Transformer 1

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thirty-third lecture 33 and what we are goingto do, is to solve the minor 2 examination problems point out the difficulties that youexperienced and the common mistakes that you made and if time permits as i hope it willwe shall do little bit more on the scattering matrix of a 2 port. minor 2 the first problem was with regardto a gyrator, this is the forward direction and the gyration constant is not given, whatis given is that the z matrix of this is given as 0 1 by n minus 1 by n 1 by n 0 this isthe z matrix that is given and then it is cascaded to a transformer ideal transformer1 is to n this is ideal. this is port 2 and this is port 1 and what is needed is scatteringparameters that is: you have to find out s11

s12 s21 and s12 with the reference registersat both ports equal to 1 ohm this was the question alright. now, some of you tried to solve it by finding out the abcd parameters, abcd parameters ofthe 2 networks multiplying them out and then from the abcd parameters go back to the voltagecurrent equation and then find out s11 s12 and so on and so forth. in the process 1 ofthe mistakes that you did was that; you assumed ab minus bc for this is equal to 1 which isnot correct i used formulas, which are true for reciprocal networks only. in fact, some of you calculated s21 and said because this is reciprocal this must be equalto s12 equal to 1 which is not correct the non reciprocity is obvious here the gyratoris a nonreciprocal network and therefore,

you cannot do that one of the other mistakesthat happened was with regard to the transformer equation you see the side that; has n theturns ratio, it is from this side that the impedance is n squared times the impedancehere not from this side and the voltage current equation many of you could not write correctlythat is: a pity is a open book open notes examination and you must be able to writethe voltage current equation. now, the simplest way that i find of doingthis is to assume the voltages and current do not make any conversion just find out justwrite down the voltages and currents at various points and you will see how easy it becomes.this is v1 this is i1 i suggest that you take this down this is; let us call this, v2 primei2 prime these are all conventions then this

is v2 and this current is i2 alright thatis: all that we need. now, if you look at the gyrator, well whatyou have to do is to calculate s11 and s21 you have to terminate the port 1 into a 1ohm resistor in series with a vg 1 a voltage a voltage source vg 1 alright. and. if i look at the gyrator then; obviously,v1, v1 you see the equation is v1 v2 prime the z matrix is given 0 minus 1 by n 1 byn 0 times i1 i2 prime this is the this is the z matrix equation v1 v2 prime these arethe 2 voltages and these are the 2 currents. therefore v1 as you can see is equal to minusi2 prime by n and then v2 prime is equal to v2 prime is equal to i1 by n that is all.so, let us find out the input impedance of

at port 1 that is z1, which is equal to v1by i1 v1 by i1. now let us, substitute for v1 minus i2 prime by n multiplied by i1 isn times v2. so, n times v2 that is equal to 1 by n squared and i can write this as v2by minus i2 alright now v2 by no prime v2 prime by minus i2 prime i missed the prime.now, if you look at the network v2 prime by minus i2 prime is the impedance looking hereand since, you have to terminate this into a 1 ohm resistor, 1 ohm resistor the impedancelooking here would be. 1 by n squared not n squared 1 by n squaredalright and therefore, i can write this as 1 by n squared multiplied by 1 by squaredwhich is simply equal to 1 the input impedance is 1 and therefore, s11 is equal to z1 minusr01 divided z1 plus r01 that is equal to 0

and in an exactly similar manner. if you goback that is terminate this in 1 ohm and find out the impedance here that will also be exactly1 ohm and therefore, s12 s22 is also equal to 0 the procedure is exactly similarly, letus calculate s21 and s12 for s21. let us go back 1 ohm vg 1 then you have thegyrator and the ideal transformer 1 is to n this is 1 ohm this is what you have to calculatev2 and to do that this is v1 i1 again v2 prime i2 prime. now, since this impedance lookinghere is 1 ohm; obviously, these are simplifying things, things are it should be v1 shouldbe equal to vg 1 by 2 and v1 is the voltage dropped across a 1 ohm resistor by the flowof a current i1, i1 flows i1 sees a resistance of 1 ohm, and therefore this must be equalto i1 is that clear? and once you is that?

you see the impedance here is 1 ohm. so, v1is the drop in a 1 ohm resistor due to a current of i1 and therefore, v1 is equal to i1 multipliedby 1. and once you recognized this it will become very simple v2 prime, which is equalto i1 by n from the z matrix equation v2 prime is i1 by n therefore, this is equal to vg1 by 2 n, because i1 is vg 1 by 2 and v2 is equal to v2 this is the voltage that; youhave to find out that should be equal to n times v2 prime n times v2 prime.so, this is vg 1 by 2 and therefore, s21 is equal to twice v2 by vg 1 square root r1 byr2, r1 and r2 are the same therefore, this is simply equal to half. this will be equal to 1 because s21 is twice v2 by vg 1. now the total device is non reciprocaland therefore, if you found out many of you

did, if you found out by mistake due to changeof sign of a particular current that; s12 is equal to 1 you should have suspected yourcalculation right then and there will be good. s12 cannot be equal to 1 because the gyratoris nonreciprocal and nonreciprocal device cascaded with a reciprocal device cannot makeit cannot make it reciprocal, it is possible the other way round you may have combinationsof nonreciprocal devices, which make the total network reciprocal, but, not the other way round. therefore, if you proceed in an exactly similarmanner and take account of the directions of current correctly and the transformer equationcorrectly you should be able to obtain s12 equal to minus 1 that is the correct resultthat is question number 1.

excuse me sir. yes. for a reciprocal network, is it necessarythat s21 should be equal to minus s 1. no for a reciprocal network. reciprocity demands that s12 is equal to s21not a negative no for reciprocity s12 should be equal to s2, for the definition of reciprocitybecause what you are doing is interchange the source and the load interchange the causeand excitation. reciprocity demands that this cross scattering parameters or the transmissionparameters, transmission scattering parameters. let us say, this is the forward transmissioncoefficient and this is the reverse transmission coefficient the 2 should be equal.this device is nonreciprocal question number

2 is a network which is a parallel combination of 2 networks 1 is r r and c is terminated in the resistance of capital r and then there is an inductance here of l equal to cr square do not tell me i did not do such a problem in the class. in factwhat i did was a i took bridge to t the exactly the same except that there is a small resistancehere this is a simpler 1 not only simpler from that consideration, but, it is also simplerbecause l is related to c and r square and even though there are 2 energy storing elementsand inductance and the capacitance it behaves as a first order network there are cancellations.now, the question says find out this impedance z, then if this voltage is v1 and this voltageis v0 you have to find out z you have to find out v0 by v1 and you have to find v0 by vithese are the 3 things to be found out let

me go systematically. to solve the networkwell the constraint was you cannot use mesh and node analysis some of you played a littlesmart i did not penalize it to the extent that i should have done, but, some of youwhat they did is assume a current here take this to be i1 take that, to be i2 alright.then again another distribution here do not you see that this is: in disguise a mesh analysisa combination of mesh and node analysis. you were asked not to resort to that alrightthose who have done so 1 or 2 1 or 2 i did not tell penalize, if this is a general phenomenonthen i will penalize heavily, but, since there are only 1 or 2 and i was in a good mood thatday i did not penalize greatly. so, the first thing we do is we find out they parameters of this the z matrix can be written

down by inspection r plus 1 over sc i didthis in the class 1 over sc 1 over sc r plus 1 over sc and the determinant of z is equalto r squared plus twice r divided by sc which i can well let me, leave it like this unnecessarysimplifications i will not do till the last moment and. so, the y parameters of the t network, y parametersof the t network therefore, is equal to r plus 1 over sc divided by r squared plus twicer by sc z this is what z11 or 22. 22. now, 22 and 11 are the same here because of symmetry and therefore, it does not make adifference the 22 y 22 parameter is the same as this, but, this z21 and z12 parameter wouldbe r squared plus twice r by sc, it would

be a negative sign and 1 should not loosesight of that that would be equal to y12 also y12 and y21 are equal and as far as the inductanceis concerned l equal to cr squared. now, here at these, right at this point of time becauseof the special relationship it should have occurred to you and there must be purposein choosing the inductance to be equal to cr squared. so, there is no point in proceeding with the general terminology l we will put cr squaredfor l and the y parameters of the l network is; obviously, 1 over sc r squared then plusor minus. plus no minus 1 over sc r squared y12 or y21 will be minus all over sc r squared minus 1 oversc r squared plus 1 over sc r squared.

therefore, the total y parameters y overallwould be equal to let me, write down z y11 and y21 or y12 that should be enough it wouldbe r plus 1 over sc divided by r squared plus twice r by sc plus 1 over sc r squared and1 2 parameter would be. minus 1 over sc minus 1 over sc divided byr squared plus twice r by sc minus 1 over sc r squared and this and this are equal.now, to be able to calculate the impedance the voltage transfer functions etcetera werequire the equivalent pi networks. so, what we do is we find out the network then boilsdown to this we have vi r, what we require is since we have found out the y parametersit is not difficult to find out this impedances ya yb and yc. then this is terminated in rohms and this is v0 and this is what we want

z and we also want this voltage v1.now obviously, because of symmetry ya and yc are equal it suffices to find out just1 of them and yb it should be obvious yb is simply minus y12 or minus y21. therefore, ya is equal to you see the simplification,it is simply r divided by r squared plus twice r by sc because the other things cancel minus1 by minus sc and minus 1 by sc r squared they cancel and this is equal to y sub c andy sub b is simply the negative of minus y21, which is 1 over sc divided by r squared plustwice r by sc plus 1 over sc r squared do you understand this. yes sir. the shunt the bridging elements should beminus y12 or minus y21 and once you have done

that the let me go back, to the network onceyou have done that; the impedance would be this impedance would be now, at this stagei would suggest that you make a simplification of this make a simplification and things shallbe obvious. what you have to do is find the admittance of this combination that wouldbe yc plus 1 over r then take the impedance of that let me write it down. z would be equal to 1 over. let me look, atthis and write it down 1 over ya this admittance comes in parallel with an impedance. so, 1 over. no i have done a mistake.it should be 1 over by 0 plus 1 by 1 over by yb plus.plus 1 over yc plus 1 over r.

now, it is looks complicated, but, if yougo systematically, you will see that this comes simply as capital r the input, inputimpedance is simply equal to r and if it that is so if that is. so, let us go back to the network. vi r yb yc r v zero and there is a ya herethis is v1, if this impedance is r then; obviously, v1 is equal to vi divided by 2 is that correct.if this impedance is r then vi divides into 2 equal resistors and therefore, v1 is viby 2 and v0 by v1 v0 by v1 does is not effected by ya and therefore, it is simply equal tor parallel zc r parallel zc which you have already found out divided by r parallel zcplus zb and f you substitute the values and do a bit of simplification, simplificationshould be obvious it simply becomes 1 by sc

r plus 1. it is as if the total network you see that, it is a first order transfer function thetotal network behaves as a first order although there are 2 energies storage elements andit is as if what you are doing is you are taking an r is not it the same transfer function1 by sc divided by 1 sc plus r. so, the total network behaves like this and; obviously,v0 by vi shall be equal to half of this. so, 1 by 2 sc r plus 1 that is the correctanswer the mistakes or other mistakes that you many of you committed in this is: nothuge the mistakes that; many of you have committed are silly 1s they are in simplification youmade a mess of a very long expression and you continued with that, you did not simplifyand ultimately there were problems there were

also problems regarding the conversion oft into phi some of you try to try to put the formula and in a za zb zb zb and zc and soon so forth not from first principles i have always told you that in such simple casesdo it from first principles and in this simplification somewhere along the line there is occurreda mistake. question 3 was to calculate 2 input impedancesand there also what i wanted the first problem was to judge whether you have understood whatis an ideal transformer? that is about all. it was an ideal transformer like this andyou will see how simple the question is 1 is to n ideal and these 2 are connected, these2 are connected these 2 are also connected, but, through a resistance r.now, the input impedances is to be calculated.

so, we say this is v1 now, the mistake thatyou many of you did was to assume this current to be i1 and it is the same as the currentflowing through the primary, which is not correct because there is a another path andsome of you draw a line here and put a current i2 here, it was as if i2 as dividing into2 parts no i have given that this is has been left open and if this current is i2 ; obviously,this current is also i2 and the input impedance. now is v1 divided by i1 plus i2 this is theinput impedance. that is all that 1 has to calculate and well you formulate a voltagehere and since, it is 1 is to n you get v2 equal to nv1 and i2 equals to minus.minus. minus i 1 by n this is the equation that manyof you faltered and this is why some of you

got incorrect answers. you see what you shouldremember among this is that this is an ideal transfer. so, it is it should be perfectlylossless v 2 i 2 plus v 1 i 1 must be equal to 0. this is the easiest way of remembering v 2 i 2 must be equal to minus v 1 i 1 yes what was your question. sir 3 term because… you cannot, you cannot because i was goingto say this because l 1 and m both are infinite. so, l 1 minus m you do not know whether itis infinity or 0 it is a different between 2 infinities and we do not know what it is.an ideal transformer has primary as well as secondary inductance equal to infinity, buttheir ratio is finite. in that case you can assume i 1 to be negligible.in that case you can assume i 1 to be negligible

no, there does flow a current there does aflow a current it is a magnetizing current. what is your logic for assuming this to be0? sir we are saying n 1 is infinity. or so. you see this is the equivalent circuit nowthis may be infinity this is also infinity and this is also infinity. and therefore,there occurs division between infinite inductances it is exactly like let me answer this question.if you have a battery and 2 capacitors let us say equal and this is a 2 volt batterythe impedance here is infinity no current flows. that does not mean that the capacitancedo not get charged they get charged instantaneously. and the voltage here shall be one volt thereoccurs a division between 2 equal infinite

impedances and therefore, a voltage does appear.similarly, here a current does flow now to to complete the question this if you can writethis 2 then; obviously, there is no other problem. you also noticed that i 2 this currenti 2 is equal to v 1 minus v 2 divided by r which means that this is equal to v 2 is ntimes v 1 therefore, v 1 one minus n divided r. and all that you required to do now isto find out v 1 divided by i 1 plus i 2. now, i 1 plus i 2 is equal to minus n i 2i 1 is minus n i 2 plus i 2 therefore, this is i 2 one minus n. and therefore, v 1 dividedby i 1 plus i 2 is equal to v 1 divided by 1 minus n. then substitute for i 2 which isv 1 one minus n divided by r. so, this equal to r divided by 1 minus n whole squared that is it.

one conclusion which i attempted to make isthe following that if you have a circuit like this, this 2 are connected and this is r asfar as the input impedance is concerned isn’t this equivalent to some resistance r prime here. no, some resist the same resistance r, but it turns ratio instead of 1 is to n it is 1 is to.n minus 1… why not 1 minus n? turns ratio cannot be negative therefore, it is 1 minus n or n minus 1 depending onwhether n is greater than 1 or less than 1. one cannot say definitely. so, you see oneminus n mod because, when you square it when you square it does not matter alright. thisis 1 equivalent which is sometimes found useful in network synthesis problem.

the last question, question 3 b was simplyif i am not mistaken i think i did this in the class. no for an infinite ladder no, but many of you most of you got it correctly.not all not all because, i tell you as far as the phrase silly mistakes are concernedthis is not a monopoly of any particular individual. there are many individuals who do commit sillymistakes. and all that you have to do here is to recognize that 1 2 and then what yousee to the right is the same as the network and therefore, all that you have to do isthis. and the input resistance then satisfies the equation r squared minus. what was it?r squared minus r plus 2. no that is not correct let me see r squaredminus r minus 2 equal to 1 and it was at this

stage that some of you made a silly mistakethis is r minus 2, r plus 1 equal to 0. and it is 2 or minus 1, minus 1 is not possiblebecause it is all positive resistors and therefore, r must be equal to 2. but some of you madea mistake here in solving this simple quadratic equation alright in the. is there any question? no. in the remaining time that is left twentyminutes we would like to look at some properties of the scattering matrix. if you recall thescattering matrix for a 2 port is s11, s12, s21, s22 and if we considered a 2 port; a2 port in which the incident and reflected parameters are a 1 and b 1 a 2 and b 2.then the power that is absorbed by the network n would be the sum of the powers fed fromthe 2 ports from the 2 ports. the power fed

to the network n from this port is given byhalf magnitude s squared minus magnitude b squared if you recall. 1 a 1 squared minus magnitude b one squaredalright why this factor half comes because we have taken a 1 and b 1 to be peak values.and similarly p 2 the power that is fed at port number 2 is half a 2 squared minus b2 squared. therefore, the total power that is observed by the network p. shall be given by half a 1 squared plus a2 squared minus b 1 squared minus b 2 squared. this is the total power that is fed to thenetwork. now, i can write this as the magnitude i can write as a 1, a 1 star alright plusa 2 a 2 star minus similarly, b 1 b 1 star

where star denotes complex conjugate plusb 2 b 2 star alright. now, let us define the vector a let us definethe vector a as the column vector a 1 a 2 and the vector b to denote that it is a vectoror a matrix, we put a we put an underline is equal to b 1 b 2. then you see that a 1a 1 star plus a 2 a 2 star can be written as a star transpose a is that this quantitycan be written as a star transpose a. if you are not convinced let us do this what is astar transpose a 1 star a 2 star this multiplies a a 1 a 2. so, you get a 1 star multiplied by a 1 plus a 2 star multiplied by a 2 is that simplematrix manipulation. similarly, this quantity b 1 b 1 star b 2 b 2 star can be written asb star transpose b alright.

so, our power absorbed by the network capitalp is equal to half a star transpose a do not forget the vector signs minus b star transposeb. and if the network is passive if n is passive then you know that this power this power hasto be greater than or equal to 0, p must be greater than or equal to 0.which means that, a star transpose a minus b star transpose b should be greater thanor equal to 0. but, you also know that the b parameters are expressed in terms of a parametersby the matrix s b is equal to sa. that is how the parameters are defined s11, s12, s21,s22 and you also know that if a matrix is transposed that is b transpose shall be equal to. do you know this? a transpose. a transpose then multiplied by s transpose

they get interchanged their positions getinterchanged and therefore, if i substitute this in this expression. then i get a star transpose a minus b startranspose this will be a star transpose or transpose star is the same thing; a star transpose.then s star transpose then s a this is b star this is b star transpose and this is b thismust be greater than or equal to 0. which i can write as follows you see a star transposepre multiplies both the quantities. so, i can take i have made a mistake here, i didnot use the underlining to denote matrix. if there is a question interrupt me if youdo not understand interrupt me. i am doing matrix manipulations a star transposepre multiplies both the terms and a post multiplies

both the terms. so, let us take a star transposeand a out this from the beginning and this from end. under bracket then what shall ibe left with? from this term i shall be left with. i should be left with the identity matrix of dimension 2 by 2 that is 1 0 0 1 minusi shall have s star transpose s, this should be greater than equal to 0. now, this formwhat is the dimension of a star transpose let us look at the dimensions what is thedimension 1 row and 2 columns and a is 2 rows and 1 column. so, this must be 2 by 2 in thismultiplication is to be compatible then this must 2 by 2. it is indeed so because i is an identity matrix 2 by 2 s is a 2 by 2 matrix and s star transposemultiplied by s will also be a 2 by 2 matrix.

this is 2 by 2 this is 2 by 2 agree such aform such a form let us say in general a star transpose a multiplied by any matrix a wouldbe a scalar quantity. this is a scalar quantity this is the power it is twice the power actuallyhalf multiplied by this is the power. so, this is twice the power it is a scalarquantity and the variables you see; obviously, these are not variables this is a matrix ofconstants it contains 0 1 1 0 and s11, s12, s21, s22. this is a matrix of constants thevariables are here the pre multiplier and the post multiplier and therefore, the variableswill be involved here there will be no linear term. all terms would be quadratic alrightyou might have a 1 squared you might have a 2 squared you might have a 1 a 2 there cannotbe a term like a 1 or a 2 is that is that clear.

no, you see a star transpose a a shall not contain any linear term with regard to thevariables alright it shall always come in product form and what does a contain a containsa 1 and a 2. so, the only terms that is allowed here a 1 a 2 a 1 magnitude squared and a 2magnitude squared because, there is a complex conjugating such a form is called a quadratic form. quadratic form, did you come across this termearlier no it is called a quadratic form; we shall come across quadratic form once morein this class itself. and that is why i am tempted to introduce this at this moment.now, it is also it can also be proved from matrix theory that if a quadratic form isnonnegative if a quadratic form quadratic

form is a scalar quantity. if it is nonnegativethe another name another name for this is positive definite. if it is greater than,no this is called positive semi definite. semi definite. positive semi definite. if this is equality sign is not there, thenit is called positive definite. since, it can be either greater than 0 or equal to 0this form is also called it is said the language is the quadratic form is positive semi definite.even if you forget the language it does not matter all that you have to remember is thatif a quadratic form is greater than or equal to 0. this is possible this is a theorem thisis it can be very easily proved, but i will not go into the proof. if a quadratic form is greater than equal

to 0 then this implies that, the determinantof the coefficient matrix here the coefficient matrix is a. the determinant of the coefficientmatrix is positive semi definite. which means, that if this is so if a transposet identity matrix minus s transpose t s a is greater than equal to 0. this implies thatthe determinant of the coefficient matrix is greater than equal to 0. a quadratic formgreater than equal to 0 implies it can be proved that the determinant of this is greaterthan equal to 0 alright. now, let us specialize the network let us specialize the networkto a lossless network. suppose, n is lossless then the sign that shall apply here is equalitysign the power absorbed by the network shall be 0. if the network itself is lossless it cannot

dissipate any power it cannot absorb any powerand therefore, p capital p should be equal to 0. which means that, 1 0 0 1 minus s11star yes next term. no it is a transpose… s21 star, s12 star then s22 star is it multiplied by s11, s12, s21, s22 this shall be equalto 0. no, the determinant of this how do i show this the determinant of this shall beequal to 0. now, how can the determinant be equal to 0 this means that, the 2 matriceshave to be identical all terms should be equal to 0 alright. which means that, if i now multiply s11 star? this is not. let us write down the total matrix total matrixis 1 minus s11 star s11; the total matrix s11.

what is it next? plus s21 star s21 correct? this is the first term second term is 0 minus which means minuss11 star s12 then plus s21 star s22. third term is minus s, s12 star s11 then plus s22star s21 is it yeah and the fourth term is 1 minus s. s12 star s12 plus s22 star s22 this determinant the determinant of this matrix is equal to0. now, what is the condition what is the condition can this 2 diagonals both of them can be equal to 0. the products of both the diagonals are thesame. product of both the diagonals is the samenow, what is this element and this element. they are complex conjugates of each otheralright. and therefore, what it means is that

if i if i call this element as let us sayf11 f12 f1 f21 and f22. then my f11 f22 minus f12 f12 star should be equal to 0. what isthe condition under which this would be valid that is the question. if this is 0 then; obviously,this is also 0 then this product should be equal to 0 alright. can this be non 0 canthis be non 0 f12 f12 star. no. why not? if f 12 is 0 then f12 star must be also be0 complex conjugate is 0 is 0, but the question is question is can we have f one 2 not equal to 0. is that a possibility? complex part of the other side. yeah this becomes 0.what becomes real?

you see this is a real quantity yeah thatis where lies the; you see this real quantity. so, is this therefore, the product of this 2. is real. is real. whereas no no this product s11 star s12, so last s21 star s 2. that is also real correct. so, it is a real quantity minus a real quantity equal to 0we will prove next time; thursday that that, each of them has to be individually 0 andi will give the logic next time.