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.
