WEBVTT
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so welcome friends we are discussing about
the some alternate forms of the riccati
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equations the the previous lecture we have
discuss about the specified degree of stability
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in which by modification into the arithmetic
riccati equation we can ensure the a stability
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of a closed loop system for a specified degree
today we will discuss about the inverse matrix
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riccati equation which normally appears if
we will have a fixed endpoint problem say
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as we know we have consider the transformation
or say our final condition will appear in
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a system as lambda t f equal to f of t f x
of t f p t f x of t f so previously we already
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have discussed this condition so if my x of
t f is zero this means my endpoint is specified
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and my system is writtening to the origin
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so my x t f will be zero if x t f is zero
so what will be my p t f from this equation
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so my p t f will be infinite so p t f infinite
this means i have to solve my riccati equation
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given as the p t f equal to f of t f if my
p t f is going to be infinite so in my backward
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solution of the riccati equation i have to
consider this p t f value to be very very
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large so this is one of the way to solve this
riccati equation so we start with a very large
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value of the p and then backward solve my
riccati equation to get the solution on the
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other hand we can have the concept of inverse
matrix riccati equation how this appears that
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we will see today
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we consider it time varying system define
as x dot t a a t x t plus b t u t with a cost
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function in which the f t f we have considered
to be the zero so it does not has the terminal
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cost and given as the x prime q x plus u prime
r u as time varying system we have so my a
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b q and r all are the time varying we consider
the boundary condition as at t equal to t
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zero it is x zero and t equal to t f this
is x f which in this particular case we are
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considering to be the zero so this equation
in a similar manner i can solve using the
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hamiltonian approach this already we have
done i can define my hamiltonian as my v plus
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lambda prime f
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so v is x prime q x plus u prime r u with
half so half of x prime q x u prime r u plus
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lambda prime f is nothing but my a t x t plus
b t u t so in this solution our first step
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is to get the optimal control which is del
h by del u equal to zero and this give me
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nothing but r t u t plus b prime t lambda
t equal to zero so by this if i will find
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out the u t this is nothing but r inverse
b prime lambda and so lambda we have to express
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in terms of a closed loop feedback
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so this means i have to express lambda in
terms of the x is state and the co state equation
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if you will drive this is nothing but by a
x plus b u is my state equation and minus
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q x minus a x is my co state equation so i
can write my hamiltonian system as a minus
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e minus q minus a prime x lambda this already
we have got up to this point in our previous
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derivation of a normal riccati equation so
with this hamiltonian matrix
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so my hamiltonian matrix is
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x dot t lambda dot t and this is equal to
nothing but my a a t minus e t minus q t minus
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a transpose t and what the a t we are considering
a t we are taking as b r inversed b prime
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and multiplied with
x of t lambda of t and if we will write the
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final condition and we know we will have lambda
t f has f of t f x of t f with this consideration
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we consider a transformation in place of the
lambda equal to p x just reverse to this
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we consider a transformation here is x t as
m times of lambda t while the previously we
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have considered the lambda in terms of the
x we have taken as a lambda equal to p x just
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we are taking the inverse of this this means
my lambda t m considering simply as m inverse
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t x t so my optimal control which we have
considered as we have taken this as r universe
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b prime lambda t so lambda i can consider
simply as m inverse t x t so this inverse
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of m is appearing while in the previous case
it was appearing as a p t x t so this is a
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basic difference here
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so this we will replace later on lambda t
first we will see how to get the value of
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my m so i will differentiate this equation
so this is x dot t as m is also a function
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of time m dot t lambda t plus m t lambda dot
t x t and lambda dot t i can place it from
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the hamiltonian system in this x dot t is
a t x t minus e t lambda t this i am replacing
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x dot t from this equation a t x t minus e
t lambda t and this is equal to i keep it
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as such m dot t lambda t plus m t and lambda
to t from here minus q t q t x t minus
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a prime t lambda t that is by lambda dot
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so in this equation i am placing the value
of x dot and lambda dot i also have x t as
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m t lambda t so substitute
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x t is sorry this is my m t lambda t so if
i will place it here so this is nothing but
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my a t x t in place of x t i will write m
t lambda t minus e t i keep it same lambda
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t again remain here this is m dot t lambda
t i multiplying to this so this negative minus
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m t q t and in place of x t again i am placing
my m t lambda t and this is say minus m t
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a prime t lambda t
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so if i can write this equation substituting
this on one side where i will have m dot t
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minus a t m t minus m t a transpose of t that
i am writing from this
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plus a t and minus m t q t m t into lambda
t equal to zero so this means by substituting
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x dot and lambda dot from hamiltonian system
we got this equation in this we will substitute
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x t is m t lambda t so the hole equation i
can write in terms of the lambda t which will
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be m dot t minus a t m t m t a transpose t
e t minus m t q t m t into lambda t equal
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to zero
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lambda t is my arbitrary variation so the
coefficient of lambda t will be zero so i
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can write
if the coefficient equal to zero m dot t equal
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to a t m t plus m t a transpose of t minus
e t in place of e t we can write b r inverse
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b transpose
plus m t q t m t so this m dot i can write
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this so i am writing m dot equal to a t m
t plus m t a transpose t m t q t m t minus
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b r inverse p prime and this we call has the
inverse matrix differential riccati equation
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so how to solve this as we have considered
the initial condition as x t zero has x zero
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and x of t f as zero so in this case because
and my transformation is x t equal to m t
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lambda t
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so x t zero is a non zero quantity so if i
will take this as x t zero so lambda t zero
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with no con conclusion but if i will write
this as x t f as m of t f lambda t f this
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will be zero so this implies i can take my
final condition m of t f equal to zero so
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i can solve this inverse matrix differential
equation in a backward form and can find the
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solution of this equation in terms of the
m t if i am solving this equation in backward
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that will see through an example so we take
an example we consider a first order system
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a of x t plus b of u t a b we are considering
to b the time invariant for example just for
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the simplification otherwise they may be time
varying also
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obtain optimal u
which will minimize j which is given as one
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by two t zero to t f q x square t plus r u
square of t d t and the terminal condition
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given as x t at zero point is x zero and x
at t equal to t f point is x f and this we
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are considering to be zero so this is my problem
for my given system i have to obtain the optimal
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value of the u which will minimize my performance
index with given initial condition has this
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i know what is the solution of this my optimal
u is r inverse b prime now we say m inverse
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x t because if we will recall what we have
taken my u was r inverse b prime lambda t
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and lambda t we are taking as or we are writing
simply as r inverse b prime m inverse t x
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t
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so my solution is sorry my optimal control
is u star is minus r inverse b prime m inverse
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x t where m is the solution of my inverse
matrix differential equation
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and what is this and my this equation is m
dot equal to a t m t
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so m dot t is a m t plus m again a so this
a of m t is giving me the a of m t m t a transpose
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this again giving me the a of m t so i can
simply write this as first two term is two
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a of m t plus m q m where q is my small q
so this is q m is square this is q m square
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of t and minus b r inverse b this is give
the b square by r and this is minus b square
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by r
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so this is my inverse matrix differential
equation and this equation i have to solve
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has my x t f is zero
so solve this with final condition as m of
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t f equal to zero now this is the simple first
order differential equation which can be solved
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the solution of this equation i will directly
take so
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the solution of this is m t equal to so a
plus beta e to the power minus beta t minus
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t f minus a minus beta and in this my beta
is a
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square plus q b square by r and this is the
value of my beta
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so solution of this equation with t f equal
to zero is has m t once m t is known then
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my optimal control
u star t is minus r inverse p prime m inverse
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of x t so i can write my optimal control as
r inverses one by r b prime is b so this is
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minus b by r m inverse this is my m so i have
to take the inverse of this so this means
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r upon b square inverse of this quantity a
plus beta minus a minus beta now numerator
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will be the denominator in this case minus
beta t minus t f
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so the overall u t we can write so directly
we can write it here so if i will cancel this
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r r will cancel so this is nothing but minus
one by b so this will be my final optimal
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control which i can find out utilizing the
inverse matrix riccati equation so in this
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wave the riccati equation can be used in a
different forms also we have seen the matrix
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differential riccati equation algebraic riccati
equation riccati equation with this specify
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degree of a stability and then inverse matrix
riccati equation
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so this class we stop here and in the next
class we will start about discussion on the
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tracking problem so till now we have discuss
about the regulator problem in which we have
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taken the reference r as zero so my also my
is states are returning to the origin so but
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if my reference point or my set point will
continuously change then this will be a tracking
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problem and this tracking problem we will
start in the from the next class
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thank you very much