 Welcome to the 39th lecture in the course Engineering Electromagnetics. The topics for discussion today are monopole antennas and then we go on to longer antennas specifically the half wave dipole antennas. You would recall that we considered short dipole antennas last time. Today we consider the corresponding monopole antennas. A monopole antenna consists of one half of a dipole, one half of a dipole antenna mounted above the earth or a conducting ground plate. That is how we define a monopole antenna. The length is usually restricted to lambda by 4 and of course if it is an electrically short monopole antenna the length is going to be even smaller less than lambda by 4. The monopole antenna is one of the simplest antennas that one may have and it has important applications for example in broadcasting in the AM band where the wavelengths corresponding to this frequency band are very large and therefore one has to use a very simple antenna. And the other important application for the monopole antennas is in land mobile communication services where the monopole antenna is used as the base station antenna. So before we consider the quantitative expressions for the monopole, let me show you how in these practical applications the antenna may look like. For the AM broadcast the antenna is going to take the shape of a tower. As we have just seen the wavelengths are of the order of hundreds of meters. So even if this tower is a quarter wavelength it is going to be a considerably large structure and to provide mechanical strength it will have to be in the form of a big tower made out of trusses etcetera. It is insulated from the ground and is fed by a coaxial conductor in this manner. Because of the size of the structure it will have to be supported with the help of guy wires. And so that induced currents on these guy wires do not corrupt the radiation or do not affect the radiation pattern and the other properties. There are insulators inserted at regular intervals on these guy wires themselves. Similarly as a base station in the land mobile communication where the antenna may be mounted on top of a high building since the earth is not available for mounting. One may simulate the effect of a conducting plane with the help of this kind of four radial wires which act as a ground screen fairly effectively. For the case of the broadcast antennas where the efficiency is very important and if the earth beneath the antenna does not have sufficiently high conductivity. So that the ground here can be approximated as a perfect conductor. One can take a remedial step such that one creates an artificial ground screen by putting in the earth just beneath the antenna this kind of radial wires. These will be large in numbers and if they have a size which is 0.25 to 0.35 wavelengths then these effectively act as a ground plane. So these are some practical measures that are taken in the use of the monopole antennas. Going back to obtaining some quantitative results you would recall that last time we considered a short dipole antenna and as long as the length of the antenna the overall length was less than a quarter wavelength. We were able to approximate the current distribution as a triangular function and then we considered what would be an effective uniform value of the current on such a length and therefore we extended the results for the alternating current element or the hertz in dipole which could be applicable to a practical radiator like this short dipole. The monopole antenna on the other hand is going to look like this a short vertical wire mounted on a ground plane fed by a coaxial cable that is alright. Now how are we going to analyze this antenna or what is the relationship between this antenna and the previous dipole antenna that we have considered? It can be shown rather simply that actually this monopole antenna is equivalent to a corresponding dipole whose total length is twice the height of the monopole antenna. How we may show that let us consider the current flowing let us say on the actual monopole antenna and let us say that at some instant of time the charges here are positive and negative like this at some two points on the actually physically existing monopole antenna and let the current flow in this direction from the positive charge to the negative charge. Now if we assume that this surface on which the antenna is mounted is a perfect conductor a ground plane then one can apply the image theory and see that this positive charge will have an image which is located at the same distance below the surface and has an opposite polarity like this and this charge similarly images into a charge which is of opposite polarity and is situated at an identical distance below the surface of the earth. Now the current which is flowing from the positive to the negative charge may once again be considered to be like this which is nothing but the kind of current that flows on the corresponding dipole antenna with having current directions which are similar in both the arms of the dipole and therefore the monopole antenna corresponds exactly to a corresponding dipole antenna. Yes Ravi well you just apply the image principle yeah no we assume that the current is flowing from the positive charge to the negative charge the same thing is applicable to the image where will be the image of this positive charge it will be here and with an opposite sign right yes or no. So similarly the image of this charge will be here and the current direction from positive to negative remains the same alternatively one could say that the fields that are radiated from this actual antenna directly and the ones that are reflected from the surface of the earth those reflected fields are simulated by considering this kind of an image of the antenna the two are identical points of view this of course is very simple to take into account. Now is there any difference in the monopole antenna and the corresponding dipole antenna since it appears that the two are identical the difference is only in the fact that the actual monopole antenna radiates fields only above this perfect conductor below this perfect conductor what exists in is an image which is a concept which helps us to find out the total fields here but actually there is no radiation below this surface right. So that is the difference between the monopole and its corresponding dipole the monopole will radiate power which is half that of the corresponding dipole because the monopole is radiating only in the half space as compared to the corresponding dipole. If we take this difference into account then we can work with the monopole just as we do with the corresponding dipole you would recall that the radiation resistance of the alternating current element was obtained to be 80 pi squared L by lambda whole square which for the dipole since the dipole a practical dipole does not support a uniform current went down to 20 pi squared L by lambda whole square and now we are saying that the monopole is identical to its corresponding dipole except that it radiates half the power as the corresponding dipole does and therefore for the monopole antenna the radiation resistance is going to be how much 10 times pi squared L by lambda whole square and since what exists physically is not L what exists physically is H and L is twice H therefore this is going to be 40 pi squared H by lambda whole square or approximately 400 times the square of the ratio of H and the faith length at which the antenna is operating. Now since we have extended the results for the hertz in dipole or the alternating current element which were derived on the basis of uniform current there is a limitation on this result that is up to what value of H this result would be we will have a sufficient accuracy and that limitation is that H should not exceed lambda by 8 so that the corresponding dipole does not exceed lambda by 4 which was the same restriction we put on the short dipole antennas with this kind of restriction or with this kind of short dipoles one can see that the radiation resistance is not going to be very large but still this kind of antennas are used because of the convenience and the cost and that kind of considerations. What do we do if we want to increase the effectiveness of the antenna as a radiator a simple step would be to increase the dimensions increase the length in terms of wave length and very important antenna that one has is the monopole antenna which has a quarter wave length height and the corresponding dipole antenna which is a half wave length height which will be more effective as a radiator that is what we want that is what we expect. So let us consider the longer antennas and for this purpose we consider again a wire antenna a linear antenna located at the origin of the coordinate system in this manner so that the origin is symmetrically located with respect to the length of the antenna the antenna has an overall length L to begin with we will keep the considerations general and later on for the sake of simplicity we will obtain results when this length is half a wave length what is the what is going to be the objective the objective is going to be to find out the fields at a distant point P which itself has coordinates are theta phi the dipole antenna is fed with the help of this kind of connections may be through a transmission line and there will be current on the two arms of the antenna which is going to be as we have seen earlier of this nature that is similarly directed in both the arms. We will see more general current directions later but right now I think this kind of current directions can be accepted now before we plunge into finding out the vector potential and the corresponding magnetic field and then the electric field and the radiated power etcetera what is it that we require we require the current distribution on the antenna itself without which we cannot find the vector potential a right in the first case for the hergene dipole we assumed a very short entity and therefore we were able to assume a uniform current distribution now this is not such a small entity in fact it has a considerable length it is at least half a wave length and may be more. So, what is the kind of current distribution that is going to exist on these antennas in general that is the first problem that one has to address when one has antennas which are longer which are electrically longer which are no longer electrically short antennas. Then we recall that the dipole antenna was evolved from an open circuited transmission line itself right we said that if we increase the separation between the two open circuited ends of a transmission line then the structure is likely to radiate. So, therefore it is nothing but an evolution a modification of an open circuited transmission line fine and what kind of current distributions exist on open circuited transmission lines it is a sinusoidal current or voltage distribution particularly for the current at the open circuit end because of the boundary condition on the current the current must be 0 right. Now the antenna the opened out transmission line is not exactly a very regular kind of transmission line but as a first order approximation to start with we can consider that the current distribution is sinusoidal even on this kind of structure. Now it is not that we have to go by an assumption like this which some of you may say that is imposed on us one can solve the Maxwell's equations subject to the actual boundary conditions for example this is a perfect conductor usually made of a very good conducting material. So, that the tangential electric field is 0 and subject to that kind of boundary condition one can find out actually calculate the current distribution right. But it turns out that as far as the purpose of the calculation the far field the radiation field or the distant field is concerned there is not a very great difference that one obtains in the fields obtained through that kind of more accurate current distributions. And therefore, since right now our purpose is restricted to obtaining the far field and the related parameters we will consider that the current distributions sinusoidal right. And therefore, we put down the current distribution on the antenna as follows i equal to i m which is the amplitude of the current times sin beta h minus z prime where h is the half length of the antenna. This is the value of the current as long as z prime is between these limits 0 less than or equal to z prime less than or equal to h. So, that at z prime equal to plus h the current drops to 0 value and beyond that point the current will have a sinusoidal distribution normally on a transmission line where we put the z coordinate equal to 0 at the open circuited end this kind of thing would not come in. But now we have shifted the coordinate and therefore, we have shifted the origin and therefore, this kind of expression is coming in. Similarly, for the lower limb of the antenna we write i m sin of beta h plus z prime as long as minus h is less than or equal to z prime less than or equal to 0. So, by taking the analogy with an open circuited transmission line we can put down the current distribution on the antenna satisfying the boundary conditions at this open ends that the current is 0 and as we go closer to the feed points the current varies in a sinusoidal manner. What is going to be the current distribution when the antenna length is exactly half a wavelength then it is going to be an exact half sinusoidal. Otherwise these two ends will remain at 0 and depending on the actual length the current will have a maximum and then drop off depending on the sinusoidal function. And therefore, one can visualize what is going to be the current distribution for longer antennas, antennas which are longer than half a wavelength. And if we increase the length beyond a certain length for example, beyond a wavelength we could have current lobes on the antenna lengths which are in opposite directions just as the current direction changes every half a wavelength on the transmission line. What is the beta that we have used here in the expression that is the phase shift constant on the parent transmission line from which the antenna is derived or this is the phase shift constant of a plane wave in free space or in the medium in which the antenna is located. So, these are familiar terms keeping the considerations general for the moment. We now recall the general method that we put down for determining the vector potential and this figure should be familiar to you. This is what we see every time when we want to consider the vector potential due to a certain source. Now, here in this general situation the source current or the charge density is distributed over a certain volume. The situation we have with us here is we have taken a wire antenna and actually the sinusoidal approximation is pretty good or gets better as the thickness of the antenna keeps on getting small. And therefore, we have taken a very thin infinitesimally thin if you please antenna. And therefore, this volume distribution of current density will change into a current distribution over the length of the antenna. This should be fairly clear and therefore, we can write the expression for the vector potential A which is mu by 4 pi j which is in general a function of r prime e to the power minus j beta r by r integrated over v prime going to in this case mu by 4 pi and this becomes I of z prime and then e to the power minus j beta r by r integrated over the length of the antenna. In this case going from minus h to plus h. Now, we have been saying that we are interested in the far field as far as the radiation pattern is concerned it is the far field which is of interest. A question can come up here that is it that we are never interested in the near field the field close to the antenna we are depending on the application. For example, if you want to find out the impedance that the antenna presents to the feeder transmission line which will depend upon the kind of field distribution close to the feed n that is the near field distribution or when we want to find out what is the effect of mutual coupling when two antennas are situated close together then we will have to find the near field the field close to the antenna. But when it is the radiation pattern of the isolated antenna then it is sufficient to calculate the far field or the radiation field because we have seen that as far as the power radiated away from the antenna is concerned it is the radiation field which contributes to that. As far as the far field is concerned one can make some simplifying approximations. We says we say that in the denominator since we are calculating the far field the field point p is at a very large distance from the antenna and therefore, it is more or less equidistant from the various points on the antenna. The antenna is going to be of the order of a wavelength in general and if the field point is sufficiently far away a few meters or even a few kilometers one can justify one can see the justification in this approximation easily. So, that in the denominator we can simply say that r is approximately small r with respect to the picture on the overhead projector we can connect that the capital r which is the distance of the field point from different points on the antenna. As far as the amplitude of the contribution of the various small small radiators on the antenna at the field point is concerned that amplitude is governed by the same factor that is 1 by r. The amplitude variation because of the minor differences in the distance of the field point p from different points on the antenna is ignored is negligible. Obviously, if the field point is brought closer if you are interested in the near field then depending on how close we bring the field point we may have to consider the variation in the amplitude because of this variation in capital r. As far as the numerator is concerned where capital r appears in the argument of an exponential factor affecting the phase of the contribution due to the different infinitesimal sources distributed over the antenna that we cannot ignore in a in such a simple manner. For that we make the following consideration we redraw the antenna here and of course, I need not repeat that whatever analysis we carry out will be applicable to a dipole antenna and also to its corresponding monopole antenna. As far as the fields are concerned there will be no difference except that the total power should be calculated appropriately and therefore, what we say holds good for both the monopole and the corresponding dipole antennas. Now, we say that the field point p is located at a large distance from the antenna and therefore, the while the actual situation is like this. This is small r and this is capital r when point p is at a large distance we can consider both these lines to be almost parallel to each other. And if the coordinate of this point is z prime and this angle is theta we can write capital r approximately as r minus z prime cos theta. In vector notation what we would have written would be r minus r cap dot r prime the projection of the position vector of this point in the direction of the position vector small r of the point p. The two are identical this is a more general notation which will be utilized even if we have a volume distribution of current fine, but for the linear antenna that we have this is effectively equal to this. Now, we like to pause here for a moment and consider the effect of this approximation have I not written z prime. I have written r prime I think it is all right you can consider this and may be we can discuss it. What is the effect of this approximation that I like to emphasize if theta is 0 or if theta is 90 degrees that is if the field point p is on the symmetry plane normal to the axis of the antenna. Then capital r is simply small r and it has the same phase. Now, we are talking about the numerator how we are going to approximate capital r in the numerator where it is the phase which is governed by the capital r term and therefore, if theta is 90 degrees if point p is located on the symmetry plane perpendicular to the antenna axis. Then the contribution from the various sources on the antenna is assumed to be arriving in the same phase which is fine as long as point p is at a very large distance. But, if point p is located at some other angle then these phase differences are taken into account. This kind of approximation where we have ignored the amplitude variation and also the phase variation for a point which is symmetrically located. Because, the point p is situated at a large distance is known as the Fraunhofer diffraction approximations approximations which lead to Fraunhofer diffraction. And therefore, the radiation pattern that we will obtain using these approximations is also many times called the Fraunhofer diffraction pattern. This is the simplest case and now one can consider the complication that will gradually need to be taken into account as a point p becomes closer. As a point p comes closer initially we will not ignore the phase differences even for a symmetrically located point p which becomes the case of Fraunhofer diffraction. And when point p comes even closer then we will not be able to ignore the differences in the amplitude also which becomes the near field pattern. So, this way there will be different regimes where different types of approximations will hold good. And as I said we are interested in the far field pattern and therefore, we are operating in the Fraunhofer diffraction region. And as some of you have done experiments in the laboratory and you would realize that there is a certain limitation on the minimum distance beyond which we can consider the pattern to be a far field pattern that arises from this kind of approximations. When a point p symmetrically located there should be no appreciable phase errors from different points on the antenna. So, that far field distance limitation comes from this kind of consideration. The effect on our analysis right now is that in the numerator we take r to be equal to r minus z prime cos theta which will consider at least 1. Considerably simplify the expression that we need to evaluate for the vector potential and therefore, the subsequent field components. Shall we proceed further? See it is this kind of approximations which one makes when one is dealing with diffraction. And the same kind of approximations will be applicable even when you consider the radiation pattern due to aperture antennas that we had shown earlier. And normally purely based on the geometrical optics the kind of field you will expect actually the radiation pattern goes beyond that and that is considered to be the diffraction of the fields at the finite aperture. And therefore, we use the diffraction term here as well. And the approximations that are made are identical in the two frameworks. Therefore, I said that the radiation pattern that we will obtain here is also many times referred to as the Fraunhofer diffraction pattern. Therefore, now we can write the vector potential and since the current that we have is entirely z directed we should have kept this vector we will get only the z component of the vector potential. And therefore, we have a z equal to mu i m by 4 pi which is a common factor and then for the two parts of the antenna as shown in the diagram we write the integrals separately. We have minus h 2 0 integral of i m sin beta h 2 0. h plus z prime e to the power minus j beta r by r d z prime as far as the lower part of the antenna is concerned plus 0 to h of i m sin beta h minus z prime e s sorry we cannot have it twice thank you. Is there anything anything else bothering you sin beta h minus z prime e to the power minus j beta r by r once again d z prime. And we make the approximations that we have discussed in fair detail already. So, that approximately this is mu i m by 4 pi small r which will come out as common and also e to the power minus j beta r which comes out as common. And then we have integral from minus h 2 0 of sin beta h plus z prime and then e to the power j beta z prime cos theta d z prime as far as the first term is concerned d z prime as far as the first term is concerned. The second term yielding 0 to h sin beta h minus z prime and once again e to the power plus j beta z prime cos theta d z prime. And it goes without saying that this kind of expression cannot be the basis of near field results. Now, this is as far as we proceed by considering the antenna to be of a general length 2 h from this point onwards it is simpler to consider the specific length where h is equal to lambda by 4 h is the half length. So, that for h equal to lambda lambda by 4 the total length becomes lambda by 2. So, now it becomes a half wave dipole. Why do we stress so much this particular length quarter wave length for the monopole and the half wave length for the dipole? It turns out that for this length the radiation resistance that we will calculate and in fact the total antenna impedance is not a very sensitive function of frequency. Actually it is a resonant length and therefore, one gets some kind of a flat behavior of the impedance particularly the real part which is the radiation resistance as a function of frequency. So, frequency variation is small. In addition the imaginary part of the antenna impedance is quite small and therefore, the antenna more or less presents a real impedance to the feeder transmission line and therefore, it is considerably easier to match such an antenna. So, from this point of view the half wave dipole or the corresponding quarter wave monopole is practically very significant provided you can mechanically or otherwise afford this kind of length. Otherwise as I said when the wave lengths become very large you may be forced to use an electrically short antennas for which we have already derived the results. When h goes to lambda by 4 then beta h is pi by 2. So, that the sine function will change into a cosine function and we are going to have a z equal to 0. Mu i m e to the power minus j beta r by 4 pi r and then we have integral minus h to 0 of cosine beta z prime e to the power j beta z prime cos theta d z plus integral from 0 to h of cosine beta z prime e to the power j beta z prime cos theta d z prime which are fairly similar integrals, but not they do not appear to be identical and they are not with some suitable change of variable. For example, here you could change the limits of integration and then if you let say z double prime equal to minus z prime. So, making that kind of simple manipulations one can see that this integral is similar to this with an important difference and therefore, these two take the following form. We get a z equal to mu i m e to the power minus j beta r by 4 pi r and the final effect of this kind of manipulation and clubbing together is that it becomes an integral which goes from 0 to h and then we have twice cosine of beta z prime cosine of beta z prime e to the power minus j beta r by 4 pi r and cos theta d z prime. With this change of variable the exponent the argument of the exponential term will change sign and when we combine them together it will turn into cosine of beta z prime cos theta which can further be written as twice the integrant as twice cosine of beta z prime into 1 plus cos theta on one hand and plus cosine of beta z prime into 1 minus cos theta on the other hand. Writing this product of two cosines as this kind of a sum which can be integrated without much difficulty and then you put in the limits and this also is lambda by 4. So, that beta h that you get after substituting limits becomes pi by 2. On this basis one can write the final result for the vector potential which only has a z component in this case and let me write the result here on the over head projector. The result that we get is as follows we have a z equal to mu i m e to the power minus j beta r by 2 pi beta r and then we have cosine of pi by 2 cos theta here in the numerator and then sin square theta here. This is the kind of expression that we get for the vector potential. It has only a z component because the current was only z directed and therefore this is the kind of expression. Some terms will always be there when we calculate the vector potential in this kind of far field approximation. Those terms are e to the power minus j beta r and 1 by r term and then of course the stress. So, this is the kind of expression that we get for the vector potential. So, this is the kind of expression that we get for the vector potential. So, this is the kind of expression that we get for the strength of the current. Then there will be an expression which will be a function of theta and or phi. Therefore, if you move on a surface let us say r equal to constant surface around the antenna the fields are going to change because of this kind of terms. So, these will be involved in what is called the radiation pattern and the other terms only govern the overall phase factor and the amplitude. That is as far as the evaluation of the vector potential is concerned. This is the first step in the evaluation of the far field. What will be the next step? The next step as you can recall in the first step is the calculation of the magnetic field intensity using this kind of an expression which will require the spherical coordinate components of the vector potential. So, that is what we will have to do next, but let us keep that for the next lecture. If you have any questions you can consider those. Otherwise, thank you.