Welcome, in this web lecture we're going to introduce vector based antenna theory. And in this particular web lecture we going to look at the general case of radiated fields by any antenna source you can imagine. So the objective of this lecture is first to derive a general formulation to determine the radiated fields from any antenna. And of course, we 're going to start with Maxwell's equations. At the end, we'll show you the expressions for the electric and magnetic field outside the source region, so outside the antenna region. Now let's first have a closer look at the configuration. We will assume that the antenna is located within the volume V naught. So any antenna, current and charges are located within this volume V naught. We have an observation points, where are we going to determine the electric and magnetic fields. P, so that's the observation point and the observation point is indicated by the position vector R. We also have source points, so points on the antenna where you have certain currents or charge distribution. And the source points are indicated by the position vector r naught. Another important position vector is indicating the distance between source point Q and the observation point P, and that is the position vector r minus r naught. In this formulation we're going to assume that the whole configuration is part of the free space medium. So we will assume that epsilon is equal to epsilon naught and mu is equal to mu naught. Now, the Maxwell's equations in the frequency domain are shown here. So we have four equations which are relevant. You observe in these equations, first the electric field and for per meter, the magnetic field and bar per meter. We have the electric current distribution on the antenna in per, per meter squared and we have a certain electric charge distribution also on the antenna in glowworm per cubic meter. What you can also observe in this set of equations is one equation which looks rather simple. And that's the third equation where the divergence of the magnetic field seems to be always equal to 0. And that is due to the fact that in our physical world, we do not have magnetic sources. Now, let us use that knowledge to see if we can solve Maxwell's equations. So let's start with this divergence of the magnetic fields is equal to 0, and note that we have used now a compact notation without using the precision vector r. We also know from effective theory that the divergence of rotation of any vector is also equal to 0. So if we combine these two equations, we apparently can write the magnetic field as the rotation of vector A e and A e is the magnetic vector potential. And you also observe that we have used a constant term 1 over mu naught, and we do that because it appears to be convenient in the notation later on. Now, the first Maxwell equation is shown here. So the rotation of the electric field is equal to minus j omega mu naught times the magnetic field. If we combine that with the notation of expression, suppressing the age field in terms of the magnetic vector potential. We apparently can write also the electric field in terms of the magnetic effect of potential minus the gradient of a lot of function phi E. And phi E is the scalar potential, which is still also unknown of course. And that is introduced because of the vector identity, that the rotation of the gradient of any scalar is always 0. And later on, we will show that we can express this magnetic scalar potential in terms of the magnetic vector potential. Now the approach that we're going to follow to solve for the magnetic vector potential and the electromagnetic fields is shown here. So we all get introduced to vector potential using that knowledge, we're going to drive the Helmholtz equation for the magnetic vector potential. We're going to apply this so called Lorenz gauge, then we will, find the so called Green's function. And using that we can find finally the solution for the electric and magnetic field. And you recognize in this flow charge three world famous scientists who contributed to this theory, Helmholtz, Lorenz and Green. Now, we already introduced the magnetic vector potential and as a next step we going to derive the Helmholtz equation and apply the Lorenz gauge. So let's see if we can work out these equations in more details. So we have the first Helmholtz equation, we have the relation between the magnetic field and the vector potential and the same for the electric field and the vector potential, and scalar potential. So if you start with the first two equations, we find that the rotation of the rotation of the vector potential can be written in terms of j omega epsilon times the electric fields plus j a. And we have to add a mu naught because it's over here. So adding the second equation. Then provides us with omega squared epsilon naught mu naught times the vector potential minus j epsilon naught mu naught times the term with the scalar potential plus mu naught times the current. And this is equal to k naught squared the wave number times the vector potential minus this term, Scalar and the current. Now this might look still rather ugly. As a next step we can work out this vector product here. Check your appendix in the book and you will find these expressions. That can be also written in terms of the gradient of divergence of a factor so this holds for any vector. But also of course for the factor potential minus Nablus squared times this vector. Now if we use that, in this expression here, we find that Nablus squared. So, we reorganized a little bit the equation all ready with the minus sign. Plus k naught times this, is equal to this term here. Plus j omega absolute mu naught, scalar potential minus because really reshuffled with the minus sign. Now, this still looks odd or rather ugly. But we still have, one choice to make. And that is we didn't put a fixed relation between the factor and the scale of potential. And, we can do that because the scale of potential can be an arbitrary function. And this was also recognized by Den scientists, Lawrence already a long time ago. And he suggested, of course, to make this particular choice. Is equal to minus. And by using this Lawrence gork, as it's called. This term here finishes equal to 0. Well, that's nice. And that brings us to the well known Helmond equation, which is the differential equation for us to solve. For the effect of potential, Is minus mu naught times the current and this equation we need to work out further. So in our approach to find the radiated fields we've now derived the Helmond equation by applying the loans gork and that resulted in this expression. What we're now going to do is to consider the response due to a point source and that brings us to the so called greens function. Now in order to derive the greens function we have to first introduce the configuration. And the configuration is that our point source is located first at the origin of our coordinate system. So we have grown distribution along a an electric dipole. So an electric dipole is a very short on antenna with a length L, and L is, much smaller than the wavelength. And because of that, the current distribution along such in point source can be written in the form as shown here. Which includes the delta function, it includes an amplitude I naught times L. That's the so called a dipole moment. And is directed in this particular case, along the z-axis. And we will generalize it further on to an arbitrary direct point source later on. Now since the current, in this case is the rector, the long, the set axis, the magnetic factor potential will only have a component also along the z axis. So this simplifies the Helmond equation. So let us look at the Helmond equation for this particular configuration where we have an Z directed point source in the origin of our coordinate system, let's say call this case one. The Helmond equation is shown here. We only have a set component and because of that we can using factor calculus formulas that you can find in the appendix of the book. We can rewrite it in terms of the spherical coordinates in this form. So we have one over r squared, left it to r and then a term r squared. Again derivative, to r of A set Plus k naught, Delta r, so the solution of this differential equation has two forms. So in fact, we have two solutions, and you can substitute the solutions in the differential equation to check it yourself. So the first solution is a constant divided by r radial expansion times exponential jk naught r. And the second solution similar constant divided by r, then with the minus sign in the exponential. Or the first solution corresponds to a wave that is propagating two words the source and the second one is a wave propagating away from the current source. So, the first solution is not a solution that can exist in the physical world. So the second solution is our solution. Now we still have to find the constancy and we can use a trick, we can look at the differential equation at 0 frequency. So we're going to look at through the static situation. So then the equation still holds, so then omega is 0, k naught is 0. And then we know from electromagnetics course that this differential equation turns into the Pocono equation. And the solution of the Pocono equation is that the factor potential can be written in this form divided by 4 pi r. Which means that the constant that we're looking for C is equal to mu naught I know l divided by 4 pi. Having said that, we have now the solution for the first case, okay? Case one, so that's our solution, There's only a component in the insert direction. And that is equal to mu naught, I know l, 4 pi r times the exponential, Directed in the set direction. Of course this is a point source in the origin. We want to generalize it, so the first step, let's generalize it to the situation where we have a point source somewhere. With a certain direction Like this at the, Position indicated by position vector i knot so let's call this case 2. And by inspection, we can generalize the first equation to this form, is mu knot I l which is then a vector divided by 4pi times the exponential but then we have to look at the difference between the position vector of the source and the position vector in the far fields. O, and then like this, Divided by r minus r knot so this is between the observation point in the far field region outside the antenna region and the source point, yeah, and the direction is indicated here. Now this is the situation of a point source in an arbitrary direction. Of course, an antenna is not a point source, but it's a volume or surface where currents can flow so in general, we would have a volume. For example, like this v knot with currents flowing here everywhere and these currents can be indicated by a precision factor again. And the solution of the vector potential can be found by superimposing all the individual point sources that we have in this antenna. So we can distribute the current distribution in terms of point sources and then use the superposition principle to find the overall solution. And superimposing means that we're going to integrate over the volume where we have the currents. So then generalization of case 2 is then the following. So we have mu knot divided by 4 pi and then we're going to integrate over the volume v knot, where we're going to integrate over the electric current distribution times this exponential term r minus r knot divided by r minus r knot and then we're going to integrate over the volume v knot. And this is the general solution for the magnetic vector potential due to general arbitrator current distribution on an arbitrary antenna. So let's go back to our general approach to the flow diagram shown here. So we now solve for the Helmholtz equation, we find the solution for the magnetic vector potential as shown here, which is an integration over the current distribution of the antenna times the so-called Green's function. So that's the function indicated here in the red box. Now the final task that we have is to find the solution for the electric and magnetic field and to express them both in terms of the magnetic vector potential. Well, the starting point here is relation we already derived. The magnetic field can be expressed in terms of the rotation of the magnetic vector potential, and the electric field could be written in terms also of the vector and scalar potentials. Now the scalar potential can be expressed in terms of the vector potential by using the Lorenz gauge again. If we do that, we arrive at this expression and we can further simplify it by using some vector identity that you can find in the appendix of the book of course. By planning this vector identity, we come to this relation where you might already recognize again the Helmholtz equation or part of the Helmholtz equation, I should say. So we can replace this part by minus mu knot times the current distribution. And by doing that, we finally get this expression for the electric field in terms of the effect of potential and the current. Of course outside the source region, so outside the region where the antenna is located, the current will be equal to zero. So to summarize the approach, we first started with introduction of the vector potential. Then we derived the Helmholtz equation, applied the Lorenz gauge and found the response due to a unit current source and by integrating over the volume we found the total solution for the magnetic vector potential as shown here. And then we related the vector potential again to the electric and magnetic fields resulting in these two equations here. And the electric field here is then considered outside the source region. Now, with these expressions, we can solve for any antenna problem. To summarize, we started with Maxwell's equations, we introduced the step by step approach to find the fields outside the source region. And in the next web lecture, we will simplify these expressions further by using far field approximations. And of course that is very relevant for antennas since we are mainly interested in the far field performance of an antenna. Hope to see you back next time.