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Good morning friends we shall continue with the topic on parts of network functions.
Yesterday, we discussed about functions r omega being given how to get z(s) okay so
we desorb this by substituting omega square equal to minus s square, we took the roots
in the left of plane and finally from the co efficient of the numerator a0, a1 and so
on.
Now today we shall be discussing of about the magnitude function, the magnitude function,
if the magnitude function is given how to calculate phase magnitude function is given
to calculate to determine phase. Now the magnitude z square can be written as z(s) into z minus
s at s equal to j omega is that all right and what is z(s) if I write in terms of even
and odd parts m1 plus n1 by m2 plus n2, this is z(s) and what is z minus s, m1 minus n1
it is only the odd part which will be having a negative sign now divided by m2 minus n2
is that all right at s equal to j omega. So what you are getting is m1 square minus n1
square by m2 square minus n2 square to s equal to j omega is that all right.
So if you given z magnitude square it is basically m1 square minus n1 square by m2 square minus
n2 square at s equal to j omega see if I make reverse substitution I will get m1 square
s minus n1 square s by m2 square minus n2 square s from here okay. So there if I factorize
and drop out the roots in the right up plane I will get m1 plus n1 and correspondingly
m2 plus n2 it is very simple. So let us take one example z j omega square is equal to 1
plus omega square by 1 plus omega square plus omega to the power 4, so that gives me if
I make reverse substitution it will be 1 minus s square by 1 minus s square plus s to the
power 4 okay.
So what are the roots of this 1 plus s into 1 minus s okay what are the roots of this
s square, see 1 plus s to the power 4 plus twice s square minus 3s square, so it will
be 1 plus root 3 s into s square minus root 3s plus 1 all right. Obviously, the quadratic
be the negative sign this is m2 minus n2, this is m2 plus n2, this m1 plus n1, this
m1 minus n1. So that gives me z(s) equal to the left up plane roots the factor corresponding
to that is 1 plus s divided by this 1 s square plus root 3 s plus 1 is that all right s1
plus s by s square plus root 3 s plus 1 realization of this you can realize this by Bott Duffin
or Brute synthesis okay we are not going to the realization part at this stage z(s) is
this much.
Let us take another example z omega square is equal to omega square plus 16 by omega
4 plus 10 omega square plus 9. So I can write z(s) into z minus s as 16 minus s square divided
by s to the power 4 minus 10 s square plus 9 which gives me 4 plus s into 4 minus s,
this gives me s square minus 9 into s square minus 1 which gives me finally 4 plus s into
4 minus s by s plus 3 into s plus 1 into s minus 3 into s minus 1. So what will be z(s)
4 plus s by s plus 3 into s plus 1 okay is this all right. Like this you can see for any function mind you z magnitude square will
be a function of omega square only there is nothing like omega, omega to the power 3 and
so on, this will be an even function of omega, this will also be an even function of omega.
Next we go to phase to magnitude, phase to magnitude, phase to magnitude okay.
Now if I see the angle z(s) if I write as even z(s) plus odd z(s) then z j omega this
will correspond to a real part
this is nothing but even z(s) at s equal to j omega and this gives me odd z(s) okay at
s equal to j omega j will come out of this, is that okay. So what is tan theta you would
give an say the angle tan theta is odd z(s) at s equal to j omega but odd z(s) will generate
j term okay mind you divided by even z(s) all right. So this is nothing but if I write
j tan theta, j tan theta you will get n1, m2 minus m1, n2 divided by m1, m2 minus n1,
n2 okay at s equal to j omega all right this is nothing but okay.
Suppose this is equal to this 1 even z(s) plus odd z(s) is say after division I am making
a reverse substitution suppose we get m by n okay m by n. So m plus n(s) if the ratio is m(s) by n(s)
if it is m(s) by n(s) then what is m(s) plus n(s) this plus this okay which will be n1,
m2 minus m1, n2 plus m1, m2 minus n1, n2 which means m1 plus n1 into m2 minus n2 okay. So
the 1 with m1 plus n1 term that is roots in the left up plane will be numerator and roots
in the right up plane will if you shift those roots in the left up plane that is m2 minus
n2 you get so you just change the sing of n2 that will give you the denominator all
right. So let us take an example it will be clear.
Suppose tan theta is given as minus omega to the power 5 okay. So you substitute omega
is equal to s by j so j tan theta you are computing j tan theta that is minus j omega
to the power 5 and that gives me minus s to the power 5 all right. So this you are calling
as m by n, so s to the power 5 by 1, is it not this will be s to the power 5 by 1 with
a negative sign.
So what will be m plus n, 1 minus s to the power 5 is that all right. So 1 minus s to
the power 5 what are the factors 1 minus s, now let me work it out here, s to the power
5, no you do it this way 1 minus s to the power 5 equal to 0 you want to find out the
roots say s to the power 5 is equal to 1 so on the unit circle, so that is equal to e
to the power j2 phi n. So what will be s e to the power j 2 pie n by 5 put n equal to
0 1 to so if I put n equal to 0 this is 1 root then 2 phi by 5, 72 degrees, this is
72 degrees 2 phi by 5 then 2 into 2 phi by 5, 144 degrees okay then 3 into 72 so 216
degrees, so the values are 1 e to the power j 72 degrees, e to the power j 144 degrees,
e to the power j 216 degrees, e to the power j 288 degrees, these are the 5 roots is that
all right, that equal to be 1. Obviously 1 into e to the power so 1 to the power 1 by
5, so that will be 1 they will not be on the unit circle.
So what are the factors corresponding to m1 plus n1, it will be this this is giving me
m2 minus n2 so m2 plus n2 will be just images of this. So m1 plus n1 will be s minus this
a to the power j 144 degrees into s minus a to the power j this one 216 degrees all
right. So that gives me cos 144 degrees okay that means this one how much is this angle
180 minus 144, 36 degrees 36 degrees, so cos 36 degrees so s plus cos 36 degrees and this
will also give me cos 36 degrees so 2 into cos 36 degrees. So s square plus 2 into cos
36 degrees into s plus the imaginary parts alpha square plus beta square is 1 it will
be 1 is that all right. Any quantity coming out of complex pair of roots on the unit circle
will be s square plus 2 into the real part which is cos 36 degrees into s, s square plus
2 into alpha into s plus 1 alpha square plus beta square is 1. So it is s square plus cos
36 degrees approximately .8, so 1.6 s plus 1, 36.8 degrees is .8, m2 minus n2 you know
so m2 plus n2 will be this is 72 degrees.
So s s square plus 2 into cosine 72 degrees into s plus 1 straight away images of these
will be on this side all right. So the real part of this will be 72 degrees, cosine 72
degrees s square cosine 72 degrees is approximately .38 degrees approximately .3, so 2 into .3,
so 0.6 s plus 1. So z(s) is this one will be the numerator s square plus 1.6 s plus
1 divided by s square plus 0.6 s plus 1 approximately this will be the function, is that okay. So
we have seen how to calculate the function z(s) from the phase.
Now from here you can calculate the magnitude, once you know the actual z(s) you can calculate
the magnitude sir, yes please, that kind of problems when when we can be in practical
use, no in the practical system sometimes we measure only the phase we are not in a
position to estimate say the gain then what will you do okay. Sometimes you try to track
both magnitude and phase all right and try to find out z(s) it is a cross check, it is
a cross check.
Suppose the magnitude function is known we take only the real part or the only the total
magnitude with frequency then you approximate it by even polynomial ratio, ratio of even
polynomials all right and then only from there you can find out just previous to this you
have found out z(s) into z minus s, is it not. So factorize it find out the right up
plane roots left up plane roots all right take only the left up plane roots and that
gives you the function z(s) you are tracking only the phase then from the phase if you
can estimate again phase function in the form of polynomials then you can estimate z(s).
In the laboratory depending on your facilities available the accuracy for measurements of
phase and magnitude may be different. So you try to find out z(s), approximate z(s) from
both sides another example tan theta equal to omega cubed minus 3 omega divided by 1
minus 3 omega square all right.
So how much is j tan theta whenever the phase function is given in the form of tan theta
we write j tan theta that will be j omega cube minus 3 omega j by 1 minus 3 omega square
make the reverse substitution s equal to j omega. So that gives you j omega cube means
minus s cubed minus 3 s okay divided by 1 minus 1 plus 3 s square all right. So this
will give me minus s into s square plus 3, correct me if I am wrong, 1 plus 3 s square
all right. So what is m plus n, what is m plus n minus s cube just add these 2 plus
3 s square minus 3 s plus 1, so that gives me s minus 11 minus s whole cube is that all
right, 1 minus s whole cube.
So where are the roots in all on the right up plane at s equal to 1, so what will be
m1 plus n1 by m2 plus n2 this corresponds to m2 plus n2 this does not give any finite
roots that is 0s, so 1 by 1 plus s whole cubed is that all right. So this is not a driving
point impedance, it is an impedance function, it is a transfer impedance function, in case
of a transfer impedance function you may have difference in the degree more than 1 but in
case of a driving point impedance or admittance function, we know the difference in the degree
should be restricted to 1.
So here you can have more than 1, so this is not a driving point function sir why do
you sir de difference is 1 and the difference is greater than1, not greater than 1 it may
or may not be 1. So transfer function is a ratio between 2 voltages or voltage and current,
so the transfer function can be anything it can have only thing the roots must be in the
left of plane that is poles must be in the left of plane yes, this is an insist able
system any stable system may have a phase function like this but this is not a impedance
function, driving point impedance function not a driving point impedance or admittance
function. I will call it immittance function all right, so this is basically a transfer
impedance about the properties of transfer impedance we will discuss later on when you
go for 2 port synthesis.
Now we shall take up 1 or 2 more examples, it is all about realization of networks from
given parts. Let us take another example real part of z j omega okay 1 plus omega square
by 1 plus omega square plus omega to the power 4 what will be z(s). So yesterday we had handled
this kind of problems, so what will be 1 plus omega square plus omega to the power 4 did,
we discuss about this particular problem yesterday, 1 plus omega square by did I take this very
example then otherwise we will skip to another problem, no I do not think so.
We took up another example, 1 plus omega square by omega to the power 4, no no not this example
okay, I do not think we did it okay. I have chosen this because we want to see 2 types
of functions given by the same expressions. So what we do here make a substitution s square
equal to minus omega square so that will give me 1 minus s square by 1 minus s square plus
s to the power 4. So what are the roots of this s 4 minus s square plus 1 gives me s
to the power 4 plus twice s square plus 1 minus 3 s square. So it is s square plus 1
whole square minus root 3 s square.
So the roots corresponding to m2 plus n2, m2 plus n2 will be having the left up, left
up plane roots or this. So s square plus root 3 s plus 1 this is the denominator, so numerator
let us assume a0(s) a0 plus a1(s) plus a2(s) square divided by s square plus root 3 s plus
1 okay. So even part is how much a0 plus a2(s) square into s square plus 1 minus root 3 a1(s)
square divided by m2 square minus n2 square which is this. See if I put s equal to j omega
here I get real z j omega okay and that gives me how much a0 sorry a2 omega to the power 4
plus a0(s) square all right plus a2(s) square minus root 3a1 is that okay. So a0, so plus
root 3 a1 minus a0 minus a2 into omega square plus a0, this is the numerator is that okay
equate the co efficients.
So a2 omega 4 omega 4 is absent so a2 is 0 next root 3 a1 minus a0 plus a2 root 3 a1
minus a0 minus a2 is equal to 1 co-efficient of omega square and a0 is equal to 1 substitute
here a2 is 0 so root 3 a1 and a0 is 1 minus 1 is 0, so a1 is 1 by root 3 therefore z(s)
will be a0 that is 1 plus a1(s), so 1 by root 3 s a2 is 0 divided by a square plus root
3 s plus 1. So that gives me s plus root 3 divided by root 3 into s square plus root
3 s plus 1 this is z(s), is that all right.
That is an interesting point to be noted in case of synthesis that if you have a 2 element
synthesis if you have a 2 element synthesis from the network function, you can identify
whether it will be a2 element synthesis or a3 element synthesis. For example z(s) if
you are having functions like s square plus alpha s square plus beta and so on s square
plus gamma, s square plus delta functions of this type and if the poles and 0s are interlaced
they are coming alternately then it is an l c function, is it not l c. You can have
Foster1, Foster2, Cauer1, Cauer2 realizations and in those realizations you have minimum
number of elements they are canonic forms if z(s) is having real roots s plus alpha,
s plus beta and so on and if they are interlaced then they can be either a r l or r c depending
on whether for z(s), whether the pole is closest to the origin. The first one starts with a
pole or not then it will be an r c with the first root is a 0 nearest to the origin is
a 0 then it will be an r l function and we can realize it again by 4 canonic forms.
Suppose it is a positive real function but roots are not alternately coming then it is
neither r l nor r c, it will be an r l c network. This is also what we saw earlier, if it is
tested for positive realness, if it is tested for positive realness and the conditions are
satisfied then it is realizable but if the poles and 0s are not interlaced then it will
not be an r l or r c then it will be r l c and if it is an r l c network then we will
find now you cannot have necessarily a Foster 1 or Foster 2 realization you have to go for
Bott Duffin or Brune's synthesis, well not that you cannot factorize it I will give an
example, you cannot factorize it so easily. Let us see, I will take a simple example,
let me see if I have that okay.
Let us take a function s plus 4 by s plus 1 into s plus 3 okay. This is a function which
we got just sometime back if you remember the first example but I worked out yes it
is the 1 s plus 4 by s plus 3 into s plus 1 okay. So how to realize this, it is a positive
real function though poles and 0s are not coming alternately there are 2 poles first
then there 0.
So it is neither r l nor r c, so I thought of trying to realize it z(s) suppose it is
k1 by s plus 1 plus k2 by s plus 3 obviously, one of them will be negative because poles
and 0s are not coming alternately. Let us see how much is k1, how much is k1 multiplied
by s plus 1 put s plus1 equal to0 so 0, 3 by 2, 3 by 2 into s plus 1, so this will give
me an r c combination.
Now here it is k2 s plus 3 here 0 if I put this will be 1 this will be 3 minus 3, so
minus 1 third, so it will not work. Suppose let us put it as s into this then how much
is k2 multiplied by s plus 3 divide by s, so it will be s plus 4 by s into s plus 1
and then make s plus 3 equal to 0. So this will be 3 divided by sorry, 1 divided by 3
into 2 minus 3 into minus 2, so minus 6 plus 6, 1 by 6 into s plus 3 into s, what is this
an r c combination, what is this an r l combination. So if someone goes by Foster1 realization
it is possible for such a simple function it is possible to get this, problem comes
when you are having quadratics and not so easily factorizable, for in a factorizable
form then you have to go for Brune's synthesis or Bott Duffin synthesis, not necessarily
all ways you will be having that kind of synthesis even by Bott Duffin synthesis also you may
be landing up in similar functions all right. Let us see what canonic means let us this
is Foster 1 synthesis all right means we are putting series elements z1, z2, z3, let us
try y s Foster 2 then it is s plus 1 into s plus 3 divided by s plus 4 and that is equal
to s square plus 4 s plus 3 by s plus 4 okay s square plus 4 s by s plus 4 is s, s into
s plus 4 s plus 3 by s plus 4.
So this is an, this is a capacitor in parallel with and s plus 4 by 3 is the impedance. So
it is 1 third 4 by 3 ohm resistance and 1 third Henry inductor is 1 Farad capacitor.
Now you see here you are having 3 elements, here you are having 4 elements. So y(s) and
z(s) you see the difference all right if someone goes for Cauer synthesis, ladder synthesis,
let us see what it gives, z(s) equal to s plus 4 by s square plus 4 s plus 3 is it not
I can write this as1 by s square plus 4 s plus 3 by s plus 4. So let us carry out the
division s square plus 4 s plus3 s into s square plus 4 s here so 3 s plus 4, s by 3
s 4, 3 3 by 4, 3 okay.
So it starts with 1 by s plus 1 by s by 3 plus 1 by 3 by 4, so 4 by 3 so what does it
give me admittance of capacitor then an impedance of 1 third Henry and 4 by 3 this is Cauer
1 okay, Cauer 1 synthesis is same as Foster 2 the same thing 1 Farad, is it not, it is
coming as Foster 2, Cauer 1 and Foster 2, Cauer 2 you reverse the order and then see
z(s) equal to 4 plus s divided by 3 plus 4 s plus s square equal to 1 by 3 plus 4 s plus
s square divided by 4 plus s. So let us divide 4 plus s 3 plus 4 s plus s square so this
will be 3 by 4, 3 plus 3 by 4 s, so 13 by 4 s plus s square 4 plus s how much is it,
16 by 13 is that all right that gives me 4, 16 by 13 s in the denominator all right.
So that gives me 4 plus 16 by 13 s now that gives me minus 3 by 13 s. So since there is
a negative sign it is not possible to realize the Cauer 2, Cauer 2 does not exist that means
you cannot have something like resistance and then capacitance and so on then an inductance
or resistance, it will feel.
So canonic forms of that kind Cauer 1, Cauer 2, Foster 1, Foster 2 all may not be applicable
here in 3 element networks, some of them may succeed, some of them may not not obviously,
that you have seen just now I have tried all the 4 methods and number of elements also
differ in some case we have got 3 element, in some case we got 4 elements in Bott Duffin
synthesis we have seen we can avoid the use of that Brune's transformer with 100 per cent
coupling but number of elements will be more all right, number of elements will be more.
Let us take 1 or 2 examples on z(s) with 3 elements that is Brune's and Bott Duffin's
synthesis. Some of the interesting networks okay sorry before we go to that there is another
function suppose real z j omega equal to 1 plus omega square by 1 plus omega square plus
omega to the power 4 already it was sorry, real z was given Suppose this is magnitude
square suppose this is equal to magnitude square then what will you do have you understood
if z(s) square is there then you make substitution z(s) z minus s will be 1 minus s square 1
minus s square divided by 1 minus s square plus s to the power 4 have I have I done it
sorry, just now, it was the real part, no I did the other problem that was the real
part. So this will be 1 plus s into 1 minus s okay and how much is this s square plus
root 3 s plus 1 into s square minus root 3 s plus1 okay so how much is z(s) choose only
the left up plane roots. So s plus 1 by s square plus root 3 s plus 1 is that all right.
Earlier while computing for the when it was given as real z equal to this we, we got the
same denominator but the numerator was remember 1 plus s by root 3 something like that know
I will just yes 1 plus1 by root 3 s ,1 plus s by root 3 that was the function.
So you can see the difference yes, 1 plus s by root 3 was that so that is the only difference
in the numerator denominator remains same okay. Now I will take general r l c synthesis
by both Brune's method and Bott Duffin's method this is just a devising what we have learnt
earlier. So realize z(s) equal to s square plus 2 s plus 16 divided by s square plus
2 s plus 4 okay.
Let us try by Brune's synthesis so what was the starting point a real z j omega we compute
that is m1, m2 minus n1, n2 by m2 square minus n2 square so s square plus 16 even part into
s square plus 4 minus 2 s into 2s, 4 s square divided by s square plus 4 whole square m2
square minus 4 s square n2 square s equal to j omega is that all right.
So that gives me s to the power 4 plus 16 plus 4, 20 minus 4, so 16 s square plus 64
divided by s to the power 4 plus 8 s square minus 4 s square plus 16 is that all right,
s equal to j omega. See if I put s equal to j omega omega to the power 4 minus 16 omega
square plus 64 divided by omega to the power 4 minus 4 omega square plus 16. Now this you
can see is omega square minus 8 whole square okay, omega square minus 8 whole square. So
this function becomes 0 at omega square is equal to 8 we have to first of all track that
frequency at which the real part vanishes, is it not.
So the real part z j omega 1 equal to 0 at omega 1 equal to root 8 that is 2 root 2.
Once you have traced that frequency omega 1 calculate z j omega 1 that will be the imaginary
part is it not that will be the imaginary part. So if I put omega equal to 2 root 2
in this how much is it 16 minus s square is minus omega square that is 8, correct me if
I am wrong, okay plus 2 into s 2 into j into 2 root 2 divided by s square plus 4. So minus
8 plus 4 plus 2 j, 2 root 2 so that gives me 8 plus 2 2's are4 root 2 j divided by minus
4 plus 4 root 2 j is that okay.
Now you can see for yourself 8 plus 4 root 2, 8 plus 4 root 2 somewhere here 8 plus 4
root 2 how much is the magnitude 8 8's are 64 plus 32, 96, square root of 96 okay tan
inverse 1 by root 2 and this is 4 4's are you see some where here 4 4's are 16 plus
32. So root 48 tan inverse root2 this is tan inverse 1 by root 2 okay and this is tan inverse
this magnitude is tan inverse root 2 but it is more than 90 degrees. So how much is this
angle, how much is this angle the difference is 90 degree, it is the denominator angle
which is more.
So if this is theta this is 90 plus theta is that all right magnitude is root 2 and
denominator is having an angle of 90 degree that means minus 90 in the numerator. So 90
degree in the numerator, so minus root 2 j is that all right, it is just a simple competition.
Otherwise also by rationalizing you will get minus root j I thought it is better to write
in the polar form and you can imagine the angles very clearly okay one is alpha the
other one is 1 by alpha tan inverse alpha and tan inverse 1 by alpha, so find out the
difference in angle is that all right. So z is known so x is minus root 2 I call it
x1 so in Brune's synthesis if you get a negative reactance to start with what will that be
it represented as l1 is equal to x1 by omega 1 which is root 2 divided by minus root 2
divided by 2 root 2, 2 root 2 is the frequency, thank you all right.
So this will be minus half Henry so to start with you will have a minus half Henry inductor,
do not get frightened in the realization in a transformer you will find that minus can
be accommodated. So once you have identified this then you go for the next impedance z1
what is z1(s), z1(s) is z(s) minus l1(s) which is nothing but what was z(s) sorry, s square
plus 2 s plus 16 divided by s square plus 2 s plus 4 minus l1(s), l1 is minus half so
minus and minus will make it plus half s is that all right. So add these what do you get
by adding this, so s cube s square plus 2 s plus 4 s square plus 2 s plus 16 by s square
plus 2 s plus 4 plus half s is that all right.
So you get 2 into s square plus 2 s plus 4 and here it is twice s square plus 4 s plus
32 plus s cubed plus twice s square plus 4 s, correct me if I am wrong. So that gives
me s to the power 3 plus 4 s square plus 8 s plus 32 divided by 2 into s square plus
2 s plus 4 is that all right. This you can write as s cube plus 8 and s see s square
plus 8 s square plus 8 into s plus 4, 4 into s square plus 8 s into s square plus 8 divided
by 2 into s square plus 2 s plus 4. By the way, why did I take s square plus 8 as a fact
factor, if you remember, if you remember in the network this function z(s) is having a
value minus half sorry root 2 j at that frequency all right and it is also the value of this
that we have equated.
So this will be a0 at that frequency if you remember this element z1(s) must be 0 at that
particular frequency omega 1 okay. So this z1(s) at omega 1 that is 2 root 2 that is
s square plus omega 1 square will be a factor in the numerator and omega 1 square is 8.
So s square plus 8 will be a factor that we know because what we try to ensure is this
impedance at that frequency is equal to the reactance at that frequency of this coil.
So what will be the impedance of this it has to be0 because this impedance is equal to
this impedance so this has to be 0, so z1(s) must have a0 at that frequency, so s square
plus 8 must be having a numerator factor, must be a numerator factor of z(s) this z1(s)
allowed obviously, we are making it equal to that inductor okay, thank you very much
we will stop here for today. We will continue with this in the next class.
Good morning friends, we will continue with the numerical problem that we are discussing
in the last class.
We took at function z(s) equal to s square plus 2 s plus 16 divided by s square plus
2 s plus 4. We started of with this function and we found that at omega 1 is equal to root
8 the real part vanishes and we also obtained the value of x1 as half s whether corresponding
impedance this was l1(s) all right and this was minus half s minus half Henry, this is
z1(s) which will be equal to z(s) minus l1 s and that give me l1 is having a negative
sign, so that give me a plus sign. So we got s square plus 8 into s plus 4 divided by 2
into s square plus 2 s plus 4 okay.
To realize a pole to realize a pole we know we can make partial fraction, so this 0 is
to be converted to a pole so we can very easily realize z1(s) in terms of admittance function
which will be 2 into s square plus 2 s plus 4 divided by s square plus 8 into s plus 4,
you write as k1(s) by s square plus 8 we know whenever there are roots 1 the imaginary axis
it will be realized in terms of an l c network k1(s) by s square plus 8 plus k2 by s plus
4, this is to be seen later.
Let us see what this means how much is k1 multiply by s square plus8 divide by s make
s square plus 8 equal to 0. So this will give me 2 into minus 4 minus 8. So minus 4 plus
2 s divided by s into s square plus 8, so s into s s plus4 so s square plus 4 s which
gives me this is minus 8 plus 4 s divided by minus 8 plus 4 s, so that is equal to 1,
there must be some frequency where it is minimum omega square is equal to x square plus 15
x plus 24 x square plus 17x plus 16, I hope this is all right okay then 2 x plus 15 into
x square plus 17x plus 16, 2 x plus 17 into x square plus 15 x plus 24. So 2 x cube 2
x cube they get cancelled 2 x into okay. Let me rewrite it any way what I wanted to stress
is you will get a real value of x that is equal to omega square.
So calculate omega omega and substitute that omega here in the real part that will give
you the minimum value that is when the real part, real part varies like this it is this
minimum value and after computing that r minimum subtract it from z(s), subtract it from z(s)
whatever is left over you start realizing that z(s), the remainder z(s) there may be
a small slip some where here we will discuss it in the next class, if time permits. Otherwise,
you work it out your self and since there is not much of time okay, thank you very much.
We will continue with this in the next class.