Elements of Realizability Theory (Contd.) - Electrical Engineering Video Lecture - Electrical Engineering (EE)

This is on elements of realizability theory
is being continued, last lecture we ended
up in a bit of discussion about realizability theory.
And to recall what we did was that the network
synthesis problem is that the excitation and
the response they are given, excitation and
the response r some relationship between them.
And the usual relationship that is given is:
the transfer function or the Laplace transform
of the output divided by the Laplace of the
input under the conditions of 0 initial conditions.
If this is given the problem is to find N
that is the network synthesis problem.
As I said the first thing that we ask is is
this is a solvable problem. In other words
is the given H of s realizable this is the
first question that we ask. And there are
various realizability constraints, for linear
time invariant passive networks the constraints
are that they should be causal and also stable
these are the basic 2 constraints that we
will make as far as this course is concerned
and H of s realizable causality and stability.
As you know causality implies that the output
cannot precede the input that is: the and
it is stated in terms of unit in pulses h
of t should be equal to 0 t less than 0 because
delta t is applied at t equal to 0 this is
the causality constraint. And we also mentioned
last time that in terms of the frequency domain
the causality constraint can be transformed
into H of j omega where H of j omega is the
Fourier transform of h of t.
And the constraint is that if we take log
of H of j omega magnitude and then take the
magnitude because log could be positive or
negative divide by 0 plus omega squared and
integrate this from minus infinity to plus
infinity d omega then this should be finite.
This is the equivalent of this simple relationship
h t equal to 0 t less than 0 in the frequency
domain that is.
No no it is only log of magnitude H.
Well, if we take the magnitude squared, it
is only multiplied by 2 and therefore, if
this quantity is less than infinity divided
by 2 also makes it less than infinity. So,
it clearly does not matter whether you take
squared or not, but we are not deriving this
we are simply saying that this is the so called
Paley Wiener criterion. But we can have a
deeper look at what it implies.
For example, if there is a frequency at which
the magnitude of H of j omega is 0. As an
example you take the parallel t network, parallel
t network or the bridged t network which has
a 0 which has a transmission equal to 0 at
some frequency that is a frequency of 0 transmission.
Now, if magnitude H of omega equal to 0 what
is log? Log is minus infinity all right.
Now, the value of the integrant equal to minus
infinity is tolerable at a particular value
of omega all right which simply means that
at that frequency this quantity shall have
a pole agreed. All network functions and analytic
functions they are differentiable everywhere
except at a finite number of singularities
all right So, singularity does not matter,
if there is singularity and we want to make
a contour integration then what we actually
do is we make a like this you know this contour
integration business. But suppose the magnitude
is 0 over a band of frequencies over a band
of frequencies.
For example, you have the ideal low pass filter.
Let us look at this, ideal low pass filter
is: it is unity let us say, between some frequency
omega c and 0 minus omega c and it is 0 elsewhere
this is the ideal low pass filter. Now, in
the ideal low pass filter you see the magnitude
of the transmission this is magnitude H of
j omega this is 0 over a band of frequencies.
Now if it is over a band of frequencies then;
obviously, infinity multiplied by any finite
non zero quantity will be infinity. And therefore,
this Paley Wiener criterion constraints that
magnitude H of j omega cannot be 0 over a
non zero band of frequency all right. In other
words if it restricts that this is not a realizable,
realizable magnitude function. An ideal low
pass filter any ideal filter is not in fact,
realizable because it is required to be 0
over a non zero band of frequency this is 1; 1 of the restrictions that this Paley Wiener criterion implies.
The other restriction is that, if magnitude
H of j omega falls with frequency as a low
pass filter should let us say, it falls like
this on the other side also it falls like
this, then this fall should not be mark my
words more rapid than an exponential fall
it cannot decay more rapidly than how an exponential
decays. Now, to make this a little more specific
let us take, the Gaussian curve the bell shaped
curve which can be represented by e to the
minus omega squared.
Suppose, my H of j omega is equal to e to
the minus omega squared then the fall with
increase of frequency is more than exponential right.
Now, if you take the log magnitude H of j
omega and magnitude of this this is simply
equal to omega squared an integral minus infinity
to infinity omega squared d omega divided
by 1 plus omega squared goes to infinity it
is not a finite quantity, it is 1 example.
But one of the interpretation one of the consequences
of the Paley Wiener criterion is that the
magnitude squared function or the magnitude function cannot decay more rapidly than an exponential.
Similarly, it cannot rise more rapidly for
high pass filter it cannot rise more more
rapidly than an exponential. So, there are
2 consequences of the Paley Wiener criterion
which we should be aware of there are many
others and you have to actually test by Paley
Wiener criterion, but the easier way is the
h of t and for us for circuit theory purposes
h of t would be good enough. On the other
hand suppose, magnitude H of j omega.
Let me ask you a question suppose, magnitude
H of j omega is square root of 1 plus omega
squared well, is it realizable this is a trivial question.
It is realizable why? Because we
know the network I think, it is obvious when
we see it we know the network what network
does realize this a simple RC where R and
C are both equal to unity or the product is
unity that is good enough.
Let see 1 by sc r plus 1 and if the product
is unity then the magnitude squared function
is this. So, you know and if you substitute
this in the Paley Wiener criterion you will
see that the integral is a finite quantity.
So, in summary we have not proved anything
we have only stated the Paley Wiener criterion
and we have seen 2 consequences of the Paley
Wiener criterion and these are good enough
for an electrical engineer that is.
Suppose why we cannot take 0 or not.
The question is why can H of j omega why is
it constrained to have not why it is constrained
not to have a band of frequencies for which
the value is 0. You see if the value is 0
then log is infinity log is minus infinity
and minus infinity magnitude is infinity.
So, of necessity this integral is going to
be infinite because the integrant itself blows
up all right. That is why; however, if it
is only at a particular frequency then infinity
multiplied by 0 because the band width is
0 integral means an area the area will be equal to 0.
Now, the second question for realizablity
then that is more important in the context
stability and the kind of stability that we
bother about in electrical engineering is
the B I B O stability. That is if the input
e of t is bounded. If, e of t is bounded less
than equal to C1, which is less than infinity
some C1 some constant C1. Then B I B O stability
says that r of t magnitude should be less
than equal to some other quantity C2 which
is finite this is the criterion for stability.
And you know that for a linear time invariant
system causal system now, r of t and e of
t they are related by the convolution integral.
So, the convolution integral is that r of
t is equal to integral 0 to infinity why 0
because we assume causal h tau r of e of t
minus tau d tau this is the convolution integral.
And the mod r of t shall be less than integral
0 to infinity mod h of tau d tau.
In general 0 to infinity because we have not
assumed e of t as causal signal in general
0 to infinity, but you if this is also causal
that is e of t does not exist for t less than
0 then the upper limit shall be equal to t
this is more general. Now, if e of t minus
tau is bounded by C1 then it is less than
C1 multiplied by this and this is required
to be less than infinity therefore, this is
a a revision of what you did in the signals
and systems course. That is stability requires that h of tau d
tau integral should be less than infinity
this is this you must have done in signals
and systems that is: the impulse response
should be absolutely integrable, the absolute
value of this integrated from 0 to infinity
should be finite. Now, this is in the time domain.
The frequency domain criterion is that the
corresponding transfer function that it is
H of j omega must have poles or the other
way around no poles in right of plane, what
will happen if we have poles in the right
of planes? Then the time domain response would
have would increase indefinitely with time.
The time domain response would have factors
of the formation of sigma t sigma 0 t, where
sigma 0 is positive and therefore, it would
grow indefinitely. No poles in righter plane.
If there are poles on the j omega axis if
there are poles on the j omega axis they must
be simple why what is the restriction?
Why is this restriction simple poles on the
j omega axis? why is it so?
If there are multiple poles then we shall
have the time domain response would be would
contain product of t and sin or cosine of omega 0 t.
If there are simple poles of the
j omega axis the corresponding impulse response
shall be either sin or cosine. It can be explained.
If there are multiple poles let us say, second
order pole that is pole of multiplicity 2
then we shall have a term of the of this form
t multiplied by this and this grows indefinitely as t increases.
So, we can have only simple poles on the j
omega axis. This also implies, if there are
simple poles on the j omega axis. And we assume
that H of s is a polynomial in s divided by
another polynomial that is: something of the form K let us say, s to the power m plus etcetera plus a 0.
And if the denominator is s to the
n plus etcetera plus b 0 let me, write it again.
If our H of s is of this form K multiplied
by s to the m plus let us say, a m minus 1
s to the m minus 1 plus etcetera plus a 0
divided by s to the n plus bn minus 1 s to
the n minus 1 plus etcetera plus b 0. Then
there is a restriction on the degrees of m
degrees of the numerator and denominator you understand how we wrote this transfer function.
This is in general a ratio of polynomials
n of s divided by d of s. And what we have
done is the living terms here and here we
have made into unity the constants, they have
observed in K all right this will be a standard
form for writing the transfer function.
Now, if that is so then what is the restriction
between m and n, which we will say that, if
there are poles on the j omega axis they must
be simple. That means how can there be pole
on the at infinity, infinity after all is
a point on the j omega axis.
And can we . M should be greater than n, but can m be greater
than n by more than 1 that is the point. No.
Therefore m should be less than or equal to
n plus 1 this is the restriction. Then the
numerator degree can be the less than the
denominator degree, it can be constant for
example, the numerator can be a constant.
For example in 1 by SCR plus 1 well the numerator
is not a polynomial it is a constant. So,
it is a zeroth degree polynomial. Now, if
you take a second order network for example, 2 RC in then numerator can still be
a constant unity the denominator will be a
quantity.
So, the numerator can the degree of the numerator
can be less than that of the denominator,
but if it is greater than, it cannot be greater
by more than 1 that is the restriction. This
is why if you take an H of s is equal to s
squared this is not realizable because the
degree the, because there are 2 poles at infinity
you can only allow a simple pole.
Now, to this realizability criterions.
Excuse me sir.
Yes. In this how did we prove that
How did we… Yes sir.
If there are multiple poles on the j omega
axis then the time domain response shall contain
terms of the form t times sin or cosine. You
see, if we have A factor of the form s plus
omega s squared plus omega not squared let
us say, whole squared then we shall have terms
of the form t times sin omega 0 or cosine
omega 0. And as t tends to infinity this term
it grows indefinitely. So, the system
becomes unstable any other question?
Excuse me sir. Yes.
Let me have m greater than n by 1 degree.
Then we will have a power of s. So, we will
have a term at quotient in s . Correct.
Sir, if we can work it to the time domain. Right.
We get a delta dashed e. Sir, but delta
H of s equal to s.
Yes sir s plus something else.
All right. And sir, if we convert this we get 1 term as.
Delta prime t.
Yes sir. And the other becomes K delta t.
Moreover, the delta prime t is not realizable
sir So, it does not mean anything.
All right it does not mean anything.
Well we know that, if we have a if we have
this is an impedance function. Yes .
S plus k I know that, an inductance realizes
s is not that right an inductance the impedance
is s times L Now, if you, if this is a transfer
function that is a voltage ratio then we shall
have problems we will come to that. But; obviously,
a network function H of s which is an impendence
function can be of this form s plus k. And
our function H of s that we wrote was of a
general form, it is not a it is not a true
transfer function, it is a network function
all right. Any other question?
Intimately related to this question of realizability, an intimately related to testing for realizabilty
is a kind of polynomial which go by the name of a gentleman by the name Hurwitz Hurwitz polynomials.
And let us, study what they are
because they are very intimately related to
the testing for realizability suppose, we
have a network function. It is N of s by D of s.
Then realizability demands that H of s poles
must be in the left of plane open left of
plane that is they can be on the j omega axis
also, but no none on the in the right of plane.
D of s therefore, and the poles of H of s
at the 0s of D of s. A polynomial which closely
obeys this condition is the Hurwitz polynomials.
So, 1 says that a the D of s the denominator
polynomial must be a Hurwitz polynomial.
Now, let us see the Hurwitz polynomials are.
Okay. Conditions at for realizabiltity of H of s
we want, it is poles to be in the left of
plane including the j omega axis that is closed
left of plane such polynomials, polynomials
whose roots are in the closed left of plains
are called Hurwitz polynomials. And therefore,
the realizability constraint can be stated
in terms of Hurwitz polynomials rather than
saying where it is poles should be if poles
occur on the j omega axis they should be simple
and so on and so forth.
Poles of H of s are the 0s of D of s.
Pardon me.
Sir poles of numerator or. Okay.
The numerator as you will see in the case
of an impedance function, the numerator also
has to be Hurwitz’s because you know for
a driving point function poles as well as
0 should be in the left of plane. So, we will
apply that to N of s also when it comes to.
But let me define first what a Hurwitz polynomial
is: a polynomial this is the definition a
polynomial P of s in general is defined to
be Hurwitz if and only if 2 conditions are satisfied.
That is it should be a real polynomial
that is P of s should be real for s real what
does that mean; that means, that since, it
is a polynomial it should be of the form ai
s to the i let us say, i is equal to 0 to n.
Then it, P of s real for s real means that
the coefficients ai must all be. Real.
Real that is all as simple as that.
And the second condition is that roots of
P of s should be or put it this way no root
in RHP P of s cannot have any roots in the
righter plane.
This statement that no root of RHP can be
amplified like this that is suppose, P of
let us say sk is equal to 0 P of sk is equal
to 0 k equals to let us say, 1 to if n is
the degree of the polynomial then the fundamental
theorem of algebra states that every polynomial
of nth degree will have.
N roots therefore, k will go from 1 to n.
So the second requirement that no root in
right of plane means that if P of sk equal
to 0 that is sk is one of the roots where
the subscript k goes from 1 to n. And sk equal
to let us say minus sigma k plus j omega k
where this is the real part and this is the
imaginary part then what does it mean. It
means: that sigma k must be. Greater than .
Greater than or equal to 0 because it is the
close left of plane. The real part of the
roots must be non negative 1 of the ways of
expressing this greater than equal to 0 is
that the real part should be non negative
all right. Any question? Our polynomial we
took as summation ak s to the k, k equals
oh no I took i is not it ai s to the i where
i is equal to 0 to n all right.
Let us look at this a little more closely
P of s is equal to ai s to the i, i is equal
0 to n. Now, I have already said that this
ai these coefficients must be real because
Hurwitz polynomial demands that the polynomial
be real for s real. If s is real obviously,
have to be real. What about sin can some of
the coefficients be negative. The answer transfer
to be no; we cannot have a negative let us
see why? The roots that are permitted the
roots that are permitted for P of s are belong
to these locations there can be a root on
the negative axis. Let us say, this point is minus gamma i. So,
this root will contribute a factor of s plus
Gamma
Gamma i is not it right this root will contribute
a factor of s plus gamma i. This is 1 possible
root another possible root could be a pair
on the j omega axis let us say, this is omega
j why 2.
There must be complexed on the way because.
Because the coefficients are real questions correct.
So, what do this factors, what do this 2 roots
contribute what factor; obviously, s square
plus omega j square all right these 2 roots
contribute to a factor like this and the only
other possibility only other possibility is
that we have roots like this: complex conjugate
ar. Let’s say this is this distance is omega
k and this distance is let us say, minus alpha
k all right. These 2 roots contribute to a
factor of s plus alpha k whole square plus
omega k square is this clear these 2 roots
contribute to a factor of this form because
what you have is s plus alpha k plus j omega
k multiplied by s plus alpha k minus j omega
k and if you multiple this out this is precisely
what you will get all right.
So, in general these are the only 3 possible
forms of way you cannot think of any other.
And the polynomial must be a product of factors
like this: that is you can have continued
product of factors like this continued product
factors like this and continued product factors
like this. In theses component polynomials
all coefficients are positive and by multiplying
a number of component polynomials in which
the coefficients are positive you can get
a you can never get a negative term is not
that correct. This is right or wrong which
means that the coefficients ai’s must be
positive ai’s are real and positive agreed.
If you see a network function in it a denominator
contains a complex coefficient immediately
you can throw it out not realizable. If you
see all coefficients are real, but 1 of them
is negative immediately you can throw it out.
Now, next question.
We are talking of denominator . Denominator yeah in general for impedance
functions the numerator has also to be tested. Sir excuse me.
Yes non negative.
That is the question I am going to ask you
now whether it is whether we should call it
as positive or non negative.
Let’s say P of s once again ai s to the
i i is equal to 0 to n.
A question that I am going to ask you is:
between a0 and an between a0 and an can there
be a missing term. Yes sir .
It can be. Now, how can a missing term be created from
product of factors like this: how can you
think of a missing term. There cannot be the
answer is there cannot be unless you have
a term like s minus gamma i when because missing
terms requires cancellation and cancellation can occur only if in one of the components
there is a negative term. Obviously there
cannot be a negative term here unless this
is minus alpha k. There cannot be a negative
term here at all these are both positive.
Okay that is a good question.
Good question what he says is: if I have terms
only of this form continued product over j
yes there are missing terms possible, but
it is a very special case what is the special
case. That it is a purely even polynomial,
it is a purely even polynomial and it is Hurwitz
because it is roots are not in none of the
roots are in the righter plane.
So, you must qualify that in general there
cannot be a missing term between a 0 at an
unless all odd terms are missing that is unless
the polynomial is purely even. And we are
being a bit bias when we are saying purely
even because such a polynomial if multiplied
by s also shares the same property. If you
multiply by s then the polynomial becomes purely.
Odd. Odd and all even parts are missing. Therefore
you should say that in general no term between
a0 and an should be missing unless all odd
terms are missing or all even terms are missing
all right that is unless i unless the polynomial
is purely even or purely odd. Now, I look
at the question I ask you a question unless
you have a question.
Sir how it makes difference that some other
How it makes a difference. If if let us say,
you have a term s cube plus s square plus
1. Then the coefficient of s is missing how
can that coefficient be missing unless there
are cancellations unless there are cancellation
you see our only term are s plus gamma i s
square plus omega j squared. And s plus omega
k squared no alpha k squared plus omega k
squared by multiplying such terms you can
never get a missing term unless either this
gamma i is negative or alpha k is negative;
that means, roots in the right of plane. Any
other question?
Now, let me ask my question.
Suppose, you have a polynomial s to the 4
plus 1 it is purely even, it is purely even.
Is it Hurwitz. No.
No because. 1 of the event terms is absent. There can
you see you can allow the polynomial to be
purely even or purely odd provided there are
no missing terms again you can allow all odd
terms to be missing but no even terms should
be missing, if it is even polynomial all even
terms should be there missing means that there
are there are roots in the righter plane can
you show this here it is very easy you can
write this as twice s squared plus 1 minus
twice s squared.
So, this is s square plus 1 plus root 2 s
s square plus 1 minus root 2 s and you can
see that this is a non Hurwitz polynomial,
because there is a.
Negative coefficient. Negative coefficient. Is that clear? We shall
have lot of fun with Hurwitz polynomials and
Hurwitz polynomials pay play a very important
role in the circuit theory in network synthesis.
Now, suppose I have a P of s which is purely
even no missing terms, where do you think
it is roots shall be, it is purely even.
Purely even with no missing they must be
Similarly, if it is purely odd then also it
is roots must be on the j omega axis only
with. And 0.
Yeah with 1 root at the origin.
Can there can be 2 roots at the origin.
No. No if it is odd then there cannot be…
1. Can you have 3 roots on the origin.
Yes . Yes you can, but if there are 3 roots at the
origin then the then the function is useless
for network synthesis why.
Multiple . Multiple poles absolutely wonderful;
that means, Hurwitz polynomial has.
Suppose, P of s in general it is a Hurwitz
polynomial with its even part denoted by small
m and it is odd part divided denoted by n
of s small n of s. Then 1 of the properties
and this is the test for a Hurwitz polynomial
the test for a Hurwitz polynomial is that.
The ratio m to n; the ratio of the even part
to the odd part or the odd part to the even
part depending on which 1 has a higher degree.
Suppose, the even suppose P of s the highest
degree highest power is even then you should
take the ratio m by n.
Pardon me.
M is the even part I think I small m is the
even part of P of s and small n is the odd
part of the P of s. For example, if I have
plus s square plus s plus 1 P of s. Then m
is equal to s square plus 1 and n is equal
to s even part and odd part all right. And
the ratio that you should take is now, is
s square plus 1 divided by s that is you should
take the higher degree polynomial in the numerator
and the lower degree 1 in the denominator.
What can be the degree difference between
these 2 m and n.
Can it be 2 why not?
Can be 3. Why not 2?
Because 1 is even and the other is the odd.
So, the difference must be difference must
be odd.
Sir it cannot be 3. It cannot be 3 it can be 3, but then we shall
not be interested all right. Now, 1 of the
properties of a Hurwitz polynomial is and
this property leads to the testing of Hurwitz
polynomials is that this ratio m by n or n
by m putting this higher degree polynomial
in the numerator. If you make a continued
fraction expansion of this or this depending
on which ever is applicable the continued
fraction expansion of this shall yield, yield
positive quotients, it can be rigorously proved,
but we are not interested in the proof.
The testing of a Hurwitz polynomial says that
the continued fraction expansion of m by n
or n by m will yield positive quotients. Now,
quotients; obviously, and I am not this is
not quite correct what will be the form of
quotients. Suppose, I divide s square plus
1 by s if I expand this into continued fraction
the first quotient will be. S. S.
Now, what do you mean by positive s no positive.
Positive coefficient. Coefficient quotient all right. The continued
fraction expansion should yield positive coefficient
quotient. Before I take an example, tell me
will this continued fraction expansion will
it always be finite or is there a possibility
that it can go on and on like a recurring
decimal continued fraction expansion will
it be finite or shall it continue indefinite.
Sir if it is realizable .
It has to be finite, how many quotients do
you expect?
1 How many quotients continued fraction expansion
well all you have forgotten what is continued
fraction expansion it would be
Difference
No sir degree of difference . Depending on the degree of the numerator which
is also the degree of the polynomial is not
that right. Therefore continued fraction expansion
1 of the properties is not a Hurwitz polynomial
any continued fraction expansion shall be
finite and the number of quotients that we
expect is equal to the degree of the polynomial.
For example, if you have a cubic polynomial
like this suppose, you give only finite only
the it is a realizable . No
Any it is realizable or unrealizable it does
not matter, Hurwitz or not it does not matter.
If you have a cubic polynomial for example,
the continued fraction expansion would be
of this form. s cube plus 2 s you shall divide
by 3 s square plus 1 all right s by 3 s cube
plus s by 3 what am I left with. 6.
5 s by 2 3 s square plus 1. s by 3 .
5 s by 3. So, this would be nine by 5 s.
So, 3 s square. 1 5 s by 3 5 s by3 and that is it continued
fraction expansion ends. I have exactly 3
quotients the coefficients of which are positive
and therefore, I conclude that this must be
a Hurwitz polynomial. The this is did only
to illustrate the number of quotients is exactly
equal to degree of the polynomial. Now, it
may. So, happen in Hurwitz testing it may.
So, happen that you do not get this required
number of quotients in other words the continued
fraction expansion may end prematurely is
that possible. It is possible only if the even part and the
odd part have a common factor. For example,
if this polynomial is multiplied by s square
plus 1 all right then what is the degree of
this polynomial degree would be 5, but you
shall get exactly 3 quotients because in forming
what shall we form here, n by m in forming
n by m n has a factor s square plus 1 m has
a factor s square plus 1 and therefore, the
number of quotients shall be exactly equal
to 3 not 5.
In other words there is a possibility that
the continued fraction expansion will end
prematurely and if it ends prematurely that
shows that even part and the odd part have
a common factor. And this common factor if
an even polynomial let me, go slow if an even
polynomial and an odd polynomial has a common
factor what can be the form of this common
factor you understand what I mean.
s plus alpha .
It cannot be s plus alpha
It has to be a s squared or a
It has to be an even polynomial and this is
why you said s square plus 1 and where how
do you detect it. Let us before taking an example.
Suppose, I have let us say, s square plus
omega j squared a number of them. Suppose,
this multiply in m of s plus n of s. Suppose,
I have a number of factors how do I detect
these factors because in the continued fraction
expansion of the even part divided by odd
part or the reciprocal of it I shall get a
number of quotients exactly equal to the degree
of m plus n these factors how do I
It is it requires a bit of thinking to convince
oneself that; these factors will make their
appearance during the continued fraction expansion.
In fact, the last divisor the last divisor
shall be proportional to these cancelling
factor the last divisor after that there is
no other divisor because the remainder is
0. Okay
Could you please explain me the last
May be I take an example, then I will come
to the theory. Let me take an example.
Suppose, I have s to the 7 plus 2 s to the
6 plus 2 s to the 5 plus s 4 clearly non trivial
example 4 s cube plus 8 s squared plus 8 s
plus 4 this is my P of s. And the question
being asked is is this Hurwitz or not. So,
what we do is we make a contribute fraction
expansion like this 2 s to the 5 plus 4 s
to the 3 plus 8 s I divide by 2 s to the 6
plus s to the 4 plus 8 s squared plus 4. So,
I get half s as the first portion I get s
to the 7 plus s to the 5 by 2then plus 4 s
cubed 4 s cubed plus 2 s agreed please do
not allow me to make a mistake is that.
So, I get 2 this would be 3 s to the 5 by
2 all right. These 2 cancel plus 6 s is that
right? all right. So, I divide 2 s to the
6 plus s 4 plus 8 s squared plus 4. Now, what
do I get? Four s divided by 3 this is the
quotient. Don’t allow me to make a mistake.
So, it is 2 to 2 s to the 6 plus 8 s square
8 s square. So, I am left with s to the 4
plus 4 all right. So, let us bring 3 s to
the 5 by 2 plus 6 s all right then what do
I get. 2 by 3 s.
3 by 2 s can I slow down of course, I can.
Where did where did you lose yourself.
Or you might all right. Prevention is better than cure.
We shall do that 3 by 2 s. So, I get 3 s to
the 5 by 2 plus.
6 s. 6 s. And you see that I expected 7 quotients
and end up with 3. The last divisor must be of degree 4, I have
got 3, I have missed 4 of them the last divisor
must be of degree 4 and the last divisor is
the common factor not the common factor there
could have been a constant. For example, 2
s 4 plus 8 the last divisor is a constant
multiplied the constant here happens to be
1 the last divisor is the common factor.
And therefore, P of s shall have a factor of s to the 4 plus 4 multiplied by P1 of s
some P1 of s. Do we have to go further? To
test whether P of s was Hurwitz or not do
we have to go further. No because this polynomial is non Hurwitz.
Yes.
Is not that right because there is a missing
term here s square term is missing. But suppose
there were no missing term. Suppose, this
was Hurwitz then you had to test P1 s for
Hurwitz character is the point clear? This
is only 1 step if it ends prematurely you
discover the common factor you take the common
factor call the rest of the polynomial as
P1, if the common factor is itself Hurwitz.
Sir; that means, you have to divide
s 4 plus 4 of P of s and find the. That
is right you have to divide this given polynomial
P of s by s 4 plus 4 find out what is P1 of
s then subject P1 of s to Hurwitz character
Hurwitz state is that clear? And then only
at the end of this exercise we will be able
to say whether, it is Hurwitz or non Hurwitz.
Now, this is a purely even polynomial, but
its roots are not all on the j omega axis
if they are all common roots; common roots
shall be discovered by Hurwitz testing all
right. And can this polynomial for example, be odd
can the last divisor be odd. No.
No it shall be with caution I make this statement
with caution you might have to disappoint
yourself later. Suppose now, another tricky
question.
Suppose, P of s is given as let us say, s
to the 8 plus 6 s squared no 6 s to the 6
plus 3 s to the 4 plus 2 s squared plus 1.
Therefore this is the polynomial that is given
that is it is a purely even polynomial how
do you test this because it is odd part is
0. On the other hand this could be a, this
could be a Hurwitz polynomial if all it is
roots are in the j omega axis this could be
a Hurwitz polynomial all right. Then the method
of testing is again without proof. If you
are given a purely even or purely odd it applies
to odd also what you do is. Take the first
differential coefficient. p prime s; obviously,
p prime s shall be purely. Odd .
Off. And then you make a contribute fraction expansion
of P of s divided by P prime of s first differential
coefficient. And if this satisfies the criterion
that is it does not end prematurely number
1 number 2 it is all its quotients are positive
then P of s is Hurwitz. Sir if it ends prematurely it does imply that
it is non Hurwitz. Non Hurwitz correct if it ends prematurely
then you shall have to subject the remainder polynomial.
Correct.
Sir it does not take
Correct correct that is what I am saying.
If ends prematurely then you have to discover
the common factors you have to discover P1
of s and then subject both of them to Hurwitz test.
If they come out to be. In this particular case, this was by inspection
you could see that this is non Hurwitz. So,
you could discover it. But suppose, you cannot
do that by inspection then you have to subject
both them to Hurwitz test all right.
For example, if I take this polynomial s to
the 4 plus s squared plus 1. Suppose, this
is my P of s then P prime of s is equal to
4 s cube plus twice s. Now, shall I divide
that is another question that I want to ask
shall I divide by 4 s cube plus 2 s can I
divide by s 2 s cube plus s.
2 s cube plus s can I do that?
Yes sir. Why can I do that?
All I want is whether the quotients are positive
or not and dividing by dividing any part by
a positive constant. Constant.
Doesn’t matter. This is the simplifying
feature otherwise we will have to handle large
numbers ar. So, I can do that I get s by 2
s to the 4 plus s squared by 2. What I am
left is s squared by 2 plus 1 then I divide
2 s cube plus s this is how continued fraction is made.
I take this as the as the what is
it called dividend dividend. So, I get how
much? 4 s 4 s 2 s cubed plus 4 s. Minus 3 s.
Minus 3 s do I have to further? No sir.
No because the next quotient will be negative
therefore, this is non HP, non Hurwitz polynomial
Now, in tomorrow’s class in tomorrow’s
class, we shall continue the discussion of
Hurwitz polynomials and their importance and their importance in testing network functions.
That is all.