 Welcome to the tenth lecture in the course engineering electromagnetic. The topic for the lecture today is impedance matching using stub lines. We take up the two parts of the topic first. The first part is the impedance matching. First we need to understand what is the necessity for impedance matching. You would be aware of the maximum power transfer theorem saying that the source delivers maximum power to a load which has the impedance which is the complex conjugate of the source impedance. And therefore, first requirement for impedance matching the first point because of its impedance matching is required is to deliver maximum power to the load. So, the first point that we could list here is for maximum power transfer to the load impedance matching is necessary. What could be the second reason? We know that if the load impedance is not matched to the source to the characteristic impedance of the transmission line then there are going to be standing waves. The incident power is going to be partly reflected back causing formation of standing waves on the transmission line. Implying that there will be build up of voltage and current at certain points on the transmission line as well as some build up of voltage and current on the source also. And this build up particularly for high power systems could damage the source on one hand or could cause dielectric breakdown in the transmission line. And therefore, the second reason because of which one should carry out impedance matching is to avoid damage to the source to the generator or the transmission line or the transmission system. Can there be a third reason also for impedance matching? Yes. We know that in practice the transmission lines are going to have some loss. There is no transmission line which is going to be ideal. The losses will depend upon the voltage and the current on the transmission line. In fact they are proportional to the square of the voltage and current depending on which loss we are calculating whether it is the dielectric loss or the ohmic loss in the resistances. Since the loss is proportional to the square of these quantities voltage or current considering that the power generated by the source and coupled to the transmission line remains constant the losses are more when there are standing waves on the transmission line because the voltage and current are non-uniform in that case compared to the case when there is no standing wave on the transmission line. And therefore, to keep losses low also we require to carry out impedance matching. Now, these losses are in the transmission line this should not be confused with the power delivered to the load. So, to keep losses in the transmission line low which are strong reasons to carry out impedance matching. The next question that follows is how to do impedance matching how to achieve the condition that the impedance that the source or the generator sees is matched. The concept of this is fairly straight forward one could either attach reactance in series at an appropriate distance from the load impedance of an appropriate value. So, that seen at this plane the input impedance offered here taking into account the contribution of the added reactance appears to be equal to that of the characteristic impedance of the transmission line. A question could be asked why are we considering only reactive components the reason is straight forward if we add some an element which has a real part in it it will add to the power loss in the system. And therefore, particularly for impedance matching we will avoid anything which has some loss some real part of the impedance in itself. Therefore, we consider only elements which have reactive values and one possibility is as we have stated is to add some appropriate reactance in series. What is equally well possible is to add something in shunt. So, the same system can be matched by adding an appropriate value of a substance in shunt at an appropriate distance from the load impedance. How to estimate these values and how to estimate their location we will be doing in a short while conceptually this is what impedance matching would require. The next question that follows rather naturally is how are we going to realize these elements the required reactance and the required susceptance. We have seen earlier we have stated earlier that at high frequencies of the order of hundreds of megahertz the familiar configuration of inductors capacitors etcetera the lumped element configuration does not work very accurately. One cannot design very precise inductance and capacitance values and as an alternative we have stated that one could use sections of transmission lines serving effectively as inductors or capacitors. That brings us to the second part of today's topic that is the stub lines. What are stub lines they are sections of transmission lines. In principle they could be open circuited or short circuited but as some of you would have experienced it is easier more accurate to realize a short circuited practice and therefore what we would consider here are stub lines which are sections of short circuited transmission lines and more commonly they are added in parallel. A parallel connection in practice is somewhat more convenient to make on various types of transmission lines and therefore these are added in parallel or connected in parallel to a transmission line for the purpose of impedance matching alright. Now we will take up two examples to show how using stub lines one can achieve the condition of impedance matching. Going to the overhead projector let us say that the problem that we pose for ourselves is of this kind. We have a transmission line of characteristic impedance Z naught with a load impedance ZL connected here and in general this load impedance is not going to be equal to the characteristic impedance that is how the problem would arise. We have to find out the location of a transmission line stub made out of a transmission line of characteristic impedance say Z naught again for the sake of illustration and the length say D2 of this transmission line stub. So that seen looking into the plane denoted as A the input impedance appears to be equal to Z naught which would be the condition of impedance matches right. The problem of course can be solved using the various formulae for transformation of impedance that we have already discussed but the problem is quite simply solved using Smith chart and the Smith chart also provides a visual perspective to what is going on on the transmission line and therefore let us try and solve this problem using a Smith chart. Let us super impose our problem on a Smith chart and let us specify the value of ZL say ZL is equal to 150 plus J50 ohms and let the characteristic impedance Z naught of the main transmission line as well as of this stub transmission line be equal to completely real and 100 ohms. The starting point would be to calculate the normalized impedance corresponding to the load which is ZL upon Z naught. So that it is 1.5 plus J 0.5 and to locate this point 1.5 plus J 0.5 on the Smith chart. So 1.5 is a constant resistance circle like this which is not shown on the Smith chart and it intersects with the 0.5 constant reactant circle at point P1 alright. So the impedance at the point the normalized impedance at the point P1 is equal to the normalized impedance of the load. The stub is connected in parallel and we know that parallel connections are more easily handled using admittances. Therefore the next step will be to calculate the corresponding normalized admittance. We have already stated that the admittance calculation on the Smith chart is particularly simple. If we consider a constant VSWR circle centered at the center of the Smith chart and with this as the radius this is the constant VSWR circle passing through the point P1. A point diametrically opposite this point P2 represents the corresponding normalized admittance. That point is P2 and the admittance at point P2 one can read it is going to be it is on the point 6 constant resistance circle and minus point 2 constant reactance circle and therefore its value is point 6 minus J point 2 which can be easily verified alright. The next step is to move towards the generator or away from the load. We have already discussed how this movement should be carried out. This should be carried out on a constant reflection coefficient magnitude circle or on a constant VSWR circle alright. In a direction which takes us away from the load or towards the generator which is marked on the Smith chart. This is the direction which is away from the load or is towards the generator clockwise direction. The question is how much to move. In fact that is one of the answers sought in the problem. So then we argue like this that the stub is going to add only some imaginary part to the input admittance seen here and therefore we must move up to a distance where the real part of the admittance seen here becomes equal to 1 in the normalized system. That is on the constant VSWR circle passing through point P2 we should move away from the load that is in the clockwise direction such that we come across the constant conductance equal to 1 circle. As one can make out there are two possibilities one leading to point P3 the other leading to point P5 I think alright. Let us consider these two possibilities separately. So let us say we reach point P3. Now at point P3 what is the admittance that one can make out the real part is of course 1 by design and the imaginary part is plus j point 58 alright. So which we put down here plus j point 58. Once we have reached point P3 two questions are answered. We know how much distance we have travelled in reaching point P3 from point P2 alright. That distance can be measured on the smith chart itself by taking a reading on the outermost scale corresponding to P2 in terms of wavelengths and by taking a reading corresponding to point P3 once again on the same scale in terms of wavelengths and finding the difference. When that is done then in this particular problem the distance D1 comes out to be point 194 lambda. Here itself one can make out that this particular choice corresponding to point P3 is going to lead to a shorter distance D1. There will be another advantage in the length of D2 as we will see later. The second question that is answered once we have reached point P3 is what should be the input susceptance seen at this point for this stub so that after the stub is added at the plane A the transmission line appears matched. The stub susceptance must be minus j point 5A so that one is able to move on a constant conductance circle from point P3 to point P4 where the conductance is 1 and there is no susceptance. This is how we say that the smith chart affords a graphical perspective. In this kind of problems the attempt will be to move towards the centre of the smith chart using steps which are permitted. And once we have reached point P4 we are sure that the transmission line is matched. We have still not found out the value of D2 the length of the transmission line stub that can be calculated since we know the formula for the input impedance or input admittance for a short circuited transmission line or it can be determined from the smith chart itself. Now instead of the main transmission line so to say we focus on the stub transmission line alright. This is the load what is the load admittance? The load admittance is infinite it is a short circuit the load admittance is infinite. The infinity admittance point can be located on the smith chart. The resistance is the conductance is increasing this way the susceptance is also increasing this way. So this point is the infinity infinity point the short circuit point. Now from the short circuit point or from the load end we have to move away from the load. So once again the direction remains clockwise away from the load towards the generator. What distance do we travel or first let us see what path do we travel on? The constant reflection coefficient magnitude path or the constant WSWR path because on this transmission line now there is no other discontinuity and therefore one will move on the constant WSWR circle that passes through this load point which is the reflection coefficient magnitude equal to one circle the outermost circle in the clockwise direction and up to what point does one move? Wherever one reaches the susceptance equal to minus j 0.58 value which can be read on the outermost on the on the circle. We are going from minus 10, minus 5, minus 2, minus 1. So somewhere here we reach minus 0.58 minus j 0.58 if you like. The distance that we have travelled is already available on the outermost scale in terms of wavelengths. Therefore the distance that we have travelled is 0.167 lambda which becomes the value of the d2 or the length of the stub that is used for matching the transmission line d2 is equal to 0.167 lambda alright. Now we were able to make out that d1 is the shorter length out of the two possible choices where we could locate our plane a is d2 also the shorter length corresponding to this point. Now one can make out that alternative to point p3 we would have reached say point p6 alright where the admittance would have been equal to 1 minus j 0.58 since this point is symmetrically located with respect to point p3 in any case one can read the value. So the required susceptance of the stub would have been plus j 0.58 and our experience of moving on the outermost circle tells us that the positive susceptance values require longer stubs. So had we chosen by mistake point p6 instead of point p3 we would have ended up with longer stub a point somewhere here which is certainly more than lambda by 4 since half the semicircular distance is lambda by 4 greater than the value we have obtained now alright. This kind of problems where we use a single stub is called matching using a single stub it has its features for example any load impedance or any load admittance can be matched using a stub of a variable length and which can slide or which can be connected anywhere on the transmission line that is an advantage of this scheme. Disadvantage of this scheme perhaps would be that on some transmission lines it is not possible to keep on sliding this stub for example in a coaxial transmission line okay. So there one would like to utilize a scheme where the stub a single stub or a double stub we would require two stubs actually so that we have two degrees of freedom to match the real part and the imaginary part. So we would like to have an alternative where we use two stubs which are fixed in location of course the lengths will have to be variable that will lead us to what is called a double stub matching problem. So unless you have any questions I will remove this. The double stub matching problem can also be introduced in a similar manner we have shown here a transmission line let us say the main transmission line with characteristic impedance Z0 terminated in a load impedance ZL and we have two stubs located at fixed points. Let us say that their distances the distance of the first stub from the load is lambda by 4 and the separation between the two stubs is lambda by 8. What is required in the problem? These are stubs of characteristic impedance Z0. What is required in the problem is the values or the lengths of these stubs that is D1 and D2 which are shortest so that at the plane A looking into the plane A just after the stub shown in green the input admittance or input impedance appears to be equal to that of the transmission line so that an impedance match condition is obtained alright. Now once again while one can use the formulae the problem is relatively simply solved using the Smith chart. Let us superimpose the problem we have set for ourselves on the Smith chart and put down some values of the load impedance say ZL is equal to 50 plus j 100 ohms and let the characteristic impedance of the transmission line be once again completely real and equal to 100 ohms. By now we are familiar with the first step that is since we are using a normalized switch chart we should normalize the load impedance so that the corresponding normalized load impedance is 0.5 plus j 1.0 which can be located on the Smith chart on the constant resistance equal to 0.5 circle and where it intersects the constant reactance equal to one circle that is point P1. So the normalized impedance at point P1 is equal to 0.5 plus j 1 the stubs are being attached or added in parallel it will be more convenient to work in terms of admittances so we determine the corresponding normalized admittance. The procedure is straight forward we consider a constant VSTBLR circle passing through point P1 and consider a diametrically opposite point on that circle that is P2 in this case. So that the admittance normalized admittance at point P2 is equal to 0.4 minus j 0.8 P2 is on the constant conductance equal to 0.4 circle and on a constant susceptance equal to minus point 8 circle alright. The next step is also fairly straight forward we move away from the load by a distance lambda by 4 where the blue stub is connected. The movement will be on a constant VSTBLR circle so the circle that we have drawn through P1 remains unchanged and the movement will be by a distance lambda by 4 in the clockwise direction since we want to move away from the load towards the generator in any case the direction is marked on the Smith chart. So from point P2 we move on a constant VSTBLR circle in the clockwise direction by a distance corresponding to lambda by 4. The distance is also marked on the Smith chart this particular value that we have taken happens to be such that we come back to point P1. Now what is point P1 that we have got? It is the impedance admittance at the plane D before the stubs are added alright and the value one can read it is equal to the admittance at point P1 itself and it is point 5 plus j1. This of course does not happen in all the problems it is a very special case where the distance chosen for illustration is lambda by 4 which is a movement of 180 degrees so we have come back to the original point just a coincidence alright. What should be the next step? The next step is not obvious if we continue considering the right hand side. So we consider the left hand side we argue in the following manner. At the plane A once these stubs have been added what is it that we expect? We expect a normalized admittance equal to 1. So just before the stub the green stub is added what should be the condition? The condition should be input admittance equal to 1 plus minus j something because the stub can only alter the imaginary part of the admittance. The real part must become equal to 1 once we have reached plane A because the stub cannot alter the real part alright which is the circle which is the path which on the Smith chart represents all these points where the real part is equal to 1 that is the constant conductance equal to 1 circle mark as circle C1 which means that if we are to be successful in achieving an impedance match condition just after plane A then just before plane A the admittance must have some value lying on this circle C1 where how have we arrived at circle C1? Where the conductance is equal to 1? The argument is that the stub cannot alter the real part fine. So just before plane A we have identified in some restricted way this situation it must be represented by circle C1. Now we transform this entire set of possible admittances towards the load by a distance lambda by 8. Lambda by 8 corresponds to an angle of 90 degrees on the Smith chart a movement towards the load now anticlockwise. So this entire circle C1 is rotated by 90 degrees in the anticlockwise direction leading to circle C2 okay it is very interesting. Now what does C2 represent? C2 represents the possible admittances just after the blue stub has been added. If after adding the blue stub the admittances lie on circle C2 then as we have argued out we will be able to match the transmission line with us a green stub of a suitable length. Now where were we? Just before the blue stub we were at point P1 with admittance 0.5 plus J1 the stub can only alter the imaginary part. So from point P1 we move on a path where the conductance is constant because the stub cannot alter the conductance it can only alter the susceptance. Just before the stub we are at point P1 just after the stub we must be on some point on circle C2 the movement must be on a constant conductance circle right in which direction that is not known that is left to us and that will come out of the requirement that this stub lengths must be as short as possible alright. For illustration or using our experience that if this stub susceptance required is of negative value then this stub lengths are shorter either way let us say that we make the choice that from P1 we go to point P3. What is point P3? It lies on a constant conductance circle passing through P1 that is the first condition it satisfies and secondly it lies on the circle C2. The admittance at point P3 can be identified it is y P3 is equal to 0.5 again by design the conductance cannot change plus j 0.14 it lies on the constant conductance equal to 0.5 circle and the constant susceptance circle that crosses it at point P3 is of value 0.14 and therefore the stub must have a susceptance which is equal to minus j 0.14 which is the difference between the admittance required after the addition of the stub blue stub and the admittance without the stub the difference is going to be the difference between these two and therefore the change that we have sought is by a value j 0.86 minus j 0.86 in moving from point P1 to point P3 we have reduced the susceptance value by this amount which must be the contribution of the blue stub. Let me repeat just before the blue stub this situation is P1 where the admittance is 0.5 plus j1 just after the blue stub the situation is P3 where the admittance is 0.5 plus j 0.14 the susceptance is less because of the addition of the blue stub the stub must have a susceptance equal to minus j 0.86 had we gone in the other direction satisfying other requirements equally well say to point P4 the susceptance there would have been more than the susceptance at point P3 and therefore the contribution of the blue stub would have been of a positive susceptance value which we have seen earlier are going to require longer short circuited transmission line steps alright therefore P3 is the correct choice and the susceptance that is required to be contributed by the blue stub is minus j 0.86 minus 0.86 corresponds to short circuited transmission line of length d2 equal to 0.138 lambda using the procedure we have just described for the single stub matching starting at the infinite admittance point moving away from the load till our requirement is satisfied we see that the transmission line stub must have a length d2 equal to 0.138 where are we now we are at point P3 we have added the blue stub just after the blue stub this is the situation now we move to plane A before the green stub is added the movement is away from the load towards the generator by a distance lambda by 8 on a constant V SWR circle there is no discontinuity in this region so reflection coefficient magnitude must remain constant so with this as the center this as the radius we draw a part of a circle starting from P3 and move in the right direction by this amount that is 90 degrees and we will automatically land on the constant conductance equal to one circle this is the way we have designed the whole thing C2 was a circle which was 90 degree which was the circle C1 rotated counter clockwise by 90 degrees so now we are moving clockwise by 90 degrees we are bound to land on a circle C1 reaching point P7 yes any question Sir the admittance is 0.5 plus j 0.140 yes the conductance is 0.5 then as we move towards plane A the conductance should remain constant why the only the reflection coefficient magnitude should remain constant that is the only requirement which we are satisfying we are moving on a constant P SWR circle and this we have seen that as the impedance transforms on a transmission line the real part or the imaginary part do not remain constant so we should not be surprised that both parts have changed could you please explain the rotation of this C1 circle to C2 okay let me just run through the solution and then the points which remain outstanding we will discuss those we were at point P3 that was just after the blue stub now we rotate counter clockwise by 90 degrees reaching point P7 and the admittance at point P7 is 1 plus j 0.73 it is on constant conductance equal to 1 circle and the susceptance of the equal to 0.73 circle intersects this at P7 therefore this is the admittance at point P7 what is this point on the transmission line this is the point at this location at plane A but before the green stub is added and naturally the green stub should have a susceptance which is minus 0.73 so that we are able to move from point P7 to point P9 on a constant conductance circle achieving our goal of matching the transmission line after the two stubs are added the stub length for the green stub so that this is the susceptance that it shows can be found out using the familiar procedure and from these considerations D1 comes out to be 0.15 lambda just as there was an alternative to point P3 in point P4 satisfying all of the requirements except for the length of the stub there will be another choice corresponding to point P7 if we start from point P4 we will end up on a different point on the circle C1 the susceptance that the green stub will be required to have at that for that choice will be plus 0.73 plus J 0.73 and will naturally require longer lengths of the transmission lines alright and this completes the solution of this double stub impedance matching problem and these are the answers for this particular problem now one question that was asked is the role of the circles C1 and C2 so let us repeat the various steps we were given an impedance value load impedance we calculated the corresponding normalized load impedance located that point as P1 on the Smith chart since we have connections in parallel we consider the corresponding admittance leading us to point P2 with value 0.4 minus J 0.8 then we move away from the load by distance lambda by 4 on a constant VSWR circle in clockwise direction in this case by coincidence taking us back to point P1 which is the admittance at plane D before the blue stub is added. Now since we are at a loss as to what to do next we start from the right hand side from the left hand side and at plane A after the green stub has been added just after it the situation should be that the normalized input admittance is 1 the green stub or for that matter the blue stub they can only alter the imaginary part of the admittance therefore just before the green stub is added the situation must be that the conductance is 1 the susceptance may have any value therefore circle C1 represents all those possibilities where conductance is 1 this is the situation of admittance that we should have if we are to accomplish impedance matching successfully using this scheme C1 represents the possibilities of admittance just before the green stub is added all these possibilities the entire circle C1 is therefore transformed towards the load by this separation yeah so that C1 is the locus of the various admittance values at plane A before the green stub is added it is required that at plane D after the blue stub is added this situation is represented by circle C2 and circle C2 is nothing but circle C1 rotated in the counter clockwise direction which is a movement towards the load by a distance corresponding to the separation between the two stubs lambda by 8 if you have any other questions we like to take those now let us have some other question yeah it hardly matters because no matter what characteristic impedance of these auxiliary transmission lines the stub transmission lines you choose you will be able to make up the required susceptance these are the normalized susceptance values the actual values can be obtained using the characteristic admittance and no matter what characteristic admittance you use you will be able to make up the required normalized admittance for the sake of simplicity and also ease in practice these will have the same characteristic impedance as the main transmission line now before we stop I want to pose a question can such say a two stub tuner as it is called match any load impedance or any load admittance or is it going to have some restriction are there likely to be some regions where double stub tuner like this with a fixed separation cannot match the load admittance or the load impedance the answer is that yes there are certain regions which turn out to be out of bounds for which the matching cannot be accomplished using this double stub tuner the region which we cannot match is going to be different for different double stub tuners for example depending on their separation between themselves and the separation from the load in this particular example let us say that we calculated the normalized impedance then the normalized admittance and then transformed to plane D before the stub is added in this problem we reached at point P1 had we reached some point let us say on this circle now any amount of movement on this constant conduct in circle would not have permitted us to intersect with circle C2 okay and the load admittance or impedance values which cause that kind of values corresponding to point P1 we will not be able to match those using such an arrangement otherwise the double stub tuner is more convenient than the single stub tuner since no movement along the transmission line of the stub lines is required we end this lecture here in this lecture we have considered how one can use the smith chart for impedance matching in particular we have considered a single stub tuner and a double stub tuner thank you.