 Welcome to the 40th lecture in the course engineering electromagnetic. We continue in this lecture with our discussion on the main topic radiation which is the last main topic in this lecture and we complete our discussion for the purpose of this course on the half wave dipole antenna which was initiated in the last lecture and then we go on to consider what are called the folded dipole antennas and then finally the radiation pattern of the various antennas that we have considered. You would recall that in the last lecture we were talking about antennas which are not electrically short. A half wave dipole antenna would fall under this category. So antennas which are not electrically short and we were considering linear antennas. So we locate such an antenna symmetrically about the origin and the antenna is oriented along the Z axis and we consider that the current distribution on these antennas which are not electrically short using an analogy with the open circuited transmission lines is sinusoidal and then we developed an expression for the vector potential for the specific length L equal to half a wavelength and we said that the half wave dipole is used very frequently because it offers an impedance at these feed terminals which is more or less real. Recalling that expression that we had got last time, the expression for the vector potential which has only a Z component in this case was obtained in this manner. You would recall that we made a number of approximations in the derivation of this vector potential expression which approximations are suitable for the evaluation of the far field and these approximations will be valid will hold good as long as the point the field point p is at a large distance from the antenna. So under those approximations this is the kind of expression that we get the factor 1 by the factor r is in the denominator the factor e to the power minus j beta r is in the numerator as far as the excitation strength and the radial variation and the phase variation are concerned those appear in this factor and the second factor includes the variations with respect to theta in this case or with respect to phi in the general case. What is the next step? The next step as you would recall is the calculation of the magnetic field intensity from the vector potential a using the expression h equal to 1 by mu del cross a which was more or less a defining relation for the vector potential a as we obtained such an expression from the Maxwell's equations for the curl of a the expansion in the spherical coordinates is written like this as we have seen before and therefore we require the components of the vector a in the spherical coordinate system alright what we have got and we will require the a r a theta and a phi components. Let us put down the expression for a z together with this alright and then write on the board the various spherical components for this case. We will have a r which is a z cos theta and a theta which is minus a z sin theta and as far as a phi component is concerned in common with the herzen dipole or the alternating current element the phi component is 0 here also such is the nature of the source that we are considering the source is entirely z directed so the related vector potential also has only a z component and it cannot have any phi component and as far as these non-zero components are concerned a r and a theta these also are not functions of phi again because of the phi symmetry of the source that we are considering and therefore as for the herzen dipole we are going to have only a phi component in the curl of a alright which means we will have only a phi component for the magnetic field which will be related to the phi component of the curl of a. So let us transfer this phi component of the curl of a on the board as follows we have del cross a and the phi component please help me out it is equal to 1 by r and then del by del r of r times a theta minus del by del theta of a r alright let us try and expand this and that gives us 1 by r and then r times del by del r of a theta that is the first term arising out of the first term here plus a theta the second term out of the derivative with respect to r and as far as a r is concerned since it has the expression in terms of a z and cos theta as a z cos theta this can be written as minus del a z by del theta cos theta plus a z sin theta. So completing this further what we get is 1 by r and then we substitute for a theta in terms of a z so that gives us r times del by del r of a theta which is minus a z sin theta plus a theta let us substitute for a theta also in terms of a z so that it is minus a z sin theta and then we have minus del a z by del theta cos theta plus a z sin theta and therefore we see that these two terms cancel out as far as this term is concerned the variation of this term with respect to r is not going to be affected even when we take the theta derivative and a z itself as you recall has a variation with respect to r which is such that it is 1 by r and therefore keeping the factor 1 by r in front in mind the variation of this term as far as the variation with respect to r is concerned is going to be 1 by r square which will not be a term contributing to average power flow because it will not be varying as 1 by r earlier we have argued out that it is the terms which vary as 1 by r only which will contribute to the average power flow and therefore based on such an argument as far as the distant field or the radiation field terms are concerned this is going to be simply or approximately del by del r of minus a z sin theta which is more simply minus del a z by del r times sin theta and therefore the phi component of h which is 1 by mu del cross a phi component of this curl is going to be minus del a z by del r times sin theta all right going back to the over a trajectory now this can be worked out what we require for the magnetic field is an h of phi which is 1 by mu del cross a phi which as we have seen is equal to minus del a z by del r times sin theta as far as the far field is concerned and therefore we require the r derivative of a z here again there are 2 terms involving r and one can make out that one of the terms arising out of this derivative will have a variation which will be 1 by r square so once again using the same logic that term will be discarded and therefore h phi as far as the far field is concerned will give us j i m e to the power minus j beta r divided by 2 pi r and then of course cosine of pi by 2 cos theta times divided by sin theta which completes the second step involved in the analysis of this kind of antennas the vector potential a the relevant component and the magnetic field we see that there is only a phi component and as far as the far field is concerned it has a relatively simple expression like this the third step is the evaluation of the electric field and once again we are going to be interested in the evaluation of the far field component of the electric field and the far field components of the electric and the magnetic fields are simply related therefore this task is relatively simple e theta which should be eta times h phi can be written as j eta i m e to the power minus j beta r by 2 pi r and then cosine of pi by 2 cos theta divided by sin theta and therefore as far as our task of evaluating the far field expressions is concerned that is now complete which is the root that we followed the vector potential the magnetic field and then the electric field and there have been considerable simplifications because we are interested at the moment only in the far field there is a simpler root which is available and that can be reached when we consider the expression for the electric field in terms of the potentials potentials v and a you would recall that we obtained e equal to the negative gradient of the potential function v minus del a by del t in general otherwise in phasor notation j omega times a now it so turns out that the terms that result from the first term on the right inside del v they do not contribute to the far field therefore as far as far field is concerned we can consider only the second term perhaps one can advance some explanation for this but right now what is important for is that the far field contributions are made by the second term even there in the far field since the average pointing vector is expected to have a radial direction what kind of terms should be involved in the far field it should be the theta and phi terms which should be involved in the far field therefore as far as the far field is concerned far field the distant field or the radiation terms which are way you like to call it for far field we can simply write e theta equal to or approximately minus j omega a theta and depending on the situation there may be a phi component in the present case it is not there e phi which would be minus j omega a phi okay and the expressions for a theta related to a z are there before you if you apply this process you will find that the expression that results will be the same as we have got through the alternative route okay which route is extremely simple and straight forward once we have got the electric field the far magnetic field can be determined in a simple manner we will have h of phi given by e theta by eta and if there is a phi component of the electric field present there will be corresponding theta component of the magnetic field given by minus e phi by eta going by the cyclic order of the subscripts okay in general as far as the far field is concerned this route is the shortest and the simplest alright but that is just an alternative method having got the expressions for the far field we will now be interested in the evaluation of the total power that is radiated by the half wave dipole that we are considering as far as the total power is concerned we will first consider the pointing vector okay for the power radiated the average pointing vector and we have already identified the fact that for the far field it will have a radial direction in fact the average pointing vector always has a radial direction even when it is evaluated close to the antenna alright so that radial direction we are not attaching it is understood and the average pointing vector can be written as half eta magnitude of h square or alternatively it could have been written in terms of the electric field and this results in the following expression for the average pointing vector gives us eta im squared by 8 pi squared r squared and then we have cos squared of pi by 2 cos theta divided by sin squared this is the average pointing vector and of course the average power radiated will be an integration of this pointing vector over a surface which encloses the antenna you would recall that in this context the area element that we have considered is the following this is the way we can consider the area element which is in the shape of a ring of a very small width like this we can consider this kind of an area element because the expression that we need to integrate is phi independent and therefore we have considered this kind of an area element and this element of area has an expression dA equal to 2 pi r squared times sin theta d theta okay so this is the element of area that can be used in the integration and carrying out this process what we get is the following w average is going to be equal to eta im squared by 8 pi squared r squared and then we have the integral of cos squared of pi by 2 cos theta divided by sin squared theta multiplied by the area of this element of integration so that it is 2 pi r squared sin theta d theta and as a result you see that some terms drop out or cancel out now this integration will have to be evaluated this turns out to be equal to eta im squared by 4 pi and then integral of cos squared pi by 2 cos theta by sin theta d theta what about the limits of integration now depending on whether we want to consider the monopole antenna or the dipole antenna the limits can be put appropriately for the monopole antenna it is enough to consider the limits from 0 to pi by 2 that will cover the region above the ground plane over which the monopole antenna is erected and for the dipole the limits will go from 0 to pi and therefore right now maybe we can use the limits 0 to pi by 2 and if you want to get the power radiated by the corresponding dipole will simply double this expression that we get now the integrand is unfortunately not so straight forward and it requires a few mathematical manipulations first of all we express the numerator in terms of twice the angle okay and that gives us let us focus our attention on the integral alone so that we get half of 1 plus cosine of pi cos theta by sin theta d theta okay next we make the following substitution which unfortunately is not obvious the substitution is let V be equal to pi into 1 plus cos theta okay so that dV is equal to minus pi sin theta d theta and pi cos theta is equal to V minus pi so that sin square theta which is going to be required a little later which is 1 minus cos square theta can be written as 1 minus V minus pi by pi whole square which simplifies to twice pi V minus V square or V into 2 pi minus V by pi square when we make the substitution in the integral then cosine of pi cos theta will become minus cos of V okay because it will be cosine of V minus pi and then we will also get a sin square theta term in the denominator so taking care of these things what we get is the integral will be minus pi by 2 and then we have 1 minus cos of V dV upon V into 2 pi minus V which is to be integrated this is further written as by expanding this in terms of partial fractions so that we get expressions which are easily integrable as minus half and of course 1 minus cos of V and then here we can write simply 1 by V plus 1 by 2 pi minus V okay actually there should be this factor should be 4 alright continuing this further we get the expression follows this gives us the integral to be equal to minus 1 by 4 and then integral of 1 minus cosine of V by V dV on one hand and integral of 1 minus cosine of V by 2 pi minus V dV on the other hand where the limits of integration considering the change of variable that had been made earlier turn out to be 2 pi to pi here as well as okay now by a suitable change of variable one can make out that this entire thing actually is equal to an integral of 1 by 4 and then 1 minus cosine of V by V dV from the using the limits 0 to 2 pi okay finally this is where we end up now is this expression easier to handle certainly as if cosine V is expressed in terms of its power series expansion and then the integrand is simply a power series in V and therefore each term can be integrated and one can put the limits and evaluate this one will have to use a sufficient number of turn terms so that one gets the acceptable accuracy so using that process this term comes out to be 0.609 okay now it is true that after so much of manipulation we been able to integrate this analytically a question could come up that what happens in general in general one can use numerical integration okay which algorithms which software is now readily available but here it was possible to do this analytically so we have taken so much of trouble now once this integral has been evaluated this integral is this one this has come out to have a value 0.609 and therefore that can be substituted here and we get for W average an expression which is 0.609 eta times IM squared by 4 pi which so that we can determine the radiation resistance can be alternatively written as 0.609 eta by 2 pi and then I RMS or I effective squared considering that IM was the amplitude of the sinusoidally spatially varying current distribution so that this part the part which is in front of I effective squared becomes the radiation resistance for this case since the limits of integration have been used as 0 to pi by 2 this will become the radiation resistance of the quarter wave monopole right and therefore we have the radiation resistance for the quarter wave monopole and recalling that eta the intrinsic impedance of free space in which we have assumed the antenna to be situated is 120 pi one can see that this comes out to be 36.5 so many ohms and by a simple extension the radiation resistance for the corresponding dipole which will be the half wave dipole will be simply the double of this that is 73 ohms it is interesting to consider what is the result we would have obtained for the radiation resistance for this half wave dipole had we continued our approximation of the approximation based on the alternating current element or the Hertzian dipole there you would recall we got the radiation resistance expression as 20 pi squared and then the length of the radiator divided by the wavelength whole squared which is approximately 200 L by lambda whole squared which for a half wave length dipole L equal to lambda by 2 is almost 50 which is quite different from the more accurate value which we have obtained taking the sinusoidal spatial distribution of the current into account alright so as far as our purpose of analyzing the half wave dipole is concerned that part is now completed we have got the radiation fields we have got the power radiated we have got a measure of the effectiveness of the half wave dipole as a radiator that is its radiation resistance then it turns out that this radiation resistance of 73 ohms for the half wave dipole which is an antenna which is frequently used in practice is not very convenient for connection with commonly available low cost transmission lines for example you would recall the parallel wire transmission line that is often used to connect the television receiving antenna to the television receiver that kind of low cost transmission lines have a characteristic impedance which is close to 300 ohms which is also closer to the input impedance of the television receiver now of course one could through some sort of impedance matching utilize this dipole with this kind of radiation resistance but everything adds to the cost and of course the performance will be affected somewhat therefore it will be very nice if through some simple modification of the half wave dipole we could increase the radiation resistance so that it meets the requirements we have just mentioned that brings us to what is called a folded dipole antenna a folded dipole antenna is nothing but a dipole antenna with an additional arm connected in parallel and these two arms are short circuited at the other end just because of the looks it is called a folded dipole antenna now how does this improve the aspect of radiation resistance one could look at it in a number of ways the argument runs like this one says that since these two straight arms are located very close to each other a distance which is negligible in terms of wavelength they support the same kind of current same amplitude same distribution which is quite obvious if we consider some signal incident on the antenna and depending on the field that is incident some current will be induced and since these two are almost co located they will have the same kind of current induced for the transmitting case also one could look at it in this manner let us say the additional arm is not connected yet then this is an open circuited transmission line that a knowledge we have already used for a line like this the ends will have zero current alternatively the ends will have maximum voltage and voltage and current will have sinusoidal or co sinusoidal distribution so with that kind of voltage distribution now if we short circuit these two ends with an additional conductor then a sinusoidal current will flow on this additional conductor at these high frequencies where time delay effects are important this does not amount to a simple short circuit. So depending on which argument appeals to you we can make out that the currents on these two arms are going to be identical with one condition that these two conductors are identical in dimensions and material alright which is usually the case where does that lead us now therefore we have two radiators located very close to each other therefore it can be considered to be a radiator having twice the current of the original radiator alright therefore the fields that will now be caused will double everything will just become double because now you have a radiator which supports double the current compared to the simple dipole case since the fields are doubling the power radiated will become four times and therefore the radiation resistance of a folded dipole is going to become four times that of the corresponding dipole from which it is derived and therefore if it is a half wavelength dipole which is the parent dipole then we will have the radiation resistance of the lambda by 2 folded dipole it will be four times that is almost 300 ohms okay and very simply one meets the requirement of increasing the radiation resistance for the purposes I mentioned earlier. There is another advantage to using the folded dipole which will not be able to discuss in detail here but the bandwidth the range of frequencies over which the folded dipole can be operated with acceptable performance is wider than that of the simple dipole from which the folded dipole is derived this fact is very useful when we want to receive television signals which have a considerable bandwidth alright therefore the folded dipole is a very popular antenna element and if you recall this is the center element of the antennas which we used for the reception of television signals alright to understand the need or the effect of the additional elements that are there in the television receiving antenna we will have to wait a little more we will have to consider the concept of radiation pattern first so that is what we take up next the radiation pattern very simply described is a graphical representation of the radiation of the antenna as a function of direction how in the three dimensional space around the antenna the radiation is distributed that is what the radiation pattern indicates for the purpose of the radiation pattern one could of course use the field strength pattern the field strength as we have calculated analytically for some of the simple antennas and that could be used for drawing the radiation pattern alternatively one could use the term which is called the radiation intensity the radiation intensity is defined as follows radiation intensity is the power radiated per unit solid angle the area associated with a unit solid angle is r square okay and the power density the pointing vector is proportional to 1 by r square since we are using the since it is the 1 by r terms only which contribute to average power flow away from the antenna and therefore this power radiated per unit solid angle in watts per unit solid angle or stear radian is independent of r and therefore it can be measured at any radial distance from the antenna and it can be plotted as a function of direction around the antenna so these are two alternative quantities which can be utilized for the purpose of plotting the radiation pattern expressed in decibels the two plots will appear identical otherwise one plot can be obtained as the square of the other plot so the two plots are simply related expressed in decibels they will just look the same and the radiation pattern is nothing but the function e for r equal to constant so that it is a function of theta and or phi as I indicated in the beginning of the lecture the conceptually what is done to plot the radiation pattern is that we locate the antenna at the origin of the spherical coordinate system and then we see what is the radiation intensity or the field strength as a function of theta and phi on a spherical surface with r equal to constant okay and that usually results in a radiation pattern which looks like this this is the radiation pattern often of the alternating current element or the herdsian dipole this is the coordinate system x y z this is where the dipole the herdsian dipole is located okay and as a function of different directions this is how the three dimensional plot will look like usually called a dov nut shape no radiation along the z direction if you look at the e theta and h phi field expressions for the alternating current element maximum radiation in the x y plane the plane normal to the antenna axis that is theta equal to 0 direction now such a three dimensional representation is of course visually very complete but is difficult to put down on the two dimensional media that we usually have to deal with paper or monitor screens and therefore we go to what are called the patterns in the principle planes principle planes will be the planes for example theta equal to 0 degree theta equal to 90 degree or some phi equal to constant plane okay these will be the principle planes therefore in this kind of a principle plane this kind of principle planes this is the kind of patterns that we will get for the alternating current element this is the pattern in any phi equal to constant plane or in any plane containing the antenna alright plane containing the z axis along which the antenna is located this is the kind of plane we will get a kind of pattern we will get for theta equal to 90 degrees plane the plane normal to the antenna axis and if we consider the orientation of the electric field which is e theta for such a radiator you can see that this can be called the e plane pattern the plane containing the far field point the antenna and the electric field vector and by the same token this can be considered the H plane pattern alright so therefore as far as this plane is concerned the plane normal to the antenna axis the radiation is uniformly distributed but in the other plane containing the antenna the radiation is not uniform it makes a figure which is just like the figure 8 the next thing will be to quantify the radiation pattern in some manner by quantification what we want to do is to find out how directional is the antenna how sharp how sharply peak the radiation pattern is and that can be done by considering by going over to the concept of beam width the beam width is the angular separation between the 3db points on the radiation pattern expressed in decibels whether we calculate the beam width from the radiation intensity pattern or the field intensity pattern it will come out the same for example for the alternating current element the 3db beam width is 90 degrees okay that can be seen in a very simple manner by considering the expression for the electric field on the other hand the other antenna that we have considered the half wave dipole antenna has a pattern which looks wise is similar but the pattern in a 5 equal to constant plane is like this it is somewhat more directional than the alternating current element and the 3db beam width is 78 degrees and therefore we say that the alternating current element has a 3db beam width which is 90 degrees but the half wave dipole has a beam width which is 78 degrees a question can come up that well we were trying to have a very good antenna when we have when we had a half wave dipole but from the point of beam width there is not much difference that is true but when we consider the radiation resistance the effectiveness as a radiator we find that half wave dipole has a much better performance 73 ohms compared to the alternating current element or even for that matter the electrically short antennas this is where we will stop if you have any question we can take it up thank you