Hello viewers. Welcome to the lecture on feedback
control. In this lecture, we will consider
linear control system and its feedback control.
Let us see what is a feedback control. Any
control which can be expressed in the form
of or as a function of the current value of
the state variable is called a feedback control.
That is at any instant of time t, if the control
function u of t is a function of the state
variable x of t, then it is called a feedback
control.
In our everyday life, we normally use feedback
control in performing many tasks. For example,
when we drive a vehicle, then the control
on the vehicle is always based on the current
position and velocity of the object and apart
from various other observations. So such controls
are called a feedback control. So we will
consider linear feedback in this lecture.
So let us see what is a linear feedback control.
Consider the control system 
x dot=Ax+Bu. So as usual we will consider
x of t is in Rn is the state space, u of t
belongs to Rm is the control space and A is
n*n matrix and B is n*m matrix. These are
constant matrices. A given linear control
system. Now what do we mean by a feedback
control, linear feedback? So if the control
function, control vector function u of t is
written as some k matrix*the x of t where
k is m*n matrix, constant matrix, then we
call it as the linear control, linear feedback
control.
Then u of t is called 
linear feedback control. So now how we use
it in a practical situation because in many
practical problems we use a linear feedback
control in the following way. For example,
let us consider x dot=2x and x of 0=x0, it
is a very simple dynamical system. So the
solution of this problem is x of t=e power
2t*x0. For example, let us say it is a population
growth of some bacteria, okay, or any insect
population or something like that.
So for a short period of time, this grows
in an exponential manner if it is a model
for the population. So if x0 is the initial
population, t is this and the x is this. At
t=0, the initial population is x0 and it grows
in an exponential manner. Now if you want
to control this population using some kind
of pesticides or some medicine, etc. So if
we control this system using some kind of
controller u of t and then want to make this
population to go to 0.
Instead of going to infinity, we want the
trajectory should travel and come towards
0. So if the initial condition is let us say
x1. At time t1, we see that the population
is this much x1 and from there it should not
grow exponentially but it should come down
and make the population 0. If you are interested
in this type of controlling the population,
then we can apply a feedback control in the
following manner.
So if you say u of t is some constant time
x of t itself, whatever is the population,
we take proportional to that the control function
u of t. That will imply that the system, the
control system will become x dot=, if you
substitute u of t in this equation 1, then
we will get this as 2+B*k of x of t with the
initial condition xt1=x1.
So if we select this k suitably so that 2+Bk
is negative, so choose k such that 2+Bk; for
example, let us say it is -1. Then that means
you have to select k to be equal to -3/B where
B is a known number already. If you put it
like this, then the system becomes 
x dot=-1*x and x of t1=x1. So this immediately
implies that x of t will be e power -t-t1*x1.
So after the time t1, you can easily see that
it tends to 0 as t tends to infinity.
So by selecting a feedback control, like this
u=k*x of t, we will be able to control the
system like this. So if without control it
was going exponentially and with control,
then it will start coming down and it will
become 0, t versus x graph. So we will be
able to change the behaviour of a system by
using feedback control in a suitable manner.
So that is the usefulness of the feedback
control.
So for the second example, consider the harmonic
oscillator x1 dot=x2 and x2 dot=-x1. Let us
say this is the equation. So this you can
easily see that it comes from the equation
x double dot+x=0. if you convert this system
into 2 first order equation, you will get
this one. This is nothing but; for example,
if x of 0=L and x dot of 0=0. So it is an
oscillator MOS1. A 
particle of MOS1 which is oscillating having
O as the center, the force of attraction is
at O and then it is oscillating indefinitely
from -L to +L.
Initially it is at L and the velocity is 0
and then it start moving in the, indefinitely
like this. So we can easily see that if we
solve this equation, so here if you convert
this x of 0 as x1 of 0=L and x dot of 0 means
x2 of 0=0. So with this initial condition,
it is a harmonic oscillator which will move
indefinitely. The solution is, if you solve
this equation using the standard way of solving
the dynamical system and substituting this,
we will get L cos of t, okay, and x2 of t
is -L sin t.
So it is a periodic solution keep on moving,
never stops. Now if we apply a force on this
particle, then it will become a forced harmonic
oscillator and then we can change the behaviour
of the motion as we wish. For example, if
you want to stop this harmonic oscillator
at one position, we can apply a suitable force,
so controlled oscillation. So if we apply
x1 dot=x2, it cannot be changed because the
rate of change of displacement is velocity
and rate of change of velocity x2 dot is here
-x1.
So now if you add another force here, u of
t on the particle, we will get the forced
equation here, controlled equation with this
initial condition. So by selecting this u
of t properly, then we can control it in the
following way. So if we select 
u of t=-2*x2 of t, for example, okay. It is
a planned selection like this, if you do it
like this, then we will get by substitution,
x1 dot will be equal to x2, x2 dot is -x1-2x2.
So this is obtained and this implies that
the matrix involved in this system is 0 1
-1 -2 with eigenvalues are -1 and -1 repeated
2 times, okay. So when the eigenvalues are
negative, both the eigenvalues are negative,
this implies that the system is asymptotically
stable. So this is a feedback control that
is what we are doing. This control which we
have obtained, it is a feedback control because
we are writing u of t as 0 and -2 multiplied
by x1 x2.
So it is like 0 -2 is the k matrix and x is
the state variable. So u=kx is the feedback
control which we have applied here. And by
applying this, the system response, it has
changed to, asymptotically stable property
has been obtained here. So here we have just
by trial and error, we have chosen a feedback
control like this. So the question here is,
if we have a general system, is a given system,
is it possible to find a feedback control
u=kx such that 
the system, resulting system x dot here equal
to Ax+Bkx which is nothing but A+Bk*x.
By using a feedback, we are getting a new
system x dot=A+Bkx. Is it possible that this
A+Bk is chosen in such a way that it has any
arbitrarily assigned eigenvalues? So the question
is, is it possible to find a feedback control
u such that the system, this has arbitrarily
assigned eigenvalues? Eigenvalues for the
matrix, eigenvalues of A+Bk.
So what is the importance of the eigenvalues
of this matrix, eigenvalues will decide the
nature of the solution because we have seen
that if all the eigenvalues are having negative
real part, the system will be asymptotically
stable or even if one of the eigenvalue is
positive, it will be unstable, etc.
So the stability of the system is decided
by the eigenvalues of the matrix involved
in the system. So if you are able to select
a set of eigenvalues which we are interested
using a feedback control, then it is an advantage,
that is we can control the given system in
a manner which we are interested in.
So we will see here that result. So if the
system is controllable, then we can find a
matrix k such that 
the matrix A+Bk has arbitrarily assigned eigenvalues.
So this is the result which we will; so before
going into this result, so let us first see
the property of the companion form. Property
of the companion matrix C. If C is in the
form 0 1 0 0, 0 0 1 0 and 1 here, -alpha n
-alpha n-1, -alpha 1.
So this is the standard form of a companion
matrix. Then it can be easily seen that, then
its characteristic 
equation that is C-lambda I determinant=0
is given by. These will be the coefficient
of the characteristic equation, that is 
lambda to the power n+alpha 1*lambda n-1 alpha
2lambda n-2 etc. alpha n=0. So the characteristic
value has the coefficient which are in the
last row of the matrix C that can be easily
verified.
If A is similar to C, A is a given matrix,
C is a matrix if they are similar, that is
TAT inverse=C which we have seen already in
this particular form, then the set of eigenvalues
of A and the set of eigenvalues of C, both
are the same. Then A and C has same set of
eigenvalues, okay. So this is a known fact
from the linear as we draw that can be easily
verified.
Now making use of this, we will see, that
is, so if let us say mu 1 mu 2 etc. mu n is
any given set, arbitrary set of numbers whether
real or complex numbers, and if we want that
this should be; and if this set is the spectrum
of a matrix, any matrix, then its companion
form, its 
C is given by the following, given by the
coefficients of the 
polynomial 
lambda-mu 1 lambda-mu 2, etc. So if you calculate
the coefficient of this polynomial and write
them in the reverse order with a negative
sign, we will get the last column of the C
matrix.
Here -alpha 1 -alpha 2 etc. are the last column.
So these numbers will be obtained by finding
this expression. So for example, if let us
say 1 and 2 are the eigenvalues required for
a matrix C, then we will calculate lambda
-1*lambda -2=0. This implies we will get lambda
square-3lambda+2=0. 
So that implies that the companion matrix
is of the form 0 1 and 2 here and -2 in the
place. Is not it?
The coefficient of the characteristic equation
is obtained here. Is not it? It is lambda
square-3lambda. So it is 3 here, okay. The
coefficient is given in this manner. So this
thing will be used for proving the particular
theorem, for proving the statement of the
theorem given here. If the system is controllable,
then it can be converted into the companion
form.
To see it very briefly, now we will consider
the system x dot=Ax+bu where A 
is n*n and b is n*1 matrices. So for the general
case, we will see in a later lecture. So now
let mu 1, mu 2, etc., mu n be an arbitrary
set of numbers. It can be real or complex
number. So the question is, is it possible
to find a k matrix so that A+bk has this set
as the eigenvalue. So consider this system.
Now the theorem is, so if you call this equation
as 1 and this set as a set S, okay, the theorem
is if the system 1 is controllable, then we
can find a matrix k such that A+b*k has the
set as S as its spectrum.
Spectrum is the set of all eigenvalues of
the matrix. So the proof is the following
way. It is given that the system is controllable.
So this implies that one can be converted
into its companion form. So there exist a
matrix T such that TAT inverse will be C and
T*b=D. So this we have already seen. System
is controllable.
We can get C and D matrix like this. So now
as usual, we multiply both side with T and
from 1, we will get Tx dot=TAx. So TAx+Tbu.
So that implies that if you substitute y=Tx,
we get y dot=TAT inverse of y+Tb of u, okay.
So now this implies that y dot=Cy+Tbu, Cy+d*u.
So if you substitute, put u=kx and which is
equal to kT inverse y, we will get y dot=Cy+dk*T
inverse of y.
So this we will get it as C+dkT inverse of
y. So now let k is k1, k2, it is kn, kn-1,
k1 is a column vector. Already we have that
d=0 0 1 and T inverse is the n*n matrix. So
we will obtain, in this case, we will get
y dot=, let us write the whole thing here.
C is 0 1 0 0, 0 0 1 0, -alpha n to -alpha
1, +d*this k1 will give 0 0 etc., the last
one, 1*kn, etc. will come, kn, kn-1, k1, *T
inverse the whole thing*the y vector.
So we can get here, now again multiplying
this matrix with T inverse, we will get all
this, -alpha n, -alpha 1, +, again we will
get all the first n-1 column, rows are 0.
Only the last row multiplied by the matrix
will give a non-0 row vector. Last row multiplied
by T inverse, we get this expression*y. “Professor
- student conversation starts” This will
be gamma n. Gamma n-1 yes. “Professor - student
conversation ends.”
And this is gamma 1 here. So this implies,
we will get the matrix as 0 1 0 0, if you
add the element wise, up to n-1 rows, we will
get like the companion matrix and the last
element will be -alpha n+gamma n. So we will
get gamma n-alpha n, etc. Last element is
gamma 1-alpha 1, this*y. Now if you are interested
in having mu 1, mu 2, mu n as the eigenvalues
of the matrix, so now we can select 
this gamma values in such a way that 
the eigenvalues of the matrix, let us call
it as some notation D*y.
The eigenvalue of this matrix D are mu 1,
mu 2, etc., mu n which is always possible
because the last row is nothing but the coefficient
of the characteristic root of this polynomial.
So the roots are given already, mu 1, mu 2,
mu n are given. Then by finding the characteristic
polynomial using this eigenvalues, we can
get the coefficients which is in the last
row. And alpha 1, alpha 2, alpha n are already
known.
Only gamma 1, gamma 2, gamma n can be obtained
from the given values of the roots here. So
this theorem shows that using the controllability
of the equation 1, we are able to get a companion
form. And using the companion form, we are
able to get the particular form D*y in the
last line and because of that, we will be
able to get all the values of gamma corresponding
to the roots given here as mu 1, mu 2, mu
n. And from here, we can calculate the control
function, that is k*u because from gamma,
we can obtain the k values.
From k value, you can obtain the control as
k*x. So this completes the proof of the theorem
on the feedback control of the system, okay.
So the next lecture, we will see some example
of how to compute the feedback control of
a system and for the general case. Here we
have seen that the result is proved for the
particular case where B is a column vector.
So the next one, we will see if B is a general
matrix, the m*n matrix, how to get this feedback
control for the system. Thank you.
