This will be lecture six on fuzzy control. The topic that we will be discussing today
is linear controllers using T-S fuzzy model. In the last class, we discussed how to design
controllers for T-S fuzzy model when the input matrix is common for all subsystems. Now,
we will consider the generic T-S fuzzy model and how do we design linear controllers for
it.
We will just revise our notion of T-S fuzzy model representation of nonlinear systems.
, The approach of Controller design linear controller design. We will give two controllers
for this, stabilizing controller design robust controller approach; we will propose controller
I and as well as controller II - two different types of controller, simulation results: two-link
manipulator, ball beam system and summary.
The T-S fuzzy model is expressed in terms of r fuzzy rules where ith fuzzy rules has
the following form: if x1t is F1 i and x2t is a F2 i and so on until nxnt is an FnI then
x dot t is a is Aix t plus Biu t. This is my ith fuzzy rule consisting of n states and
each state is fuzzy variable and where the fuzzy variable FjI is the jth fuzzy set of
the ith rule. Then, the fuzzy index muI associated with the ith fuzzy rule is given by this formula
where mui j xj is the membership function of the fuzzy set Fj i. i equal to 1 to r.
Given an input output pair x t u t the fuzzy variable around this operating point is constructed
as the weighted average of the local model as x dot t equal to summation j equal to 1
to r sigma i Ajx t plus Bju t. This is jth local model and with jth local model associated
membership function is actually sigmaj, so sigmaj is a normalized fuzzy index associated
with jth local model. The summation of such local model multiplied with fuzzy index over
j equal to 1 to r gives me the complete fuzzy dynamics in terms of T-S fuzzy model. A T-S
fuzzy model approximates a nonlinear system as a cluster of a linear system. This is my
linear system; this is my jth linear system, sigmaj is the normalized fuzzy index associated
with jth linear system. When the cluster of such a linear system if there are there, then
if I fuzzy cluster these r number of linear systems then, I get the approximation of a
non linear system. The advantage of the fuzzy system is more informative in terms of local
dynamics because I can look at a nonlinear system in terms of linear system. Dynamics
is governed by subsystems fired at each operating point. In this class we will talk about variable
gain controller using single nominal plant.
This is the first type of controller will be talking today and in this we expressed
the fuzzy systems as a linear system with non linear disturbance. Our fuzzy system which
is x dot sigma j equal to 1 to r sigmaj Ajx plus Bju, this is my T-S fuzzy model approximation
of nonlinear system using T-S fuzzy model you can easily see that this has a very convenient
form. It looks as very convenient form easy to handle and I can write this as Ax plus
Bu plus disturbance term. This is my disturbance term, this is worst part I am saying here;
express the fuzzy system is the linear system with nonlinear disturbance. Then design a
controller to stabilize in the linear system in the presence of disturbance. The original
plant was x dot is sigmaj Ajx plus Bju j equal to 1 to r. So, this is my original plant.
I can always write this plant you see as: A x t plus B u t this is the nominal plant.
Once I separate from this as a nominal plant how do I write that this sigmaj Aj x in that
I have to subtract A x t which is here. So sigmaj Aj minus A x t, I subtract that and
j equal to 1 to I. Similarly, if I subtract b because, Bj is there I subtract b here.
How can I do? You must know sigma j equal to 1 to r sigmaj A is A; because, j equal
to 1 to r sigma j equal to 1. You may worried that I should have only subtracted A outside
no, this is one and this is equal to A so I can subtract this quantity here. Similarly,
about B so x dot t is written in the particular form; this is my original dynamics; this can
be written in this particular form. This can be finally written A x t plus B u t F x t
u t and this is my disturbance term around this nominal plant where, A x t B u t is the
linear system and F x t is a nonlinear disturbance given by. I represent this nonlinear system
in terms of three different components. So, this is simply F x t B h1 x t B h2 u t. You
can easily see that, B x t u t is obviously taken into account of this. This one is actually
B h2 u t you can easily see that this is a nonlinear term and that is given by B x2 u
t, we can express that. Similarly, these two terms combined represents this one. If I say
this is the second term this is my second term. This term is represented by this. I
wish that you understood what we are trying to do my original plant was given by this
particular thing x dot is sigma Ajx plus Bju this one I am representing this same dynamics
as this. There is no difference between this and this. I am representing this whole thing
as A x t plus B u t is the first plant nominal plant and this F x t is disturbance and this
has three component which the first component here I is this one. Sigmaj Bj minus B u t
sigmaj equal to 1 to r and the second component is F x t plus B h1 x t this component here.
Always in controller design whenever we say any disturbance, we are not interested in
exact represent of disturbance rather an upper bound. This is the principle of robust control;
we want to know the upper bound of this disturbance. That means if I am designing controller for
the worst case naturally the controller also will be a stable for the other cases. This
is the robust control design principle is design the controller for worst case.
That is why we will represent first of all disturbance terms more elaborately, we will
define the upper bound. Let us see the first disturbance B h2 u t the component one as
I said which is: sigmaj Bj minus u t j equal to 1 to r. We can write this equation as,
I have to write in terms of B here multiplied sigmaj Bj bar t u t and this Bj bar is obviously
such that the b Bj bar is v j minus v. This identity has to be satisfied, so B Bj bar
is Bj minus B
Similarly, F x t plus B h1 x t is this term, the second term, j equal to 1 to r sigmaj
Aj minus x t and then this can be written in terms of two. Looking at this you can easily
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
as B h1 x t. So, here Aj minus A has to be written in terms of A1j plus BA2j where, the
A1j represents the unmatched disturbance, unmatched means you see that, this disturbance
is not magnified by u, means this is not with there with the control input. This is matched
you see the b in to this and you see our normal plant is A x x dot is Ax plus Bu. So, anything
with B means you can say that, this term is like an excitation because anything multiplied
with B is kind of excitation to the system and this term is not with b. It is separate
term unmatched disturbance and this is matched disturbance means this is disturbance excitation
that is exciting system because, we say b is the control matrix. In that sense, A1j
is the unmatched disturbance and this is the matched disturbance.
Now, we will talk about the control problem. The T-S fuzzy model approximates a nonlinear
system as a fuzzy cluster of r linear subsystems. Since each sub system is linear, linear control
theory can be applied to design fixed gain controller for each system design fixing controller
for each subsystem. Since the desired system output traverses a specific trajectory system
states traverse across different fuzzy zones. It is thus expected that, the controller will
be characterized by variable gain instead of fixed gain which I have already discussed.
The control problem is given a T-S fuzzy model representation design a variable state feedback
controller u t equal to minus kx t where, k is the variable such that, the T-S fuzzy
model is Lyapunov stable. Here, the matrix k represents variable state feedback gain
it is not a constant gain as in case of a linear system. Now, we will be talking about
this disturbance measure. Since the nominal plant is linear while disturbance term is
nonlinear one can possibly solve the control problem using the principles of robust control
theory.
For that the norm bounds of uncertainties have to be computed first. The norm bound
of unmatched state disturbance which is h1x t. I will show you which is h1 x t is 1 this
one is matched state disturbance. Matched state disturbance h1 x t is alpha j you see
h1 x t we have represented h1 x t is this is h1 x t. You can easily say sigmaj A2 x
t sigma j equal to 1 to r. So that is what we have written here: h1 x t is sigmaj A2
j x t known from j equal to 1 to r overall norm. I can represent this particular term
using a triangular in a quality as less than equal to sigmaj and this is the induced norm
of A2j is alpha h x j, alpha h x j represent the maximum singular value of A2j and the
norm of x2 x t separately. This is a triangular in equality and what we are saying is that
the kind of this disturbance is represented in terms of a major disturbance which is less
than this quantity.
The norm bound of input disturbance is, which is h2 u t you see h2 u t given here is this
one and this sigmaj bar Bj bar u t. This was actually sigmaj Bj bar u t j equal to 1 to
r. This can be written in triangular inequality as j equal to 1 to r sigmaj and norm of B
bar j into u t norm and we can write now B bar j norm the maximum norm. I can put here
the maximum norm of induce norm of B bar j is alpha u j alpha u j is the norm of B bar
j that means this is maximum singular value of B bar j. This inequality gives a disturbance
measure for this quantity where we already know alpha u j, we know sigmaj. Similarly,
the norm bound of unmatched state disturbance which is F x t the previous one is unmatched.
This is unmatched one F x t and F x t is sigmaj A1j x t, so this whole norm can be written
as again less than equal to this A1j alphaf is norm of A1j induced second norm A1j this
is alphaf.
I wish that you have understood now what we have been talking about the disturbance measure
so once the disturbance measure I define, we will be now telling a theorem which says
that: if I design a state feedback controller, if I have a state feedback controller for
the system, what is the system now, my system is: x dot is Ax plus Bu plus f x plus B h1
x plus B h2 u. This is my system which my original system is simply sigmaj Aj x plus
Bj u sigma j equal to 1 to r. This is same as this quantity we said we approximated and
then we found out the measure upper bound of effects h1 x and h2 u and then, we are
saying that, the system will be stable if this u control input u is given by minus sigma
j equal to 1 to r sigmaj gammaj B transpose B x t this is my control law and gammaj satisfies
this particular condition where alpha h x j and alpha u and alpha m alpha f they are
all disturbance measure as we define just recently.
Now, this can be only valid if the two conditions are satisfied and that is alpha f is less
than equal to lambdamin the minimum eigen value of Q and lambdamax is the maximum eigen
value of P where, A transpose P PA is equal to minus 2Q. You know that is my A is nominal
plant model and for a nominal plant model A, I can always find out AQ for such that,
I have also P which satisfies A transpose P plus PA equal to minus 2Q and so given this
P and Q the alphaf which is here, this alphaf is the induce norm.
You see alphaf is the induce norm of A1j and this A1j is coming from our term Aj minus
A this can be written as A1j plus A2j this is what we have shown earlier Aj minus A is
A1j plus BA2j. So Aj minus A A1j plus so this A1j is the induce norm is alphaf means the
maximum singular value A1j. This is the theorem and we will just prove this theorem again
repeat what is this theorem implies that means, if I propose a control law u t where the gammaj
satisfies this condition then, the system will be stable provided alphaf is less than
this identity as well this identity is true. Now, consider the Lyapunov candidate v equal
to x transpose Px this is the theorem one proving, we trying to prove the theorem one.
Now, the time derivative of v is given by v dot is two x transpose is P x dot.
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
is we have already given u to be so u is we just define u is here this is our u and if
I write down that u is sigma minus j equal to 1 to r sigmaj gammaj B transpose P x. This
is my u so before I introduce u inside what I will do is that, this is my V dot is 2 x
transpose P x dot and I can write this expression by introducing this x dot inside here. If
I do that what I will get 2x transpose P x dot. So, there is A x so what I get is that
2 x transpose P A x so this two and we know that A transpose P plus P A is minus 2Q this
combined with the knowledge that 2x transpose P A x this can be written as x transpose A
transpose P plus P A x. You see that A transpose P plus PA is equal to minus T Q then, I can
write 2x transpose P A x transpose A transpose P plus P A into x and this quantity is now
minus 2Q. We have already said that given a we achieve this A transpose P plus B transpose
P A equal to minus 2Q if you can write. This writing that this first term 2x transpose
P A x can be written as minus 2x transpose Q X, so, this is minus 2x transpose Q X, the
first one. So, from x dot I took care of it AX. Now, let me take care of Bu so Bu is this
quantity, so how do I write it is B u if I put it here, so 2x transpose P Bu, so 2x transpose
P B and u has B transpose P x this one and the other quantity is that sigma j equal to
1 to r sigmaj gammaj.
So because, we have already j inside so j we have k equal to 1 to r sigmak gammak, so
this quantity is given by 2X transpose P B u and again we have multiplied here f x here
x dot has also another component f s, so that is 2 x transpose P into f x. Similarly, B
h1 x 2 transpose P B h1 x similarly 2 x transpose P B h2 u. So, all that here what we have done
instead of x dot, we have replace this and we can write the equation like this. Here,
further what we can do we can rewrite this term as kind of a using the properties of
matrices that is x transpose Q x. It can be bounded by two quantities the lower bound
is lambda minimum that means the minimum eigen values of Q x norm square and the upper bound
is lambdamax singular value of lambda eigen value of Q norm square because, Q and P there
are symmetric matrices. Hence, the singular value maximum singular value is same as maximum
eigen value and therefore minus x transpose Q x, if I write this minus I can write minus
x Q x transpose Q x is less than equal to minus lambdamin Q x norm square.
For the symmetric positive definite matrix P that induces 2-norms is P norm is we can
say induce norm is lambdamax P. So, this is my maximum value of P and further more if
I look at this expression That this is not capital x this is small x bold x, x is a vector
small x. So, x transpose P B x transpose P is 2 x transpose 2 x transpose P. This is
not capital x. So, x transpose P B transpose B x is this quantity, this can be written
as x transpose P B again x transpose P B transpose which is B transpose x. In this I can write
this is a norm noun x transpose P B norm square, this is matrix theory. Taking all the relation
to this account and using norm bounds on certain elements. We get, norm earlier V dot is this
so using the norm bound using norm bound means this is less than this quantity. Similarly,
we can find this is less than this quantity, this less than its norm bound; this is less
than its own norm bound; this is less than own norm bound.
V dot can be written as a less than some quantity and that quantity is here V dot is less than
minus 2 lambdamin Q x norm square. This is the norm bound of the second term; this is
the norm bound of the third term; this is norm bound for the fourth term and this is
norm bound of the fifth term, we have five terms. By putting that, this particular expression,
so what we did is that, we first derived what is the direct derivative of Lyapunov function
and then we expressed that in terms of norm bound using the sine less than equal to and
then this particular quantity can be written as: minus 2 x bar transpose Q bar x bar where,
x bar is this two quantity x norm and x transpose P B. This you can say one element and another
element this x bar is a vector norm, this is first element and this is a second element
and Q bar has four element q11 q12 q21 q22 where q11 is given by this quantity that is
minimum eigen value of Q minus alpha f lambdamax of P the maximum eigen value of P. Similarly,
q12 q21 is half of this quantity and q22 is this quantity and you may be wondering how
we got this but, I will just explain to you in a very simple understand you can easily
see that, I can now add all x norms square together so that, if r that you can see that,
this x norm square if I take common, I get here two lambdamin 2 and 2 alpha x lambda
x P. So, if I combine them I can get lambdamin Q minus alpha x lambda x and lambda alpha
f lambdamax. This is my q11 so you see that, if I write this expression as this and I have
taken 2 out, 2 is common here. So, if I take 2 common out so I have also taken negative
outside, so this becomes lambdamin Q minus alpha f lambdamax P which is this quantity
q11 into so the point is that, you can easily see this quantity and with definition x bar
and this and this is we can write out q11 x norm square plus q22 norm x transpose P
B square and you can easily see q12 plus q21 x norm and x transpose P V.
What you saw that, V dot is less than this five terms and I am trying to represent five
terms in terms of minus 2 x bar transpose Q bar x bar. If I define x bar is this two
terms vector of two terms and Q bar is vector matrix of 2 by 2 then, this x bar transpose
Q bar x bar is this quantity q11 x bar x norm square q22 x transpose P V norm square plus
q12 q21 x norm into x transpose P V norm. So, you can easily see here that, obviously
I have to find out this is x bar transpose Q bar x y is this quantity. So, minus 2 if
I take common here, what I get is that easily by comparing the coefficient of qc q11 will
be the total coefficient of x1 bar. That is lambdamin x minus alpha f lambdamax p which
is here q11.
q22 is the coefficient of x transpose P V norm square, so this is one term x transpose
P V square and another term is x transpose P V square. If I take minus 2 common here
I get here minus 2 common k equal to 1 to r k equal to 1 to r sigmak gammak sigmak gamma
k. This is the first term and second term is here where, this is sigmaj alpha u j j
is equal to 1 and sigmak gammak k equal to 1 to r. So, this is q22 and then q12 plus
q21 is the coefficients half x bar and x transpose P V you see that, this all one term. Since
we have q12 and q21 the coefficient is simply j equal to 1 sigmaj alpha H x j, so this is
my coefficient and I have two term, so I can easily do that by dividing them equally and
making q12 equal to q21 so q12 to equal to q21 which is minus half because, here if I
take minus 2 common I am getting minus half here minus j. So, minus half sigmaj alpha
H x j equal to 1 to r, so I wish that you understood how we finally wrote V dot in terms
of a quadratic function x transpose Q bar x bar and instead V dot what is the advantage
of this is we can write V dot is x bar transpose Q bar then, if this is Q bar is positive definite
if Q bar is positive definite since there is negative sign here, V dot is negative definite
hence the system is stable. We can find the positive definite of Q bar using Sylvester
criterion all principal minor should be positive. So, lambdamin Q minus alpha lambda x mean
P is greater than zero and determinant of Q bar which is this quantity determinant of
Q bar is greater than 0. So, this quantity the first minor for first alignment for this
q11 actually, so this q11 has to be greater than 0, so this quantity gives you if you
go back to theorem one this identity and by equating this identity determinant of Q bar
has to be greater than zero.
Then by doing some manipulation Q bar; this is the Q bar quantity and then if satisfied,
you get this quantities greater than this, comparing the coefficient both sides, you
can write: gammak is greater than this for all k. The above equation gives constraint
on the controller parameter for kth subsystem as the controller parameter is the positive
one it results in the second constraint in theorem one which is gammak.
We actually prove that, the theorem is correct and in the controller I the salient points
are: the linear system considered as a nominal plant, may not fire at all operating results
as system is traversing from one point to another point x1 to xf x to xf. Then all the
nominal plant, we have selected one of the system matrices a and b of specific system
to be nominal plant. It may not fire all the time as it moves. Unmatched disturbance is
measure for the entire fuzzy system considering the above fact controller II is designed such
that, the nominal plant changes with the operating region thus reducing the norm bound on unmatched
disturbance. What we are trying to say here in the controller this one this desired criteria
before that we can implement the controller, the alphaf is less than lambdamin Q by lambdamax
P because, this nominal plant is the norm bound on the nominal plant.
Because, distance between the nominal plant and the actual plant where, the system rule
is fire a specific rule is fired corresponding to that in a plant and this alphaf kind of
measure distance measure between the nominal plant and actual fuzzy the plant associated
with fuzzy rule that has been fired. Hence this condition becomes little too harsh. To
make that relaxed what we are doing is we are now talking about second controller that,
the nominal plant changes with the operating result. As operating zone changes so which
ever rule is fired from that rule we take the plant if I have two rule fire so I consider
each of them as a nominal plant and so what I am trying to do in the second controller
that, let us think that two rules are fired: rule i and rule j. So, rule j is ů rule i
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
second controller what we are aiming is that, we consider all of them to be nominal plant.
This is my nominal plant and also this is my nominal plant, so I design a controller
u around this plant and then the fuzzy blending of the controller for both the plants is the
overall controller gain. That is the idea for second controller which we did not do
for the first controller considering kth subsystem to the nominal plant the T-S fuzzy system
can be written as this particular one where x dot t is represented around kth plant associated
with the kth rule. Similarly, where we did the disturbance term, so Fk can be written
has in terms of three disturbance term as we saw for the controller I. All the approaches
are same only thing little bit difference will be there.
A controller problem given a set of r representative dynamics compute uk such that, each represented
dynamics is locally stable so that, fuzzy blending of these individual actions defined
as u makes T-S fuzzy model Lyapunov stable. That is what I said here, r represented dynamics
mix two represented dynamics mix if two rules are fired or all rules can be fired actually
in principle. That is why, r represented dynamics compute vk for each subsystem and then fuzzy
blending of u k equal to 1 to r sigmak uk makes the T-S fuzzy model Lyapunov stable.
This is the idea which comes from our the first class the last class we discussed and
again the disturbance measure the way it has to be computed for h1k x t u t and fk x t.
Then theorem two the controller II is suppose that, Ak is the asymptotically stable and
Pk is the positive definite matrix that is fine Ak transpose Pk plus PkAk is minus 2Qk
for some symmetric positive representation Qk. Suppose also that alphafk is less than
this quantity and alphau is less than one then the state feedback controller u t is
equal to minus gamma k equal to 1 to r sigmak B transpose Pk x t where, gamma is greater
than this quantity asymptotically stabilizes the uncertain fuzzy model. So, here you see
that, this is called k equal to 1 to r sigmak B transpose Pk x t and this is called fuzzy
blending of the controller where, we find the gamma has to be greater than for this
system to be stable. We also relaxed the minimum condition that is required for implementing
this controller where alphafk is the distance between kth plant and the corresponding jth
plant which is also fired and normally the distance would be less. The proof is similar
to the theorem one so I will not explain that in this class you can this is an exercise
for you that, how this theorem can be proved an exercise. It is similar just like we moved
to theorem one theorem two can be also proved.
This is our theorem two we define two controllers and the salient point difference between controller
I and controller II: An arbitrary subsystem is selected as a nominal linear plant from
the set of all linear subsystems. The nonlinear disturbance system at each operating point
is computed derivation of actual dynamics from the selected nominal point. As the dynamics
moves from one operating point to another operating point the disturbance also varies
accordingly; whereas, in controller II, each linear subsystem is considered as a nominal
plant. The disturbance is modeled for each nominal plant by considering the effects of
its neighbor subsystems. The implementation constraint is relaxed in case of controller
II.
This is our controller I where the structure is minus gamma B transpose P x gamma is given
by this and this gamma makes this controller time varying the gain is time varying, x is
my state so minus gamma B transpose P is the time variable quantity it is not a constant
quantity because gamma is varying. In controller II if you look at here gamma is a constant
quantity but here sigmai Pi makes this again variable quantity but, the design principle
between Q1 controller I and controller II are different; whereas, the same principle
of robust control has been used to design controller I and controller II.
This is our two link manipulator and this is our dynamics we have already discussed
a lot about this. I will not discuss a lot about how we find a two link manipulator dynamics.
Theta1 double dot and theta2 double dot these are the two link angular accelerations and
tau1 tau2 minus v1 v2 where v1 is given by the quantity and v2 is given by this quantity
and theta1 and theta2 are shoulder and elbow angle, tau1 and tau2 box applied to shoulder
and elbow manipulator and these are the m11 m12 m21 and m22.
Here we do a little changes little transformation that is two link planar mechanism needs finite
torque at x equal to 0 0 0. That means if I have a robot manipulator like this and if
I want to keep this robot manipulator; this is my 0 0 0 position or due to gravity it
will fall down so hangs at every joint; this joint and this joint I have to keep some finite
torque I have to apply so that the manipulator remains stable. But, you see that, other case
the vertical position if I keep the manipulator here the torque required is zero. That is
when the no torque requirement the torque is not require for balancing at vertical operating
point. If I keep two link one above the other this is called vertical operating position
which is state wise pie by 2 0 0 0 then, you see the system does not require at equilibrium
in any control action. The control input if I assume my u is minus Kx so you see that,
we will give zero input at origin and because it will give zero input around origin. If
I define this as origin I require finite torque it is not possible but here, I require zero
torque hence I can define this to be origin and I can implement u equal to minus Kx. To
do that the origin is shifted to vertical upright position by co ordinate transformation
phi1 is equal to pie by 2 minus theta1. So, theta1 has been transferred to theta phi1
by pie by 2 and we make this as the origin, this reversed.
Doing that transformation what we do so is we have this transformation and then the rule
base: Considering the states as: x1 equal to phi1 x2 equal to phi1 dot this is transformed
theta1 and transformed theta1 dot x3 is theta3 and x4 is theta2 dot, two did not change that.
The system is linearized around the operating point with zero input both x1 x3 are fuzzified
into seven equally specified reasons in range minus phi by six the operating point of the
state x2 and x4 are always considered as 0. Thus we have 49 fuzzy rules and a linear subsystem
corresponding to each rule. So, one rule is given as follows: this is just taking an example
so we have 49 fuzzy rules you understand because what I am trying to do is that we are keeping
here because we have 4 states.
But what we are trying to do is that linearizing x2 and x4 we are always making a 0. So, hence
x1 and x3 they are varying and x1 is fuzzy partition into seven as well as x3 is fuzzy
partition into seven equally specified reasons. By doing that we have 49 rules, so if x is
around 0 0 0 by linearizing using taylor series expansion you had x dot equal to this quantity
A x plus v tau. Similarly, I can linearize using this dynamic I am giving this dynamics.
I can use this dynamics to linearize and then I get this, so you can just do it given the
plant model. Once done that, this is my linear subsystem around the equilibrium point around
the vertical upright position.
Done that, you see that in the beginning I have to stabilize the nominal plant. To stabilize
the nominal plant, we place the poles at minus 2 minus 3 minus 2 minus 3 and we got these
are the state feedback. You can use any mat lab program or formula or pole plus technique
then you get this gain. So, for controller I upper norm bound is the disturbance alphahx
is bound to be 23.3854 and for controller II the input matrix for rule one taken as
common input matrix. The closed loop poles for all nominal subsystem are selected as
minus 2 minus 3 minus 2 minus 3 minus 2 minus 3 and the preliminary feedback is given accordingly.
Then maximum norm bound of matched state disturbance alphahx computed as 8.44536 this is for controller
II and the constraint on parameter gamma is found to be greater than 27.9 and gamma is
selected as 30.
Controllers are proposed for stabilization. To achieve to tracking those the overall control
input tau is given as: tau is minus K x plus u where, minus K is the stabilizing control
input and u equal to u1 and u2 is the tracking controller yet to be designed. We can easily
design the stabilized fuzzy system dynamics has a form this particular form and we have
to give now the tracking controller u1 u2 so that, it tracts any desired trajectory.
The output equation y is my theta1 position and theta2 position and this using this equation
the y double dot which is x2 dot and x4 dot if x1 dot and x4 dot by y then x2 dot and
x4 dot you know that x2 is x1 dot and x4 is x3 dot. Using that principle I can write y
double dot is a1y plus a2y dot plus bu where, a1 is given by the matrices given by a21 a22
and a41 a43 a21 a22 a41 a43 a21 a23 a41 a43. Similarly, a2 is a24 a42 a44 and b is similarly
here b21 b22 b41 b42. So, let the decided output vector be yd and the error vector is
defined as e equal to y minus yd.
Then I can say u is this is my tracking control and if I design this tracking controller I
get the closed loop error dynamics as this. If I take kp and kd and this is a stable dynamics
and tracking is possible. Two link manipulator trajectories tracking you can easily see that,
desired trajectory and controller I and controller II they are very much following; whereas,
this other one which is not following is actually proposed by Jack which is a fixed gain controller
for T-S fuzzy model and it is performing very badly. Similarly here also at a tracking at
this joint 1 and this is joint 2. This joint 1 position tracking for both control one and
control two is very good and here also for both controller I and controller II is very
good. But, Jack which is we compare with another algorithm given by Jack as I said is not able
to do properly.
Control input this is the control input tau1 and tau2 which is very smooth. Controller
parameter you see that, how the controller is varying at different operating zones theta1
theta2. If you see that is not a flat surface it is constant it is varying so variation
in gain K11 controller II you see that, how it is varying over a range.
Performance comparison controller I RMS error is 0.016 and 0.049 and controller II 0.013
and 0.036 the performance has been improved. Similarly we can simulate the system for ball
beam system. What is the ball beam system is, I have a beam and on this I place a ball
and the ball like it is like this beam and I place a ball here and the ball will roll
over the beam and the controller is that I do up and down this beam. That means this
beam is made up or down such that, the ball always remains in the center point and this
is the dynamics and for that also we have designed the controller. I will not go in
details of this. You see that, this controller for ball beam system we could not implement
controller I as well as Jack controller. We could implement only controller II because
of the relaxation that this controller provided.
This is called the relaxation and this was not satisfied for ball beam system whereas
it is satisfied for controller II for ball beam system. None of the system satisfies
the non beam system non bound for controller I. Hence, it cannot be implemented for the
subsystem. What I am trying to show is I am trying to show you another system for which
the controller I cannot be implemented but controller II can be implemented; control
two satisfied norm bound condition. Hence, it can be implemented for the system.
You see that for ball position the desired and actual very perfect tracking and this
is the tracking error. Similarly beam angle you see that and controller input. Summary
in this lecture, we have covered the following topics: T-S fuzzy model representations of
nonlinear systems. T-S fuzzy model is represented as linear plants with nonlinear disturbance
terms. Two variable gain controllers have been designed using robust control approach.
Simulation results are represented for two nonlinear systems showing the comparison.
For the references you can see that, we have two papers here, the first paper is Zak paper
that is there in IEEE Transactions Fuzzy Systems in 1999. The second is our paper which is
in IEEE Transaction Systems Man Cyber net in 2006 which is called variable gain controller
nonlinear system using T-S fuzzy model. Thank you very much. 411