WEBVTT
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This will be lecture six on fuzzy control.
The topic that we will be discussing today
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is linear controllers using T-S fuzzy model.
In the last class, we discussed how to design
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controllers for T-S fuzzy model when the input
matrix is common for all subsystems. Now,
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we will consider the generic T-S fuzzy model
and how do we design linear controllers for
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it.
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We will just revise our notion of T-S fuzzy
model representation of nonlinear systems.
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, The approach of Controller design linear
controller design. We will give two controllers
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for this, stabilizing controller design robust
controller approach; we will propose controller
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I and as well as controller II - two different
types of controller, simulation results: two-link
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manipulator, ball beam system and summary.
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The T-S fuzzy model is expressed in terms
of r fuzzy rules where ith fuzzy rules has
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the following form: if x1t is F1 i and x2t
is a F2 i and so on until nxnt is an FnI then
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x dot t is a is Aix t plus Biu t. This is
my ith fuzzy rule consisting of n states and
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each state is fuzzy variable and where the
fuzzy variable FjI is the jth fuzzy set of
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the ith rule. Then, the fuzzy index muI associated
with the ith fuzzy rule is given by this formula
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where mui j xj is the membership function
of the fuzzy set Fj i. i equal to 1 to r.
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Given an input output pair x t u t the fuzzy
variable around this operating point is constructed
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as the weighted average of the local model
as x dot t equal to summation j equal to 1
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to r sigma i Ajx t plus Bju t. This is jth
local model and with jth local model associated
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membership function is actually sigmaj, so
sigmaj is a normalized fuzzy index associated
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with jth local model. The summation of such
local model multiplied with fuzzy index over
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j equal to 1 to r gives me the complete fuzzy
dynamics in terms of T-S fuzzy model. A T-S
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fuzzy model approximates a nonlinear system
as a cluster of a linear system. This is my
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linear system; this is my jth linear system,
sigmaj is the normalized fuzzy index associated
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with jth linear system. When the cluster of
such a linear system if there are there, then
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if I fuzzy cluster these r number of linear
systems then, I get the approximation of a
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non linear system. The advantage of the fuzzy
system is more informative in terms of local
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dynamics because I can look at a nonlinear
system in terms of linear system. Dynamics
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is governed by subsystems fired at each operating
point. In this class we will talk about variable
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gain controller using single nominal plant.
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This is the first type of controller will
be talking today and in this we expressed
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the fuzzy systems as a linear system with
non linear disturbance. Our fuzzy system which
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is x dot sigma j equal to 1 to r sigmaj Ajx
plus Bju, this is my T-S fuzzy model approximation
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of nonlinear system using T-S fuzzy model
you can easily see that this has a very convenient
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form. It looks as very convenient form easy
to handle and I can write this as Ax plus
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Bu plus disturbance term. This is my disturbance
term, this is worst part I am saying here;
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express the fuzzy system is the linear system
with nonlinear disturbance. Then design a
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controller to stabilize in the linear system
in the presence of disturbance. The original
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plant was x dot is sigmaj Ajx plus Bju j equal
to 1 to r. So, this is my original plant.
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I can always write this plant you see as:
A x t plus B u t this is the nominal plant.
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Once I separate from this as a nominal plant
how do I write that this sigmaj Aj x in that
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I have to subtract A x t which is here. So
sigmaj Aj minus A x t, I subtract that and
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j equal to 1 to I. Similarly, if I subtract
b because, Bj is there I subtract b here.
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How can I do? You must know sigma j equal
to 1 to r sigmaj A is A; because, j equal
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to 1 to r sigma j equal to 1. You may worried
that I should have only subtracted A outside
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no, this is one and this is equal to A so
I can subtract this quantity here. Similarly,
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about B so x dot t is written in the particular
form; this is my original dynamics; this can
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be written in this particular form. This can
be finally written A x t plus B u t F x t
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u t and this is my disturbance term around
this nominal plant where, A x t B u t is the
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linear system and F x t is a nonlinear disturbance
given by. I represent this nonlinear system
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in terms of three different components. So,
this is simply F x t B h1 x t B h2 u t. You
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can easily see that, B x t u t is obviously
taken into account of this. This one is actually
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B h2 u t you can easily see that this is a
nonlinear term and that is given by B x2 u
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t, we can express that. Similarly, these two
terms combined represents this one. If I say
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this is the second term this is my second
term. This term is represented by this. I
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wish that you understood what we are trying
to do my original plant was given by this
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particular thing x dot is sigma Ajx plus Bju
this one I am representing this same dynamics
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as this. There is no difference between this
and this. I am representing this whole thing
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as A x t plus B u t is the first plant nominal
plant and this F x t is disturbance and this
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has three component which the first component
here I is this one. Sigmaj Bj minus B u t
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sigmaj equal to 1 to r and the second component
is F x t plus B h1 x t this component here.
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Always in controller design whenever we say
any disturbance, we are not interested in
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exact represent of disturbance rather an upper
bound. This is the principle of robust control;
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we want to know the upper bound of this disturbance.
That means if I am designing controller for
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the worst case naturally the controller also
will be a stable for the other cases. This
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is the robust control design principle is
design the controller for worst case.
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That is why we will represent first of all
disturbance terms more elaborately, we will
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define the upper bound. Let us see the first
disturbance B h2 u t the component one as
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I said which is: sigmaj Bj minus u t j equal
to 1 to r. We can write this equation as,
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I have to write in terms of B here multiplied
sigmaj Bj bar t u t and this Bj bar is obviously
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such that the b Bj bar is v j minus v. This
identity has to be satisfied, so B Bj bar
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is Bj minus B
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Similarly, F x t plus B h1 x t is this term,
the second term, j equal to 1 to r sigmaj
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Aj minus x t and then this can be written
in terms of two. Looking at this you can easily
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sigmaj A1 j x t is F x t and sigma Bj plus
x t is I can take of B and I can write this
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as B h1 x t. So, here Aj minus A has to be
written in terms of A1j plus BA2j where, the
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A1j represents the unmatched disturbance,
unmatched means you see that, this disturbance
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is not magnified by u, means this is not with
there with the control input. This is matched
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you see the b in to this and you see our normal
plant is A x x dot is Ax plus Bu. So, anything
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with B means you can say that, this term is
like an excitation because anything multiplied
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with B is kind of excitation to the system
and this term is not with b. It is separate
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term unmatched disturbance and this is matched
disturbance means this is disturbance excitation
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that is exciting system because, we say b
is the control matrix. In that sense, A1j
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is the unmatched disturbance and this is the
matched disturbance.
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Now, we will talk about the control problem.
The T-S fuzzy model approximates a nonlinear
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system as a fuzzy cluster of r linear subsystems.
Since each sub system is linear, linear control
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theory can be applied to design fixed gain
controller for each system design fixing controller
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for each subsystem. Since the desired system
output traverses a specific trajectory system
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states traverse across different fuzzy zones.
It is thus expected that, the controller will
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be characterized by variable gain instead
of fixed gain which I have already discussed.
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The control problem is given a T-S fuzzy model
representation design a variable state feedback
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controller u t equal to minus kx t where,
k is the variable such that, the T-S fuzzy
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model is Lyapunov stable. Here, the matrix
k represents variable state feedback gain
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it is not a constant gain as in case of a
linear system. Now, we will be talking about
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this disturbance measure. Since the nominal
plant is linear while disturbance term is
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nonlinear one can possibly solve the control
problem using the principles of robust control
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theory.
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For that the norm bounds of uncertainties
have to be computed first. The norm bound
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of unmatched state disturbance which is h1x
t. I will show you which is h1 x t is 1 this
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one is matched state disturbance. Matched
state disturbance h1 x t is alpha j you see
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h1 x t we have represented h1 x t is this
is h1 x t. You can easily say sigmaj A2 x
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t sigma j equal to 1 to r. So that is what
we have written here: h1 x t is sigmaj A2
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j x t known from j equal to 1 to r overall
norm. I can represent this particular term
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using a triangular in a quality as less than
equal to sigmaj and this is the induced norm
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of A2j is alpha h x j, alpha h x j represent
the maximum singular value of A2j and the
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norm of x2 x t separately. This is a triangular
in equality and what we are saying is that
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the kind of this disturbance is represented
in terms of a major disturbance which is less
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than this quantity.
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The norm bound of input disturbance is, which
is h2 u t you see h2 u t given here is this
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one and this sigmaj bar Bj bar u t. This was
actually sigmaj Bj bar u t j equal to 1 to
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r. This can be written in triangular inequality
as j equal to 1 to r sigmaj and norm of B
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bar j into u t norm and we can write now B
bar j norm the maximum norm. I can put here
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the maximum norm of induce norm of B bar j
is alpha u j alpha u j is the norm of B bar
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j that means this is maximum singular value
of B bar j. This inequality gives a disturbance
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measure for this quantity where we already
know alpha u j, we know sigmaj. Similarly,
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the norm bound of unmatched state disturbance
which is F x t the previous one is unmatched.
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This is unmatched one F x t and F x t is sigmaj
A1j x t, so this whole norm can be written
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as again less than equal to this A1j alphaf
is norm of A1j induced second norm A1j this
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is alphaf.
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I wish that you have understood now what we
have been talking about the disturbance measure
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so once the disturbance measure I define,
we will be now telling a theorem which says
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that: if I design a state feedback controller,
if I have a state feedback controller for
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the system, what is the system now, my system
is: x dot is Ax plus Bu plus f x plus B h1
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x plus B h2 u. This is my system which my
original system is simply sigmaj Aj x plus
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Bj u sigma j equal to 1 to r. This is same
as this quantity we said we approximated and
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then we found out the measure upper bound
of effects h1 x and h2 u and then, we are
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saying that, the system will be stable if
this u control input u is given by minus sigma
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j equal to 1 to r sigmaj gammaj B transpose
B x t this is my control law and gammaj satisfies
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this particular condition where alpha h x
j and alpha u and alpha m alpha f they are
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all disturbance measure as we define just
recently.
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Now, this can be only valid if the two conditions
are satisfied and that is alpha f is less
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than equal to lambdamin the minimum eigen
value of Q and lambdamax is the maximum eigen
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value of P where, A transpose P PA is equal
to minus 2Q. You know that is my A is nominal
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plant model and for a nominal plant model
A, I can always find out AQ for such that,
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I have also P which satisfies A transpose
P plus PA equal to minus 2Q and so given this
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P and Q the alphaf which is here, this alphaf
is the induce norm.
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You see alphaf is the induce norm of A1j and
this A1j is coming from our term Aj minus
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A this can be written as A1j plus A2j this
is what we have shown earlier Aj minus A is
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A1j plus BA2j. So Aj minus A A1j plus so this
A1j is the induce norm is alphaf means the
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maximum singular value A1j. This is the theorem
and we will just prove this theorem again
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repeat what is this theorem implies that means,
if I propose a control law u t where the gammaj
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satisfies this condition then, the system
will be stable provided alphaf is less than
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this identity as well this identity is true.
Now, consider the Lyapunov candidate v equal
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to x transpose Px this is the theorem one
proving, we trying to prove the theorem one.
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Now, the time derivative of v is given by
v dot is two x transpose is P x dot.
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My x dot, as I have already told you is Ax
Bu plus f x plus B h1 x plus B h2 u and u
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is we have already given u to be so u is we
just define u is here this is our u and if
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I write down that u is sigma minus j equal
to 1 to r sigmaj gammaj B transpose P x. This
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is my u so before I introduce u inside what
I will do is that, this is my V dot is 2 x
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transpose P x dot and I can write this expression
by introducing this x dot inside here. If
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I do that what I will get 2x transpose P x
dot. So, there is A x so what I get is that
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2 x transpose P A x so this two and we know
that A transpose P plus P A is minus 2Q this
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combined with the knowledge that 2x transpose
P A x this can be written as x transpose A
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transpose P plus P A x. You see that A transpose
P plus PA is equal to minus T Q then, I can
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write 2x transpose P A x transpose A transpose
P plus P A into x and this quantity is now
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minus 2Q. We have already said that given
a we achieve this A transpose P plus B transpose
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P A equal to minus 2Q if you can write. This
writing that this first term 2x transpose
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P A x can be written as minus 2x transpose
Q X, so, this is minus 2x transpose Q X, the
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first one. So, from x dot I took care of it
AX. Now, let me take care of Bu so Bu is this
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quantity, so how do I write it is B u if I
put it here, so 2x transpose P Bu, so 2x transpose
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P B and u has B transpose P x this one and
the other quantity is that sigma j equal to
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1 to r sigmaj gammaj.
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So because, we have already j inside so j
we have k equal to 1 to r sigmak gammak, so
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this quantity is given by 2X transpose P B
u and again we have multiplied here f x here
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x dot has also another component f s, so that
is 2 x transpose P into f x. Similarly, B
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h1 x 2 transpose P B h1 x similarly 2 x transpose
P B h2 u. So, all that here what we have done
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instead of x dot, we have replace this and
we can write the equation like this. Here,
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further what we can do we can rewrite this
term as kind of a using the properties of
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matrices that is x transpose Q x. It can be
bounded by two quantities the lower bound
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is lambda minimum that means the minimum eigen
values of Q x norm square and the upper bound
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is lambdamax singular value of lambda eigen
value of Q norm square because, Q and P there
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are symmetric matrices. Hence, the singular
value maximum singular value is same as maximum
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eigen value and therefore minus x transpose
Q x, if I write this minus I can write minus
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x Q x transpose Q x is less than equal to
minus lambdamin Q x norm square.
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For the symmetric positive definite matrix
P that induces 2-norms is P norm is we can
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say induce norm is lambdamax P. So, this is
my maximum value of P and further more if
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I look at this expression That this is not
capital x this is small x bold x, x is a vector
29:00.860 --> 29:07.860
small x. So, x transpose P B x transpose P
is 2 x transpose 2 x transpose P. This is
29:09.909 --> 29:16.909
not capital x. So, x transpose P B transpose
B x is this quantity, this can be written
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as x transpose P B again x transpose P B transpose
which is B transpose x. In this I can write
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this is a norm noun x transpose P B norm square,
this is matrix theory. Taking all the relation
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to this account and using norm bounds on certain
elements. We get, norm earlier V dot is this
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so using the norm bound using norm bound means
this is less than this quantity. Similarly,
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we can find this is less than this quantity,
this less than its norm bound; this is less
30:00.999 --> 30:05.179
than its own norm bound; this is less than
own norm bound.
30:05.179 --> 30:12.179
V dot can be written as a less than some quantity
and that quantity is here V dot is less than
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minus 2 lambdamin Q x norm square. This is
the norm bound of the second term; this is
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the norm bound of the third term; this is
norm bound for the fourth term and this is
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norm bound of the fifth term, we have five
terms. By putting that, this particular expression,
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so what we did is that, we first derived what
is the direct derivative of Lyapunov function
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and then we expressed that in terms of norm
bound using the sine less than equal to and
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then this particular quantity can be written
as: minus 2 x bar transpose Q bar x bar where,
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x bar is this two quantity x norm and x transpose
P B. This you can say one element and another
31:03.740 --> 31:09.419
element this x bar is a vector norm, this
is first element and this is a second element
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and Q bar has four element q11 q12 q21 q22
where q11 is given by this quantity that is
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minimum eigen value of Q minus alpha f lambdamax
of P the maximum eigen value of P. Similarly,
31:28.590 --> 31:35.590
q12 q21 is half of this quantity and q22 is
this quantity and you may be wondering how
31:36.549 --> 31:43.549
we got this but, I will just explain to you
in a very simple understand you can easily
31:44.490 --> 31:51.490
see that, I can now add all x norms square
together so that, if r that you can see that,
31:53.610 --> 32:00.610
this x norm square if I take common, I get
here two lambdamin 2 and 2 alpha x lambda
32:03.899 --> 32:10.899
x P. So, if I combine them I can get lambdamin
Q minus alpha x lambda x and lambda alpha
32:24.019 --> 32:31.019
f lambdamax. This is my q11 so you see that,
if I write this expression as this and I have
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taken 2 out, 2 is common here. So, if I take
2 common out so I have also taken negative
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outside, so this becomes lambdamin Q minus
alpha f lambdamax P which is this quantity
32:51.159 --> 32:58.159
q11 into so the point is that, you can easily
see this quantity and with definition x bar
32:59.009 --> 33:06.009
and this and this is we can write out q11
x norm square plus q22 norm x transpose P
33:15.869 --> 33:22.869
B square and you can easily see q12 plus q21
x norm and x transpose P V.
33:38.379 --> 33:45.379
What you saw that, V dot is less than this
five terms and I am trying to represent five
33:45.450 --> 33:52.369
terms in terms of minus 2 x bar transpose
Q bar x bar. If I define x bar is this two
33:52.369 --> 33:59.369
terms vector of two terms and Q bar is vector
matrix of 2 by 2 then, this x bar transpose
34:01.830 --> 34:08.830
Q bar x bar is this quantity q11 x bar x norm
square q22 x transpose P V norm square plus
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q12 q21 x norm into x transpose P V norm.
So, you can easily see here that, obviously
34:18.169 --> 34:25.169
I have to find out this is x bar transpose
Q bar x y is this quantity. So, minus 2 if
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I take common here, what I get is that easily
by comparing the coefficient of qc q11 will
34:39.179 --> 34:46.179
be the total coefficient of x1 bar. That is
lambdamin x minus alpha f lambdamax p which
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is here q11.
34:52.240 --> 34:59.240
q22 is the coefficient of x transpose P V
norm square, so this is one term x transpose
35:01.730 --> 35:08.730
P V square and another term is x transpose
P V square. If I take minus 2 common here
35:08.850 --> 35:15.850
I get here minus 2 common k equal to 1 to
r k equal to 1 to r sigmak gammak sigmak gamma
35:21.470 --> 35:28.470
k. This is the first term and second term
is here where, this is sigmaj alpha u j j
35:34.460 --> 35:41.460
is equal to 1 and sigmak gammak k equal to
1 to r. So, this is q22 and then q12 plus
35:44.200 --> 35:51.200
q21 is the coefficients half x bar and x transpose
P V you see that, this all one term. Since
35:54.390 --> 36:01.390
we have q12 and q21 the coefficient is simply
j equal to 1 sigmaj alpha H x j, so this is
36:07.490 --> 36:14.490
my coefficient and I have two term, so I can
easily do that by dividing them equally and
36:14.710 --> 36:21.710
making q12 equal to q21 so q12 to equal to
q21 which is minus half because, here if I
36:24.710 --> 36:31.710
take minus 2 common I am getting minus half
here minus j. So, minus half sigmaj alpha
36:32.600 --> 36:39.600
H x j equal to 1 to r, so I wish that you
understood how we finally wrote V dot in terms
36:40.450 --> 36:47.450
of a quadratic function x transpose Q bar
x bar and instead V dot what is the advantage
36:49.260 --> 36:56.260
of this is we can write V dot is x bar transpose
Q bar then, if this is Q bar is positive definite
36:58.480 --> 37:03.920
if Q bar is positive definite since there
is negative sign here, V dot is negative definite
37:03.920 --> 37:10.920
hence the system is stable. We can find the
positive definite of Q bar using Sylvester
37:12.810 --> 37:19.810
criterion all principal minor should be positive.
So, lambdamin Q minus alpha lambda x mean
37:20.340 --> 37:27.340
P is greater than zero and determinant of
Q bar which is this quantity determinant of
37:27.720 --> 37:34.720
Q bar is greater than 0. So, this quantity
the first minor for first alignment for this
37:36.790 --> 37:43.790
q11 actually, so this q11 has to be greater
than 0, so this quantity gives you if you
37:44.360 --> 37:51.360
go back to theorem one this identity and by
equating this identity determinant of Q bar
37:58.560 --> 38:01.530
has to be greater than zero.
38:01.530 --> 38:08.530
Then by doing some manipulation Q bar; this
is the Q bar quantity and then if satisfied,
38:10.720 --> 38:16.000
you get this quantities greater than this,
comparing the coefficient both sides, you
38:16.000 --> 38:22.580
can write: gammak is greater than this for
all k. The above equation gives constraint
38:22.580 --> 38:27.560
on the controller parameter for kth subsystem
as the controller parameter is the positive
38:27.560 --> 38:34.560
one it results in the second constraint in
theorem one which is gammak.
38:36.170 --> 38:43.170
We actually prove that, the theorem is correct
and in the controller I the salient points
38:48.270 --> 38:54.090
are: the linear system considered as a nominal
plant, may not fire at all operating results
38:54.090 --> 39:01.090
as system is traversing from one point to
another point x1 to xf x to xf. Then all the
39:07.070 --> 39:14.070
nominal plant, we have selected one of the
system matrices a and b of specific system
39:17.730 --> 39:24.730
to be nominal plant. It may not fire all the
time as it moves. Unmatched disturbance is
39:25.040 --> 39:31.190
measure for the entire fuzzy system considering
the above fact controller II is designed such
39:31.190 --> 39:36.180
that, the nominal plant changes with the operating
region thus reducing the norm bound on unmatched
39:36.180 --> 39:43.180
disturbance. What we are trying to say here
in the controller this one this desired criteria
39:45.000 --> 39:52.000
before that we can implement the controller,
the alphaf is less than lambdamin Q by lambdamax
39:52.780 --> 39:59.780
P because, this nominal plant is the norm
bound on the nominal plant.
40:00.620 --> 40:07.620
Because, distance between the nominal plant
and the actual plant where, the system rule
40:11.040 --> 40:18.040
is fire a specific rule is fired corresponding
to that in a plant and this alphaf kind of
40:18.220 --> 40:25.220
measure distance measure between the nominal
plant and actual fuzzy the plant associated
40:25.840 --> 40:32.840
with fuzzy rule that has been fired. Hence
this condition becomes little too harsh. To
40:34.530 --> 40:41.530
make that relaxed what we are doing is we
are now talking about second controller that,
40:46.020 --> 40:52.930
the nominal plant changes with the operating
result. As operating zone changes so which
40:52.930 --> 40:59.930
ever rule is fired from that rule we take
the plant if I have two rule fire so I consider
41:03.350 --> 41:10.270
each of them as a nominal plant and so what
I am trying to do in the second controller
41:10.270 --> 41:17.270
that, let us think that two rules are fired:
rule i and rule j. So, rule j is ů rule i
41:20.640 --> 41:27.640
is x dot is Ai x plus Bi u and rule j is x
dot is Aj x plus Bj u so this is i. In this
41:38.990 --> 41:45.990
second controller what we are aiming is that,
we consider all of them to be nominal plant.
41:46.380 --> 41:51.670
This is my nominal plant and also this is
my nominal plant, so I design a controller
41:51.670 --> 41:58.670
u around this plant and then the fuzzy blending
of the controller for both the plants is the
42:03.300 --> 42:09.560
overall controller gain. That is the idea
for second controller which we did not do
42:09.560 --> 42:16.560
for the first controller considering kth subsystem
to the nominal plant the T-S fuzzy system
42:18.280 --> 42:25.280
can be written as this particular one where
x dot t is represented around kth plant associated
42:31.480 --> 42:38.480
with the kth rule. Similarly, where we did
the disturbance term, so Fk can be written
42:41.130 --> 42:47.690
has in terms of three disturbance term as
we saw for the controller I. All the approaches
42:47.690 --> 42:52.370
are same only thing little bit difference
will be there.
42:52.370 --> 42:59.370
A controller problem given a set of r representative
dynamics compute uk such that, each represented
42:59.760 --> 43:04.990
dynamics is locally stable so that, fuzzy
blending of these individual actions defined
43:04.990 --> 43:11.990
as u makes T-S fuzzy model Lyapunov stable.
That is what I said here, r represented dynamics
43:12.360 --> 43:19.360
mix two represented dynamics mix if two rules
are fired or all rules can be fired actually
43:20.490 --> 43:27.490
in principle. That is why, r represented dynamics
compute vk for each subsystem and then fuzzy
43:28.290 --> 43:34.070
blending of u k equal to 1 to r sigmak uk
makes the T-S fuzzy model Lyapunov stable.
43:34.070 --> 43:41.070
This is the idea which comes from our the
first class the last class we discussed and
43:42.920 --> 43:49.920
again the disturbance measure the way it has
to be computed for h1k x t u t and fk x t.
43:52.380 --> 43:59.380
Then theorem two the controller II is suppose
that, Ak is the asymptotically stable and
44:00.990 --> 44:07.240
Pk is the positive definite matrix that is
fine Ak transpose Pk plus PkAk is minus 2Qk
44:07.240 --> 44:12.810
for some symmetric positive representation
Qk. Suppose also that alphafk is less than
44:12.810 --> 44:18.470
this quantity and alphau is less than one
then the state feedback controller u t is
44:18.470 --> 44:25.470
equal to minus gamma k equal to 1 to r sigmak
B transpose Pk x t where, gamma is greater
44:25.800 --> 44:32.660
than this quantity asymptotically stabilizes
the uncertain fuzzy model. So, here you see
44:32.660 --> 44:39.490
that, this is called k equal to 1 to r sigmak
B transpose Pk x t and this is called fuzzy
44:39.490 --> 44:45.920
blending of the controller where, we find
the gamma has to be greater than for this
44:45.920 --> 44:52.920
system to be stable. We also relaxed the minimum
condition that is required for implementing
44:53.560 --> 45:00.560
this controller where alphafk is the distance
between kth plant and the corresponding jth
45:05.580 --> 45:12.320
plant which is also fired and normally the
distance would be less. The proof is similar
45:12.320 --> 45:18.890
to the theorem one so I will not explain that
in this class you can this is an exercise
45:18.890 --> 45:25.890
for you that, how this theorem can be proved
an exercise. It is similar just like we moved
45:26.060 --> 45:29.390
to theorem one theorem two can be also proved.
45:29.390 --> 45:36.390
This is our theorem two we define two controllers
and the salient point difference between controller
45:37.250 --> 45:44.250
I and controller II: An arbitrary subsystem
is selected as a nominal linear plant from
45:44.680 --> 45:49.840
the set of all linear subsystems. The nonlinear
disturbance system at each operating point
45:49.840 --> 45:55.460
is computed derivation of actual dynamics
from the selected nominal point. As the dynamics
45:55.460 --> 46:00.560
moves from one operating point to another
operating point the disturbance also varies
46:00.560 --> 46:06.620
accordingly; whereas, in controller II, each
linear subsystem is considered as a nominal
46:06.620 --> 46:11.530
plant. The disturbance is modeled for each
nominal plant by considering the effects of
46:11.530 --> 46:16.500
its neighbor subsystems. The implementation
constraint is relaxed in case of controller
46:16.500 --> 46:19.240
II.
46:19.240 --> 46:25.100
This is our controller I where the structure
is minus gamma B transpose P x gamma is given
46:25.100 --> 46:30.720
by this and this gamma makes this controller
time varying the gain is time varying, x is
46:30.720 --> 46:37.220
my state so minus gamma B transpose P is the
time variable quantity it is not a constant
46:37.220 --> 46:44.220
quantity because gamma is varying. In controller
II if you look at here gamma is a constant
46:44.620 --> 46:51.620
quantity but here sigmai Pi makes this again
variable quantity but, the design principle
46:54.150 --> 47:01.150
between Q1 controller I and controller II
are different; whereas, the same principle
47:02.860 --> 47:08.830
of robust control has been used to design
controller I and controller II.
47:08.830 --> 47:14.730
This is our two link manipulator and this
is our dynamics we have already discussed
47:14.730 --> 47:20.190
a lot about this. I will not discuss a lot
about how we find a two link manipulator dynamics.
47:20.190 --> 47:27.190
Theta1 double dot and theta2 double dot these
are the two link angular accelerations and
47:27.630 --> 47:34.630
tau1 tau2 minus v1 v2 where v1 is given by
the quantity and v2 is given by this quantity
47:35.230 --> 47:39.400
and theta1 and theta2 are shoulder and elbow
angle, tau1 and tau2 box applied to shoulder
47:39.400 --> 47:46.400
and elbow manipulator and these are the m11
m12 m21 and m22.
47:49.710 --> 47:56.050
Here we do a little changes little transformation
that is two link planar mechanism needs finite
47:56.050 --> 48:03.050
torque at x equal to 0 0 0. That means if
I have a robot manipulator like this and if
48:04.940 --> 48:11.940
I want to keep this robot manipulator; this
is my 0 0 0 position or due to gravity it
48:14.750 --> 48:20.660
will fall down so hangs at every joint; this
joint and this joint I have to keep some finite
48:20.660 --> 48:26.590
torque I have to apply so that the manipulator
remains stable. But, you see that, other case
48:26.590 --> 48:33.590
the vertical position if I keep the manipulator
here the torque required is zero. That is
48:35.540 --> 48:42.540
when the no torque requirement the torque
is not require for balancing at vertical operating
48:43.280 --> 48:49.110
point. If I keep two link one above the other
this is called vertical operating position
48:49.110 --> 48:56.110
which is state wise pie by 2 0 0 0 then, you
see the system does not require at equilibrium
48:59.970 --> 49:06.970
in any control action. The control input if
I assume my u is minus Kx so you see that,
49:10.250 --> 49:17.250
we will give zero input at origin and because
it will give zero input around origin. If
49:17.340 --> 49:24.340
I define this as origin I require finite torque
it is not possible but here, I require zero
49:25.610 --> 49:32.580
torque hence I can define this to be origin
and I can implement u equal to minus Kx. To
49:32.580 --> 49:37.520
do that the origin is shifted to vertical
upright position by co ordinate transformation
49:37.520 --> 49:43.790
phi1 is equal to pie by 2 minus theta1. So,
theta1 has been transferred to theta phi1
49:43.790 --> 49:50.790
by pie by 2 and we make this as the origin,
this reversed.
49:52.410 --> 49:59.410
Doing that transformation what we do so is
we have this transformation and then the rule
50:06.440 --> 50:12.940
base: Considering the states as: x1 equal
to phi1 x2 equal to phi1 dot this is transformed
50:12.940 --> 50:19.940
theta1 and transformed theta1 dot x3 is theta3
and x4 is theta2 dot, two did not change that.
50:20.140 --> 50:27.020
The system is linearized around the operating
point with zero input both x1 x3 are fuzzified
50:27.020 --> 50:34.020
into seven equally specified reasons in range
minus phi by six the operating point of the
50:34.440 --> 50:40.890
state x2 and x4 are always considered as 0.
Thus we have 49 fuzzy rules and a linear subsystem
50:40.890 --> 50:46.290
corresponding to each rule. So, one rule is
given as follows: this is just taking an example
50:46.290 --> 50:53.290
so we have 49 fuzzy rules you understand because
what I am trying to do is that we are keeping
50:53.570 --> 50:55.810
here because we have 4 states.
50:55.810 --> 51:02.130
But what we are trying to do is that linearizing
x2 and x4 we are always making a 0. So, hence
51:02.130 --> 51:09.130
x1 and x3 they are varying and x1 is fuzzy
partition into seven as well as x3 is fuzzy
51:09.600 --> 51:16.430
partition into seven equally specified reasons.
By doing that we have 49 rules, so if x is
51:16.430 --> 51:23.430
around 0 0 0 by linearizing using taylor series
expansion you had x dot equal to this quantity
51:24.530 --> 51:31.530
A x plus v tau. Similarly, I can linearize
using this dynamic I am giving this dynamics.
51:35.460 --> 51:42.460
I can use this dynamics to linearize and then
I get this, so you can just do it given the
51:43.080 --> 51:50.080
plant model. Once done that, this is my linear
subsystem around the equilibrium point around
51:53.460 --> 51:55.930
the vertical upright position.
51:55.930 --> 52:02.930
Done that, you see that in the beginning I
have to stabilize the nominal plant. To stabilize
52:03.400 --> 52:09.980
the nominal plant, we place the poles at minus
2 minus 3 minus 2 minus 3 and we got these
52:09.980 --> 52:16.980
are the state feedback. You can use any mat
lab program or formula or pole plus technique
52:17.820 --> 52:24.820
then you get this gain. So, for controller
I upper norm bound is the disturbance alphahx
52:25.400 --> 52:32.400
is bound to be 23.3854 and for controller
II the input matrix for rule one taken as
52:32.610 --> 52:38.110
common input matrix. The closed loop poles
for all nominal subsystem are selected as
52:38.110 --> 52:42.780
minus 2 minus 3 minus 2 minus 3 minus 2 minus
3 and the preliminary feedback is given accordingly.
52:42.780 --> 52:49.780
Then maximum norm bound of matched state disturbance
alphahx computed as 8.44536 this is for controller
52:49.960 --> 52:55.760
II and the constraint on parameter gamma is
found to be greater than 27.9 and gamma is
52:55.760 --> 52:59.930
selected as 30.
52:59.930 --> 53:03.600
Controllers are proposed for stabilization.
To achieve to tracking those the overall control
53:03.600 --> 53:10.190
input tau is given as: tau is minus K x plus
u where, minus K is the stabilizing control
53:10.190 --> 53:16.050
input and u equal to u1 and u2 is the tracking
controller yet to be designed. We can easily
53:16.050 --> 53:22.680
design the stabilized fuzzy system dynamics
has a form this particular form and we have
53:22.680 --> 53:29.420
to give now the tracking controller u1 u2
so that, it tracts any desired trajectory.
53:29.420 --> 53:36.420
The output equation y is my theta1 position
and theta2 position and this using this equation
53:38.510 --> 53:45.510
the y double dot which is x2 dot and x4 dot
if x1 dot and x4 dot by y then x2 dot and
53:46.820 --> 53:53.820
x4 dot you know that x2 is x1 dot and x4 is
x3 dot. Using that principle I can write y
53:55.990 --> 54:02.990
double dot is a1y plus a2y dot plus bu where,
a1 is given by the matrices given by a21 a22
54:05.030 --> 54:12.030
and a41 a43 a21 a22 a41 a43 a21 a23 a41 a43.
Similarly, a2 is a24 a42 a44 and b is similarly
54:22.700 --> 54:29.700
here b21 b22 b41 b42. So, let the decided
output vector be yd and the error vector is
54:30.100 --> 54:33.260
defined as e equal to y minus yd.
54:33.260 --> 54:40.240
Then I can say u is this is my tracking control
and if I design this tracking controller I
54:40.240 --> 54:46.930
get the closed loop error dynamics as this.
If I take kp and kd and this is a stable dynamics
54:46.930 --> 54:52.230
and tracking is possible. Two link manipulator
trajectories tracking you can easily see that,
54:52.230 --> 54:59.230
desired trajectory and controller I and controller
II they are very much following; whereas,
54:59.280 --> 55:06.280
this other one which is not following is actually
proposed by Jack which is a fixed gain controller
55:06.540 --> 55:13.510
for T-S fuzzy model and it is performing very
badly. Similarly here also at a tracking at
55:13.510 --> 55:20.510
this joint 1 and this is joint 2. This joint
1 position tracking for both control one and
55:24.360 --> 55:28.300
control two is very good and here also for
both controller I and controller II is very
55:28.300 --> 55:35.300
good. But, Jack which is we compare with another
algorithm given by Jack as I said is not able
55:36.610 --> 55:39.320
to do properly.
55:39.320 --> 55:46.320
Control input this is the control input tau1
and tau2 which is very smooth. Controller
55:47.030 --> 55:54.030
parameter you see that, how the controller
is varying at different operating zones theta1
55:55.750 --> 56:02.750
theta2. If you see that is not a flat surface
it is constant it is varying so variation
56:03.000 --> 56:10.000
in gain K11 controller II you see that, how
it is varying over a range.
56:11.040 --> 56:18.040
Performance comparison controller I RMS error
is 0.016 and 0.049 and controller II 0.013
56:23.120 --> 56:30.120
and 0.036 the performance has been improved.
Similarly we can simulate the system for ball
56:31.190 --> 56:38.190
beam system. What is the ball beam system
is, I have a beam and on this I place a ball
56:40.280 --> 56:46.480
and the ball like it is like this beam and
I place a ball here and the ball will roll
56:46.480 --> 56:53.390
over the beam and the controller is that I
do up and down this beam. That means this
56:53.390 --> 56:59.710
beam is made up or down such that, the ball
always remains in the center point and this
56:59.710 --> 57:04.620
is the dynamics and for that also we have
designed the controller. I will not go in
57:04.620 --> 57:11.620
details of this. You see that, this controller
for ball beam system we could not implement
57:11.790 --> 57:17.360
controller I as well as Jack controller. We
could implement only controller II because
57:17.360 --> 57:23.390
of the relaxation that this controller provided.
57:23.390 --> 57:30.390
This is called the relaxation and this was
not satisfied for ball beam system whereas
57:31.280 --> 57:37.050
it is satisfied for controller II for ball
beam system. None of the system satisfies
57:37.050 --> 57:41.030
the non beam system non bound for controller
I. Hence, it cannot be implemented for the
57:41.030 --> 57:46.100
subsystem. What I am trying to show is I am
trying to show you another system for which
57:46.100 --> 57:50.510
the controller I cannot be implemented but
controller II can be implemented; control
57:50.510 --> 57:55.320
two satisfied norm bound condition. Hence,
it can be implemented for the system.
57:55.320 --> 58:01.190
You see that for ball position the desired
and actual very perfect tracking and this
58:01.190 --> 58:08.190
is the tracking error. Similarly beam angle
you see that and controller input. Summary
58:10.350 --> 58:14.870
in this lecture, we have covered the following
topics: T-S fuzzy model representations of
58:14.870 --> 58:19.680
nonlinear systems. T-S fuzzy model is represented
as linear plants with nonlinear disturbance
58:19.680 --> 58:25.410
terms. Two variable gain controllers have
been designed using robust control approach.
58:25.410 --> 58:32.370
Simulation results are represented for two
nonlinear systems showing the comparison.
58:32.370 --> 58:39.370
For the references you can see that, we have
two papers here, the first paper is Zak paper
58:39.610 --> 58:46.330
that is there in IEEE Transactions Fuzzy Systems
in 1999. The second is our paper which is
58:46.330 --> 58:52.450
in IEEE Transaction Systems Man Cyber net
in 2006 which is called variable gain controller
58:52.450 --> 58:59.450
nonlinear system using T-S fuzzy model. Thank
you very much.
411