Practice English Speaking&Listening with: Lecture - 34 Robot Dynamics and Control

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It is start with the lecture number 04 in these module robotics dynamics and control

okay, Robot dynamics and control so let us what we have seen in the last lecture so we

had look at the actuators okay, How to module the actuators and how to integrate actuator

dynamics with the robot dynamics and finally we come with a liner equation which are learn

in the system dynamics okay. it had a leaner form with a non leaner disturbances acting

on it okay, you recall that form so that is first thing.

In that we did and after that we took that form and like we develop some controllers

algorithms basically for slow point to point regulation okay,So those controllers algorithms

like we studied them what is standard state error we did some analyses for that okay,

now in these class we will see suppose you want track a trajectory what kind of controller

algorithms we can come up with using the same form for the equations of a robot dynamics

along with the act voters okay.

We will use the same form.

But now will concentrate developing some control algorithms trajectory tracing kind of a objects

okay, And then we will have to see it will give you some improved performances over the

privies algorithms which are regulation algorithms but still to improve the performance further

you will have to go for a non linear control okay, the linear form of the equation whod

no longer give you very high kind of performances of a control algorithms okay.

You have to study some part of non linear theories.

Okay, specially leopna theory that we are going to see okay, So for that we will just

build up some mathematical pulmonary understanding for leopna in the todays class okay, So

let us start with a trajectory tracking problem you remember these is our linear form with

a disturbance for robot actuator, robot and actuator.

So you have these effective inertia and effective damping terms okay these effective inertia

again recall that it is not a constant we just assumed it to be some value over the

range of the motion manipulator okay, So it is a function of Q okay, Then B effective

is the consent damping kind of a term and then these our control input which is going

into the system and then these is the disturbance term DK which accounts for Colas cardiologist

centripetal ollas acceleration or inertia terms and like coupling terms in the dematrix.Okay,

now we will look at what is trajectory tracking problem again so we had our standard example

of tolling Manipulator okay , now trajectory problem is to go from if you take the end

effecter from these point PS to these point PF okay.

In some predefined trajectory faction okay, So in this case for example you are having

a liner trajectory like going in a straight line from one point to other point okay, so

that is a trajectory tracking problem okay, So in earliest given these trajectory in terms

of a end effector cordite in a global cordite you have given a path along which we have

to go okay, then that is your trajectory tracking problem.

Now you do the inverse analyses inverse kinematic analyses to get the disregard joint angels

okay.

So this D is nothing but D1= this J okay, and still now our conditions that D. -2C is

Q symmetric matrix will be preserved because this J is constant in ratio, because you take

derivative that will go to zero okay, So now ill substitute in this equation your control

law that we have presented the last slide so this D time stay because QD..

? * E. +C+B* this term new Kd * R somewhat there is R we have seen here to be

E. + ? E okay.

So this gives us this equation.

Now we simplify that by bringing this terms on other sides so you get here Q..-QD..+ ? E,

and similarly these terms which finally results into these equation okay, So we term this

variable R, i mean we write the equations in terms of this variable R ,and then will

become as simple as this okay.

This R is E. + ? E. okay .so, this is the form of a equation which we have to analyze

for asymptotic stability with the error dynamics okay.

R is basically representing the error dynamics okay .

it has both error derivative and actual error E and E. is function of E and E. okay so in

terms of these of new functions R will have to we have this equation of dynamics equation

of error dynamics appearing over here , now we need to prove by lyapunov stability theory,

that this indeed gives a error and error derivative both going to zero time times to infinity

. so , how do we do that so we have to select first the lyapunov function.

so, selection or choice of lyapunov function will be usually in the D base okay.

In this case since R is a new term or new variable for your dynamics we choose the lyapunov

function being in terms of R okay.

So, this is just not dependent on error derivative it is also dependent on error okay these terms

, and in additions this kind of a spring energy that is there because of the this part of

the term.

So KD* l ? E is basically spring constant when it is multiplied by E . you can recall

your standard spring mass system equations and compare this term and then you will see

that is kind of spring term in the spring mass system .okay with corresponding to that

this spring energy is getting added here okay.

Now we will have to first see whether this positive definite function or not okay.

So, that is expand this N terms of error . now see we want to see this is as a function of

error dynamics as a positive definite function okay .so our all conclusions are basically

about error error dynamics will have to conclude . so, we have to see in terms of error whether

this is a positive definite function . see in terms R its clearly a positive definite

function here and here now in terms of error also we can prove that is true okay.

So we just expand this as substituting for R at E + l ? E so this E + l ? E transpose

D times E.+ l ? E okay.

So suppose this R is E. + l ? E so you take transpose inside you get this equation, and

then you multiply this E. transpose times DE the first term then this E transpose l

? transpose in to D l ? E this multiplied by these and now this cross multiplication

terms , E. transpose D l ? E and E transpose l ? transpose De.

Okay, now this both of this terms is this mat races are symmetric projective definite

mat traces D and l ? because of that this both this terms will cannot be the same okay.

So, you can just write these as twice one of these terms , so how does V look like now

so it has now I have put these subsided d as per our previous expansion and then you

will find at this can be simplified in this fashion like you collect the terms corresponding

to E. square or E transpose DE.

E. transpose DE.

And then this is a term corresponding to E transpose something in to E so this part is

also added over here okay.

And this terms is represented in this okay.

And then this two terms are twice one of this terms and multiplied by half is somewhat here

, now again we can put it.

In a quadratic form in this fashion in terms of these matrix okay.

So here this is half l ? transpose D l ? + l ? KD matrix this is l ? transpose see this

are two equal terms , and then this is half D which is term corresponding to these, so

this is a quadratic form and now we will have for this form to be positive definite we know

that we need this matrix to be positive definite this first matrix and then we need the determinant

of this entire matrix to be positive and then if you multiplied and see it is indeed a case

this term is getting canceled.

This term multiplied by this term okay, because of that this is a positive definite we can

conclude about the positive definiteness of this entire matrix okay.

Now let us see how we can take a derivate and then the what is a property that V.

Satisfy okay so, let us have a look at versions again so V. V is the form okay now will you

use form for taking the derivative so it is half two times R transpose the R. to see this

and this R they both are differentiated and see these are the this is a three product

of three term so we have to differentiate one at a time okay.

And then again using as= symmetric property you can write this as two times R transpose

the all dot okay then this is R transpose D. R okay , and here it will become two times

i am sorry two is missing in this part this will become two times D. transpose l ? KD

E.E transpose l ? KD E. or E. transpose l ? KD E both are the term at all symmetric

property okay, now this R. or D times R. You recall from the previous case or basic

equation this just negative of C+B +KD and R okay , so I have substituted that in this

equation to two times R transpose C+B+KD times R for DR. and negative sign of here and other

terms are as here now you can see that this D. -2 C can be separated out from this equation

you have two C here and here you have D. and R transpose and R their same R transpose and

R. I have separated out this terms D.- 2C multiplied

by this half or here okay.

An then all the other terms are continuing as here okay, now we know that this R transpose

D.-2C R = zero because of the Q symmetric property of this matrix that we have seen

in the earlier classes okay, this D. -2C Q symmetric so quadratic form with the Q symmetric

matrix will have zero, if you evaluate that form it will be zero okay!

So , we use that property here so that is by now again For more higher and higher values

of X the function will take higher and higher value okay, it is strictly increasing function

so, these in did positive definite okay, and this is a positive definite not just a locally

positive definite because this is valid what are the conditions we have seen their valid

for all X okay, so, now let us come to these function here.

We have x2 x32 x1 then x4 okay, now similar to these part these part is positive definite

now these part x1 and x24 is it positive definite or not ,It is right, you can see that so,

and then is x here is given as x1 x2 x3 okay these x vector is having all these x1 x2 and

x3 term and this function also having x1 x2 x3 term okay, I will tell you why I am saying

that is typical in this case so, we have these reason or this function again that it is a

positive definite now let us come to these function were are x is specified as these

okay.

So it contain x1 it contain x2 but it is not containing x3 so, you see which condition

now gets spoiled here then certain condition okay ,we have 3,0 will be 0 but it will not

be that it will be 0 like if you have x other than 0 still we will have we took the 0 which

is not valid okay, for any value of these nom of x some over here also in this diagram

you have we to be go to 0 because if you have some vector like this 0 x2=0 and x3 = some

value the nom will turn out to be this some value over here and the function be is 0 which

is violating.

The condition on the this function condition on we this is the third condition okay, the

third condition will get spoiled okay this is not positive definite okay, you see this

is the very important understanding vey very important you should never forget these understanding

will find at your lay up now of function when you do the analysis you will use this PDF

function for lay up now theory analysis.

So, will see at that time also I will tell you but it is very important understanding

for when a function does not contain some terms okay then it cannot be a PDF function

okay , it should contain all the like compounds of vector x okay, if one of the compounds

of vector x is missing okay then the function can be a PDF function.

You see that now let us see some other definitions to see the first definition is some called

digression the function is said to be digression if there exist a constant small or grater

then 0 and function beta of class k such that these V is now lesser or equal to is these

beta you remember the pervious conditions for alpha function V was greater than or equal

to alpha and now these V is lesser than or equal to is beta function of class and again

now is either local condition or it can be a global condition.

If you respect x2 this small of radius R then it will became a local condition and if you

do not resent x when it global digression or locally digression.

Now function b will radial unbounded if again our condition are on alpha with alpha function

is satisfied okay, B is lesser than or equal to b is greater than or equal to alpha of

class A and this condition are valid for all the domain at we are talking about global.

It should be valid okay all X belong to RN presentation should be valid and than for

some continues for alpha with additional property that alpha for will tent to infinite.

So, the function alpha should go on increasing as R goes on increasing will infinite okay

that how it is called readily unbounded function these is the really unbounded function now

another definition v is totally definite function.

If V is lpdf function and negative definite function if v is a pdf okay so, locally

when like negative function again locally and globally follow just a negative function

pdf okay, so, this is another definition so, let us we see will later how we use this definition

for a analysis purpose but just have patient to ready this definition okay later on we

will use them for robotic function.

Now this quadratic function let us study little bit more about this quadratic function why

we have to study them because they are very popular function used in the lyapunov theory

okay, especially theory applied to robotic kind of application okay, this we will find

at this function quadratic lyapunov function they are very popular okay and then now let

us see the quadratic function this quadratic function can be written in this form.

Okay you can understand this form you have power or degree only two here not more than

two okay polynomial kind of function and then you have these terms coming in cross coupled

way then square way okay, so like these can be represented in the matrix form in this

pattern if you see that remember here the X is the vector o or here state okay x will

consist of x1 x2 x3 up to X in compounds were N is a like typical degree of freedom number

of degrees of freedom or number of degree of rate equation.

So, X as and N compounds okay X belongs to RN state we can fear like mathematical terms

okay and then this A is a matrix given by this expression so, A is a symmetry matrix

you can see that the element one two is same as the element A to one okay, so, because

of these term 2A12 expanders to sums here.

Okay so given this kind of matrix you should be able to write this quadratic formation

or give an quadratic expression you should able to convert it into these form X transpose

to AX okay so, it is very easy to do that actually we just need to look at the co efficient

multiplying x12X22 and coefficient multiplying x1 x2 x3 and x-3 and these kind of terms and

then you like divided these terms there will be coefficient of this cross terms by 2 and

put them in a appropriate place in these matrix and then you Construct this matrix A okay,

this is how you go about the matrix okay Now next we will see suppose we want this

quadratic function be a positive definite function what are the condition okay, so ,this

the theorem will not see the proof of this theorem but I will just tell you the theorem

and explain you this theorem .the quadratic function are positive definite function if

the determinants of successive minors of A are all positive okay,

You have this matrix A if take its first minor okay it will be form by only the first element

A11 determinant of each is a magnitude of these element it should be positive so this

element should be positive certain magnitude is the determinants is this element itself

okay, So let us take the next minor okay if

you look something of these slot so now you take the determinant

of these it will be A11 A12 ,A11 *A22 A122 okay.

You check whether that

is positive like that all you find all these successive minors and see wheaten these are

positive if they all positive then it is guaranteed that your quadratic form is positive definite

form okay, this is how you verify whether the function is positive definite

are not oaky, there is other condition also for verifying the weather function

is

positive

definite are

not based on the same matrix in terms of

the eigenvalues you can

see this theorem it is quadratic function if

the ideal value are

of all positive okay.

The Description of Lecture - 34 Robot Dynamics and Control