 38th lecture in the course Engineering Electromagnetics. You would recall that in the last lecture we initiated discussion on a Hertzian dipole or what is also known as the alternating current element. We continue our discussions on this Hertzian dipole or the alternating current element today and the topics for discussion today are the following. We will consider the power radiated by the Hertzian dipole and then we will go on to define what is called the radiation resistance. And then you will recall that we said that the results that we obtained for the Hertzian dipole are going to be applicable to some electrically short antennas as well. So that is what we will consider next, application of these results to electrically short antennas. So that is our agenda for today's lecture. Going back to the Hertzian dipole that we have been considering. This is the way we considered a very short radiator and we located it at the origin of the spherical coordinate system and then following the procedure lay down earlier we obtained the following field expressions. We saw that the radiation field due to the Hertzian dipole has the H5ER and the Etheta field components which are non-zero and the other three field components turn out to be zero. This is not always the case, this is the case only for this particular case and as you would recall there are many simplifications that we have built in this particular case. In the last lecture we considered the explanation or the significance for the various terms that appear in the radiation field due to the Hertzian dipole and today we go on to consider the power that is going to be radiated by the Hertzian dipole. Any amount of information that can be transmitted through an antenna will require that some power is received at the other end that is by the receiving antenna and therefore power transfer is an essential aspect involved when we use antennas. So what is the power radiated that is a question of very great significance. As far as the power radiated is concerned we will use the simple expressions that we have developed earlier and let us first consider the complex pointing vector. What is the complex pointing vector? It is obtained as half E cross h star where star stands for the complex conjugate of the phasor magnetic field intensity expression. Let us develop this further for the spherical coordinates in which we have obtained the field components for the Hertzian dipole. What we will have is the following expression the various components are going to be half and then we will have since we have the r component of the electric field it will be E r times h phi star which will be a direction which is theta cap and then we are also going to have E theta times h phi star which will be the radial direction. That is what are going to be the components of the pointing vector in the spherical coordinates. We do not have any phi component from the field components one can see that and we have a radial component and a theta component that is indicated. So let us try to calculate the theta component P theta which is half and then minus E r h phi star alright. The field components are available here the E r and the h phi field components and let us write down the expression for this and this is going to be please check that we do not go wrong it will be j eta i l square and then sin theta cos theta divided by 8 pi squared beta right and then within the brackets we are going to get beta squared by r cube the product of these two terms and the product of these two terms gives us minus j beta by r to the power 4 why minus j beta because we are taking the complex conjugate of the phi component of the magnetic field. Similarly the product of this term with the other two terms gives us plus j beta r to the power 4 and finally plus 1 by r to the power 5 and what happens to the exponential term they get cancelled there is a half outside so let us include that here okay thank you. Now as far as these two terms are concerned they cancel each other and therefore they simply drop out of consideration. Then yes Ravi well you are forgetting I think that it is h phi star so once that is taken into account the sign comes out plus any other problem okay now once we recall that as far as the average power flow is concerned we are going to require the real part of the entire expression and since there is a j term right in front the real part is 0 for all distances for all conditions right and therefore we note that the real of p theta which is going to be the p theta average is 0 which means that the terms which contribute to the theta component of the point in vector the theta component of the point in vector may have an instantaneous value and it may just indicate a surging back and forth of the power in the theta direction in one part of the cycle some power flows in one direction but in the rest of the cycle the same power is transferred back in the opposite direction and therefore there is no net average transfer of power in the theta direction fine. Next therefore in a similar manner let us consider the r component of the pointing vector which is half of e theta h phi star these are the three components you will have to keep these in mind and okay let us try it this way I think that should be alright and this comes out to be minus half and then j eta il squared sin squared theta divided by 4 pi squared beta alright and then we are going to get six terms in the brackets and let us write down at least some of these first let us consider the product of this term with the rest of the terms and what we get is plus j beta cube by r squared okay then the product of these two terms which will give us again plus beta squared by r cube and finally these two terms which give us minus j beta by r to the power 4 similarly let us take up this term which gives us minus beta squared by r cube and plus j beta by r to the power 4 and finally plus 1 by r to the power 5 alright six terms as a product of these two terms and these three terms and here once again we find that two pairs of terms cancel each other and what we are left with is simply j beta cube by r squared and plus 1 by r to the power 5 and once again if we consider the average value of this pointing vector the radial component of the pointing vector it is only the first term which is going to contribute to the average power flow the second term once again is going to drop out when we consider the real part of this component right and therefore if we note it down here itself the average PR average is going to come out going to be IL whole squared maybe an eta here and then beta squared sin squared theta by 32 pi squared and what else and r squared okay in the radial direction now this is very interesting we see that at large distances it is these 1 by r terms which are going to be significant from the point of view of magnitude the other terms will become less and less significant we have seen earlier where the induction term becomes starts becoming less than the distant field term or the radiation term right but even very close to the antenna these other terms do not contribute to average power flow which are the terms which are contributing to the average power flow finally it is the 1 by r squared term in PR which is contributing to average power flow and which are the component terms in the magnetic field components which are contributing to this term those are this one and this one only right and therefore from one point of view we call these terms the distant field term from this point of view these are also called the radiation terms or the radiation field terms because these are the ones which contribute to average power flow away from the antenna right and therefore the average power flow which is in the radial direction varies as 1 by r squared and of course then it has other terms and other types of variations now is this just an accident that the power flow has come out like this and it is the 1 by r terms which are contributing to the average power flow that is not so actually one can advance a very simple argument for making out logically that it should be this kind of terms which should contribute to average power flow for this purpose let us consider a point radiator at large distances from an antenna the antenna can be considered to be fairly similar to a point radiator particularly the kind of radiator that we are considering then let us consider two spherical surfaces enclosing the antenna spherical surfaces of different radii let us say one surface is like this it is completely symmetric concentric with the source and let there be another spherical surface let us say this is of radius r 1 this is of radius r 2 now let the antenna radiate some average power we call this p average is flowing radially away from the antenna this power that is radiated has one unique value and therefore whether we consider this power coming out of this surface or power coming out of this surface the total value must remain the same unless there is some regular dissipation in the medium or there is regular generation or amplification of power which is really the case and therefore the power that comes out of this surface must equal the power that comes out of this surface which clearly implies that the average power density divided by area must vary as 1 by r square otherwise the power conservation will be violated now we recall that p average or the pointing vector magnitude is proportional to magnitude e square or magnitude h square and since this quantity must vary as 1 by r square each one of these quantities either e or h they must vary as 1 by r at least the terms which contribute to average power flow if it is a temporary instantaneous surging back and forth of power we do not bother about that because that is not going to violate conservation of power but the average power flow must satisfy this kind of conditions and therefore we see that based on this very simple argument it is possible to make out which kind of terms in the radiation field will contribute to average power flow and that is precisely what we are getting here right and therefore in work with the antenna antennas most of the time it will be this distant field term or the radiation term which will be of significance the other terms which are significant at close distances also play in a point of role but depending on the application and in most practical applications will be concerned about this distant field term this setup has become at least it appears quite complicated and if we can identify a simple relationship at least between the radiation field terms the task will be considerably simplified and that can be done in a simple manner if you consider the radiation terms in the magnetic field and in the electric field okay let us say that we are considering the a E theta term which is proportional to 1 by r divided by the h5 term which also is proportional to 1 by r these expressions are long but we are considering only the terms which vary as 1 by r in that case what do we get okay if you look at the expression closely then what we get is simply and if you consider the magnitude well not even magnitude is required you get simply the value which is eta the intrinsic impedance of the medium in which the antenna in this case the hertz in dipole is located this is simply the intrinsic impedance of the medium that we have been considering earlier which holds good for plane waves also and for the radiation field terms in the antenna also this relation is holding true this is equal to square root of mu by epsilon considering that the medium has no conductivity if it does then the intrinsic impedance will be suitably modified so there exists a very simple relationship between the radiation field terms and if we get one the other one can be simply calculated okay which means that our calculation of the electric field from the curl of edge is unnecessary as long as we are interested only in the radiation field term so this kind of simpler procedures exist and they will be pointed out as we proceed the expression that we have got for the average value of the pointing vector is only the power density power density per unit area right but we would be interested in the total power that the antenna radiates how are we going to calculate the total power radiated we know the power density radial power density away from the antenna and therefore if we integrated on some closed surface we should be able to get the total power radiated for this purpose we consider the following diagram we say that the antenna is located at the center of the at the origin of the spherical coordinate system and then we consider a closed surface around it while in principle any closed surface would do a spherical surface which is concentric with the antenna is most convenient okay because then one can exploit the symmetry of the radiation field if any which exists for example in this case we notice that the power density that we have to work with has no phi variation and therefore we can consider an element of area for the purpose of integration which is symmetric with respect to phi and therefore we consider this element of area which is in the form of a thin band a thin circular band okay what is the thickness of the width of the circular band the width is r times d theta and what is the circumference of this band the radius is r sin theta and therefore circumference is 2 pi r sin theta okay so that circumference times the width gives us the area of this element of area for the purpose of integration. Now we have got the power density and therefore we can get the total power radiated total power radiated wr is going to be p.da in general integrated over a surface which encloses the antenna and the units would be watts. Now the pointing vector that we have got has only one radial component one component the radial component and the element of area that we have selected also if we associate a direction with it will be a radially outward direction and therefore the dot product will become fairly simple and we can write this as and to calculate the average power it is the average power that we are interested in we can add the subscripts average here and therefore it becomes p r average times the area element integrated over the closed surface. So many watts will be the result the element of area that we have got okay we will put it down next this becomes eta i l whole squared beta squared sin squared theta divided by 32 pi squared r squared that is the expression for p r average and then the expression for the area element as we pointed out is 2 pi r sin theta and then multiplied by the width that is r d theta therefore this becomes r squared and then d theta right what is the limit of integration we must cover the entire spherical surface that that is how the limit of integration should be and the entire surface is covered if we let theta vary over what limits 0 to pi that will cover the entire spherical surface and putting the limits accordingly that is 0 to pi one can integrate this the expression is fairly straightforward what we get is eta times beta squared i l whole squared by 16 pi that is the factor that comes out of the integral sin and then we have the integral of sin cube theta d theta integrated from 0 to pi and already as we had argued out earlier you find that the average radiated power is independent of r no matter at what value of r you calculate it you will get the same value fine so we are being consistent as far as this integral is concerned and I hope most of you will try it out yourself this value is 4 by 3 and therefore we get the result for the average power radiated as eta beta squared i l whole squared by 12 alright so this is the average power radiated it can be rewritten in a slightly different manner by recalling that beta is 2 pi by lambda and eta is 120 pi if you are dealing with free space if the antenna is radiating in free space then eta is 120 pi so based on this we get the expression for average radiated power as 40 pi squared l by lambda whole squared into i squared right in terms of the current that is fed to the antenna and the size of the antenna relative to the wave length this relatives with respect to the wave length will always appear in antenna work okay and you will notice it at subsequent stages also we can rewrite this average radiated power by using not the amplitude of current amplitude of the sinusoidal time varying current but using the rms value of the current and in that case it becomes equal to 80 pi squared l by lambda whole squared times what we may call i rms or i effective squared since i effective is i by root 2 okay now of course it says that the longer the antenna in terms of wave length the more power it can radiate for the same current fed to it right of course the approximation that we made that the antenna is much shorter than a wave length that should not be violated too badly otherwise this result will become inaccurate now at this point we are in a position to introduce the concept of what is called the radiation resistance which becomes parameter in terms of which one can assess the performance of the antenna the performance of the antenna as a radiator how effective the antenna is as a radiator that can be made out if one can specify its radiation resistance we introduce the concept of radiation resistance by equating this average radiated power to the following expression we say that let wr average be equal to i effective squared times the radiation resistance of the antenna what will be this radiation resistance this will be a fictitious resistance which when connected in place of the antenna and when the same current is fed to it it will dissipate the same amount of power as the antenna radiates and therefore if we know the radiation resistance of different antennas we will be able to compare their efficiency effectiveness as a radiator the higher the radiation resistance the more efficiently more effectively the antenna radiates fine so in this case for the herzen dipole what is the radiation resistance the radiation resistance is simply a t pi squared l by lambda whole squared since l must be much less than a wavelength the value of this radiation resistance is not very large and that is going to happen whenever the antenna is much shorter than a wavelength in fact when we introduced the concept of the mechanism of radiation we said that the antenna is nothing but something which is made deliberately large in terms of wavelength so that the separation between the charges or the currents increases and if you are keeping the antenna short naturally it will not be very effective as a radiator that part that constant that thing apart this is the radiation resistance for the herzen dipole or the alternating current element we can note down this fact on the board the radiation resistance for the herzen dipole which is a t pi squared and l by lambda whole squared so many ohms alright now when we were providing the motivation for studying the herzen dipole we said that first of all any antenna sustaining some current can be looked at as a combination of this kind of building blocks and then in addition we said that some simple antennas which are of great practical importance can be considered to be similar to this herzen dipole. So, this is what we come to next where can we apply in practice the results that we have got for the herzen dipole or for the alternating current element whenever we have electrically short antennas mind you it is not the physical length that we are worried about whenever we have electrically short antennas electrically short wire antenna to be more precise. So, that the various simplifications that we were able to make in the derivation of the radiation fields and the subsequent results are not violated then these results would be applicable electrically short antennas now there may appear a seeming contradiction in what we have been saying on one hand we said that antenna is nothing but something which is made deliberately large in terms of wavelength to increase it is a possibility of radiation on the other hand we are talking about electrically short antennas. So, at the outset we can keep this in mind that the if the antenna is short in terms of wavelength it will not be very effective as a radiator, but there may be circumstances where we are forced to use electrically short antennas or by improving some other parameter in the communication system we are able to make to with an electrically short antenna. I will give you a very simple example consider the AM broadcast signals amplitude modulated broadcast what is the frequency range used for this kind of broadcast signals the frequency range is 550 to 1650 kilohertz. Let us take a typical frequency in this frequency range in this frequency band which is used for this purpose very commonly say 1000 kilohertz. So, which is 1000 kilohertz or 1 megahertz what is the corresponding wavelength some typical frequency we have taken the corresponding wavelength here is 300 meters. Now, if you want to build an antenna which is significant in terms of wavelength it will have to be hundreds of meters which structure will have its own practical problems. And therefore, you have to make to with the antennas which are tens of meters long. So, that they are much less than a wavelength and therefore, they will become although physically large, but electrically short antennas. Also antennas which are required for consumer applications or which have to be mounted on vehicles they have to be light and small. And then the size becomes an important consideration and therefore, in many such situations we do use electrically short antennas. In fact, if we consider the total number of antennas which are used the electrically short antennas may out number any other type of antennas. Because of their simplicity and ease of operation. So, this is a very important class of antennas and one of the most simple ones to deal with also. There are two types of antennas which are used in this category one is the dipole antenna and the other is the let us write it here monopole antenna. What do these antennas look like? The dipole antenna is nothing but a system which looks like this. There is a definition of this antenna which is been given in the eye triply handbook. The dipole antenna is a straight radiator usually fed in the middle. It is made of a conducting wire. The length that we talk about for the antenna is the total length from end to end. And the direction of maximum radiation is in the plane normal to the antenna axis. All dipoles share this property. We may have dipoles of different lengths, but here we are talking about the dipoles where L is much less than a wavelength. In fact, we can extend the results of the Hertzian dipole for application to the dipole antennas if L does not exceed lambda by 4. Otherwise, there will be too much of an error in such an extension. Now, how do we extend the results? Do we simply replace the L in the expression for the radiation resistance by the actual length? That is not possible because the practical dipole although it is electrically short does not support a uniform current. The radiation resistance expression we got earlier was assuming that the current is uniform over the entire length. And we were able to make that assumption because it was a very short antenna. For any physically realizable length, the current distribution will not be uniform. In fact, at the ends the current must drop to 0 from the point of view of continuity. And therefore, actually the current distribution may be approximated by a triangular kind of current distribution which is maximum at the center and then goes to 0 towards the ends. As long as the length is small, we can make this kind of approximation for the current distribution. What is the direction of current on the two arms of the dipole? That can be put down. For example, if this is the transmission line feeding the antenna, then the current in both arms is in the same direction which is very interesting. And therefore, the radiation can interfere constructively particularly in the plane normal to the antenna axis. While the actual current distribution is more like this triangular shape that we have shown, we can say that if this is I naught, we can approximate this by an effective uniform current which is say I naught by 2. We are just considering what is the area under the current distribution curve and we are equating that in the two cases. So, on that basis, a uniform current of I naught by 2 can be assumed to be flowing on the antenna for a current at the antenna terminals which is I naught. Compared to an antenna which had a uniform current I naught everywhere, which is how things would be if we just extended the results we got earlier, the fields due to the antenna which sustains a current I naught by 2 over its entire length will be reduced by a factor of 2 everywhere. I naught, I will go to I naught by 2 or I naught will go to I naught by 2. Since the field components are reduced by a factor of 2, the power radiated will reduce by a factor of 4 and therefore, for the radiation resistance we can write no that is not so. Power radiated is proportional to I square field components are proportional to I and therefore, on that basis the radiation resistance can be written as one fourth of the radiation resistance that we got on the basis of the uniform current distribution. So, that it is 20 pi squared L by lambda whole squared that many ohms or approximately 200 times L by lambda whole square. Even if we go to the upper limit of the applicability of this consideration, we find that the radiation resistance is just a little more than 10 ohms or so, which is a fact one has to live with as long as one is using an electrically short antenna. I think we will take up the monopole antenna in the next lecture, if you have any questions we can talk about those now. Okay then, thank you.