 Welcome to the 9th lecture in the course Engineering Electromagnetics. In the lecture today we shall learn about the evolution of the Smith chart, how the Smith chart evolves from the simple relationship between the impedance and the reflection coefficient. Then we will go on to demonstrate how the Smith chart can be utilized for the impedance and the reflection coefficient calculations. And finally we will see an actual Smith chart that one uses in practice. Considering the first part of today's lecture we see how the Smith chart evolves. We have seen that there is a simple relationship that exists between the normalized input impedance and the reflection coefficient that is Z is equal to 1 plus rho L upon 1 minus rho L, where the variable Z is equal to the normalized input impedance normalized with respect to the characteristic impedance of the transmission line. And the symbol rho L stands for the reflection coefficient with appropriate phase shift at different distances at a distance L from the load impedance that is it is rho times e to the power minus 2 j beta L L being the distance from the load impedance. Both Z and rho L are complex quantities and they can be written as Z equal to r plus j x and rho L equal to u plus j v. And we have seen in the last lecture that the points in the complex Z plane particularly are greater than or equal to 0 points in the complex Z plane map or transform onto points on the complex reflection coefficient plane within the unit circle. We have considered different possibilities for points on the complex impedance plane normalized complex impedance plane and we saw that all such points as long as r is greater than or equal to 0 map or transform within or on to an area which is inscribed in a unity radius circle in the reflection coefficient plane. We consider next how we may manipulate these relations for useful purposes. We write the relation as r plus j x equal to 1 plus u plus j v upon 1 minus u minus j v and then we rationalize so to say the denominator by multiplying by 1 minus u plus j v and dividing also by the same quantity this may be put in brackets. Giving us the result 1 minus u whole squared plus v squared as far as the denominator is concerned and in the numerator we get 1 minus u squared minus v squared plus 2 j v. After this step it is easier to separate out or compare the real and imaginary parts on both sides of the equation. Thank you that is right and comparing the real and imaginary parts we get r equal to 1 minus u squared minus v squared upon 1 minus u whole squared plus v squared on one hand and x equal to 2 v upon 1 minus u whole squared plus v squared on the other hand. Next we pose the question what will be the locus of points in the complex reflection coefficient plane which have a constant value of the resistance or similarly what will be the locus of points in the complex reflection coefficient plane which have a constant value of the reactance. Now this can be seen rather easily by manipulating these mathematical expressions in a suitable manner. One can take the u and v terms on one side and try to put them in the form of perfect squares and then it will be possible to see what is the locus of such points. So quoting this result skipping the manipulation we see that we get u minus r upon r plus 1 whole squared plus v squared which is equal to 1 upon r plus 1 whole squared on one hand and similarly for the other equation we get u minus 1 whole squared plus v minus 1 upon x whole squared equal to 1 upon x whole squared. It is fairly straight forward to come to these relations starting from these and immediately one sees that assuming that r is constant here it will be a circle in the complex reflection coefficient plane with the center r upon r plus 1 and 0 coordinate and the radius as 1 upon r plus 1 and as we give or consider different values of r we will obtain circles centered at different points and with different radii in general. So constant resistance paths will be a family of circles. What happens on the other side is also something similar we will have a constant reactance paths also a family of circles with the centers 1 and 1 by x and the radius equal to 1 by x taking care that we take the positive square root so that the radius comes out to be of a positive value. Now one could consider different illustrative values of r and x and see what will be the corresponding centers and the radii and as we shall see that is the way this mixture evolves. So let us consider different values of the resistance r and the reactance x and we put r here and x here. We start with r equal to 0 which will be a constant which will be a circle with center at 0 0 and of radius 1 implying that if for example the load has 0 resistance load is only reactive the reflection coefficient magnitude will be 1. So this kind of interpretations can always be offered. Similarly for r equal to half we have center at 1 by 3 and 0 and the radius equal to 2 by 3. For r equal to 1 the center is at half 0 with radius half for r equal to 2 it is 2 by 3 0 and radius is 1 by 3 and if r becomes very large say goes to infinity then we have the center at 1 0 and the radius equal to these are some typical values of the normalized resistance. Similarly what can complete the other side for x values taking on similar values for x equal to 0 the center is at 1 infinity and the radius is also infinity. For x equal to half the center is at 1 2 and radius is 2 for x equal to 1 the center is at 1 1 and the radius also is 1 for x equal to 2 it is 1 1 half and half for x equal to infinity it is 1 0 with radius equal to 0. Now these illustrative values of r and x are going to be enough for us to demonstrate what a Smith chart looks like and actually a Smith chart is nothing but families of constant resistance and constant reactance circles. When these are plotted methodically then we get Smith chart. A Smith chart is a plot of constant resistance and reactance circles in the reflection coefficient plane that is what a Smith chart is. Now before we show this family of circles we see a very simple relationship between impedance Smith chart and the admittance Smith chart. So far the way we have described it might appear that the Smith chart is an impedance chart. We are talking about the resistance and reactance. However we have the relation for impedance as z equal to 1 plus rho L upon 1 minus rho L and if we consider the corresponding admittances and realize that the normalized admittance is equal to the reciprocal of the normalized impedance. Then we see that why the normalized admittance is equal to 1 minus rho L upon 1 plus rho L and if we substitute for minus rho L a new quantity say rho L prime equal to minus rho L then y is simply 1 plus rho L prime upon 1 minus rho L prime which is an expression which has an identical form to the expression for the normalized impedance. And therefore is going to have similar behavior that is if one plotted the constant conductance paths they will come out a family of circles and similarly if one plotted the constant susceptance paths they will come out a family of circles with identical paths for identical numerical values. And therefore the same Smith chart serves as an impedance chart usually normalized impedance chart and a normalized admittance chart. All that one needs to do is to transform the reflection coefficient by this amount that is by 180 degrees. In the complex reflection coefficient plane minus 1 is equivalent to a rotation by 180 degrees. So, if one has considered some impedance point in the on the Smith chart to obtain the corresponding admittance all that one needs to do is to consider the diametrically opposite point with the same value of the magnitude of the reflection coefficient. So, that aspect is extremely simple on the Smith chart and therefore the same chart serves as an impedance chart or as an admittance chart. Now let us show you on the OHP a plot of these illustrative values. We first consider the R equal to 0 circle and you see that that has a center at 00 this is the complex UV plane. So, center at 00 radius is 1. So, with some scale you draw a circle with radius 1 and all points on the circle will represent R equal to 0. What will be different at different points on this circle? The difference will be in the reactance values. Similarly one can see that R equal to half circle is this it is obtained by considering the center at 1 by 3 0 somewhere here and the radius equal to 2 by 3. So, this way you get the R equal to half circle and so on R equal to 1 R equal to 2 and R equal to infinity is just a point here with radius equal to 0 essentially the open circuit point. Similarly the X equal to 0 point is obtained or path is obtained by considering a center at 1 infinity. So, that is beyond this slide and the radius also is infinity and therefore this is the straight line U axis is the X equal to 0 part of the constant X equal to 0 circle alright. Next one can consider the X equal to half value and that has a center at 1 2 with radius equal to 2 and therefore X equal to half plots like this. One can easily see mathematically that if one considered X equal to minus half that will have a center at 1 minus 2 with radius equal to 2 and therefore it will have a corresponding symmetric arc of the circle representing X equal to minus half and similarly one goes on to X equal to 1, X equal to 2 and symmetrically X equal to minus 1 and X equal to minus 2. And this is this can be considered to be the skeleton of the actual Smith chart. We will see the actual Smith chart a little later, but this will be somewhat more convenient for us to consider the various applications to which the Smith chart can be put. There are a few other things that are noted down on this as we proceed we will try to explain their meaning. The first application that we consider is that we are given the impedance given Z okay and the question that is posed to us is find the reflection coefficient or more specifically find rho n. That is the perhaps the simplest application for a Smith chart. Given the impedance or the normalized impedance find the reflection coefficient. We know how it can be done using the formula, but how it can be done on the Smith chart that is the question. So we consider the example let Z be equal to 1 plus J and please bear in mind these are normalized values usually the Smith chart that is available in practice is also a normalized Smith chart. Now how would one locate this impedance point on the Smith chart? Not difficult the real part is 1 so R is equal to 1. So it should be some point on the R equal to 1 circle alright and what is the reactance part? X is also equal to 1 and therefore it is the intersection of the R equal to 1 and the X equal to 1 path. We can mark it here okay relatively thick and call this point A. This is how one would mark various impedances on the Smith chart alright. Now what is the reflection coefficient? The Smith chart is already a plot of the constant R constant X circles in the reflection coefficient plane. The distance from the center will be the magnitude of the reflection coefficient and therefore we say that the reflection coefficient magnitude for this example this is rho or rho L. We have seen that the difference between rho L and rho is simply that of phase therefore this is the magnitude of reflection coefficient the distance of point A from the center. Taking the scale into account the scale is that this distance is equal to 1. So once that scale is known it is obviously available on any Smith chart this distance can be measured and it will be some quantity less than 1. For this example this quantity is point 4 alright. The phase can be read by considering this angle with respect to the positive u axis okay. This is the angle rho or angle rho L to be more specific okay. The angle measured in the counter clockwise direction with respect to the positive u axis that is how angles are measured in the complex plane alright. Now in this example this angle comes out to be 1.11 radians or the equivalent value in degrees and most Smith charts are graduated on their periphery in terms of degrees and wavelengths and therefore actual measurements are not required one just reads the values of the Smith chart okay. The corresponding value in degrees will be roughly 63, 64 degrees alright. So this is one straightforward application of the Smith chart that is we are given the impedance value. We can convert it into the normalized value if necessary and then read the magnitude of the reflection coefficient and the phase of the reflection coefficient alright. We take up the next application of the Smith chart and that is given let us say the load impedance alright and we are required to find say z at different distances from the load impedance or in other words how the impedance or the load impedance transfers along the transmission line at different distances from the load impedance what will be the impedance value that is a typical problem that one encounters in transmission lines. How the Smith chart may be utilized for this purpose. Now once again we consider some typical value say ZL is equal to 1 plus J considering the same example seeing how convenient it is. So this is the load point say represented once again by point A. Now comes a very important question depending on the question posed that is find the input impedance at such and such distance from the load impedance we have to move from the load point A is the load point representing the load impedance. Now we have to move on the transmission line away from the load point how this movement is to be accomplished okay that is the natural question. Now we realize that rho L which is V0 minus e to the power minus J beta L upon V0 plus e to the power J beta L as we move away from the load it is L that is increasing and that only affects the phase of the reflection coefficient rho L it does not affect the magnitude of the reflection coefficient and therefore unless one encounters some discontinuity or some other change as we move on a transmission line say away from a load point or some particular discontinuity we will move on a constant magnitude of reflection coefficient circle that is the center will be this okay and this becomes the radius and we will move on a constant magnitude of reflection coefficient circle okay that answers one question about how we are going to move or what kind of movement is going to take place. Since the magnitude is not affected as we move away from a load impedance the movement will be on a constant reflection coefficient magnitude circle right that tells us the path. Now if it is the load impedance that is given to you depending on different problems you will consider different load impedance values so they will plot appropriately at different points on the Smith chart that is quite fine but how are you going to move the movement is going to be on a constant reflection coefficient magnitude circle alright this is point number 1 the second point obviously will be how much do we move how much is answered here itself the angle is twice beta L normally L will be specified in wavelengths okay and therefore beta L will be easily calculated and depending on this value you move by that much amount on this Smith chart on the constant reflection coefficient magnitude circle. So the amount is clear then comes the direction one could move clockwise or counter clockwise in which direction would one move okay then realizing that rho L is equal to rho e to the power minus 2j beta L as L decreases the angle becomes less negative or in other words the angle becomes more positive alright and decreasing L takes you towards a load and therefore the increasing values of the angle will take us towards a load and this is clearly marked on the Smith chart the direction in which one has to move on the path earlier identified and the amount earlier identified the direction is this this movement takes you towards a load this movement therefore by the same argument the opposite movement takes you towards the generator and these two directions are clearly marked from a particular point if one wants to move towards the generator one will move on a clockwise direction if one wants to move towards a load supposing it is not a load point it is some other point then one will move in the counter clockwise direction alright so it is already shown here the clockwise movement is towards the generator now we know how to move how much to move and in which direction to move alright coming back to our example that it is a load impedance which is given to you given to us 1 plus j and we require to find out z at a distance L equal to lambda by 4 let us see so what is the input impedance seen for such a load at such a distance let us say that is the question posed to us then we mark point a locate the load impedance move with this as the center and this as the radius on a constant reflection coefficient magnitude circle move by how much amount to beta L for this particular example it will go to L is lambda by 4 beta is 2 pi by lambda so this will go to pi radians or 180 degrees so once again a setup problem for the sake of simplicity and one will move in the circular fashion and reach a point like this on a constant reflection coefficient magnitude circle by 180 degrees in which direction towards the generator this was given to be the load impedance so we have to move away from the load towards the generator and therefore it is a clockwise movement as identified earlier what is this value that we get one can read it here for the simple example it is on r equal to half and x equal to minus half and therefore the input impedance in this case comes out to be z i is 0.5 minus j 0.5 once the application is understood this is a very fast process as compared to the complex calculations on the calculator. Now the converse is also equally feasible that is given the input impedance at a certain distance from the load determine the load by similar arguments one will start at this point and then retrace this path and reach this point and therefore transformation of impedance is possible in a fairly straight forward manner on the Smith chart. Let us take the third application of the Smith chart and also you would have noticed that the peripheries marked in terms of wavelengths so that different distances in terms of wavelengths can be easily marked can be different amounts of movements can be easily carried out. The third application we say is that we are given the load impedance and the way we have written it it should be clear that it is the normalized load impedance obtaining which is no problem at all. We need to find out let us say the standing wave ratio the locations V max V min those of you who have attempted these calculations would recall that these are not so easy they take some effort but on the Smith chart this can be handled very easily and one actually gets a feel of what is happening on the transmission line this was one of the advantages that we quoted for these graphical aids Smith chart etc. The standard example we have been taking is that let Z L be 1 plus j which certainly is only for illustration one can use the Smith chart for all other values also alright now how do we find the standing wave ratio no that is all right given Z L you can calculate the reflection coefficient and calculate the standing wave ratio those mathematical formula there they have their own value using the Smith chart how does not do it we use the simple fact that at V max the input impedance is completely resistive and is equal to S times Z naught where S is the standing wave ratio therefore the normalized Z or Z in at all locations of V max is going to be equal to the standing wave ratio S so the simple procedure will be to move from the given load point on the transmission line we have already seen what this movement should be like it should be on a constant reflection coefficient magnitude circle the direction also we have made out whether it is towards the generator or towards the load alright so this is the load point we start moving towards the generator so in the clockwise direction and let us say this is the movement that takes place something like this and we stop at a point where the impedance is completely real that is wherever we intersect the u axis okay in this case we stop here where the reading here is if you look at a detailed Smith chart 2.6 so for this problem we see that S is equal to 2.6 this Smith chart very clearly brings out the fact that this is going to be periodic phenomenon okay if we continue to rotate by another 360 degrees which amounts to a movement in terms of wavelength by half a wavelength once again we will come to a voltage maximum therefore unless something else changes on the transmission line there will be these periodic maximum voltage maximum formed on the transmission line everywhere the input impedance seen at these locations will be S times Z now now you could come up with the argument that for this purpose we could have used the voltage minimum locations and the information that at V min the input impedance is equal to 1 by S which is perfectly valid equally valid and for that what one would do is one would continue movement towards the generator from the load point and not stop here because this corresponds to the location of a voltage maximum and continue on the same path and come somewhere here on the negative u axis where the values are less than 1 and therefore the reciprocal of that will once again give us an identical value for the standing wave ratio no 2.6 so see when the value itself was the standing wave ratio and that corresponded to the voltage maximum location for the voltage minimum it has to be the reciprocal alright so the standing wave ratio can be read more or less read on the smith chart S is never less than 1 that is one thing see now let us read the markings on the u axis 0 r equal to 0 r equal to half 0.5 r equal to 1 2 and so on it will go on to infinity so from 0 to the right of this 0 point the center of the complex reflection coefficient plane the values of r normalized values of r are greater than 1 and to the left they are less than 1 and perhaps the reason is there before us now this information should actually be enough to make out the locations of voltage maximum as well the distance that has been moved from the load impedance up to this location of the voltage maximum will give us the location of the voltage maximum as I mentioned the peripheries graduated in terms of wavelengths so this movement that we made in coming from the load impedance to the location of the first voltage maximum will tell us the location of the voltage maximum and in this case v max is located at a distance which is 0.088 lambda from the load and obviously if one adds another lambda by 4 one will get the first voltage minimum from the load and then you keep on getting adding lambda by 2 distances and you will keep on getting more maximum and minimum alright so these are the three representative very simple applications of the smith chart and we restricted ourselves to this limited form of smith chart for the sake of convenience actually what one gets to use is the following and we will try to look at various features of this chart one by one and I suggest that you look at copy of the smith chart and look at various regions as we go along first of all you see that there are 1, 2, 3, 4 circles on the outermost peripheries okay and these are marked as movement towards the generator clockwise and similarly movement towards the load okay the requirement for this we have just seen these outermost two circles are marked in terms of wavelengths so that the movements can be easily read in terms of wavelengths so we start at 0 go to a quarter wavelength here on the other side and complete the circle by coming back to this point that is half a wavelength since the angles are 2 beta l half a wavelength completes 360 degrees so there are two circles since there can be a clockwise movement or a counter clockwise movement depending on the requirement in addition the circle inner to these two is marked in terms of degrees okay so there are graduations of 10 degrees each 150, 140, 130, 120 etc. starting from 0 okay 0 to plus 180 degrees on this side and 0 to minus 180 degrees on this side so plus pi to minus pi radiance then of course we can consider the family of constant resistance circles as you would recall this the innermost circle here okay this represents the r equal to 0 constant resistance circle and its intersection on this real u axis is at 0 alright similarly one goes on in the on the right hand side on this axis so 0.1, 0.2 and so on up to 1 which is the center of the reflection coefficient plane and then 1.2, 1.4, 1.6, 2 then 3, 4 and this way going on to infinity and there are all these circles complete circles representing the constant resistance family of circles alright similarly these arcs of circles on this side or on the other side they represent the parts of constant reactance circles so starting from 0.1, 0.2, 0.3 you go up to 1 and then values greater than 1 going up to infinity here and similarly negative values on the other side and of course there are a large number of curves some full circles some arcs so that one can make finer readings on the Smith chart okay this is where we stop this lecture in this lecture we have seen how the Smith chart evolves what are the various simple applications of the Smith chart and what a practical Smith chart looks like thank you.