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Practice English Speaking&Listening with: Lecture - 32 Synthesis of 2-port Network

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Good morning friends, today we will be discussing about 2 port network syntheses.

You know in case of a 2 port you have various specifications now, for a 2 port network Y11,

Y12, Y22 these are the parameters or may be A, B, C, D or Z parameters and so on. These

are the 3 commonly used sets sometimes you may be given only a gain function G12 which

you know is given by this or minus Y12 by Y22, so if this voltage by this voltage this

is the gain function only the gain function is given then you are asked to realize a network

to give that particular gain function all right.

So these specifications can be many of many types unlike 1 port network where we are given

either Z or Y(s), Y(s) is just inverse of Z(s). So you have 1 specification here the

specifications can be in terms of say Y11 and Y12 or Y22 and Y12, any 2 can be given

or may be just the gain function all right. So how to realize a network to meet these

specifications? So before we go into that we just see the network functions how they

are related to this 2 port parameters network elements.

Let us take a t network okay or a phi network from a t, you can always find a phi, this

is start star to delta or delta to star conversion all right. Normally if you are given the specifications

in terms of Z parameters you go for an equivalent t this is easier. Similarly, for Y parameters

you go for phi network suppose this is given as YA, YB and YC. Let us take this first what

is Y11, YA plus YC very good Y22, YB plus YC and Y12 okay minus Y12 is YC okay.

Now these elements these elements YA, YB, YC they can be any RLC combinations okay.

Let us see Y11 what are the poles of Y11, the poles associated with YA and with YC will

be the poles of Y11. Similarly, poles of Y22 will be the poles of YB and YC so YC is the

common element for the poles of Y11 and Y22 and also minus Y12, so the common poles in

these 3 will be the poles of YC.

So in Y11 there can be some poles of YA which are not included in the other 2 functions

all right. So poles of YA will be present only in Y11 nowhere else, similarly poles

of YB will be present only in Y22. So I can remove these poles still have such poles can

be realized as separate elements, I am not given YA and YC, YB, YC separately, I am given

Y11, Y22 and Y12 so out of which I find out the poles which are not present in Y12 but

present in Y11.

Similarly, the poles which are present in Y22 not in Y12, I call them private poles

of the 2 ports private poles of 1, 1 dashed and 2, 2 dashed okay poles of Y11 and Y22

which are not present, which are not present in Y12 okay. So the poles which are present

in Y12 will be present in Y11 and Y22 but the converse is not true okay. So if I have

a network which is having Y11 dashed Y22 dashed and Y12 dashed as the parameters then Y11

as seen from this side will be Y11 dashed plus YA, any addition of an admittance here

will be adding to this value only, is it not. If I have another element another additional

admittance here that will be changing only Y11 that will not be affecting these 2, is

it not because it is only YA which is getting modified. So Y11 will be this additional element

plus whatever is a Y11 dashed here. Similarly, Y22 will be Y22 dashed plus YB but Y12 will

be Y12 dashed is it not that remains unaffected.

Similarly, for the Z elements Z11 will be suppose this I call as Z1, Z2 and say Z3 then

Z11 will be Z1 plus Z3, Z22 will be Z2 plus Z3 and Z12 is Z3. So here also you see the

poles of Z3 that is Z12 poles of Z12 will be present in Z11 and Z22 but the converse

is not true Z11 and Z22 will have additional poles coming out of Z1 and Z2 okay. So if

I have a series addition of such elements this is some Z1, Z2 suppose this is having

Z11 dash Z22 dash and Z12 dash then overall Z11 from this side I should write small z

small z11 will be Z1 plus z11 dash, is it okay.

Similarly z22 will be Z2 plus z22 dash but z12 remains same. So once again the privacy

poles of an impedance, so this is the series and shunt separation of private poles. Now

let us see what are the requirements for the impedance functions? All right, what are the

requirements for the impedance functions to be realized? Now let us take a y11, y12, y22

once again I draw the, now what is the y11, YA plus YC, so this YA, YC, YB, y22, YB plus

YC and y12 minus y12 is YC okay.

So y11, y11, YA minus y12 is it all right or YA you can write as y11 plus y12 okay similarly,

YB will be y22 plus y12 and YC is minus y12. Now given any network, given any network you

can have a very complex structure of a 2 port network and so all right that can be bridge

elements also whatever be the type I can always reduce by repeated star delta conversion,

repeated star delta conversion this entire set to either t or a phi, is it not. Now in

this repeated star delta conversions what are the elements? How do you calculate the

elements of a star delta set? Say in terms of impedances Z1, Z2 plus Z2, Z3 plus Z3,

Z1 divided by something all right divided by Z1 or Z2 or Z3 the operations are all positive,

there is no subtraction okay.

Now Z1, Z2, Z3 these will be R plus L, S plus 1 upon SC or their parallel combinations series

combination, so there all having positive signs and these after multiplication also

the signs do not change then again you are having additions. So all the elements will

be finally added there is no subtraction, there is no chance of a negative sign, is

it all right. So all these elements finally Z(s) or Y(s) elements here it is YA, YB, YC

or ZA, ZB, ZC these will be having the forms some a0 s to the power a0 plus say a1(s) plus

a2(s) square and so on divided by say an s to the power n divided by b0 plus b1(s) and

so on.

So bn s to the power n okay these elements YAA these will be all positive real functions,

so far as YA, YB and YC are concerned these 3 elements are positive real after all they

are obtained after reduction of some positive real functions, star delta reductions. So

this will be having expressions like this. Similarly, y12 minus y12 will be YC, so y11

sorry YA, YB and YC are all having positive coefficients of polynomials in the numerator

and the denominator. So let us see y11, what was was our y11, YA minus y12 okay.

So suppose you are given minus y12 as a0 sorry otherwise I have a small slip, there was a

small slip check y11 is YA plus YC okay y22 is YB plus YC all right and y12 minus y12

is YC okay therefore y11 it may be given, let us write the denominator may be some polynomial,

numerator can be any polynomial that will be a common polynomial if there is no private

poles okay. We can consider a common polynomial for the denominator then this can be say a0

plus a1(s) plus a2(s) square and so on, y22 is b0 plus b1(s) plus b2(s) square and so

on and y12, c0 plus c1(s) and so on okay.

So one of the conditions is a0, a1 these must be greater than equal to 0 that is what we

have seen all of them should be positive similarly b0, b1 must be greater than equal to 0, c0,

c1 etcetera must be greater than equal to 0 all right, what is y11 plus y12? YA, is

it not, y11 plus y12 is YA is it all right now YA has to be with all positive coefficients.

So if I substitute here y11 plus y12 means a0 minus c0, a1 minus c1 all right into s

so that gives me a0 minus c0 must be greater than equal to 0 a1 minus c1 must be greater

than equal to 0, a2 minus c2 must be greater than equal to 0 and so on okay.

Similarly, by the same logic y22 plus y12 is how much, YB and that also we discussed

after a repeated reductions YA, YB, YC must have all positive coefficients in the numerator,

so y22 plus y12 also must be having numerator coefficients as 0. So b1 minus c1 is greater

than 0, b0 minus c0 is greater than 0 greater than equal to 0 and so on, all right. So these

are the coefficient conditions to be satisfied when you are given the specifications is it

all right. So y11 suppose is given 2s plus 5 divided by some denominator, y22 is given

s plus 2 divided by same denominator and y12 minus y12 is given s plus 7 by the same denominator.

Is it possible to realize this, let us see y11 and y12 the coefficients first coefficient

5 minus 7 is negative, so it is not possible all right.

So this must be less than 2 if it is s plus 1 yes, it is possible suppose it is 2s plus

1 then this is all right but this is not satisfied okay. So if you are having the coefficients

of the numerator given like this the polynomials are given then by applying this rule you can

check whether it is realizable or not, all right whether the network can be realized.

So this is known as Fialkow Gerst Condition.

So before going for any synthesis we must see that these coefficient conditions are

satisfied. There are 2 ways of realizing a 2 port network, one is Ladder Synthesis that is you try to go for realizing the 0s

of transmission in steps from one end or there is a there is another one Lattice Synthesis

which will be taking up later on that is a very interesting synthesis. It is a structure

like this okay if you want you can have repeated block surface, so this is a Lattice network

they are Za, Zb, Zb, Za it is like a bridge basically this is Za, Zb and you take the

output from here, this is 1, 1 dashed and 2, 2 dash this 1 as 1, 1 dashed, 2, 2 dash

okay.

We shall see what are the 0s of transmission and how to realize them okay. Whenever you

are having a0 somewhere what do you mean by 0s of transmission transmission 0s means what

I give a signal at the other end nothing is received, is it not. Now in how many ways

that can be ensured I give a signal you take a water pipe line all right. In what are different

possible ways, the say delivery of water at the other end can be disrupted, very good

there is a leakage so water is grounded basically the pipe is grounded the entire pressure is

grounded is there any other way of blocking.

So if you have some blockade in the series path or a short circuit in the shunt path

then you have a 0s of transmission that means how do you create a 0s of transmission by

short circuiting at some frequency suppose at some frequency I find s squared plus 9 in the numerator that

means at s equal to J3, this s square plus 9 will be vanishing that is a0 at that particular

frequency. So at the frequency I want there should be no transmission of signal and I

want short circuit at that particular frequency, no, short circuit means what it should basely

is resonance if have a series resonance at s equal to J3 and that element is put in shunt

okay. So you are having an element like this there are many such elements in the ladder

I create this element is say an lc combination.

Suppose this is resonating at omega equal to 3 then s squared plus 9 will be giving

me omega is equal to 3 this is resonating frequency, this will be the numerator okay.

So in z12 if I have a numerator of this kind s squared plus 9 my target will be to create

a0 in the shunt element such that the elements are having a resonance frequency at omega

equal to 3, is it all right or alternatively I open any of these that means there is no

transmission after this. So how do I create that, how do I create that at omega equal

to 3, now how to create that parallel resonance, anti-resonance.

So if I have an anti-resonance element any of these at that frequency and then there

are others then also it can be blocked. So a0 of transmission can be created 0 of transmission

means because G12 is z12 by z11, so it is the 0 of this which will be the 0 of this

transfer function okay, so output will be 0 at that frequency. So that can be created

either by a parallel resonance here or a series resonance is here in the shunt element, is

it okay. So you keep on repeatedly removing all the 0s by adjusting these elements okay.

So this is the technique of 2 port synthesis when you are given the transmission 0s you

convert them into poles and then try to realize the pole any 0 can be realized in terms of

a pole, is it not? How to do that?

Suppose you are given z12 how do remove that, how do, how do I remove a pole a0 you take

inverse of that function if it is in the admittance form then you convert it into impedance and

then make partial fraction corresponding to that pole remove that okay.

Let us take before we go further let us take a function Z(s) equal to say any function

you tell me it is a 2 into s into s squared plus 4 by s square plus 1 okay I can make

partial fractions K1(s) by S square plus 1 plus K2 into S okay, K1 comes out as if I

multiply by S square plus 1 divided by s and make s square plus 1 equal to 0. So it will

be 3 into 2, 6, 6S by S squared plus 1 plus K2 into S, how much is K22 into s all right.

So that means 2 Henry, how much is this 1 by 6 farad and 6 Henry okay. Now if I remove

2 s from here from here if I remove this K2 totally, if I remove 2s whatever is left over

will content only the other pole all right. If I remove this I am left with only 2s now

in this the 0s are at 2, 1 and this is a pole, this is a pole 0 configuration. 0 then pole

at 1, 0 at 2 and pole at infinity okay.

Now suppose in this I, when I remove this I remove the pole at this one, what I am left

it only 2S, a 0 here and a pole here okay for this function if I remove this what will

be the pole 0 configuration for 2, s0 here and a pole here at infinity if I remove this

2s then 0 here, pole at 1 and a 0 here okay. So if I remove this pole totally this 0 also

gets eliminated all right. If I remove okay, if I remove this pole, this pole okay means

this is having a pole at infinity, is it not, if I remove the pole at infinity then infinity

becomes a 0 and this pole remains where is this pole. So this pole remains at 1 but the

other pole has created a 0 here okay.

Now if I do not remove the poles or the poles of either this or that totally if I retain

some part of it what happens that means instead of 6S plus s square plus 1 suppose I take

3S by s square plus 1 okay. Then I call that balance admittances Z dashed S as 3S by s

square plus 1 plus 2S okay and some Z removed S is 3S by s square plus 1 that means 50 percent

have removed 50 percent have retained. So if I remove 50 percent of this means what,

3S by s square plus 1 means allow 3 Henry, 3 Henry and 1 third Farad. Suppose I take

out an element like this whatever is left over is Z dash into S, is it not what is Z

dash into S, what will be the poles and 0s of this, poles will remain same I am not disturb

the pole what about 0s.

Now this is not totally removed, it will have a new 0 all right, what I have done is I have

not totally removed this pole, I have weakened the pole, I have removed the part of that

residue residue was 6S all right. So I have taken a part of it, so if I remove a part

of that pole that is called part removal of the pole then what do I get in the oral distribution

of poles and 0s poles remains as they are they were but what about 0s. So this will

be 3S plus2 S cubed plus2 S all right that means divided by s square plus1 that means

2 s cubed plus 5 S in the numerator so S into if I take 2S common then S squared plus 5

by 2, correct me if I am wrong okay.

So the new distribution of pole and poles and 0s will be 0 at the origin is remaining

intact all right, pole at 1 is again appearing because I have not removed it completely,

what about this 0, this 0, it is now root over of 5 by 2, it has earlier it was at 2,

now it has drifted towards this. Pole at infinity that also remains as it is okay. So this 0

at 2 has now shifted to root over of 5 by 2 okay approximately 1.58 okay. So if I weaken

a pole I can shift a0 to a desired position I have taken out only 50 percent of this 6S,

this I said you I could have taken any percentage and hence I could have shifted this to a desired

location, is it all right. Therefore, by partial removal of a pole I can create a0 at a desired

position some of the 0s can be shifted, had there been many other floating 0s means between

0 and infinity, had there been some more 0s they would have all drifted towards the pole

all right. The drift will be by different amounts all right.

I am not bothered about the drift of all the 0s, I may be concentrating on one 0 specially

then nearest one, the one which is nearest to this preferably to locate it in a new position

and this is not the only pole there could have been other poles and my desire would

have been to shift this to this side then I would have weakened this one, is it all

right. So it all depends on what is the 0 wanted that means in Z12 in the transmission

I am creating artificial 0s all right by removing the poles partly this is known as partial

removal of poles and shifting of 0s okay. So you can weaken either the pole corresponding

to this or this that means you weaken the pole here that also you could have seen. Let

us see now, instead of 2S if I take S what happens I remove only one hand inductor and

rest of it, I retain.

Now here I have created a 0 in Z1 dash into S which is at root 5 by 2 so this element

is having a 0 let me complete this then will go to partial removal of this pole at infinity.

So I have got a network like this and then Z dash into S is a function which is having

a 0 at S square plus that is at root 5 by 2 it is having a 0, how do I realize this

any 0 you convert it to a pole. So I correspondingly I will take Y dash into S which will be giving

me S square plus 5 by 2 in the denominator okay then there are other factors okay and

this one I will write as K1(s) by s square plus 5 by 2 plus other factors that means

Y dash the admittance is taken as some Y1, Y2, Y3.

So this entire Z dashed which was appearing here is now taken as some Y1, Y1 which will

be corresponding to this all right that means it will be having, this is an LC series element,

is it not if Y corresponds to K(s) by S square plus omega square that will give you a series

LC element okay and then rest of it I will again switch over to Z(s). I am not bothered

about this part, I am interested only rem in removing that 0, so convert it to a pole

that is go to an admittance function then realize that and then again the balance you

again invert you will get rest of the elements 1 by 1 you take out the 0s in the shunt elements

okay.

So this will be resulting at root 5 by 2 so that will be creating the 0 all right by partially

removing the pole we can realize this. Let us see if we partially remove the pole at

infinity that is let us take again 50 percent removal just take S. So we have got 1 henry it is Z dash into S

now, is s plus 6 S by S square plus 1 and Z removed is just s okay. Now what is Z dash

into S, s cubed plus S plus 6S, so s cubed plus 6S divided by s squared plus 1 sorry

7S. So s into s squared plus 7 by s squared plus1 now what will be the pole 0 location,

earlier it was a pole sorry a 0, a pole a 0 and a pole.

Now I have weakened this so under the new situation Z dash into S this was for Z(s),

Z dash into S is having 0 once again at the same point, pole at the same point, this pole

is also at infinity but here it is earlier it was 2, now it is root 7, now it has drifted

to this side. So 0 has drifted towards this pole which has been weakened, so wherever

you are weakening the pole the 0s will drift towards that okay. Earlier it was in the other

side so the 0 drifted to that side, so depending on the 0 that is given in the transmission,

this is corresponding to the transmission 0.

So the 0s are transmission where they are located and what is given in z11 if you know

that z11 is at 2 and the 0s required 0 of transmission is at root 7 then I will be weakening

this pole first, so that this drift safe all right. If I weakened this pole then it will

be drifting to this side I cannot realize this 0 all right. So this is the technique

that we follow for ladder development and we want 0 at that particular frequency so

what should be the value of that residue that is we have seen only 50 percent removal, what

should be the percentage removal? What should be the percentage removal of the residue,

percentage removal of the residue. So that we can get the desired location of the 0 okay.

So that is a next question, let us take up an example it will be clear say z11(S) is

given as s squared plus 9 into s squared plus 25 divided by s into s squared plus 16 okay.

Let us make a plot of this poles and 0s first and then decide about the shifts, so there

are no private poles z12 and z11 are having the same poles all right, if there are private

poles remove that first. So poles and 0s are for z11 is it has is the original 0 pole then

3, 4 okay then not to the scale any way 5 and then at infinity there is a pole, this

is z11, z22, z12 pole, poles are same because there is no private pole. So the poles are

identical and then this one is at 1 and then this is at 2.

Now in the transfer admittance, transfer admittance or transfer impedance, there is no hard and

fast rule that poles and 0s should alternate, transfer admittances z12, how much is z12

in terms of z1, z2, they need not be okay we will discuss about it later on, see the

poles and 0s need not come alternatively, so they need not be positive real functions,

they need not be positive real function, is that clear. Any question? They need not be

positive real functions how follows us z12 by repeated star delta conversions, you got

z12 as this impedance but this need not be a positive real function, this need not be

a positive real function.

Only thing is the numerator and denominator coefficient must be always positive when you

go for repeated star delta conversions you must get all the coefficients as positive

but repeated conversion of star and delta that need not guaranty positive real functions

as elements that means the element may not be realizable, element may not be realizable.

For example you are having an inductor and capacitor, an inductor, an inductor and resistor

and so on. If I reduce it to star or a delta this may give you some Z(s) it may not be

realizable it will be having a function all right with all positive coefficients a1(s),

a2(s) squared and so on divided by b1(s) b2 square but it may not be a positive real function

they may not be able to realize it by RLC, okay it is a very interesting point.

So this is very clear from z12 itself here s squared plus 1 into s squared plus 4 by

s into S squared plus 16. So you can see the 0s are coming consecutively coming 2 consecutive

0s and then pole comes, so what should I do I can shift this 3 so these are the values

1, 2, 4 and 5 either I can shift this 0 here or this 0 here. So either you can weaken this

or weaken this, is it not. So if you want to shift this here you have to weaken this

that means a part of this residue has to be removed okay so z11(s) let us break up z11(s)

if you write as K1(s) plus K2(s) by S squared plus 16 plus K3 by S, I am not really concerned

about all these, I am interested only in a partial removal of this.

Suppose I write this as some K1 dashed S plus z11 dashed S that means this is the part removed,

this is the part removed and this is what is remaining all right. Now z11(s), so z11

either at j equal to omega sorry omega equal to1 that is S equal to j1 or s at j2 this

will be vanished all right sorry, the impedance at that point, impedance sorry impedance at

that point j1 or impedance at j2 I want this to be 0 for z1, z12 for z12 this will be equal

to 0, is it not. So z11 dashed, z11 dashed at j1 if I take it as 1 because I am shifting

it here, should be equal to 0 is it not this must be 0. So what is this if I put on this

side z11 at j1 should be equal to K1 dashed into j, S equal to j1. So how much is K1 dashed

calculate from here, is it okay z11 that was given as s square plus 9 into this thing,

so put S equal to j1, so that is 9 minus 18 into 25 minus 1 into 24 divided by 15 into

j okay and this j is there already.

So that gives me K1 dashed into j1 all right K1 dashed into j1, so calculate this how much

is K1 dash comes out as minus 64. Now check whether it should be minus z11 j omega plus

K1 dash into j1, j1 is equal to 8 into 24 divided by S into s squared plus 1 or 15,

is it all right. We are trying to make it at 1 it is a 0, so 1 okay K1 dash it is let

us see whatever be that value minus 64 by now j into j will be so it will be plus minus

64 by 5 so that means we will have to add that seems to be some out funny this there

made some slip okay.

Let us see the other one I come back to this you also, also think over it why we should

get this or what we are trying as K2 dashed S by S square plus 16 into some z11 double

dashed S either remove this part partly, remove this partly then it will be shifting to 2

sorry, sorry, sorry, sorry 3 will be 2, thank you very much that is why this scaling was

very very important. So in any case you can weaken this to shift it to either 1 or 2 okay,

thank you.

So in that case what will be this one, z11 double dashed at j2 should be equal to 0,

this is a0 which gives me yes, sorry we are weakening the pole at the origin

that is what I am putting j2, I am going to write that, that is what

I am going to do, this will be z11. Okay, let us write it here, okay

we will continue with this in the next class. Now time is over we

will continue with this in the next class, thank you.

The Description of Lecture - 32 Synthesis of 2-port Network