certain properties of the dynamic model of therobot stated in Chapter 4

Lemma A.1 Consider a vector x ∈ IR n Its Euclidean norm,
x
, satisfies

i,j {|a ij |} for all k = 1, · · · , n

Lemma A.3 Consider a symmetric matrix A = A T ∈ IR n ×n and denote by

a ij its ijth element The spectral norm of the matrix A,
A
, induced by the vectorial Euclidean norm satisfies

A
= λMax{AT A } ≤ n

max

We present here a useful theorem on partitioned matrices which is takenfrom the literature.

Theorem A.1 Assume that a symmetric matrix is partitioned as

A.2 Vector Calculus

Theorem A.2 Mean-value

Consider the continuous function f : IR n → IR If moreover f(z1, z2, · · · , z n)

has continuous partial derivatives then, for any two constant vectors x, y ∈

where ξ ∈ IR n is a vector suitably chosen on the line segment which joins the

vectors x and y, i.e which satisfies

ξ = y + α[x − y]

= αx + (1 − α)y for some real α in the interval (0, 1) Notice moreover, that the norm of ξ

Theorem A.3 Mean-value theorem for vectorial functions

Consider the continuous vectorial function f : IR n → IR m If f i (z1, z2, · · · , z n)

has continuous partial derivatives for i = 1, · · · , m, then for each pair of

vec-tors x, y ∈ IR n and each w ∈ IR m there exists ξ ∈ IR n such that

for some real α in the interval (0, 1).

We present next a useful corollary, which follows from the statements ofTheorems A.2 and A.3

Corollary A.1 Consider the smooth matrix-function A : IR n → IR n ×n

As-sume that the partial derivatives of the elements of the matrix A are bounded functions, that is, that there exists a finite constant δ such that

Define now the vectorial function

[A(x) − A(y)] w, with x, y, w ∈ IR n Then, the norm of this function satisfies

[A(x) − A(y)] w
≤ n2 max

where a ij (z) denotes the ijth element of the matrix A(z) while z k denotes the

kth element of the vector z ∈ IR n

Proof The proof of the corollary may be carried out by the use of Theorems

A.2 or A.3 Here we use Theorem A.2

The norm of the vector A(x)w − A(y)w satisfies

A(x)w − A(y)w
≤
A(x) − A(y)

w

Considering Lemma A.3, we get

A(x)w − A(y)w
≤ n

max

i,j {|a ij (x) − a ij (y) |}
w
(A.3)

On the other hand, since by hypothesis the matrix A(z) is a smooth

func-tion of its argument, its elements have continuous partial derivatives

Conse-quently, given two constant vectors x, y ∈ IR n, according to the mean-value

Theorem (cf Theorem A.2), there exists a real number α ij in the interval

where for the last step we used Lemma A.1 (
x
≤ n [max i {|x i |}]).

Moreover, since it has been assumed that the partial derivatives of the

elements of A are bounded functions then, we may claim that

|a ij (x) − a ij (y) | ≤ n

'max

Truncated Taylor Representation of a Function

We present now a result well known from calculus and optimization In thefirst case, it comes from the ‘theorem of Taylor’ and in the second, it comesfrom what is known as ‘Lagrange’s residual formula’ Given the importance

of this lemma in the study of positive definite functions in Appendix B theproof is presented in its complete form

Lemma A.4 Let f : IR n → IR be a continuous function with continuous

partial derivatives up to at least the second one Then, for each x ∈ IR n , there exists a real number α (1 ≥ α ≥ 0) such that

f (x) = f (0) + ∂f

∂x(0)

2x T H(αx)x where H(αx) is the Hessian matrix (that is, its second partial derivative) of

where

[t − 1] ˙y(t) T x dt + y(t) T x[t − 1]1

0

=

 10[1− t)] ˙y(t) T x dt + y(0) T x (A.7)

Now, using the mean-value theorem for integrals2, and noting that (1−t) ≥

0 for all t between 0 and 1, the integral on the right-hand side of Equation

(A.7) may be written as

 10(1− t) ˙y(t) T x dt = ˙y(α) T x

 10(1− t) dt

= 1

2y(α)˙ T x

for some α (1 ≥ α ≥ 0).

Incorporating this in (A.7) we get

1 We recall here the formula:

2 Recall that for functions h(t) and g(t), continuous on the closed interval a ≤ t ≤ b,

and where g(t) ≥ 0 for each t from the interval, there always exists a number c

such that a ≤ c ≤ b and

 b a

h(t)g(t) dt = h(c)

 b a

g(t) dt

 10

y(t) T x dt = 1

2y(α)˙ T x + y(0) T x

and therefore, (A.4) may be written as

f (x) − f(0) =1

2y(α)˙ T x + y(0) T x. (A.8)

On the other hand, using the definition of y(t) given in (A.6), we get

where for each x ∈ IR there exists an α (1 ≥ α ≥ 0) Specifically, for

x = 0 ∈ IR any α ∈ [0, 1] applies (indeed, any α ∈ IR) In the case that x

−100 −50 0 50 100

0.00

0.25

0.50

0.75

1.00 α

x

Figure A.1.Example A.1: graph of α

A.3 Functional Spaces

A special class of vectorial spaces are the so-called L n

p (pronounce “el/pi:/en”)

where n is a positive integer and p ∈ (0, ∞] The elements of the L n

p spaces are functions with particular properties

The linear spaces denoted by L n

2 and L n ∞, which are defined below, are

often employed in the analysis of interconnected dynamical systems in the theory of input–output stability Formally, this methodology involves the use

of operators that characterize the behavior of the distinct parts of the inter-connected dynamic systems

We present next a set of definitions and properties of spaces of functions that are useful in establishing certain convergence properties of solutions of differential equations

For the purposes of this book, we say that a function f : IR n → IR m is

said to be continuous if

lim

x → x0

f(x) = f(x0) ∀ x0∈ IR n

A necessary condition for a function to be continuous is that it is defined at

every point x ∈ IR n It is also apparent that it is not necessary for a function

to be continuous that the function’s derivative be defined everywhere For

instance the derivative of the continuous function f (x) = |x| is not defined

at the origin, i.e at x = 0 However, if a function’s derivative is defined

everywhere then the function is continuous

The space L n

2 consists in the set of all the continuous functions f : IR+→

IRn such that  ∞

0

f(t) T f(t) dt =

 ∞

0
f(t)
2dt < ∞

In words, a function f belongs to the L n

∞ space consists of the set of all continuous functions f : IR+→ IR n

such that their Euclidean norms are upperbounded as3,

sup

t ≥0
f(t)
< ∞

The symbols L2and L ∞ denote the spaces L1 and L1∞ respectively.

We present next an example to illustrate the above-mentioned definitions

Example A.2 Consider the continuous functions f (t) = e −αt and

g(t) = α sin(t) where α > 0 We want to determine whether f and g belong to the spaces of L2 and L ∞.

Consider first the function f (t):

 ∞

0 |f(t)|2 dt =

 ∞0

f2(t) dt

=

 ∞0

e −2αt dt

= 1

2α < ∞ hence, f ∈ L2 On the other hand, |f(t)| = |e −αt | ≤ 1 < ∞ for all

t ≥ 0, hence f ∈ L ∞ We conclude that f (t) is bounded and

square-integrable, i.e f ∈ L ∞ ∩ L2 respectively

Consider next the function g(t) Notice that the integral

 ∞

0 |g(t)|2 dt = α2 ∞

0sin2(t) dt

does not converge; consequently g 2 Nevertheless|g(t)| = |α sin(t)|

≤ α < ∞ for all t ≥ 0, and therefore g ∈ L ∞. ♦

A useful observation for analysis of convergence of solutions of differential

equations is that if we consider a function x : IR+ → IR n and a radially

unbounded positive definite function W : IR n → IR+ then, since W (x) is continuous in x the composition w(t) := W (x(t)) satisfies w ∈ L ∞ if and

only if x ∈ L n

3For those readers not familiar with the sup of a function f (t), it corresponds

to the smallest possible number which is larger than f (t) for all t ≥ 0 For

instance sup| tanh(t)| = 1 but | tanh(t)| has no maximal value since tanh(t) is

ever increasing and tends to 1 as t → ∞.

We remark that a continuous function f belonging to the space L n

2 may nothave a limit We present next a result from the functional analysis literature

which provides sufficient conditions for functions belonging to the L n

2 space

to have a limit at zero This result is very often used in the literature ofmotion control of robot manipulators and in general, in the adaptive controlliterature

Lemma A.5 Consider a once continuously differentiable function f : IR+→

IR n Suppose that f and its time derivative satisfy the following

Then, necessarily lim t →∞ f(t) = 0 ∈ IR n

Proof It follows by contradiction4 Specifically we show that if the conclusion

of the lemma does not hold then the hypothesis that f ∈ L n

2 is violated.

To that end we first need to establish a convenient bound for the function

f(t)
2= f (t) T f(t) Its total time derivative is 2f(t) T f(t) and is continuous˙

by assumption so we may invoke the mean value theorem (see Theorem A.2)

to conclude that for any pair of numbers t, t1 ∈ IR+ there exists a number s laying on the line segment that joins t and t1, such that




f(t)
2−
f(t1)
2 ≤ 2f(s) T f(s) |t − t˙ 1|

On the other hand, since f , ˙ f ∈ L n

∞ it follows that there exists k > 0 such

that 
f(t)
2−
f(t1)
2 ≤ k |t − t1| ∀ t, t1 ∈ IR+ . (A.9)Next, notice that

f(t)
2=
f(t)
2−
f(t1)
2+
f(t1)
2for all t, t1∈ IR+ Now we use the inequality|a + b| ≥ |a| − |b| which holds for all a, b ∈ IR, with a =
f(t1)
2and b =

f(t)
2−
f(t1)
2

to see that

4 Proof “by contradiction” or, “by reductio ad absurdum”, is a technique widely used in mathematics to prove theorems and other truths To illustrate the method consider a series of logical statements denoted A, B, C, etc and their negations, denoted A, B, C, etc Then, to prove by contradiction the claim, “A and B =⇒

C”, we proceed as follows Assume that A and B hold but not C Then, we seek for a series of implications that lead to a negation of A and B, i.e we look for other statements D, E, etc such that C = ⇒ D =⇒ E =⇒ A and B So we conclude that C = ⇒ A and B However, in view of the fact that A and B must hold, this contradicts the initial hypothesis of the proof that C does not hold (i.e C) Notice that A and B = A or B.

f(t)
2 ≥
f(t1)
2−
f(t)
2−
f(t1)
2

for all t, t1∈ IR+ Then, we use (A.9) to obtain

f(t)
2≥
f(t1)
2− k |t − t1| (A.10)Assume now that the conclusion of the lemma does not hold i.e, eitherlimt →∞

each T ≥ 0 there exists an infinite unbounded sequence {t1, t2, }, denoted {t n } ∈ IR+ with t n → ∞ as n → ∞, and a constant ε > 0 such that

f(t i)
2> ε ∀ t i ≥ T (A.11)

To better see this, we recall that if limt →∞ f(t) exists and is zero then,

for any ε there exists T (ε) such that for all t ≥ T we have
f(t)
2 ≤ ε Furthermore, without loss of generality, defining δ := ε

2k, we may assumethat for all i ≤ n, t i+1− t i ≥ δ —indeed, if this does not hold, we may always

extract another infinite unbounded subsequence {t 

Notice that in the integrals above, t ∈ [t i , t i + δ] therefore, −k|t − t i | ≥ −kδ.

From this and (A.13) it follows that

We see that on one hand, the term on the left-hand side of Inequality (A.15)

is bounded by assumption (since f ∈ L n

2 ) and on the other hand, since{t n }

is infinite and (A.14) holds for each t i the term on the right-hand side ofInequality (A.16) is unbounded From this contradiction we conclude that itmust hold that limt →∞ f(t) = 0 which completes the proof.

♦♦♦

As an application of Lemma A.5 we present below the proof of Lemma 2.2used extensively in Parts II and III of this text

Proof of Lemma 2.2 Since V (t, x, z, h) ≥ 0 and ˙V (t, x, z, h) ≤ 0 for all

x, z and h then these inequalities also hold for x(τ), z(τ ) and h(τ ) and all

τ ≥ 0 Integrating on both sides of ˙V (τ, x(τ), z(τ), h(τ)) ≤ 0 from 0 to t we

obtain5

V (0, x(0), z(0), h(0)) ≥ V (t, x(t), z(t), h(t)) ≥ 0 ∀ t ≥ 0

Now, since P (t) is positive definite for all t ≥ 0 we may invoke the theorem

of Rayleigh–Ritz which establishes that x T Kx ≥ λmin{K}x T x where K is

any symmetric matrix and λmin{K} denotes the smallest eigenvalue of K, toconclude that there exists6 p

m > 0 such that y T P (t)y ≥ p m {P }
y
2 for all

y ∈ IR n +m and all t ∈ IR+ Furthermore, with an abuse of notation, we will

denote such constant by λ min {P } It follows that

5 One should not confuse V (t, N, z, h) with V (t, x(t), z(t), h(t)) as often happens

in the literature The first denotes a function of four variables while the ond is a functional In other words, the second corresponds to the function

sec-V (t, x, z, h) evaluated on certain trajectories which depend on time Therefore,

V (t, x(t), z(t), h(t)) is a function of time.

6 In general, for such a bound to exist it may not be sufficient that P is positive definite for each t but we shall not deal with such issues here and rather, we assume that P is such that the bound exists See also Remark 2.1 on page 25.

which, using the fact that V (0, x(0), z(0), h(0)) ≥ V (T, x(T ), z(T ), h(T )) ≥ 0

yields the inequality

V (0, x(0), z(0), h(0)) ≥

 T

0

x(τ ) T Q(τ )x(τ ) dτ ∀ T ∈ IR+ Notice that this inequality continues to hold as T → ∞ hence,

V (0, x(0), z(0), h(0)) ≥

 ∞0

x(τ ) T Q(τ )x(τ ) dτ

so using that Q is positive definite we obtain7 x T Q(t)x ≥ λmin{Q}
x
2 for

all x ∈ IR n and t ∈ IR+ therefore

V (0, x(0), z(0), h(0))

λmin{Q} ≥

 ∞0

x(τ ) T x(τ ) dτ

The term on the left-hand side of this inequality is finite, which means that

x ∈ L n

2.

Finally, since by assumption ˙x ∈ L n

∞, invoking Lemma A.5 we may

con-clude that limt →∞ x(t) = 0.

Another useful observation is the following

Lemma A.7 Consider the two functions f : IR+ → IR n and h : IR+ → IR with the following characteristics:

Proof According to the hypothesis made, there exist finite constants k f > 0 and k h > 0 such that

 ∞0

h(t)2f(t) T f(t) dt

≤ k2

h

 ∞0

where x ∈ IR m is the system’s state u ∈ IR n , stands for the input, y ∈ IR n for

the output and A ∈ IR m ×m , B ∈ IR m ×n and C ∈ IR n ×m are matrices having

constant real coefficients The transfer matrix function H(s) of the system is then defined as H(s) = C(sI − A) −1 B where s is a complex number (s ∈ C) The following result allows one to draw conclusions on whether y and ˙ y

belong to L n

2 or L n ∞ depending on whether u belongs to L n2 or L n ∞.

Lemma A.8 Consider the square matrix function of dimension n, H(s) ∈

IR n ×n (s) whose elements are rational strictly proper8 functions of the complex

variable s Assume that the denominators of all its elements have all their roots on the left half of the complex plane (i.e they have negative real parts).

numer-Corollary A.2 For the transfer matrix function H(s) ∈ IR n ×n (s), let u and

y denote its inputs and outputs respectively and let the assumptions of Lemma

Lemma A.2 appears in

• Marcus M., Minc H., 1965, “Introduction to linear algebra”, Dover

Publi-cations, p 207

• Horn R A., Johnson C R., 1985, “Matrix analysis”, Cambridge University

Press, p 346

Theorem A.1 on partitioned matrices is taken from

• Horn R A., Johnson C R., 1985, “Matrix analysis”, Cambridge University

The definition of L p spaces are clearly exposed in Chapter 6 of

• Vidyasagar M., 1993, “Nonlinear systems analysis”, Prentice-Hall, New

Jersey

The proof of Lemma A.5 is based on the proof of the so-called Barb˘alat’slemma originally reported in

• Barb˘alat B., 1959, “Syst`emes d’´equations diff´erentielles d’oscillations

non-lin´eaires”, Revue de math´ ematiques pures et appliqu´ ees, Vol 4, No 2, pp.

267–270

See also Lemma 2.12 in

• Narendra K., Annaswamy A., 1989, Stable adaptive systems, Prentice-Hall,

p 85

Lemma A.8 is taken from

• Desoer C., Vidyasagar M., 1975, “Feedback systems: Input–output ties”, Academic Press, New York, p 59.

1

f (t)

Figure A.2. Problem 1

Hint: Notice that f2(t) ≤ h2(t) where

Support to Lyapunov Theory

B.1 Conditions for Positive Definiteness of Functions

The interest of Lemma A.4 in this textbook resides in that it may be used

to derive sufficient conditions for a function to be positive definite (locally orglobally) We present such conditions in the statement of the following lemma

Lemma B.1 Let f : IR n → IR be a continuously differentiable function with continuous partial derivatives up to at least second order Assume that

• f(0) = 0 ∈ IR

• ∂x ∂f(0) = 0∈ IR n

Furthermore,

• if the Hessian matrix satisfies H(0) > 0, then f(x) is a positive definite

function (at least locally).

• If the Hessian matrix H(x) > 0 for all x ∈ IR n , then f (x) is a globally

positive definite function.

Proof Considering Lemma A.4 and the hypothesis made on the function f (x)

we see that for each x ∈ IR n there exists an α (1 ≥ α ≥ 0) such that

f (x) = 1

2x T H(αx)x

Under the hypothesis of continuity up to the second partial derivative, if

the Hessian matrix evaluated at x = 0 is positive definite (H(0) > 0), then the Hessian matrix is also positive definite in a neighborhood of x = 0 ∈ IR n,

e.g for all x ∈ IR n such that
x
≤ ε and for some ε > 0, i.e.

H(x) > 0 ∀ x ∈ IR n:
x
≤ ε

Of course, H(αx) > 0 for all x ∈ IR n such that
x
≤ ε and for any α

(1≥ α ≥ 0) Since for all x ∈ IR n there exists an α (1 ≥ α ≥ 0) and

f (x) = 1

2x T H(αx)x , then f (x) > 0 for all x n such that
x
≤ ε Furthermore, since by hypothesis f (0) = 0, it follows that f (x) is positive definite at least locally.

On the other hand, if the Hessian matrix H(x) is positive definite for all

x ∈ IR n , it follows that so is H(αx) and this, not only for 1 ≥ α ≥ 0 but for

any real α Therefore, f (x) > 0 for all x n and, since we assumed

that f (0) = 0 we conclude that f (x) is globally positive definite.

♦♦♦

Next, we present some examples to illustrate the application of the ous lemma

previ-Example B.1 Consider the following function f : IR2→ IR used in the

study of stability of the origin of the differential equation that modelsthe behavior of an ideal pendulum, that is,

f (x1, x2) = mgl[1 − cos(x1 )] + J x

2 2

and is positive definite at x = 0 ∈ IR2 Hence, according to Lemma

B.1 the function f (x1, x2) is positive definite at least locally Notice

that this function is not globally positive definite since cos(x1) = 0

for all x1 = nπ2 with n = 1, 2, 3 and cos(x1) < 0 for all x1 ∈

func-Example B.2 Consider the function f : IR n → IR defined as

f (˜ q) = U(q d − ˜q) − U(q d ) + g(q d)T q +˜ 1

ε q˜T K p˜q

where K p = K T

p > 0, q d ∈ IR n is a constant vector, ε is a real positive

constant andU(q) stands for the potential energy of the robot Here,

we assume that all the joints of the robot are revolute

The objective of this example is to show that if K p is selected sothat1

λmin{Kp } > ε2k g then f (˜ q) is a globally positive definite function.

To prove the latter we use Lemma B.1 Notice first that f (0) = 0.

The gradient of f (˜ q) with respect to ˜q is

Clearly the gradient of f (˜ q) is zero at ˜q = 0 ∈ IR n

On the other hand, the (symmetric) Hessian matrix H(˜ q) of f(˜q),

According to Lemma B.1, if H(˜ q) > 0 for all ˜q ∈ IR n, then the

function f (˜ q) is globally positive definite.

To show that H(˜ q) > 0 for all ˜q ∈ IR n, we appeal to the

follow-ing result Let A, B ∈ IR n ×n be symmetric matrices Assume

more-over that the matrix A is positive definite, but B may not be so If

λmin{A} >
B
, then the matrix A + B is positive definite4 Defining

3 LetB, C : IR n → IR n,N, O ∈ IR n andN = C(O) Then,

NT [A + B] N > 0 ,

which is equivalent to matrix A + B being positive definite.

We present next a set of definitions and properties of spaces of functions that are useful in establishing certain convergence... proof.

♦♦♦

As an application of Lemma A .5 we present below the proof of Lemma 2.2used extensively in Parts II and III of this text

Proof of Lemma 2.2 Since... belonging to the L n

2 space< /sup>

to have a limit at zero This result is very often used in the literature ofmotion control of robot manipulators