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Good morning friends, last time we were discussing about LC synthesis we got struck in a little
problem. See we had z11(s) those are slip we made, we had something like this S squared
plus 16 plus K3 by S.
So while removing the pole partially we are concentrating on K1(s) instead of that it
should have been in K3(s) okay and from there we will get the value of K1 dashed corresponding
to the one that is to be removed K3 dashed okay and we will get the pole shifted. So
I would leave it to you for completion of this after that it is a very straight forward
method. We will take up today RC networks all right, RL is not used so often because
of the imperfection in the inductance L.
So RC is a very commonly used network so will concentrate on RC network before you go to
RC network in details you would like see the properties of Z(s) and Y(s), some of the properties
of Z(s) and Y(s) which we did not study earlier. We will see what would be their plot in the
S domain. Suppose you consider the Z(s) function it may be s plus 2, let me take a very simple
example, S plus 5 divided by s into s plus 4 the pole is at the origin so 2, 4, 5, what
would be the impedance diagram, Z, sigma, if I put S is equal to sigma that is minus
2, minus 4, minus 5 these are the poles in 0s and if I put any other value of sigma what
would be the nature of variation at least near this poles and 0s. It will be a pole
here passing through 0, this is a pole. So it will be from infinity passing through 0
will be going to infinity okay again from here will be going like this at S equal to
infinity it is finite okay.
Let us see if you put S is equal to infinity it is a s squared by s squared that is 1 in
this particular case it is 1 and had there been a constant here it would have been that
constant, so this is Z infinity at S equal to infinity, why should the slope be negative,
slope should it be negative. Let us find out d, d sigma of see this can be written in a
general term some Ki by S plus pi okay summation of this pi may be 0, in that case the pole
is at the origin otherwise, the poles are at points like4 there can be many more poles.
So this is a general term, so it will differentiate any of the general terms and see what will
be the nature S equal to sigma okay we are putting S equal to sigma.
So we will find this will be giving me minus Ki by sigma plus pi whole squared which will
be always giving me a negatives slope all right, this is a respective of the value of
sigma. So here it has to pass through 0 with a negative slope, so the curve has to be this
way all right. So this is the nature of Z in case of an RC network and there been a
pole somewhere other than the origin and so on, what will be the nature of this? This
is a pole all right, this is a pole, this is a 0, this is a pole okay, this is a 0,
what about this one? So it may cross at a point Z at 0 all right. Let us have a function
like Z(s) equal to so this a minus 1 here so S plus 1 into S plus 2, S plus 3 I am just
obviously taking some points okay.
So this is 1, this is 3, this is 2, this is 5, you may have a have a function like this,
what will be Z0 if I put a S equal to 0, so it is 10 by 3, what is Z infinity. So Z0 is
always greater than Z infinity are you convinced this is there any logic for it when I put
S is equal to 0, see these are poles and 0s starting with a lower value a pole first that
means 1 then the 0 corresponding 0 will be at higher point so 1, 2 the ratio is always
greater than 1, 0 is always above the pole all right here it is 5 by 3 and so on.
So Z0, Z0 will be always more than Z infinity it can be under certain situation okay may
be equal to, s plus 3, s plus 2, s plus 6, s plus 1 okay let us see S plus 3 into S plus
2 then the poles and 0s are not interlacing it will be an RLC circuit it will not be a
RC circuit. So because of the property that RC networks will have poles and 0s coming
alternately starting with a pole that means a lowest value is a pole then a 0 then a pole
a 0, 0 is always at higher point than a pole. So the ratio is always greater than1 so the
overall ratio is more than 1 and if you take a S10 to infinity that gives me Z infinity
which will be a S squared by s squared that is 1 at the most all right otherwise it will
be 0, no for an RC network it cannot be otherwise, it cannot be otherwise.
So Z infinity is 1 in this particular case otherwise if you have a K both of them will
be multiplied by K okay. So this is an important conclusion Z naught is always greater than
Z infinity in case of an RC network. For Y(s) it will be just the other way round, so for
Y(s) for an RC network Z0 sorry Y0 is less than Y infinity and the slopes will be positive
okay. So if we take a take an admittance function it will look like this. So Y(s) is once again
I can put it s minus 1 minus 3 okay s plus 4.
So this is an admittance function of an RC network okay, what will be Y0, 3 by 10, 3
by 8, sorry 3 by 8, 3 by 8 so will be somewhere here then from here and so on then at S equal to infinity it is
1, this is much above this point. So how will it move, it will be going to infinity like this asymptotically
it will go to Y infinity that is equal to 1 is it all right.
Now suppose there are these following steps of removal of poles okay part removal I can
have a partial removal of a constant from Z if you are given the specification in terms
of Z(s) you remove a constant Z(s) which will be some Kp times Z at infinity okay. So for
Z, for Z, Z infinity was like this okay.
So some fraction of it I can remove, I do not remove Z infinity totally, what is Z infinity?
Z infinity what will it give me here. Now in this particular example it is 1 if I make
partial fraction of this, it is K1 by S plus K2 by S plus 4 plus a constant K3, so at S
equal to infinity this will vanish, this will vanish, it is this term all right, it is this
term which is partly removed that means whatever is that constant I do not remove it totally
I retain a part of it all right. So if I remove a part of it what happens to the 0s, what
happens to the 0s. So let us go back to that very example, Z(s) is s plus 2 into S plus
5 divided by s into s plus 4. So if you make a partial fraction K1 by S plus K2 by S plus
4 plus K3 that gives me 1 plus 5 by 2S plus 1 by 2 into S plus 4, you can calculate K1,
K2, K3 as 1, 5 by 2 and half all right.
Suppose I remove only a part of it half of it 50 percent then what will be my new Z(s)
is half plus some Z1(s), what will be the Z1(s), it will be the balance half plus 5
by 2 S plus 1 by 2 into S plus 4 okay. Now can you see the result and that gives me just
check up your calculations 1 by 2S, so S into S plus 4, S squared plus 4, S plus 5 into
S plus 4, so5 S plus 20 plus S, so that gives me 5 plus 4, 9 plus 1, 10S, S squared plus
10S plus 20 by 2 S into S plus 4 okay. So that gives me minus 10 plus minus root over
10 square minus 80, so by 2 so that gives me minus 5 plus minus root 20 by 2, root 20
is how much root 5, so 2.24 okay.
So it is s plus 7.24 into approximately s plus 2.76 divided by 2s into S plus 4. So
what will be the pole 0 configuration earlier we are having a pole okay at 2 there was a
0, at 4 there was a pole, at 5 there was a 0. Now this is for Z(s) I am not drawing the
entire sketch I am just showing the pole 0 configuration and at Z1, now we have got a
pole here a 0 at 2.76 here, pole here and a 0 at 7.26 you see the shift in the 0 positions,
is it all right. So if you weaken the what you have done is, you have partly removed
this, what is this part removal part removal of Y at infinity, a Z at infinity, Z at infinity,
Z at infinity this has been partly removed.
So if I remove it partly this is the intersection of this is now taking place here, intersection
of this is taking place here. So this is the new shifted position of the 0s all right if
I remove it partly then if I draw line these are the new 0s okay, not convinced Z1(s),
let me write Z(s) minus Kp into Z at infinity you all agree. So we are suppose this is the
new location of the 0 at that location Z at that location new 0, I call it z1(s), z1,
z2 okay and so on this is 0, so this is 0, so Z at that point should be equal to Kp into
Z at infinity. So Kp into Z at infinity, Kp into Z at infinity
is this that value should match with this function at various points at those 0s, so
this is the shift wherever this condition is satisfied those 1s will be z1 or z2 or
z3, so these are the z1’s z2’s etcetera, is it all right. So this is the condition
to be made to determine what should be the value of Kp to get a particular location of
z1 or z2 that means the desired 0 the shift in the position of the 0 at the desired position
at z1, z2 we can achieve by equating z at that point equal to Kp into Z at infinity.
So that gives me the values of Kp the fraction of this to be removed okay if instead you remove okay we had once again
I will start from the same function Z(s) is equal to 1 plus 5 by 2S plus 1 by 2 into S
plus 4.
Suppose we get this one partly this pole what happens so Z1(s) will be 1 plus suppose we
remove 50 percent of it. So 5 by 4 retain plus 1 by 2 into S plus 4 okay if you take
50 percent of it 5 by 4 S and remaining 5 by 4S is here. So it will be 4S into S plus
4, 4 S plus5 plus 2, correct me if I am wrong 4 into S plus 7 by 4 divided by 4 into S into
S plus 4, so 4 get cancel, cancelled. So once again if we sketch the poles and 0s, 4S plus
sorry this totally wrong yes, this is 4S into S plus 4 thank you, this is all right, now
nothing is right okay. Let me work it out 4S into S plus 4, so 1 is 4S into S plus 4
so 4 S squared plus 16 S plus 4 S squared plus 16 S plus 5 into S plus 4 so 5S plus
20 plus 2S, is it all right.
So that gives me 4S squared plus 21 plus 1 to 2, 23 S is that so plus 20 divided by 4S
into S plus 4 then why have you left the factorization for me tell me the values it will be minus
23 plus minus root over 23 square 529 minus 4 into 4 into 20, 320 okay divided by 8, so
minus 23 plus minus 529 minus 320, so 209, 14.5 divided by 8 approximately. So 23 and
14 how much 37.85 by 8 so this is one route, so 4.775 approximately and the other one is
23, so this is approximately 8 okay say little more than 1. Okay, so let us take as minus
1 and minus 1, .1 or 2 whatever it is this was 1, 2 then 4 then 5. Now it is minus 1
minus 4.75 we have weakened this pole, pole at the origin, is it not, so this 0 is drifting
to this side this 0 is also drifts towards this side that means all the 0s will be gravitating
towards the 0, towards the pole which you are weakening all right. Here both of them
will be shifting towards origin, let us weaken the third1 and then see what happens Z1(s)
is equal to 1 plus 5 by 2 S plus1 by 4 into S plus 4 half of this is removed, so14th I
have taken out, so14th is retained.
So this will be how much 4S squared plus 16 S plus 5 into 25 into 2S so10S plus 40 plus
S, so that gives me 4 S squared plus 26 plus 27S plus 40 divided by 4S into S plus 4. So
that gives me minus 27 plus minus 27 square 729 minus 4 into 4 into 4, 640 okay divided
by 8 that is sure 729 and 40, so this is 81, 89, so this will be 9 point how much 9.4 divided
by 8. So 36.4 by 8 say 4.5, 4.5 to this is one route, the other one is 27 minus 9, 18
by 8 little less than 2 little more than 2 okay approximately minus 2, is in approximation,
little less than 2. So it will be s plus
the poles and 0s these are somewhat approximate, this is the minus 2 point you should tell
me this is a little less than 27 minus9 18 okay 2.2 minus 2.2.
Let me see I hope this calculations are all right then what was the pole at minus 4 then
it was at 5, this was Z and Z1, this is a pole, this is a pole, so 2 becomes 2.2 and
this 5 becomes 4.5 you see now both are them are coming towards this pole, so the shifts
are in opposite direction but they are towards the pole this is what I wanted to do stress
that means the pole that you are weakening will be attracting the 0s from their earlier
positions towards that particular pole itself . So the movement of the 0s are known, so
once you know this then the problem is simplified you know which 0 is to be created by weakening
which pole yes please any question, no well 0s become poles means 0s will be merging with
a pole and they will vanish.
So that is the total removal that is what you precisely do, when you take out that pole
totally there is no trace of that pole in the remaining part all right. So it is like
a positive charge and negative charge when they joined together that becomes neutral.
So when the 0 merges with the pole means the pole is totally removed. So we have seen the
3 cases of removal of partial removal of a pole at at the origin or at any definite location
or removing the constant.
In case of Y what would be the effect on partial removal of say any pole or constant in case
of Y where will be this value, this value is more or less than this infinite value sorry
that can be many more such elements and so on. This value is always greater than this
value, so here you remove a part of this part of this this constant all right. Earlier case
you removed part of this constant and then again you will get 0s shifted. Let us take
up an example, you are given y22, y22 as s plus 1 into s plus 3 divided by s plus 2 into
s plus 4 okay minus y12 is given as s into s plus half divided by s plus 2 into s plus
4 all right.
So if you look at the poles and 0s I will just sketch it here only the poles and 0s
at 1 you have 0 this is for y22 then at 2 you have a pole at number 3 you have a 0 at
minus 4 you have a pole and you want y for y12 there is a 0 at the origin, there is a
pole sorry 0 at half there is a pole at this point, there is a pole at this point okay
in the original one, this will be the nature of pole 0, this is the impedance diagram is
it not admittance diagram for y22.
So it is part of this that I am going to remove part of this may be this much I remove, if
I remove this much then this will be the new 0 location, is it all right, if I remove it
totally then this 0 will come here, is it true mind you there is no pole I am weakening
it is removal of a constant. I can remove part of a constant or weaken any pole all
right that pole depending on the network function that pole can be at a finite point or at infinity
or at 0 depending on whether it is an admittance function or an impedance function. For an
impedance function RC, impedance function the pole can be at the origin but there cannot
be any pole at infinity. Similarly for an admittance function whole can be at infinity
but not at 0, first one has to be a 0, so I can remove a part of that constant value
okay whenever it hits the S equal to 0 line all right a part of that is removed. So if
I remove that wherever it hits this curve that is the shifted 0 position I can remove
it totally and then this 0 can be brought here.
So let us take up this particular problem, so y22 if you make a partial fraction, if
you make a partial fraction it gives me 3 by 8 plus dot dot dot dot plus y1(s), let
us call it y1(s) the remaining part i am interested only in the constant but I am playing with
this portion. So what will be y1(s), y1(s) is y22 minus 3 by 8, so that gives me 5 by
8, y22 is this much s plus 1, s plus 3 by s plus 2 into s plus 4. So I am subtracting
3 by 8 from here 3 by 8 is actually the constant value suppose I remove that constant then
what is left y1(s) becomes 5 by 8 into you can make partial fraction and then add also
rest of the terms or subtract directly s into s plus 14 by 5 divided by just, what it out
here s plus 1 into s plus 3 by s plus 2 into s plus 4 and I am subtracting that y at 0
value which is equal to 3 by 8, so minus 3 by 8.
So that gives me 8 into S plus 2 into S plus 4 okay, so 8 how much is it if I subtract
it by 8 are you getting the minus sign 8 into S plus 1 into S plus 3 minus 3 into S plus
2 into S plus 4 so that gives me 8S squared minus 3S squared 5S squared plus 3 plus 1,
4 into 8, 32, 4 plus 2, 6 into 3, 18, 32 minus 18, 14 S plus 3 into 8 and 3 into 8 okay divided
by 8 into S plus 2 into S plus 4 which gives me this 15 by 8 if it is a common it will
be S plus 14 by 5, is it okay. So what will be the 0s for this y1(s) I have realized 10
I will removed it completely. So this will be if I remove it completely this will be
hitting here only, is it not. So I will get a 0 here and that is one of the desired 0s
I could have realized partly and got into this 0 okay either you realize this 0 first
or this 0 first you take 1 at a time okay. So if I remove it partly I could have got
at minus half this one it I remove it totally that is this part 5 by 8, 3 by 8 then you
can go here. Can you tell me what will be the value of that partial removal of this
constant 3 by 8 such that I realize this 0 then you all try, let us see that also.
So now I have got 0 here,0 here and 0 at minus 14 by 5 and a pole at 2 then pole at 3 minus
2 minus 3 and sorry 0 pole at minus 4 no 0 here okay this is my new distribution of the
pole 0 for the balance function y1 if instead of realizing y1(s) like this suppose we take
50 percent of it or a certain percentage of it so that we get the 0 at S equal to half
minus half then what would be that value of the removed material okay. So y22 minus some
Kp times this is partially removed constant which is less than 1 times the actual value
is 3 by 8 that should be equal to my new one y1(s) and this must be equal to 0 at S equal
to minus half. So y22 at minus half how much is it that means y22 at minus half minus Kp
into 3 by 8 should be equal to 0 so Kp is y22 at minus half divided by 3 by 8. So how
much is that y22 at minus half y22 at minus half is half into 5 by 2 divided by 3 by 2
into 7 by 2 divided by 3 by 8.
So 22 goes so 5 by 21 into 8 by 3 how come here 40 by 63 this is Kp, is it all right.
So if this fraction of 3 by 8 that means 5 by 21 could have been removed I would have
got a 0 created at minus half, once you have realized the 0 then what would have been the
factor of y1, y1(s) would have been s plus half as a factor 0 created there multiplied
by are you know that you have to calculate out but I am sure s plus half will be one
of the factor and the denominator as it is s plus 2 into s plus 4 okay. So after you
have realized the 0 at particular location, how to realize the network elements, how to
get it of it whenever there is a 0, how do you remove that 0 invert it create a pole
make partial fraction and remove that is it not so from Y(s) you have to go to corresponding
Z(s).
So Z1(s), so let us go back to the first realization when you have shifted the 0 to the origin
that is totally removed that factor 5, 3 by8 all right, 3 by8 is removed from y what does
it mean you have started from the terminal 2, 2 dashed okay, you have started from the
terminal 2, 2 dashed and then we have started from here. We have started with the first
element which is a resistor 3 by 8 is a constant all right that I have removed y22 is equal
to 3 by 8 plus y1(s), so 3 by 8 is admittance so 8 by 3 ohms resistance I have put and this
is the balance y1(s) okay, yet to be realized. Now you have seen that there is a 0 at the
origin so from y1 we go to the corresponding z1 because we want to realize that 0.
So what will be corresponding z1 this was our after removal of that constant this was
our y1, so z1 corresponding z1 will be 8 by 5, s plus 2, s plus 4 by s into s plus 14
by 5 okay. So remove this as K1 by S plus whatever is left over I will call it z2 okay.
So get it of this 0 which has been realized as a pole, so how much is K1 multiply by S
make S equal to 0, so 64 divided by 5 into 14 by 5, so 14, 64 by 14, 32 by 7 all right.
So 32 by 7 this y, I have converted to a Z all right I am writing in the form a Z whose
series element is 32 by 7 what, sorry this is K1, so 32 by 7S, K1 by S means 32 by 7S,
so what is it a capacitor 7 by 32 farads okay whatever is remaining I will call this as
Z2 all right this Z2 I have to finally calculate z2. So how much is z2 once you have written
32 by 7S from z1.
So 8 by 5 into s plus 2 into S plus 4 divided by s into s plus 14 by 5 minus 32 by 7S. If
I remove this I will get the balance other alternative is you can make partial fraction
since there is only 2 poles you can make partial fractions remove that one of them will be
32 by 7S, so remove that whatever is remaining will be our z2 okay. So how much is it, how
much is it coming to okay. Let us compute I have got 8 by 5, Z1 as 8 by 5 into S plus
2 into S plus 4 divided by s into S plus 14 by 5 I could have written as K1 by S this
is the one which I have removed K2 by S plus 14 by 5 plus K3 okay and that gives 14, 14
by 5 so 8 by 5, 8 by 5 S plus 14 by 5 so 4 by 5 minus 4 by 5, correct me if I am wrong,
14 by 5 and 4, 20 minus 14, 6 by 5 divided by minus 14 by 5. So that gives me 8 by 25
is that so 24 by 14.
So 112 by 175 okay so it is 100 and this K2 is this much how much is K3, K3 make S10 into
infinity so it will be 8 by 5. So basically 112 by 175 into S plus 14 by 5 plus 8 by 5
this total sum will be your Z2 is it not this will be equal to Z2, is it all right. So from
Z2 what should I do from Z2 what you do again go back to Y2 then again weaken a pole so
as to realize the other 0, so as to realize the other 0 that is at minus half.
So if you keep on doing it so you have to remove partly one of the poles so that the
0 shifts to minus half again realizing it by taking a factor 1 by S plus half by inverting
it and making the partial fraction. So if you keep on doing it you finally get I will
give you only the final result you can work it out this procedure is same alternately
you have to go for imp impedance and admittance. So you get next stage you will get Z3 as 37
by 3 after removal of this s plus 22 by 7 by s plus half okay means in the next stage
after realizing that new 0 invert it you get the pole and then again make partial fractions.
So the final values will be like this which is 296 by 161 okay then a capacitor in the
final stage you may have since it is going to be realized as S plus half, K by S plus
half, it is an RC combination all right. Earlier, we are getting a factor like K by S, we remove
32 by 7S is it not. So that was a capacitor now it is an RC combination so you will get
something like this okay that will give you S plus half then because that Z this one will be this RC combination
and a constant that is all Y22 is an open circuit parameter or short circuit parameter,
it is a short circuit parameter, it will be shorting this. So measuring the impedance
from here we will get the same impedance that I will be doing so okay is this point clear the last1 last one is
from Y3, I have got z3 so from Y3 where we have got some factors here divided by some
factors here and you want to remove that much of a constant. So that you get in the next
stage say Y4 something like S plus half realized here.
Okay so you have removed Y3 as Y4 plus a constant which was 296 by 161 this constant was decided
so as to realize S plus half in Y4 is it not the other 0. Once you have realized that that
factor became if I invert it that factor became this is it all right, this part this was basically
S plus half into S plus 22 by 7 into 3 by 37 it is this one which was inverted and you
got this.
So after realizing this resistance whatever is left over you invert it that will be the
Z and that is just a simple RC and R combination. So leave it there, so this will be your Y22
and Y12 specification met. Obviously, Y11 can be anything there is no unique choice
I could have removed that 0 at minus half earlier. So I could have got something else
this RC combination could have been at the beginning itself do you get my point. So we
will stop here for today next time we will take up Lattice synthesis okay.
Good afternoon friends, today we shall be discussing about symmetrical network, symmetrical
Lattice network for two port synthesis.
Let us take simple network function in this form Za is an impedance, Zb is this impedance
and similarly you have Za here and Zb here. Now what will be for this network what will
be Z11 and Z12 in terms of Za and Zb you shall be see the symmetry, you shall be denoting
this henceforth by an impedance Za and Zb okay and it is understood by this dotted lines
you denote the other 2counterparts, this is Za and this is Zb so we shall be showing only
the dotted lines and what will be Z11 and Z12. If you look at this network see one side
is Za and Zb, so I can show from one say I can show it like this Za and Zb okay. The
junction is one Za and Zb okay from Za the junction of Za and Zb, this is 2 that is another
branch Zb and Za and that junction is 2 dashed.
So basically this is 1 dashed if you look at it 1, 1 dashed by 4 is equal to 1 plus
zb minus za squared divided by 4 equate za squared zb squared get cancelled then you
are left with twice za, zb by 2 okay if I bring to this side then a plus b whole square
plus a minus b whole square is equal to 4ab that divided by 4 is just ab okay thank you
za, zb comes out to be therefore 1. So if you want that network, we are planning to
have a network which if terminated by a register R, let us take a normalized situation 1 ohm,
then the impedence seen from this side also 1 ohm, what will be that za and zb. Now that
za and zb must be at this relation za, zb equal to 1. So in the next class we will start
from here this is for a normalized relation if it is equal to any R, so in general za
zb equal to R squared once you set za automatically zb sets but such lattice networks terminating
impedence R will generate an impedence of R from the looking side, thank you very much.