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Good
morning friends, we will be continuing with driving point impedance function or immittance
function.
We have seen the properties to be satisfied are poles and 0s of the immittance function
must lie in the left half plane, left half of s plane then poles and 0s on the imaginary
axis must be simple on the imaginary axis must be simple. These are 2 very important
properties and then to ensure that poles and 0s lie in the left half of s plane actually
all the coefficients of the polynomials all the coefficients of numerator and denominator
polynomials we were using p (s) and q (s) okay p (s) and q (s) must be positive all
the coefficients of p (s) and q (s) must be positive okay and real then the difference
in the degree difference in the degree of p (s) and q (s) the degrees of p (s) and q
(s) must be restricted to 1 alright, difference in the degrees must be restricted to 1 and
as a corollary to this that means I will not write as an extra point it is to ensure this
Rowth Hurwitz criteria must be satisfied.
The criteria all of you know we discussed last time anyway we will work out 1 or 2 simple
problems later on in the tutorial and lastly what was the last point we discussed, real
part of Z or Y for all positive values omega must be greater than 0, for all real omega
okay. We will see where are the exception is coming, where do you think some of these
conditions may not be strictly satisfied, poles and 0s must lie in the left half plane
alright and the image axis this must be simple all the coefficients of p (s) and q (s) must
be positive and real, is it necessary that all of them should be positive that can be
a situation when alternate terms are present that is you have an even polynomial or an
or an odd polynomial. This situation can come when you are having say purely inductance
capacitance circuits L C circuits where there is no dissipative element when there is no
dissipation then there can be a sustained oscillation.
For example I can have a functions by S squared plus 9 suppose z (s) is this. So in the numerator
polynomial you are having s the real term may or may not be present the S to the power
0 term here, s to the power 1 term is missing will it be a positive function yes, it will
be this is the only exception that is if alternate terms are present, this is an even polynomial,
this is an odd polynomial. So this you can see this is nothing but an l c parallel combination.
We will see how to realize these very soon okay.
Now if the numerator if the numerator or denominator that is p (s) or q (s) contains only even
or odd terms it may be possible to have a realizable immittance function okay other
conditions are to be satisfied alright. Now let us see 1 or 2 simple function we will
take and then we will see the properties of L C networks. We will take the typical structures
of 2 element networks that is either L C or R C or R L okay so what would be what would
be the impedance or admittance function lie for such networks, networks consisting of
only 2 types of elements that is L and C or R and C or R and L, what would be the network
function like. Let us take L C combination, let us have a simple series circuit L and
C so what would be z (s) L S plus 1 by c s if you simplify this LCS squared plus 1 divided
by c s okay.
Let us take the admittance corresponding to this y (s) corresponding to this will be c
s by LCS squared plus 1, where the poles and 0s located. For the impedance function there
is a pole at the origin alright and there are 2 0s where 1 by L C, 1 by root L C and
plus minus 1 by root L C j j. So here and here this is s plane, so roots are all imaginary
okay this is 1 by root L C, this is imaginary, this is minus 1 by root L C okay, if you consider
this is z, if you consider y the admittance function where the poles and 0s located it
will be 0 now where ever we had the 0 earlier for Z will become poles. Let us take another
example l and c in parallel what would be zl (s) into 1 by c s divided by l s plus 1
by c s okay if I multiply through out by c s it becomes LS by LCS squared plus 1 so for
this l c parallel combination this is for series for the parallel combination what will
be z (s) the 0 at the origin pole at 1 by root LC on the major axis okay. This is Z
and what will be the corresponding y, what will be the corresponding Y, it will be just
reverse.
So this will be a pole this will be 0s okay. Now can you draw any conclusion from this
the structures of poles and 0s poles and 0s are all on the imaginary axis number 1, number
2 poles and 0s are coming alternately will have a little higher order functions and this
then see and then the origin is always either a pole or a 0, is there any pole or 0 at infinity
if I put s equal to infinity in the first case if I put s equal to infinity z (s) will
tend to infinity this is higher order s square s square by s, so it will be LCS and s tending
to infinity. So z (s) will be infinity and when a functions blows up to a very high value
tending to infinity it is a pole corresponding s will be a pole so will show a break and
at this is at infinity it is a pole here also at infinity there is a pole similarly here
at infinity y (s) becomes 0. So at infinity this will be a0 at infinity this will be a0
similarly, here we can see for z s this will be a0 okay similarly this 1 will be a pole
at infinity.
Let us take another example, say this is 1 Henry 1 farad, 1 Henry 1 farad what will be
z (s) like LS plus 1 by c s plus LS into 1 by c s by LS plus 1 by c s and l is equal
to LC is equal to 1. So this become s plus 1 on 1 by s plus s by s squared plus 1, is
it alright? So that gives me s by s squared plus 1, this is s squared what will be the
numerator like S square plus 1 whole squared plus S squared. So S to the power 4 plus 3 S squared plus
1 okay. Simplify this s square plus 1 whole squared plus s square, so is 4 plus 3 s square
plus 1 okay. There are 2 real roots, what are the roots s to the power 4 okay S to the
power 4 plus 3 S squared plus 1. I can write S square as x so it is basically a quadratic
index. So what are the roots of x minus 3 plus minus root over 9 minus 4 by 2 so minus
3 root 4 so 2.24 divided by 2 alright. So 3 minus 2.24, 2.76 by 2.38 approximately and
this is 5.24 divided by 2 is it alright 5.24 divided by 2, so 2.62 sorry S squared plus
2.62 divided by S into S squared plus 1.
So lets us sketch the poles and 0s for this origin is a pole it is in the denominator,
1 is a pole then root over of .38 approximately .62 .63 so that is a 0 and root over of 2.62 that
is 1.63 or so 1.64 approximately and this is say .62 you have got 0s and then at infinity
I am not showing the negative values of omega it is identical okay.
At infinity, this is infinity so this is a pole so you see pole, 0, pole, 0, pole again
they are coming alternately okay. So LC network, LC network if we have similar to this if I
take y(s) an admittance function the poles and 0s will be just replacing interchanging
their positions this will be a 0, this will be a pole and so on so there also pole 0 poles
and 0s will be appearing alternately. Hence forth, we will draw a diagram.
So this is omega because now we can take the frequency as the axis and corresponding value
of x omega, x omega how will x omega vary with omega let us go back to the first problem
suppose, we have just an inductance what will be its x j omega l. So if I write this as
jXL, jXL then XL will be varying linearly with frequency is that alright with the slope
of l with the slope of l. So this is XL what about xc if I have a capacitor then xc will
be minus j by omega c is j xc, so xc is minus 1 by omega c is that alright xc is all ways
negative. So it is inversely proportional to frequency it will be a rectangular hyperbola
varying like this. So this is xc if I put this is tending to infinity if I put a series
combination of l and c series combination of l and c then what will be the net effect
it is xL plus xc which is minus 1 omega c so this magnitude minus this this magnitude
that will be the net result so initially this is 0 this is minus infinity.
So it will the net value I will show by dotted line will go like this and when this magnitude
is equal to this it will be 0 so the net value will pass through the origin at this point
when omega l is equal to 1 by omega c and then final this is tending to 0 but this is
increasing. So it will be asymptotically catching up with this okay. So this will be XL plus
Xc characteristics total x is that alright, if I take a series combination the reactance varies like this what about
the admittance.
Let us see the admittance of this, it will be 1 over this quantity okay 1 by j omega
l minus j by omega c. So how much is it if I multiply by omega c divided by omega square
LC minus 1 j can be taken out so minus j by omega c by omega square LC minus 1 okay if
you if you take small value of omega where will it tend to small value of omega, infinity
this is minus minus will go this will tend to infinity, this will tend to infinity because
omega is there omega will get cancelled and omega omega is tending to 0 it will that omega
will be cancel each other that omega will be in denominator 1 omega will be in the numerator
in the denominator is 1 denominator is 1 omega what will be the admittance function like?
What will be the admittance function like? When omega tends to 0 that is y j omega omega
tending to 0 infinity, omega tends to 0 means this this is negligible.
So this will be 1 and omega tending to 0, 0 this will be tending to 0. So it will be
0 and plus or minus will it be plus or minus plus j a small quantity plus j epsilon okay.
So it will be starting from here and as omega gradually increases it will be having a slope
very close to omega c neglect this compared to 1. So it will have a slope pretty close
to omega c I mean slope of c it will be varying like omega c and at omega tending to LC omega
square tending to 1 by LC a little less than that this will be a little less than 1 okay
say .99 omega l square LC is .99 what happens to that this will be 10ding to 0 from the
left hand side that is say minus .001.
So this will tend to and what about this this is minus j omega into c so minus j omega into
c divided by a very small quantity negative say .001 when I am approaching this frequency
1 by LC under root from the lower side that means this quantity is less than 1 in magnitude
say .9 or .99 then this is a negative very small quantity so this will be tending to
plus infinity, is not it is a large quantity. So at that frequency it will tend to a very
large quantity when I just cross over that frequency 1 by root LC what happens to this
magnitude this becomes suddenly plus and over all quantity becomes minus infinity. So it
will start from minus infinity and then when this quantity gradually increases from 1.001
to 2345, so this will be gaining higher and higher values so over all quantity will be
reducing in magnitude and at a very large value of omega this will tend to 0 so it will
1tend to 0 like this okay.
You can have 1 or 2 more combinations of this alright a series parallel combination you
will find, for example if we take a combination of 1 inductance, 1 capacitance in parallel
and then a series l c combination pardon 1 thank you. So if you take such a combination
this will be the nature of function if you put s equal to j omega if you put s equal
to j omega and then take out only the j you will get X, if you sketch X, if you sketch
the x against omega you will get at those poles the function will try to blow up to
infinity okay.
It will be like this it will be going up to infinity and when you just cross over that
pole it will start from negative infinity. So this will be the nature of variation in
both the cases I have taken a very simple order function this can be prove for any other
function the slope is always positive whether it is y or z, the slope is always positive,
this is nothing like going down like this the function is always increasing from a low
value to high value if it is a minus in the minus region, negative region then from a
very negative value it is gradually going to less negative value that means it is always
increasing whether it is y or z okay alright.
So one conclusion we can draw the admittance, the reactance or admittance versus omega characteristic
if we study the slope is always positive, poles and 0s are appearing alternately. So
hence forth we shall be indicating on the x axis which is the frequency axis when we
are making this x omega plot, this poles it was starting from a 0 value it was a pole
here then again see if it is only for xL then again a pole if you take the over all function
over all function for this was a pole this was a 0, this is a pole. So poles and 0s will
be coming alternately on the omega axis say if it is starting from 0 then a pole a 0 a
pole and so on then at infinity this will be a pole then a 0 then this is infinity okay.
This can be one configuration if I am giving the configuration for x, can you sketch the
variation of x it is start from a 0 and go to infinity with a positive slope. So what
can be the variation can it go this way, it cannot it has to go this way.
So this is for x or y either of them then from a pole, from a pole it has to go through 0
and approach again infinity from minus infinity it can go to plus infinity passing through
a 0 it cannot go this way because then it will ensure a negative slope and we have seen
the slope has to be positive. Similarly from this infinity to 0 it will asymptotically
approach 0 this can be the possible x omega characteristics for a given set of pole 0
configuration is that alright.
Let us take a function z (s) equal to s into s squared plus 9 by s squared plus 4 into
s squared plus 16 what will be the pole 0 configuration for x and how will the impedance
vary, sketch it free s is in the numerator. So this one will be 0 then 2 is a pole then
3 is a 0, 4 is a pole and then infinity has to be a 0 because last one was a pole.
So from 0 on the poles we draw dotted vertical lines and then it is very simple it will go
to infinity like this from infinity, it will go through 0 to this, then from infinity it
will go to 0 asymptotically. See this will be so this will be the impedance characteristic
of this function. Now my question is is it possible to have a function like this s square
plus 4 by s square plus 9 into s square plus 16, now you see 0, 0 then pole, then pole.
So first 0, 2, 3, 4 is not the corresponding roots are 2 3 4 can you have 0 0 pole pole,
0 0 pole and pole they should come alternately alright.
So you cannot have this as a LC network function. Now suppose you are given a function z (s)
equal to 10 into s into s square plus 9 by s square plus 4 into s square plus 16 okay.
Now how many information are given to you one is this frequency 9, the other one is
that is 3, 2, 4, 3 and one constant. So information given up 4 so I can realize this network with
minimum number of elements 4 alright. This function can be realized by at least 4 elements
if I have a network of say 1 Henry and 2 Henrys, this is not a canonic form a canonic form
means which requires minimum number of elements to represent a function how much is this total
is s plus 2s, 3s. So 3s I can realize by a simple 3 and inductive why should I have 2
inductances.
So this is not a canonic form this is a canonic form for that function 3s if I am given z
(s) is equal to 3s realize it I will go for this this is a canonic form, canonic form
means minimum number of elements required to realize that network function. So how many
minimum number of elements will be required here the information that you are giving will
be 4.
So if you are given 4 information is 4 minimum number of informations then I can have only
4 minimum number of elements alright. So this will be a network a canonic network with 4
minimum number of elements if I make partial fractions what will be the form k1 s by s
squared plus 4 plus k2 s by s square plus 16 can I have anything like k3 s plus k4.
Let us keep a general form like this. Since, the roots are all on the imaginary axis poles
and 0s are all on the imaginary axis there is no scope for any resistive element. So
z (s) with a k4 a resistance so that is ruled out okay.
If I add these what will be the numerator s square plus 4 into s square plus 16 into
k3 (s) highest order will be s to the power 5, s to the power 5 is not given here. So
this is also ruled out had I had a term like s square plus 25 then the numerator was s
to the power 4 maximum power. So then k3 (s) would have been possible but in this case
you are having a maximum degree of s to the power 3 in the numerator. So here if I add
up it will be s to the power 5, so this is not possible so k3 (s) is also not possible.
Now let us find out k1 (s) and k2 (s) what will be the residue k1 multiply by s square
plus 4 okay and divide by s. So this s will go then make s square plus 4 equal to 0, so
it will be 10 into 5s square plus 4 is 0 means 4 divided by 12 is that alright. So 50 by
12, so 25 by 6, 4.16, k2 similarly multiply by s square plus 6 divide by s make s square
plus 6 equal to 16 equal to 0. So this will become minus 7 this will become minus 12,
so 10 into minus 7 by minus 12 so that gives me 70 by 12, 35 by 6, 5.83, 5.83 okay.
So I have got z (s) equal to 4.16 s by s squared plus 4 plus 5.83 s by s squared plus 16, what
is this combination it mean an impedance function you get some thing like k (s) by s squared
plus alpha what does it mean if I make s10 into very very small value 10 into 0, this
will be absent. So it will be k (s) by alpha k (s) by alpha means k by alpha Henry alright.
Similarly, if I make s10 into infinity alpha will be knocked off.
So it will be k (s) by s square s will go so k by s, so 1 by k farads. So here I will
get 4.16 s by s square plus 4 will give me an inductance of how much when s tending to
small value 4.16 by 4. So that is 1.04 Henry is that alright that means at a low frequency
capacitor blocks the current the entire current flows through the inductance. So it becomes
an inductive circuit at a very high value of capacitor frequency, capacitance is a bypass.
So the entire current flows through this it becomes capacitive. So it will be 4.6 by s,
so s will so 1 by 4.16 farads is that alright. Similarly 5.83 s by s squared plus 16 could
you please tell me the values 1.04 by 16 when s is small it is 5.83 by 16 Henry and this
one will be 1 by 5.83 farads okay. So if I put these 2 in series I get the total impedance
function this plus this.
Okay, so whenever you start realizing the function from z (s) that is take the impedance
function break it up into forms like this k1 s by s square plus alpha plus k2 (s) by
s square plus beta and so on plus a free term k3 (s) plus a term k4 by s. In the previous
case I forget to include another term k5 by s is there any possibility of k5 by s coming
because s is in the numerator not in the denominator. So in this case that was also ruled out but
suppose in the general function s is in the denominator it is possible then you will have
a function like k4 by s also. So you have to see all possible elements of this type
either an inductor a capacitor or LC combinations if I can break it up in this form and realize
each element of this partial fraction separately put them together in a string that is called
coir sorry foster 1 synthesis, foster 1 synthesis.
Let us take another example z (s) equal to s squared plus 1 into s squared plus 3 divided
by s into s square plus 2 into s squared plus 4 and s squared plus 5 okay. So numerator
is s to the power 6 denominator is s to the power 4 so what will be the form like K1 (s)
by s square plus 2 plus K2 (s) by s square plus 4 will there be any K3 (s) term, will
there be any K4 (s) term yes, K4 by s will be present because s is in the denominator
K3 into s will be present because here it is s squared, s squared, s into s to the power
6s to the power 6 is present here.
So for this type of function all the 4 elements will be present realize after getting the
value of K1, K2, K3, K4, what will be the network like a capacitor of value 1 by K4
farad. Okay an inductor of K3 Henry and then terms like this K2 (s) by s squared plus 4
will give me a tank circuit a parallel LC combination, how much is this when s tends
to 0, K2 by 4 Henry is that alright and when s tends to infinity it will be 1 by k2 farads
okay. Similarly, this one k1 by 2 and 1 by k1 farad. This will be the network is that
okay any question if I have z (s) like this all the possible terms are there can I have
another inductance or another capacitance in series that will not be a canonic form
because in series you cannot have more than one capacitor in the canonic form, another
capacitor put up and add it together. I could have got an equivalent value another inductor
could have been added but these cannot be added.
So each one of them will give rise to 1 pole corresponding to each pole like s square plus
2, s square plus 4 you will have one LC parallel combination. Since they are distinctly different,
so you will have distinctly different combinations but series combinations there can be only
one K3 and K4 I cannot have another k4 by s then I could have put them together k4 by
s and k4 by s k4 by s together alright. So 0s can be different in the network no I did
not mean that whatever be the given network function in the canonic form you can have
1 series element of inductance at the most 1 capacitance at the most and parallel combinations
of LC okay, each 1 giving rise to 1 pole and depending on the order of the numerator polynomial
how heavy it is compared to the denominator, if the denominator is 4, s to the power 4,
numerator is s to the power 6 and if you divide you will always get a free s term that means
there is a series inductance, if s is in the denominator there is a series capacitance
also alright. So either this will be absent or this will be absent or both may be absent
or both may be present, so what will be the number of elements of l and c how much can
be the difference, see how many inductance is at there 1, 2, 3 how many capacitances
are there 1, 2, 3. Suppose this is not there this is also going to give me another function
z (s) this is one possibility. So how many inductance is at there 1, 2, 3
capacitances 2, difference is 3 minus 2, 1. I can have the inductance absent then how
many capacitances are there 3, how many inductances is at there 2, can you have 2 inductance and
4 capacitances, 2 inductances and 4 capacitances, no sir pole will be increasing, no let us
see the structure can be always like this I can be all get for any z (s) function a
structure like this now suppose 4 capacitances 1 I can put here 4 induct 2 inductances so
one inductance I can put like this another inductance I can put like this. Now 3 capacitances
are already gone where can I put the forth capacitance still it will be a canonic form
it is not possible if I put a capacitance here I could have added them together I cannot
put a capacitance here then I could have added this together.
So I cannot have more than 3 capacitances if there are 2 inductances to get a canonic
form that will not be a unique function if I put a capacitance here I find the overall
z (s) and that z (s) I can realize suppose you find out, suppose this is 1 farad this
is 2 farads and so on you find out z (s) and again from there if you try to realize you
will be able to get a function like this that means I can realize the same network function
that you are giving me by less number of elements. So that was not a canonic form so the difference
in the number of elements l and c cannot be more than 1 they can be equal, l can be greater
than c or c can be greater than l but then the difference can be only one, is that alright.
This is foster 1 synthesis if instead of that we would have started with an admittance function.
So if the same function z (s) is given I can realize it in terms of admittances then what
happens so I will just invert it y (s) will be s into s square plus 2 into s square plus
4 divided by s square plus 1 into s square plus 3 into s square plus 5 again I make parallel
okay I make a partial fractions so it will be a 1 s by s square plus 1 plus a2 s by s
square plus 3 plus a3 s by s square plus 4 okay will there be a4 s and a4 by s. Since,
s is not in the denominator this is ruled out since it is already s to the power 6 and
numerator is s to the power 4. So if I put this then it will be s to the power 7, so
this is also ruled out okay so I will get a1 s by s squared plus 1. Let us call these
as Y1 plus Y2 plus Y3 okay so what will be Y1, Y1 is say 2 s by s square plus 1, suppose
it is 2 s by s square plus 1 what does this mean if I make s10 into 0 it will be going
to 2 s so when is the admittance becoming 2 s admittance is c s. In case of admittances
c s is the admittance, so it is 2 farads, so it will be a 2 farad capacitance at that
time at a very low frequency inductance is 0 it is a series combination, you see if Y1
is this corresponding z1 is how much s square plus 1 by 2 s that means s by 2 plus 1 by
2 s. So half Henry inductor and 2 farads capacitor will given me this impedance whose admittance
is this. So each admittance you just invert and then separate you will get the corresponding
LC values is that alright. So similarly, you get another LC combination another LC combination
okay interestingly this values will be different from the foster realization foster 1 realization
but how many inductances were there 1, 2, 3 and 3 capacitances you go for this realization
you will get the same number of elements type of elements will be of the same number 3 inductances,
3 capacitances. So if we go from impedance it is foster 1 if you start from admittance
function this is foster 2 realization okay we will stop here now, we will continue with
this in the next class.
Okay good morning, last time we are discussing about foster 2 realization of network functions. Suppose you are given
z (s) and you have obtained y (s), y (s) you say s into s squared plus 4 divided by s squared
plus 1 into s squared plus 16 okay. Then I can write this as k1 by k1 (s) by s square
plus 1 plus k2 (s) by s square plus 16 how many elements will be there, can you guess
how many elements will be there, how many informations are given to you I could have
add a constant here 10 or so.
So this is one information 1, 2, 3, 4 so you can have only a 4 elements synthesis, now
each of these factors will take away 2 elements. So 2 plus 2, 4 elements are already gone alright.
So blindly you can write there is no other term k4 s or k5 by s okay. So how much is
K1 multiply by s square plus 1 divided by s make s square plus 1 equal to 0. So 10 into
3 by 15 so that is equal to 2 K2 so 10 into 12 by 15 make a square plus 16 equal to 0,
so this will be 12 into 10 by 15, 8. So I can write this as 2s by s squared plus 1 plus
8 s by s squared plus 16 okay. So this is Y1 (s) plus Y2 (s) so how much is Y1 (s) what
are the values of l and c corresponding Z1 (s) Y1(s) gives me Z1 (s) which is just inversion
of this s squared plus 1 by 2s.
So that is s by 2 half Henry 1 by 2s, so 2 farads is that okay another combination Y2
(s) how much will be the element values of y2 (s) invert it s square by 8, s that is
s by 8, 1 by 8 Henry and 16 by 8s, so 2 by s that will give me half a farad parallel
combination of r l elements, is that alright 17,16, 17, 7 by 32, 16 into 232, a 7 by 16,
7 not 17 sorry, not 17, thank you very much, it is 7, 7 by 16 ohms resistances and 7 by
32 Henry inductance. So if you are having in the impedance function the first one, first
factor as 0 then it will be an r l network that means admittance function of rc network
and impedance function of rl networks are identical, conversely impedance of rc network
and admittance of rl network are identical, poles and 0s thank you poles and 0s will be
alternately coming.
If in the impedance function it starts with a pole it is an rc network if it is in the,
if it starts with the 0 it will be an rl network and then the method of partial fraction is
identical alright. The partial fraction that you have made for admittance of rc network
will be similar to impedance of rl network, is it not. So in the next class we will be
taking up a some tutorial exercise on synthesis by different techniques of different rl, rc
and rc, lc networks. Thank you.