So, now we will see the Lyapunov Stability
theorem results for the discrete time case.
So, consider now the following discrete time,
linear time varying system where now this
t belongs to the set of integers. So, x t
plus 1 is equal to A of t into x of t plus
B of t u of t. So, here A B C D matrices are
time varying matrices in the discrete time
domain.
So, the result what we had discussed for the
continuous time case, all those results almost
remains the same for the discrete time case.
So, the first result says is the system is
the system is set to be marginally stable
in the sense of the Lyapnov or the system
is internally stable whenever for every initial
condition x naught the homogeneous state response
x of t is uniformly bounded. So, the first
definition we know already that the solution
of the state based system must be bounded.
And when we speak about the uniform boundedness
we actually mean to say that the norm of the
signal x of t is always less than equal to
some constant c, ok. Now, the second definition
says the system is asymptotically stable occur
in the sense of the Lyapunov, whenever in
addition to the first statement for every
initial condition x naught we have x of t
approaching to 0 as t tends to infinity.
The third definition says that the system
is exponentially stable whenever in addition
to the above two statements there exists constants
c and lambda both positive constants, but
here with respect to the continuous time system,
we have an additional condition on lambda
that it should be less than 1. Such that for
every initial condition x naught the norm
of the signal x is less than equal to c into
lambda to the power t minus t naught and the
norm of the vector at t is equal to t naught.
The system is set to be unstable whenever
it is not marginally stable in the sense of
Lyapunov.
So, almost all the definitions remains similar
to their counterpart of the continuous time
system, there is a minor difference only in
the third statement. Again, here the B C D
matrices played no role in this definition.
So, therefore, we will mostly speak about
the stability of the homogeneous system that
is without considering the input. So, for
the continuous time system, we studied two
conditions, basically 2 tests to determine
the stability, one is the eigenvalue test
and another is the Lyapunov test. So, the
first eigenvalue test for the discrete time
system or homogeneous LT I system; so, this
x plus denotes the x t plus 1, ok.
So, this LTI system is marginally stable if
and only if all the eigenvalues of the matrix
A have magnitudes smaller than or equal to
1. Basically, all the eigenvalue should be
inside a unit circle and all the Jordan blocks
corresponding to the eigenvalues with magnitude
equal to 1 are 1 cross 1, it is similar to
what we had seen in the for the continuous
time system. The LTI system is asymptotically
and exponentially stable, because we already
knew that the asymptotic stability and the
exponential stability for the LTI system basically
remains the same.
So, the system is stable if and only if all
the eigenvalues of a have magnitude strictly
smaller than 1 or the system is unstable if
and only if at least 1 eigenvalue of the matrix
A has magnitude larger than 1 or magnitude
equal to 1, but the corresponding Jordan block
is larger than 1 cross 1. So, we will not
see the proof of this theorem since we had
already gone through a detailed proof for
the continuous time system, the Lyapunov test.
So, all these following five conditions are
equivalent, the system homogeneous-discrete
time linear time invariant system is asymptotically
stable is equivalent to saying that the system
is exponentially stable, is equivalent to
saying that all the eigenvalues of the matrix
A have magnitude strictly smaller than 1,
is equivalent to say that for every symmetric
positive definite matrix Q there exists a
unique solution P to the following Stein equation.
So, this Stein equation is also known as the
discrete time Lyapunov equation. So, this
Lyapunov equation reads A transpose P into
a minus P is equal to minus Q. Moreover, P
is symmetric and positive definite. Again,
it is equivalent to say that there exists
a symmetric positive definite matrix P for
which the following Lyapunov matrix inequality
holds. So, the change here is in the Lyapunov
equation basically in the fourth and fifth
statement with respect to the continuous time
system, otherwise the first three statements
are remains the same as we had seen there.
So, again you could do the proof of this theorem
in accordance to what we had done for the
continuous time system. So, if we summarize
the overall results of the Lyapunov stability
test for the LTI systems. So, here on the
left column, we define all the definitions
of the unstable, marginally stable, asymptotically
stable and the exponentially stable systems.
Here, we are giving a summary of the continuous
time test and the discrete time test, again
it is further divided into eigenvalue test
and the Lyapunov test similarly for the discrete
time.
So, let us see for the first for the unstable
system for some t naught and the initial condition
x of t naught, the signal x of t can be unbounded.
It means, if we see the eigenvalue test in
the continuous time that for some lambda i
of A, the real part of the eigenvalue of the
matrix A is greater than 0 or the real part
of the eigenvalue is equal to 0 with Jordan
block larger than 1 cross 1. For the discrete
time case, the eigenvalue is outside the unit
circle or on the unit circle itself, but with
that Jordan block larger than 1 cross 1.
The system is marginally stable, if the signal
x of t for a given initial condition is uniformly
bounded which according to the eigenvalue
test that all the real part of the eigenvalue
should be strictly less than 0, and if it
is lying on the 0 it should have a Jordan
block of 1 cross 1 only. For the discrete
time systems, the eigenvalue should be inside
the unit circle or if it is lying on the unit
circle, it should have a Jordan block of 1
cross 1.
The asymptotic stability defines that the
signal itself should approach to 0 whenever
t tends to infinity and for the eigenvalue
test the real part of the eigenvalue of the
matrix A should be strictly less than 0. For
if we see the Lyapunov test so, for the given
Q which is positive definite symmetric matrix
Q, there exists a matrix P which is symmetric
the symmetric condition is given by this one
and the positive condition is given by this
one.
So, there exists a matrix P which is symmetric
and positive definite such that it satisfies
this equation, the Lyapunov equation. For
the discrete time case all the eigenvalue
of the matrix A should lie inside the unit
circle. Here, for the given positive definite
matrix Q there again exists a matrix P which
is symmetric and positive definite, such that
it satisfies this discrete time Lyapunov equation.
The system is exponentially stable, if there
exists a symmetric positive definite matrix
P such that it satisfies this linear matrix
inequalities, again similarly for the discrete
time case. So, this table gives you the entire
stability test specifically for the LTI systems
both the eigenvalue test and the Lyapunov
test.
Let us consider one example with which we
had started this lecture or this module. So,
here we are considering the inverted pendulum.
So, there we. So, initially we considered
the simple pendulum, but you can also consider
it as an inverted pendulum where theta is
now the axis the sorry the angle made from
the vertical axis, ok.
So, the control input to this inverted pendulum
is the torque applied to the base and y or
the state is the angle this pendulum makes
from the vertical axis. So, we very it is
well known that from the Newton’s law, we
obtain this particular equation which is you
could imagine that it is a non-linear equation.
So, while considering the case of the pendulum
we identified 2 equilibrium points, one is
at 0 and another is at pi. So, if we see the
equilibrium point theta equals 0 means that
the pendulum is standing still upwards, ok.
So, let us see what happens.
Now, if the equilibrium point theta is equal
to pi is considered then linearizing this
non-linear equation gives us this A B C matrices.
Now, if I compute the eigenvalue of the matrix
a by this formula determinant of lambda I
minus A where lambda are the eigenvalues;
so, after simplification I compute this two
eigenvalues.
Now, notice here there that this part which
is inside the under root part would always
be less than or equal to this part. Meaning
to say that all the eigenvalues would be strictly
on the left hand side. So, it means that the
system is asymptotically stable and also exponentially
stable. Now, we can also compute the P matrix
also. So, if we compute the you can do it
by yourself for computing the P matrix. So,
this is the P matrix for this system we have
obtained.
So, now, if you compute or you can do the
self test that if you compute the eigenvalues
of this P matrix, you would see that all the
eigenvalues of this matrix P are positive,
the symmetric definite can be strictly seen.
So, if you compute the transpose of this matrix
P, you would see it is actually equal to the
P transpose, ok.
Now, considering at the equilibrium point
theta is equal to 0, these are the set of
state space matrices we have obtained. Now,
the eigenvalues of this matrix A is given
by this one and it is quite straightforward
to see that these eigenvalues specifically
with the positive sign is greater than 0.
So, one of the eigenvalues is lying on the
writing side meaning to say that the system
is unstable and this we can also see while
computing the P matrix.
So, if we compute the P matrix though it is
symmetric, but it is not positive definite,
ok. So, we see that the system or the homogeneous
system without applying any control signal
is a stable. However, one can certainly make
it move up by applying some torque or some
control input u, which we will study in the
next module about the controllability and
the state feedback, ok.
