So, the next test is the PBH test which is
Popov Belevitch and Hautus Test for the stabilizability.
So, for stabilizability one can also reformulate
the in fact, this is what we have defined
the PBH test as an elegant restatement of
the eigenvector test. That the continuous
time system is stabilizable if and only if
this rank condition is satisfied for all lambdas
belonging to the set of complex number such
that the real part of those lambda is greater
than equal to 0.
So, if you recall the result for the controllability
this condition was not there while testing
the rank condition, but for stabilizability
as we had seen in the eigenvector test that
we are only concerned with the eigenvalues
which are on the right hand side. So, again
for stabilizability we are concerned with
the eigenvalues which are onto right hand
side. So, for the discrete time system we
both have the same result, but those eigenvalues
should be outside the unit circle or on the
boundary of that unit circle.
So, the proof of this theorem is analogous
to the earlier proof that is the controllability
proof except that now, we need to restrict
our attention to only the unstable portion
of the set of complex numbers, where the eigenvalues
lie. So, we would not be going through the
proof this PBH test. But, if we want to determine
the stabilizability directly from the given
AB pair first of all we will compute the eigenvalues
of the A matrix, and then only for the eigenvalues
which are on the right hand side we will carry
out this test, .
The last test is the Lyapunov test for stabilizability.
It says that the system is stabilizable if
and only if there is a positive definiteness
solution P to the following Lyapunov matrix
inequality. So, this is in the continuous
time and this is in the discrete time domain.
If you are typing this pair AP coming from
the discrete time system and we term this
as an LMI which is a linear matrix inequality.
So, let us see the quick proof for this result.
So, the first implication in that the LMI
has positive definite solution matrix P which
implies that the pair AB is stabilizable.
So, again we would be using this eigenvector
test to prove this part. So, there are two
assumptions here which the first one is already
given here that the LMI has a positive definite
solution P, it means that disc that LMI holds.
, the second part is that there is nonzero
vector x which happens to be the eigenvector
of A transpose associated with the unstable
eigenvalue lambda that is A transpose x is
equal to lambda x, .
So, forming the quadratic form of the left-hand
side of the Lyapunov equation so, this was
we were having on the left-hand side and then
using that eigenvector. We form this x star
this matrix into x which should be less than
or in fact, is less than x star BB transpose
x and using the property of the norms I can
write this as the squared norm of b transpose
x. So, we have starred we already know that
it denotes the complex conjugate transpose.
So, see now proceeding in a similar way what
we had done for the Lyapunov test in the controllability,
this right hand side part we have represented
by A star, A transpose x star transpose which
would be nothing, but your x star and AP x
. And on this side since A transpose x star
is would be equal to lambda star x star. Similarly,
here we would have A transpose x is equal
to lambda x and since lambda being scalar
I can take him onto the left to the I can
compute them which finally, I would have the
twice of real part of lambda into x star P
x.
So, see this part which is pretty much important.
So, since P is positive definite, and we know
that the real part of lambda is either greater
than or equal to 0. We can conclude that this
twice of real part of lambda x star P x would
definitely be less than of squared norm because
squared norm is positive; P is a positive
definite already.
So, this part would be positive definite,
but since 
lambda we are having onto the right-hand side,
which is also positive, but from this inequality.
This linear matrix inequality we would have
this less than part and which would always
be greater than or equal to 0. And therefore,
x must not belong to the kernel of B transpose,
if x happens to belong to the kernel of B
transpose, we would have b transpose x equal
to 0, but it is not equal to 0.
The other direction of this implication that
the AB is stabilizable implies that LMI has
a positive definite solution. So, we already
know that the here AB is stabilizable and
if AB is stabilizable, we can decompose it
into the controllable part and the uncontrollable
part. And stabilizable meaning to say that
all the eigenvalues are on the left-hand side.
So, we will go step by step. So, first of
all we from the controllability of the pair
Ac and Bc. We have this the Lyapunov equation
in the context of the controllability this
is one of the results where Qc, where we have
this Bc Bc transpose and Bc is a positive
definite solution satisfying this equation.
Now, on the other hand since A u is a stability
matrix. So, it is already assumed in in this
implication that the pair AB is stabilizable.
So, all the eigenvalues are already onto the
left-hand side and Au is a stability matrix.
Now, using the Lyapunov stability theorem
we can this write this simply we can write
this LMI, in terms of A u and P u and P u
is a positive definite solution of this equation.
So, this equation we basically have written
from the stability result and this equation
we have result from the controllability result
with different A c P c and A u P u pair.
So, now we defined another matrix P bar which
is given by the elements P c and row P u on
the diagonals and 0 on the half diagonal so,
where row is some positive scalar or parameter.
So, let us see this part for the complete
equation which is in the result. In fact,
So, this is in the result. So, we need to
show that this element has a positive definite
solution P. So, let us take the left hand
side of that equation and start putting those
matrices what we had obtained A bar is basically
this one if I write all these matrices explicitly,
P bar we are introducing we are defining this
matrix P bar.
And similarly, B bar I can write this B c.
Now simplifying them I would have A c into
P c plus P c into A c transpose minus B c
into B c transpose, which is nothing, but
minus of Q c. So, here I would have Q c and
I take the minus sign outside of this matrix.
So, similarly see this part we would have
A 1 2 row P u here and the rest would be 0
and with the negative sign because we have
taken the negative sign outside. Another half
diagonal element we would have A u rho into
P sorry, this one. So, here we would have
this would be 0 and this would be rho P u
into A transpose 1 2 and seeing the last element
we would have rho is a scalar. So, I can write
A u P u plus P u into A u transpose, which
is nothing, but minus Q u. So, I obtained
this matrix with rho multiplication.
Now in order to see this matrix, this matrix
first of all is a symmetric matrix because
the off diagonal elements are equal or let
us say in fact, if I take the transpose of
this matrix I would obtain in fact, this equivalent,
.
Now, the second part Q c we have defined already
is a positive definite matrix. Q u which we
have obtained from the uncontrollable part
is a positive definite matrix. Now, I can
choose rho u a very small value which is already
a positive such that the diagonal elements
would be positive. If it is a positive and
it is symmetric then the entire matrix would
be a positive definite matrix, and this positive
definite matrix can be written which was given
in the main result being this matrix a positive
definite matrix we would have the right hand
side the negative definite.
Now, since this P bar is the transformed matrix
of the transformed system. So, in order to
obtain the P matrix of the original pair A
B, again I using this T transformation matrix
I could obtain this P and you can verify that
this P matrix satisfy that linear matrix inequality
which is given in the result. So, this completes
the proof.
So, additional result we have on the controllability.
That so far, we have; so, in the first week
we discussed the continuous to discrete time
transformation given a continuous time system.
So, we need to sample our system sampling
time T to finally, obtain this discrete time
system if the discrete time system is already
given to us. Then we had studied different
test to carry out whether the system is controllable
or whether the system is stabilizable.
Now, we had separate test for the continuous
time system also, but now the here the problem
is if I use the continuous to discrete transformation
given a continuous time system. So, what can
I comment on to it controllability? So, if
you recall from the first week. So, we had
studied two ways of doing the discretization;
the first one is using the Euler method and
the second one by assuming that u k is some
piecewise constant signal between this time
interval.
So, but when we assume that u as piecewise
constant given by this, one we obtain this
discrete time representation where A bar and
B bar. We have explicitly computing using
this A bar is given by e to the power A into
the sampling time and B bar is given by M
into B, where m is this part.
So, now the problem here is which we want
to address that if the pair AB is controllable
will its sampled equation which is given by
after doing the C2D continuous discrete conversion
is controllable. So, we will see the answer
to this problem. In fact, this problem is
quite important in designing so called dead-beat
sampled-data systems in the computer control
of continuous-time systems.
So, let lambda i and lambda i bar be respectively
the eigenvalues of A and A bar. So, here note
that when. So, this A bar I should have written
this A d which is the discrete time matrix
of the continuous time matrix. So, here A
bar is not the transform matrix algebraically
equivalent transform matrix. So, whenever
I am speaking of this A bar and the context
of the controllability after sampling, this
A bar is the discrete time counter part of
the continuous time matrix A. So, lambda i
is the eigenvalue of this and lambda i bar
is the eigen value of A bar.
So, this is the result like suppose that the
continuous time pair AB is controllable. So,
a sufficient condition for the discretized
equation with sampling time T to be controllable
is that, that the imaginary part of the difference
of two eigenvalues lambda i. And, lambda j
of the continuous time system is not equal
to 2 pi into m by T where m is a positive
integer, whenever the real part of the difference
of two eigenvalues is 0.
So, for the single input case this condition
is necessary as well, but for the multiple
input case this condition is only a sufficient
condition. So, we would not be going through
the proof of this result, but we will see
various implication of this result.
So, first of all notice here that if the eigenvalues
of the A matrix are only on the let’s say
what would be the possible cases. So, this
is my S plane, now if all the eigenvalues
are on to the real axis. Let us say we have
the eigenvalues here or it we have the eigenvalues
anywhere. If we have the eigenvalues here,
then it means that the system would always
be controllable because there would not be
any binary part.
Now, this imaginary condition is for those
eigenvalues which have a complex conjugate
let us say I have one eigenvalue here and
one eigenvalue here or one eigenvalue here
and another eigenvalue here. Because only
in this and in this case, we could have the
difference, the real part of the difference
of two eigenvalues equal to 0. And in only
those cases we would check the imaginary part
of that difference, if that imaginary part
of the difference is not equal to 2 pi m by
T where T is a sampling time then we say that
the system is the discrete time system is
also controllable, .
So, it is straightforward to verify that if
the matrix A has only real eigenvalues then
the discretized equation with any sampling
period T greater than 0 is always controllable.
So, let us see we have complex conjugate pair
of the matrix A. Now, if the sampling period
T does not equal to any integer multiple of
pi by beta. Let us recall which is the imaginary
part of lambda i minus lambda j should not
be equal to 2 pi m by T.
Now, the difference here the imaginary part
we would have twice beta from here and this
should not be equal to 2 pi m by T which is
equivalent to same that t is not equal to
pi by beta or for m is equal to 1, . So, if
it is not equal to the integer multiple of
pi by beta then the discretized state equation
is controllable. But if it becomes equal to
pi by beta for some integer m then it only
says that the, for the discretized equation
may not be controllable. this is the because
of that this is only a sufficient condition.
So, if it satisfies, if the results satisfy
it means that the discrete time version is
also controllable, if it does not satisfy
it does not mean that the system is not controllable,
; it means that the system may not be controllable
or be controllable.
So, we can see the implication of the second
point more precisely. So, first of all is
note that since A bar the discrete time state
matrix is equal to e to the power A into T
where A is the continuous time matrix and
T is the sample time. If lambda i is an eigenvalue
of A then lambda i bar which is equal to e
to the power lambda i into T e is an eigenvalue
of A bar. So, you can take this as an exercise
if you have any difficulty in visualizing
this because both the matrices A and A bar
are related by this one. So, the lambda i,
the eigen values are also written in a similar
way.
So, the see if T becomes equal to m pi by
beta for some integer m or and given these
two distinct eigenvalues alpha plus j beta
and alpha minus j beta. In fact, become a
repeated eigenvalue either of minus e to the
power alpha T or e to the power alpha T of
A bar. So, we can quickly visualize this say
for example, take lambda is equal to alpha
plus j beta. So, the eigen value of the discrete
time would be e to the power alpha plus j
beta into the sample time.
So, this i can write as e to the power alpha
T, sorry it should be j e to the power alpha
T times e to the power j beta T, and T i already
know is a multiple of m pi by pi by beta.
So, this would become e to the power alpha
T times j m pi and this part is nothing, but
either it would be plus 1 or it would be minus
1. So, that is why we it becomes a repeated
eigenvalue of the discrete time state matrix
and this will cause the discrete time equation
to be uncontrollable.
So, the last result we have that if the continuous
time system is not controllable then its discrete
state equation with any sampling period will
never be controllable matrix. If the system
the LTI system is controllable then we had
a test or we, you could compute the sampling
time also; so, that is your discrete time
version of the state equation becomes controllable.
Now, in the original pair is not controllable
then for any sampling period your discrete
time version would never be a controllable.
