So, in the last lecture we have seen one of
the method for computing the State feedback
gain so, that method was known as the eigenvalue
assignment. So, we stressed on to one fact
that if the system is controllable then given
a set of eigenvalues we can place the eigenvalues
of the closed loop system. Now, the another
question arises that how do we get the information
about the set of desired eigenvalues.
So, there are different ways of computing
the set of desired eigenvalues. So, you have
studied in your, possibly you have studied
in your UG control course some methods from
the root locus diagram or from the bode plot
by taking into account the transients and
the steady state characteristics.
So, first of all it depends on the performance
criteria that what performance criteria we
are targeting, meaning to say what are our
control objective, it could be the rise time
and the overshoot or the settling time if
we speak these criterion in terms of the time
domain. In terms of the frequency domain also
the responses not only depend on the poles,
but it also depends on the 0s. So, so far
we have focused onto the location of the poles
and also placing the location of the poles
by using a state feedback, but the overall
response is also affected by the location
of 0s.
The factors which affects the selection of
the poles first of all is the zeros, so in
the last lecture we had seen that if by using
the state feedback you are able to place or
you might happen to place the poles at those
location, where the zeros of the plant are
already there, then it may lead to the state
of unobservability which we will discuss in
the coming weeks.
So, we need to take care that first of all
whatever the poles we are we want to place
for the closed loop it should not overlap
with the zeros. Second point here is the magnitude
of u which takes into accounts the saturation
or burn out. So, this point is more from a
practical point of view; say for example,
if you have certain plant then those plants
are basically actuated by some actuators and
we take some measurements with the help of
some sensors.
Now depending on some actuator limits we need
to put some constraints on the signal u. So,
it might happen that you have designed a controller
in such a way that the control signal always
leads to the saturation or it leads to the
maximum value to what the actuator is capable
of. So, we need to take care of that because
this is not very desired condition or scenario.
The rise time, settling time, overshoot is
pretty much obvious, the most importantly
is the bandwidth of the closed loop, which
takes into account the frequency domain characteristics
and there is a very strong correlation between
the time domain characteristics and the frequency
domain characteristics which possibly you
have studied during your UG control course.
So, all in all it involves some compromises
among many conflicting objectives. So, in
the sense that when we started this week.
We put different, two different objectives
to synthesize the control log, meaning to
say when we discuss about the open loop minimum
energy control. So, the first objective was
of the controllability and the second was
of the and there was a second objective also.
Now, there are two good references which I
would like to highlight here.
So, the first one is by Stephen Boyd on linear
controller design limits of performance. So,
this is about so many chapters of this book
discusses about what performances or what
are the limits, on the performance of the
closed loop system you can achieve. The second
is paper by Cal Astrom on limitations on control
system performance which was published in
European Journal of Control in 2000.
So, this paper is also quiet good which discusses
an overall overview of the performance of
the control system. So, we won’t go into
the details of computing the set of desired
eigenvalues, but we will see some guidelines
that how one can specify the set of desired
eigenvalues.
So, just some guidelines say for example,
on the left hand side this one is basically
the s plane in the continuous time domain
and this one is the z-plane in the discrete
time domain. So, the first of all, the first
objective that all the eigenvalues of the
closed loop system should be on to the left
hand side.
Now, here we are specifying an additional
control objective that all the eigenvalues
should be inside this region C which is defined
by this region. So, when you see this region
in the s plane by drawing the damping coefficient
line, you can assess some of the parameters
let’s say the sigma which is the real part
of the s plane; the theta the angle which
it makes from this axis and also the radius
of this semi-circle.
So, if we have larger the sigma the response
of the closed loop system would be faster
if we have closed loop poles happens to be
having the larger value of sigma. So, this
implication you could see pretty much straight
forward; say for example, if you say the larger
the sigma let’s say in terms of the transfer
function. Let’s say if we have a transfer
function some s+σ, ok. So, s+σ into the
sense I can also write this is equivalent
to (1/σ)/ ((1/σ)s+1) and replacing 1/σ
by τ so, I can write this τ /(τ.s+1) where
τ is the time constant.
So, if we have larger the σ the value of
the time constant τ would be smaller meaning
to say the response would automatically be
faster. If we have larger the theta, larger
the overshoot, if we have larger the r the
response of the system would be faster. With
the faster response as we had noted in some
tutorial problems when we if you recall some
of the problems we discussed with the controllability
as a part of the tutorial on controllability.
So, if we want to reach to some desired state
value in t1 amount of time the lesser the
time t1, the faster is the response and faster
is the energy. So, we; so this is obvious
larger the r faster the response would be
and u would also be larger in that particular
case. In the same time bandwidth will also
be larger and the resulting system will be
more susceptible to noises, ok. So, this is
in the s plane equivalently is drawn in the
z plane.
So, most of the time in this week we would
discuss the design of the controllers in the
continuous time domain, all the results whatever
we discuss we directly applicable on the discrete
time systems as well, right.
So, we will now discuss another method of
computing the state feedback gain for the
eigenvalue assignment problem. So, in the
last lecture we studied one way of computing
the state feedback gain using the controllability
matrix. So, this method has some restriction
that the selected eigenvalues cannot contain
any eigenvalues of the plant itself or of
the A ok.
So, here first we will discuss the algorithm
for computing the switch feedback gain and
then we will try to justify that algorithm
by computing certain conditions which are
necessary and sufficient for this algorithm
to work. So, the data which we supply to this
algorithm is the pair (A, b) which supposed
to be a controllable pair where A is a n-dimensional
matrix and for the moment we are taking u
as a scalar.
So, in that case we would be having n cross
1-dimension vector and a set of desired eigenvalues.
So, based on these three information the result
we want to compute, 1cross n real vector k
such that the eigenvalues of this matrix of
the closed loop matrix A minus b k has the
set of desired eigenvalues that contain no
eigenvalues of A.
So, the first step is we select an n cross
a square matrix F of dimension n that has
the set of desired eigenvalues. Now, the form
of F can be chosen arbitrarily and we will
be discussing it after the justification that
what F matrix you can select. The second step
is select an arbitrary 1 cross n vector k
bar such that the pair F transpose and k bar
transpose is controllable.
Now, if you recall in the last algorithm based
on the information of the coefficient of the
characteristic polynomial of the A matrix
and of the set of desired eigenvalues we pre
specified this k bar. Now, here we got some
freedom that once we have a set of desired
eigenvalues first of all the form of F matrix
could be anything such that it is having the
those eigenvalues.
Now, here we can select any arbitrary k bar
vector, the condition which we need to satisfy
is that the pair F bar F transpose and k bar
transpose is controllable. Third, we solve
for the unique matrix T in the Lyapunov equation
given by AT minus T times F is equal to bk
bar. The final step is once we have obtained
the T matrix we compute the feedback gain
vector k as k bar into T inverse ok.
So, once we have selected a k bar matrix such
that this pair is controllable then we can
use the MATLAB Lyapunov function this lyap
function to solve this Lyapunov equation.
So, you could use this function in a similar
way what we have discussed during the stability
week. And once this k bar matrix or k bar
vector is selected we can solve this Lyapunov
equation to solve for T and after putting
T here we would obtain the state feedback
vector.
So, how do we make sure that once we that
whatever the k we have computed would going
to yield the set of desired eigenvalues. So,
here we will see the justification of all
these four steps. So, that we can ensure that
once we plug in this k state feedback vector
we would obtain the closed loop as this one
which would contain the all the set of desired
eigenvalues that contains no eigenvalues of
A.
So, the first point is at if T is nonsingular,
if the matrix T is nonsingular then I can
write k bar is equal to k into time, k into
T from here and the Lyapunov equation can
be replaced in which the k bar can be replaced
by kT as, say for example, if I put k bar
is equal to kT here and then take it on the
left hand side I would have A times T minus
b k times T, and after taking the T matrix
common the rest of the thing is A minus bk
and we take this part on to the right hand
side so, it would give me TF. Now again taking
this T matrix on to the right hand side I
would have A minus b k is equal to TFT inverse.
Now, if you pay close attention to this equation,
the same equation we have used when we discuss
about the simulate transformation. Where T
could be any nonsingular matrix and here T
happens to be transformation matrix, meaning
to say that and we know that under the transformation
the eigenvalues do not change. So, whatever
the eigenvalues we would specify in the matrix
F, it would definitely be of the matrix A
minus b k, and this is what we expect the
result of this algorithm.
Now, thus the eigenvalues of this matrix A
minus b k can be assigned arbitrarily except
those of A as well. Now, if A and F have no
eigenvalues in common that is the plant and
the closed loop system have no eigenvalues
in common, then a solution matrix T exist
in this Lyapunov equation for any k bar and
would be unique.
So, this is the most important point here
which we would formulate into a result and
we will see the proof of this equation as
well. Because the most important thing in
all these four steps, that T because here
we solve for the unique T. So, when we should
know first as a preliminary step that whether
that T exist or not, because F is the user
defined or is completely based on the objectives,
k bar is also user defined based on this condition
which can be satisfied.
Now, using this k bar and F we solve for this
T; so, we need to ensure that whether that
such T exists or not first of all. On the
other hand if A and F matrix have common eigenvalues
a solution T may or may not exist depending
on bk bar. So, in order to remove this uncertainty
we require A and F to have no eigenvalues
in common. So, this particular point in the
first statement we have seen this while if
you recall the result on the Lyapunov stability
for linear time systems; so, this was one
of the point which we have highlighted.
So, what remains to be proved is the non singularity
of T so, at the same time if T happens to
be a nonsingular meaning to say that it would
be unique, ok.
So, this is one of the important result that
if A, if these matrices A and F have no eigenvalues
in common then the unique solution T of this
Lyapunov equation is nonsingular, if and only
if the pair (A, b) and F transpose and k bar
transpose are controllable pairs. So, we need
the controllability of the pair A comma b
and also of this one which is an additional
condition to solve for this Lyapunov equation.
So, we will see a detailed proof of this result
so that by doing the proof we want to recall
many of the basic concepts so that it would
also help you to carry out your own proof
of your own results. So, we shall prove the
theorem for n is equal to 4, but it would
also be applicable for any n-dimensional system.
So, recall that the characteristic polynomial
of the plant or of A matrix is given by this
polynomial for n is equal to 4, where alpha
1 to alpha 4 coefficients are already known,
ok. We also know from Cayley Hamilton theorem
that if I replace s by the A matrix this would
also be equal to 0 ok.
So, let’s see the proof. So, here we define
a matrix which is Δ(F) given by this one.
Now, this is a matrix we can say let’s say
F tilde, and we know Δ(a) would be equal
to 0, but Δ(F) would never be equal to 0,
ok. So, we define one matrix by replacing
s by F in the characteristic polynomial. So,
if λi bar is an eigenvalue of this matrix
F, then Δ(λi bar) is an eigenvalue of this
matrix.
So, this you can take it as an exercise for
a generic n dimensional system, but we can
see through one example that how this statement
is true say for example, we take A as an identity
matrix, if A is an identity matrix then we
know that system is an unstable system, ok.
The next step is to compute the determinant
of sI - A which is Δ(s) and it would be given
by (s-1) of square or (s square – 2s + 1).
Just for the confirmation we can verify that
delta of A would definitely be equal to 0.
So, if I substitute S by A matrix we would
have A square. So, A square again would be
identity minus 2 it would be twice into I
plus an identity so, it would definitely be
0.
So, now define another matrix let’s say
we want to place the eigenvalues of the closed
loop matrix to let’s say minus 1 and minus
2. So, I form another matrix let’s call
it F tilde which is Δ(F) and is given by
the square of this matrix would be [1 0, 0
4] minus so, I write plus and it would become
2 4 plus identity, ok. So, this matrix happens
to be 
F tilde.
Now, it says if λi bar is an eigenvalue of
F, so λi λi bar are -1 and -2 of this matrix.
So, we need to compute and verify that Δ(λi
bar) is an eigenvalue of this matrix F tilde.
So, delta of, so once we compute for Δ(-1),
Δ(-1) would be equal 4, and again Δ(-2),
Δ(-2) would be 9.
So, it is clear now that Δ(λi bar) are 4
and 9 which happens to be the eigenvalue of
this F tilde matrix ok. So, here we are taken
one example to verify this statement, but
you can try take it as an exercise to show
in generic for an n dimensional system that
this statement would hold true, ok.
Now, taking this statement further because
A and F matrix have no eigenvalue in common
we would have Δ(λi bar) is not equal to
0 for all eigenvalues of F ok. Then we compute
the determinant of this matrix we have so
we know that determinant of any matrix is
the multiplication of all its eigenvalues.
The eigenvalues of this Δ(F) is Δ(λi),
Δ(λi) and the product of this Δ(λi bar)
over all i would also not be equal to 0 because
this is not equal to 0. So, this Δ(F) is
a nonsingular matrix.
Let’s go further then we substitute taking
from Lyapunov equation AT = TF + bk bar into
another matrix which we are defining by as
A square T minus TF square, ok. So, this matrix
we are defining and we want to substitute
the Lyapunov equation into this equation which
yields that A square T minus TF square is
equal to.
So, here we would have I can take A matrix
common so, the rest matrix would be AT and
replace TF plus b k bar minus TF square. And,
then clubbing this A(TF) matrix with this
matrix and taking F matrix common I would
have AT minus TF and Abk bar the rest of the
part. And this AT minus TF is nothing, but
bk bar, so I can write A square T minus TF
square as equal to Ab k bar plus bk bar F,
ok. The relevance of doing this procedure
is would be clear in the next slide.
So, preceding forward we can obtain the following
set of equation so we start with IT - T into
I would definitely be equal to 0. Then AT
- TF we know it is already a Lyapunov equation,
further (A square T - TF square) this is what
we have computed. Now, going forward for power
3 and power 4 we would you can also simplify
by yourself, the right-hand side to whether
they are coming equal to this or not.
Now, we multiply the first equation by alpha
4; so, we multiply by this alpha 4 multiply
this equation by alpha 3 both sides this is
by alpha 2, this is by alpha 1 and this remains
as it is ok or by 1 or by I ok and we will
sum them up. So, you would notice that T matrix
we can take it common and the rest of the
elements are or in the bracket would be alpha
4 plus A times alpha 3 plus A square alpha
2, A cube alpha 1 and A 4, ok.
So, we would obtain this Δ(A).T – T.Δ(F)
would be equal to -T.Δ(F). Why because from
the Cayley Hamilton theorem this part would
reduces to 0 because Δ(A) is nothing, but
equal to 0 ok. Now, looking at the right hand
side I can rearrange the right hand side by
this matrix in times this matrix, times this
matrix, it is just a simplified version of
taking all these form by multiplying each
and every equation by alpha i’s, ok.
Now, here this is the most important thing.
So, Δ(F) we have already shown that it would
be a non-singular matrix this matrix being
upper triangular matrix it would always be
a nonsingular matrix. This matrix would be
nonsingular if and only if (A, B) pair is
controllable, and this matrix is nonsingular
if and only if F transpose and k bar transpose
is controllable.
So, in that case all these matrices this,
this, this and this matrix are nonsingular
matrices so T would definitely be a nonsingular
matrix, and this only happens if the matrix
A and the matrix F has have no common eigenvalues.
Now, if either of the matrix is uncontrollable
if this matrix or this matrix or either of
the pair is uncontrollable, then your T matrix
would be a singular matrix. So, this is establishes
the if and only if condition, ok.
Now, commenting on the selection of F given
a set of desired eigenvalues there are infinitely
many F that have the set of the eigenvalues.
So, we can take two forms, the first form
would be if w form of polynomial which we
had seen earlier from the set we can use its
coefficient to form a companion form matrix.
So, that all the coefficients I can arrange
into this way where all these coefficients
are of the F matrix, if I write the characteristic
polynomial of the F matrix this would be the
coefficients. And, arranging all these coefficients
into this form would yield me a companion
form matrix whose eigenvalues would definitely
be the eigenvalues of the required eigenvalues.
Now, this particular F matrix we have also
seen in some of the tutorial problems. Let’s
say if there are some complex eigenvalues
and we want to specify a real matrix then
we can write the complex eigenvalues as this
block. And the complex eigenvalues should
definitely occur in it’s in its conjugate
pair, ok.
So, here it contains 5 eigenvalues and it
is a block diagonal matrix; so, one block
is this one, then another block is this one,
another block is this one, ok. So, two eigenvalues
are here, two are here and one is here. So,
there are different F matrix which you can
form so that it could contain the required
set of eigenvalues.
