Hello Viewers. Welcome to the lecture on Feedback
Control. In this lecture, we will continue
with the concept of feedback control which
was discussed in the previous lecture. Now,
I will demonstrate the computation of feedback
control using an example. So first let us
summarize the concept of feedback control
which was seen in the last lecture.
So let x dot = Ax + bu be a control system
given to us where A is n cross n matrix and
b is a column vector n cross 1 matrix. Now
let the; okay. Let u the control is a feedback
control of the form kx, where k is a unknown
column; k is a unknown row vector. And so,
when we substitute the control in this equation
we will get Ax+b kx so that will finally become
(A+bk) of x. So let us S be the set Mu 1,
Mu 2 etcetera Mu n be an arbitrarily set of
numbers, so n number are given arbitrarily.
So now the question is, is it possible 
to find the matrix k such that the eigenvalues
of A+bk are exactly the set S, Mu i, where
i=1, 2, 3 up to n. So is it possible to find
a k in this manner. So the answer is we can
find provided the system is controllable.
So let us call this system equation as 1,
is the system 1 is controllable, or the rank
of b Ab A square b etcetera A-1b, if the rank
is n then we can find a k like this. So if
you are able to find how to find this k.
That is; for this we follow this procedure,
the procedure is; earlier we studied that
there exists because the system is construable,
there exists a non-singular matrix T such
that TAT inverse is the companion form 0 1
0 0, last is 1 and then - alpha n etcetera
–alpha 1. And T*b is 0 0 1 this is the d
matrix. So we can find a unique non-similar
matrix T so that this happens is it not. So
here can easily see that alpha 1, alpha 2,
alpha n these are the coefficient of the characteristic
equation; characteristic polynomial of the
matrix C.
In other words, it is the coefficient of characteristic
polynomial of matrix A because A and C are
similar matrices, they will have the same
eigenvalue so characteristic polynomial is
the same for both of them, so the coefficient
of characteristic is a last row of the matrix
C. Now we are interested in A+bk, we want
that the eigenvalues of A+bk are Mu 1 Mu 2
Mu n. So we can write this as A is T inverse,
CT. And b is T inverse d, okay. So we can
write this as T inverse Tb.
So this can be written as T inverse Tb, b
can be written like this then we have k then
T inverse T can be written like this, so that
the final thing is we can take T inverse outside
the left side and T can be taken out in the
right side so the remaining things are C+Tb=d
and then K*T inverse, so this expression is
there. So it is easy to see here A+bk and
C+dkT inverse, these are the similar matrices.
So the eigenvalues of C+dkT inverse and A+bk
are same.
So now if you calculate this; therefore, the
last column last row which will appear in
this matrix should be the characteristic coefficient
of the characteristic polynomial of A+bk,
so that is the meaning of this expression.
Now if you calculate C+dkT inverse so C is
0 1 0 0 etc., –alpha n, - alpha 1 and d
is 0 0 1 and k is kn kn-1 k1*T inverse. So
this will appear like this, plus the product
of this two it will be 0 0 etc., kn k1 * k
inverse.
Further if we calculate we will get this expression
to be the first matrix is c+, the second one
is 0 0; first n-1 column, the last column
we will multiply this row with this T inverse
matrix we will get gamma n, gamma n-1, gamma
1.
So adding these two matrices finally we will
get the matrix to be 0 1 0 0; 0 0 1 then we
will get –alpha n + gamma n and –alpha
1 + gamma 1. So this should be the coefficient
of the polynomial. So the characteristic equation
of A+bk, what is required is lambda – Mu
1; lambda –Mu 2; lambda –Mu n = 0. So
this will be some polynomial like this + beta
n=0. So what we have is the last column of
this matrix; last row of this matrix should
be –beta n, -beta n-1 etc.
So this implies that –alpha i + gamma i
that should be equal to –beta i, i=1 to
up to n. Or we want to calculate this gamma
i that is equal to alpha i – beta i. So
–beta i, i=1 to n. And the previous slide
we have seen the relation here, this expression
we want the k matrix only. But what we have
calculated is this expression. So the relation
between k matrix and this gamma kn kn-1 k1
that is the last row. And this last row is
gamma n etc.
So the relation between this two is k multiplied
by T inverse is given by this gamma matrix.
So from the previous page we know that kn
kn-1 k1*T inverse that is gamma n, gamma n-1*gamma
1 up to that. So this implies the k matrix
is gamma matrix multiplied by T. So if we
use this k matrix as the feedback matrix here
this one k, u=kx we will get Mu1 Mu2 Mu n
as the eigenvalue of the matrix A+bk. So that
is what we will see.
The computation as an example we will see
here. So if you consider the matrix A to B
say 1 2; 2 1 and the matrix B as 2 1, okay.
And let S be the matrix, the set -2 and 1.
So the question is, so is it possible to find
a matrix k which is called K2 K1 such that
A+bk has eigenvalues -2 and 1. The arbitrarily
given set are -2 1 here. So is it possible
to find k? Now we can first check whether
the system is controllable or not.
So the system is controllable, so the set
U, the matrix U is, b is 2 1 and Ab is 4 and
5 if you compute this one. So this implies
the rank of U is 2 therefore it implies the
controllability. So according to the principle
which we have seen earlier we have to find
the matrix T. So we have to find the matrix
T first, so T is, the first row is alpha,
the second row is alpha*A, where the companion
form we have studied that the alpha matrix
can be computed by using 0 1 and U inverse,
okay.
So this implies alpha = 0 1, U is this matrix
and U inverse if you calculate that is 5/6,
-4/6 and -1/6, 2/6. So if you multiply this
two we will get the alpha value to be -1/6
and 2/6 as the alpha value. The second row
of T is alpha*A. So this is alpha and A matrix
is there, if you calculate this one we will
get 3/6 and 0, okay. So this implies the T
matrix is -1/6 and 2/6, 3/6, 0. So T matrix
is obtained.
Then it is clear that TAT inverse will be
equal to C. So here C is nothing but the companion
form. And the last row of this will be having
the coefficient of the characteristic polynomial
of the matrix A, okay. So we will get the
matrix C to be; because if you see A the eigenvalues
of A matrix can be calculated and from there
we can calculate or directly we can calculate
the characteristic polynomial that is lambda
square -2, lambda -3 = 0.
So the last row of C will be 3 and 2 is the
C matrix and Tb we can easily check it will
come out to be 0 1 here, okay. So we got the
T matrix and C d matrix etcetera all these
are available. Now to find the K matrix which
K2 K1, we have to follow the procedure that,
we have to follow that K2 K1 it is equal to
gamma 2, gamma 1 multiplied by the T matrix.
K matrix = gamma*T, gamma 2, gamma 1 * T matrix.
And gamma values are obtained by alpha i-beta
i. Now we have to see what should be the value
of beta i. Beta i are calculated by the characteristic
equation of A+bk, it is given by because we
require the set -2 and 1 are the eigenvalues,
so the characteristic equation can be calculated
as lambda+2*lambda-1=0, so this implies lambda
square + lambda-2=0. So beta 1 is 1 here and
beta 2 is -2.
Therefore, we get the value gamma 1 is alpha
1 – beta 1 which is equal to; alpha 1 is
here -2 and beta 1 is 1 here so the value
is 3, gamma 2 is alpha 2-beta 2 alpha 2 is
-3 and –beta 2 is +2=-1. So we got the value
of gamma and gamma multiplied by T matrix.
K2 K1 that is equal to gamma 2, that is -1
-3 * T matrix -1/6, 2/6, 3/6, 0 so by multiplying
this two we will get the value k matrix is
gamma 2 gamma 1 * T. So we will get the value
– that is +1/6 -8/6 and -2/6 that is the
value of the matrix k. So the feedback control
u(t) is given by -8/6 -2/6 * the Xn X2 are
the solution of the problem. The given control
system, so that is nothing but -8/6 x1, -2/6
x2 is the feedback control of the problem.
And substituting this control; for finding
the control, we have to find x1, x2 but in
real life situation at each instant of time
we can measure the x1 and x2 value the position,
the coefficient; the current position of the
dynamical system can be measured at each instance
of time. So depending on those values the
control should be applied on the system, so
that it proceeds the control is working on
the system.
And it will drive the system as we desired,
because the eigenvalue are selected for a
particular purpose. In this case just to demonstrate
the problem we have taken -2 and 1. So it
may not serve any physical purpose of any
particular problem. But normally this procedure
is applied for stabilizing a system. If you
have a control system x dot = Ax+bu to make
the system stable the eigenvalues of the A+bk
matrix it should be negative that is; for
asymptotically stable solution we need that
the; the eigenvalue of the matrix A+bk all
of them should be having negative real part.
So if we select the set S in such a way that
all the eigenvalues have negative real part
and then find the feedback control then we
are sure that the system becomes asymptotically
stable, so that is the purpose of this particular
result. That is the main purpose for stabilizing
a control system.
So in this lecture we have seen a demonstration
of how to find the feedback control for the
system where A is a n cross n matrix but b
is a column n cross 1 matrix. But the general
case is we have to do a similar work for the
general system where A is n cross n and B
is a n cross m matrix. So in this case how
to find a feedback control, so here we want
to find a feedback control u=k times x same
procedure but only difference is k is here.
So we want to find a feedback control u=kx
in such a way that A+B*k has desired eigenvalues.
So this is the; so far we have seen the method
of finding the feedback control for the system
in which the matrix b is n cross 1, that is
a column vector matrix. And the next lecture
we will consider the system X dot = Ax+bu
in which b is a general n cross m matrix.
So we want to find a feedback control u=kx,
where k is a n cross m matrix so that A+Bk
has desired set of eigenvalues. Okay, thank
you.
