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Good afternoon friends. Today we shall be discussing about synthesis of RLC network,
when you say synthesis you are still restrict to ourselves to 2 terminal network that is
1 port network. Now before we go to the synthesis will define some functions special type of
functions, minimum function, what is a minimum reactance function? First, we define a minimum
reactance function. A minimum reactance function we get when an impedance function has got
no poles on the imaginary axis that is an impedance function having no poles on the
imaginary axis that means in Z(s) factors of this type will be ruled out, this type
of factor there are other factors here this type of factor is not present then in it is
a minimum reactance function.
Similarly, if you have in the admittance function factors of this type absent that is there
are no poles on the imaginary axis for the admittance function then we call it minimum
susceptance function okay. If we have the real part of Zj omega equal to 0 at some frequency
at some omega equal to omega 1 then we call it minimum resistance function okay, minimum
resistance function.
Similarly, if you have real part of Yj omega equal to 0 at some omega equal to omega 1
we call it minimum conductance function, minimum conductance function. If a function is simultaneously,
minimum reactance, minimum susceptance, minimum resistance minimum conductance, then we call
it a minimum function.
Now there are certain specific properties of the minimum functions, we will just mention
about the essential features of minimum function the first one is say Z(s) is finite at s equal
to 0 and infinity at s equal to 0 and infinity, it has no poles or 0s on the imaginary axis,
no poles or 0s on the imaginary axis because it is finite at 0 and infinity obviously if
I write Z(s) equal to P(s) by Q(s) both of them will be of the same order, same degree,
say this is S cubed plus twice S square plus something this also has to be S cubed if it
is S to the power 4, if it is of higher degree then S tending to infinity at s equal to infinity
this will be 0 and Y(s) will tend to infinity. So similarly if this is of higher order at
s equal to 0 this will tend to 0 but at S equal to Y(s) at s equal to 0 will tend to
infinity.
So it will be of the same degree these 2 polynomials then thirdly a real part of Z(s) is equal
to 0 at some omega, say omega 1 that means if I sketch the real part
may be like this. So at some omega 1 this will touch 0 value what does it physically
mean that means the real part shows the active power consumed I square into the real part
is that active power consumed at some frequency that active power will be 0 for a minimum
function.
Let us take an example, let Z(s) be given by say twice s squared plus s plus 1 by twice
s square plus 2s plus 4 then what would be a real part of Zj omega we take m1, m2 minus
n1 n2 if you remember twice s squared plus 1 is the even part multiplied by the even
part twice s squared plus 4 minus the odd parts product divided by even square, even
part squared for the denominator minus odd part square okay.
So that gives me twice s square plus into twice s square 4 s to the power 4, so 4 omega
to the power 4 okay 4 2's are 8 plus 2, 10 minus 2, so plus 8 s squared means minus 8
omega squared plus 4 divided by 4 omega 4 twice s squared so 4 s to the power 4, 4 omega
4, 4 2's are 8 minus 4 so sorry, 4 2's are 8, 2's are 16 minus 4, so plus 12 s squared
that gives me minus 12 omega squared plus 4 4's are 16.
So this can be written as 4 can be taken out so omega to the power 4 minus twice omega
squared plus 1 by omega to the power 4 minus thrice omega squared plus 4. So clearly you
see this is equal to omega square minus 1 whole squared, so this will vanish at omega
squared equal to 1 okay.
So when omega square is equal to 1 real part will be 0, real zj omega equal to 0 at omega
square is equal to 1 that is omega equal to 1, what would be Z(s) like then then Z at
omega equal to 1 how much is this. Let us see 2 into s squared means 2 into minus 1
plus j1 plus 1 divided by minus 2 twice s squared s square is minus 1, so minus 2 plus
2 into j1 plus 4 correct if I am wrong so minus 2 plus 1 minus 1 plus j divided by 2
plus 2 j is that okay so that gives me minus 1 you see minus 1 plus j that is magnitude
of root 2 and angle of 135 degree is it not and 2 plus 2 j2 plus 2 j is 2 root 2 and an
angle of 45 degrees so 135 minus 45 so that gives me 90 degrees.
So j, so that gives me 1 by 2j, so it is purely reactive the impedance function is purely
reactive that was what we expected also the real part will vanish there is no real part
at this frequency now this will be the starting point of a realization.
Now we shall follow first the method suggested by Brune, Brune's synthesis starts from this
point that means Z(s) I can conceive as at that particular frequency it is half j is
that okay so this is plus half j at omega equal to 1, so if this is the reactance can
we suggest an inductor inductor value which will give me half j as the reactance value
at omega equal to 1. So let us call that as L1, L1 into j1 is equal to half j j omega
so L1 is half Henry. So let us take a half Henry inductor out of Z(s), Z(s) I am trying
to realize as a reactance reactance which is inductance half of half Henry plus another
reactance say Z1(s) how much is Z1(s) therefore Z1(s) will be Z(s) minus this L1(s) which
is Z(s) is twice S squared plus S plus1 divided by twice S square plus2 S plus 4 minus half
s is that okay
Now at this point you just see what would be the property of Z1(s) what would be the
value of Z1(s) at s equal to j1 that is at omega equal to 1 what would be the value of
Z1(s) Z(s) is purely reactive half j and this is also half j. So this will be 0 at that
frequency so Z1 (s) let us work it out let us see what it gives if you write it this
is S squared plus S plus 2 okay simplify this twice S squared plus S plus 1 minus twice
S cubed minus s cubed s into s cube minus s square minus twice s okay.
So that gives me twice s and s that will give me minus s, minus s cube plus s square minus
s plus 1 which gives me finally you can factorize this s cubed and s if you take together that
will give me s square plus 1 as a factor, so minus s plus1 okay. So obviously s square
plus 1 this factor at s equal to j omega at s equal to j omega [Noise] s square plus 1
will be equal to0 okay so that was what we expected Z1(s) will vanish and that particular
frequency how do you remove this particular 0 then, you create a pole if you remember
in a earlier synthesis you create a pole and then remove it this becomes a 0 at that particular
frequency s equal to j
So we take the corresponding admittance Y1(s) as 2 into S squared plus S plus 2 divided
by s square plus 1 into 1 minus s at this moment let us not bother about the sign it
does not represent a positive real function at this moment but we will take care of that
in course of time we will see what it means.
So Y1(s) can be I can make partial fractions written as K1(s) by S square plus 1 plus some
K2 by 1 minus S, let us see what K2 will represent how much is K1, K1 comes out as let us calculate
the value of K1 multiply by s square plus 1 make s square plus 1 equal to 0, so that
give me 1 plus s and divided by s divided by s, so s minus s square so s minus s square
and s square is equal to minus 1 so that gives me 1 plus s divided by s plus 1, so that is
equal to 1, there is a 2 here know. So there will be a 2 here, so it is half.
So K2 similarly if you multiply by 1 minus s and then make 1 minus s equal to 0 that
is s equal to1 so this will be 1 plus 1, 2 plus 2, 4 into 2, 8 divided by 2 that is equal
to 4. So you can write Y1 S as s by 2 into s square plus 1 plus 4 by 1 minus s. So what
does it mean I have got first L1 removed then I have got a Y1 and a Y2 sorry these I will
call it Y2 plus Y3 this whole thing is Y1, so this is Y2 this is Y3 okay.
So what is corresponding Z2 this shunt term will be just inverse of this 2 into S squared
plus 1 by s that gives me twice s, check the value of K1, K1 I worked out 2 into s squared
plus s plus 1 how much is this is Z1, how much is Y12 into s square plus s plus 2 by
s square plus 1 is that so I forgot to put the values correctly K1 is equal to if you
remember K1 I multiplied by s square plus 1 make s square plus 1 equal to 0 so 2 should
be in the numerator. So it will be sorry 2 I was by mistake I was looking at Z1 S so
this will be 2, so this will be 2 s so S by 2 plus 1 by 2 S that means this branch will
be nothing but an inductor and a capacitor okay. So this is half Henry this is 2 Farads
and this was also half Henry okay, this is what we got.
Now obviously whatever be the part here you see this will start resonating at that particular
frequency s equal to j1 that is at omega equal to 1 this will resonate, so that will be a
short. So the entire impedance Z1(s) that is what we are saying Z1(s) will be 0, so
Z(s) the impedance values of Z(s) which was half j is the impedance of this itself, this
impedance is 0 that time because it is resonating at that frequency.
So Z1(s) is having a 0 at that point and that is been realized like this what is this Y3,
so from there let us see Z3 what will be Z3, 1 minus s by 4 so that is 1 by 4 ohm resistance
minus s by 4 that is I am having an inductance of minus 1 by 4 Henry and a resistance of
1 by 4 ohms resistance okay, is that all right what is this minus1 by 4 Henry, how do you
realize minus 1 by 4 Henry?
Now Brune suggested you have an inductor of half Henry another inductor in the form of
a t, half Henry and this 1 is minus 1 by 4 Henry okay these 3 can be replaced by a transformer
an ideal transformer where the primary inductance is these keep it open. So that will give me
the primary inductance half plus half 1 Henry okay then this side half and minus 1 by 4
half minus 1 by 4 is 1 by 4 Henry. Add a mutually inductance of half Henry and then have the capacitor, resistor, how
do you realize this what is the co-efficient of coupling that is equal to m by root over of LP, LS
or L1, L2 and that is here half Henry divided by 1 into 1 by 4 that is equal to 1.
So if you are having an ideal coupling, 2 transformer coils are ideally coupled and
they are having the values1 Henry and 14th Henry then you can realize this network. So
connect the 2 transformer terminals here and then connect the capacitor to this end and
the other ends are taken out like this. So this is the configuration of a Brune's network
where a negative inductance has been incorporated in the coupled coils. So if one of the terms
is negative do not get frightened it can be realized I could have got right in the beginning
in instead of plus jx at that frequency where the resistance part the resistive part was
vanishing the real part was vanishing at that time the impedance of Z(s) where we got half
j could have been negative.
Then we will start off with a negative inductor a minus half j it will not try to realize
it by a capacitor you can try it at home if you try to realize it with a capacitor, what
kind of difficulties you face later on, you can try with another example, we have many
examples from in a many problems given in the book or we will take up in the tutorial
class while realizing some of them we may find a negative sign here.
So there also the same logic we got. Now there is another method of realizing this that is
Bott and Duffin synthesis we shall be discussing about the synthesis techniques and then we
will take up a large number of problems in the tutorial class. So Bott and Duffin synthesis
is a very interesting synthesis technique
before we go to that we define another interesting positive real function Richard function which
is that if Z(s) is a positive real function then for positive real values of K you take
any value of K positive and real values of K, K into Z(s) minus s into Z(k) divided by
KZ(k) minus s Z(s) is also a p r f, R(s) is also a positive real function, this is called
Richard function.
See if R(s) is positive real then let us extract Z(s) from here write Z(s) in terms of R(s)
you multiply the denominator by R(s) and then take Z(s) terms on one side so Z(s) comes
out as KZ(k) into R(s) plus s Z(k) divided by s Z(s) R(s) and K Z(s) it will be K plus
s R(s) okay. Now we can break it up into 2 parts KZ(k), R(s)divided by K plus s R(s)
plus s Z(k) by K plus s R(s) okay. If you divide the denominator by the numerator then
K by KZ(K) so I can write this as1 by Z(k) into R(s) plus s by KZ(k), R(s) will get cancelled
plus similarly here K by SZ(k) plus s will get cancelled R(s) by Z(k), correct me if
I am wrong. Now let us see what will be the structure of this impedance function this
is 1 by 2 admittances, so this is admittance, this is an admittance. So I can write like
this what is1 by KZ(k) R(s), so the first 1 can be written as 1 by 1 by Z1, so Z1(s) plus
what is this admittance see Z(S) is 1 by this admittance this is 1 Z1, this is Z2 and Z1
I am writing as some of 2 admittances inverse of that. So this is an admittance of what
a capacitive element so a capacitor of value 1 by KZ(k) Farads is that all right.
Similarly, this one K by SZ(k) so this can be written as 1 by Z(l) plus 1 by Z2(s) where
Z(l) will be this inductance. So Z(k) by K Henry that will give me K by SZ(k) as the
admittance, is it not and 1 by Z2 means this will be Z2 okay where, Z2 means I will write
Z1(s) means Z(k) into R(s), Z2(s) equal to Z2(s) is R(s) by Z(k) okay, Zc we have already
seen Zc is KZ(k) by s and Z(l) is s into Z(k) by K okay s into Z(k) by K.
So let us start from this point where we stopped at we initiated for Brune's synthesis, we
start from there for Bott Bott Duffin synthesis also that is Z(k) by K is that inductance
value for ZL2, so if it is positive, if it is positive then Z(k) by K we take as that value of L, what was that value jx at
omega 1 by omega 1 okay. If you remember we will not put j because that will be divided
by j omega 1, this was taken as L1 is it not. So we chose that value of L1 as Z(k) by K
because Z(k) by K will give me that inductance value okay, if it is positive, if it is positive
then we will start realization from here otherwise we will start from here.
So x omega 1 by omega 1 is L1, so that gives me K if you get say half L1 is equal to in
the last example we got half then solve for K. So let us take up that same example we
will try from there how much was L half that was equal to half, so how much is Z(k) by
K, Z(k) will be K twice K square the polynomial was twice s square plus s plus1 divided by
twice s squared see I am putting in place of s, I am putting K, so twice K squared plus
twice K plus 4 all right this is Z(k) ,so Z(k) by K so this divided by L is equal to
half okay. So Z(k) by K equal to this divided by K and that has to be equated to half all
right.
So let us solve twice K cubed plus twice K squared plus 4 K equal to 2 into 2 K square 4 K square plus
2 K plus 2 okay. So let us bring all of them to this side twice K cubed, twice K square
minus 4 K square, so minus twice K squared plus 2 K minus 2 equal to 0. So K minus 1
gets common to 2 K squared plus 2 okay equal to 0, one root will be imaginary, the other
one will be K equal to 1 so we will take the real value of K that is K equal to 1 so once
you know K equal to1 how much is Z(k), Z(k) by K is equal to half, so Z(k) K equal to
1.
So Z(k) equal to half okay so if you know Z(k) is equal to half K is equal to 1 calculate the Richard function
because we need for calculation of Z1 and Z2 this R(s) function all right. So let us
calculate R(s) what was Richard function? R(s) was given as K into Z(s) minus s into
Z(k) divided by KZ(k) minus SZ(s) okay K equal to 1, so 1 into Z(s) we will substitute Z(s)
afterwards minus s into Z(k) is half divided by K into ZK1 into half minus s into Z(s)
okay, you have to expand put Z(s) and expand you will get R(s).
So let us put those values Z(s) twice S squared twice S squared plus S plus 1 by twice S squared
plus 2S plus 4 minus s by 2 divided by half minus s into twice S squared plus S plus 1
by twice S square plus twice S plus 4. So this is R(s), once you have got R(s) Z(k)
into R(s) is Z1 SZ(k) into R(s) is Z1(s) and Z2(s) is R(s) by Z(k), so how much is Z1(s),
Z1(s) is R(s) into Z(k) and Z(k) we have got already the value of Z(k) as half, so into
half Z2(s) R(s) by Z(k) am I correct Z2(s), Z2(s). Let us see Z2(s) I hope I have not
made any slip one upon R(s) by Z(k) there was a small slip Z1(s), Z1 is Z(k) into R(s) and Z2(s) should
be actually this is 1 by Z2(s), so Z(k) by R(s), is it not Z(k) by R(s). Now this is
all right, so Z2(s) is is Z(k) by R(s) and that is equal to half divided by R(s).
So R(s) let us complete this 2 into S squared, so 4 S squared plus 2S plus 2 minus twice
S cubed minus 2S sorry, it should be 2 goes so S cubed minus S squared minus 2S divided
by 2 into this and sorry, divided by this and this will also get cancelled. So let me
work out here, here it will be S, S squared plus S plus 2 minus twice S cubed minus S
squared minus S if you simplify this R(s) will come out as minus S cubed minus S square
plus 4 S square. So minus 3S squared sorry, plus 3S squared 2S and minus 2S will go is
that so plus 2 I am getting some complicated figures please check there should be 2S square plus,
I have made a slip here while computing this it should be 2S squared because 2 is a common
factor 2S squared plus S plus 1 in a, you complete this you can complete this any way R(s) will be
whatever is the simplified form I will just work out separately there is a lot of mix
up, what I wanted to stress is once you have found out R(s) you have already got Z1(s),
Z1(s) will be R(s) by 2 and Z2(s) will be half R(s). So the impedance function here
and the admittance function here will look alike okay.
We will complete this problem in the next class time is running out the other 2 parts
are there is a inductor here and is a capacitor here. So this is a kind of network you are
getting these 2 are to be realized let me work out the value of R(s) here. R(s) twice
S squared plus S plus 1 minus s into S square plus S plus 2 s into S square plus S plus
2 divided by so S squared plus S plus 2 minus s into twice S squared plus S plus1 is that
all right. Check twice minus s cubed minus s squared plus twice s square plus s squared
minus 2 s plus s minus s plus 1 divided by minus twice S cubed S squared and minus S
squared gets canceled plus S and minus S that will also get canceled, so plus 2. So that
gives me in the numerator 1 minus s into s square plus 1, check numerator and this is,
this is 2 into 1 minus s cubed which is nothing but 1 minus s will get cancelled. So it will
be 1 plus S squared divided by 2 into S square plus S plus 1 okay so this is R(s) all right.
What is therefore, what is our Z1 R(s) by 2 if you see Z1 is R(s) by 2, so Z1(s) is
1 by 4 s square plus 1 divided by S square plus S plus 1 if I invert it corresponding
Y1(s) it will be S square plus S plus 1 by S square plus 1 into 4 this is nothing but
S square plus 1 by S square plus 1 will give me 1, so 4 into 1 plus s by s square plus
1. So that is 4 mho plus 4 S by S square plus 1.
So this is our familiar elementsY1 plus Y2, Y1 is 4 means 1 by 4 ohm resistance and this
is Y2. So if I invert it how much is this, if I invert it 1 by 4 Henry and 4 Farads,
1 by 4S so 4 Farads what about Z2(s) Z2(s) was R(s) half Z(k) by R(s) that is half divided
by R(s) so half divided by RS so we can write R(s) was 1 plus S squared, so it will be inverted
2 into S square plus S plus 1 divided by 2 into S square plus 1 so 2 will get cancelled
it will be 1 plus S by S square plus 1, Z2 is this so z1 plus Z2 I mean Z2 can be written
as 2 sum sum of 2 components 1 is 1 ohm, 1 ohm resistance other 1 is 1 Henry and 1 Farad,
capacitor, so this is Z2 okay.
So now you come to the final realization Z1(s) was in parallel with the capacitor okay. So
you have final Z(s) has a capacitor in parallel with that Z1(s) that is one forth ohm resistance
and network like this and then an inductor and this one is resistance. So this is the
structure of Z(s) the number of elements required here will be more you see it is no more canonic
in that sense because we are getting a negative inductor in case of Brune's synthesis we coupled
it with a transformer the number of elements there we got much less here it will be 1,
2, 3, 4, 5, 6, 7, 8 never the less this is easier to implement there is no coupling involved
okay. So thank you very much we shall continue with this with a few more examples in the
next class.
Good afternoon friends. Today we shall be discussing about parts of network functions.
Now before we go to that I will just briefly summarize what we did yesterday in the Bott
Duffin synthesis then we will go to parts of network functions. So the Bott Duffin synthesis
we defined the Richard function which is a positive real function in terms of Z(s) as
KZ(k) minus SZ(s) if Z(s) is a positive real function then all for all positive values
of K, R(s) will also be a positive real function from there we wrote Z(s) in terms of R(s)
as KZ(k), R(s) plus SZ(k) divided by K plus SR(s) and this we broke up in this form if you remember in the form of 2 parallel
elements put in series with another set of this type K by SZ(k) plus R(s) by Z(k) okay.
So this was shown to be like this this is the admittance of a capacitor is that all right. So this is this admittance
of a capacitor so 1 by KZ(k) Farads will be the value of this capacitance this is say
this is some Z1 then in parallel with this capacitor this impedance Z1 will come.
Similarly this one is the admittance of a reactor whose value is Z(k) by K so many inductance
in parallel with that we have Z2 which means Z(k) by R(s) is that all right. So this is
what we wrote yesterday now Z1 to be calculated Z1 is Z(k) into R(s), R(s) being a positive
real function Z(k) is having a constant value for a particular value of K, real value so
this is an impedance function, realizably impedance function. Similarly, Z(k) by R(s)
is the other impedance function you can see one is Z(k) into R(s) the other 1 is Z(k)
by R(s).
For example, if you have a complex pair like this for m1 plus sorry m2 plus n2, 1 pair
of roots is like this then for m2 minus n2 correspondingly you have roots here mirror
images. So you get quad of points for each complex root set you get actually quad of
points for the function into m2 square minus n2 square okay.
So problem is simplified you are given m2 squared minus n2 squared in terms of omega.
So make a substitution omega square is equal to minus S squared then factorize it then
chose the factors corresponding to the left of plain roots that will give you m2(s) plus
n2(s) why left of plain roots because original function Z(s) is a positive real function
which should satisfy Rowther witz criteria that means all the roots of the numerator
and the denominator must lie in the left of plain. So m2(s) plus n2(s) corresponds to
the function Z(s) it is the denominator of Z(s) is it not which is a realizable function,
so the roots are here are laying in the left of plain so you select out the roots corresponding
to the left up plain.
Once you have done that then m2(s) plus n2(s) is defined m2(s) plus n2(s) is selected. Now
you chose Z(s) equal to suppose this equal to some b0 plus b1(s) plus b2(s) square suppose
it is like this may be b0(s) to the power 3 b0 plus b1(s) plus b2(s) square plus b3(s)
to the power 3 then what will be the numerator like it can have a 0 plus a1(s) plus a2(s)
squared plus a3(s) cubed okay. Any question? The condition for a positive real function
is the power of S should differ at the most by 1, can I take a4 S to the power 4 see by
division there will be a free S term that means some reactance element, additive reactance
element will come, it can be anything that becomes indeterminate, is it not. As I was
telling you right in the beginning I can identify from that only x corresponding to the minimum
value any additional x will also give me the same real part, real part remains unchanged
if you keep on adding reactances.
So we will consider only that reactance function which is just just required to realize that
S, n1 m2 okay this part divided by m2 square minus n2 square, m2 square minus n2 square
is already known that is 1 plus omega to the power 6 and so on like that. So it is this
part which has to be computed okay from that a0, a1 all right in terms of a0, a1, a2 we
can calculate this and substitute once you have calculated this co-efficient substitute
there we will get that all part.
Otherwise, in Z (s) put S equal to j omega take out the real part and the imaginary part
that will giving you x okay. So we will stop here for today, we will take up a few more
problems in the next class along with problems of RLC synthesis, thank you very much.