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### ELEC510108Lect9

Course: ELEC 5101, Fall 2009
School: Allan Hancock College
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Word Count: 1573

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- ELEC5101 08 &quot;Antennas and Propagation&quot; Lecture 9 25/09/07. Notice the emphasis here in that one pair defines FORCES while the other pair defines FIELDS. In everything that we do UNDERLINED field quantities are VECTORS. The difficulty with using these integrals is in providing the actual limits of integration when tackling a real practical problem generally an impossible task! To get...

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- ELEC5101 08 "Antennas and Propagation" Lecture 9 25/09/07. Notice the emphasis here in that one pair defines FORCES while the other pair defines FIELDS. In everything that we do UNDERLINED field quantities are VECTORS. The difficulty with using these integrals is in providing the actual limits of integration when tackling a real practical problem generally an impossible task! To get around this we make use of simple transformations which convert this integral form into a differential form and these new differential form equations apply at each and every point in space. The necessary general vector transformation theorms are due to Stokes and Gauss: Applying Gauss to the first integral equation above both sides of the equation become identical volume integrals so the contents of the integrals must be equal. I will get you to carry out the remainder of these transformations in the next tute so that you can see for yourself how easy it really is! In free space = 0 = 4*10-7 Henries/m and = 0 =10-9 /36 Farads/m. Impedance of free space = ( 0 /0 ) = 120 ohms. It is also worth mentioning that (the charge density in equation (1)) is almost always zero in a field solution situation unless there is a specific charge source present! Let me just emphasize again that these `differential' forms apply at every point in space. Most of you have probably experienced the solution of these equations for a rectangular waveguide: And that is `relatively easy' because you can separate the `Z' solution from the X Y one and then concentrate on satisfying the boundary conditions in the X Y cross section which are E must be normal at a wall (Etan = 0 at a wall!) and H must be tangential at a wall (Hnormal = 0 at a wall!). The solution that we need for antennas is very different because we are talking about fields in 3D open space!! To solve MEs for antennas we make use of the properties of the vector operator `DEL". The gradient of a SCALAR field gets you the greatest rate of change of a scalar field (eg mountain contours) and its direction. It is a linear vector result so it cannot have any curl: ie curl(gradient) = 0. Similarly the curl measures the vortex (rotational) content of a vector field so it cannot have any divergence: ie div( curl) = 0 So if we now consider Maxwell (3): This means that we can set B = the curl of an arbitrary vector function: If we now substitute this into Maxwell 4: And substituting this back in to Maxwell 2: The A and are auxiliary functions and we can happily choose them so that this (hugely simplifying!!) Lorentz condition is achieved and then: So you can see that A is directly related to J and is not needed! So if we can find a solution for A in terms of J we can proceed to evaluate E from (11) and then H follows. We could now follow a rather convoluted mathematical development to find the shape of A OR we can make an intuitive/intelligent guess and proceed from that. We will go this much simpler way! We start this by considering simple DC voltages and currents and then extend these `DC' results to our high frequency antennas situation. What this simply says is that the potential (voltage) due to a charge q at a distance is equal to the force that would be exerted on a unit charge, Same applies to a charge doublet: If we then proceed to consider a region of distributed charge which we can divide into infinitesimal volumes Volumei having charge density i (coulombs/m3 ) we can then write the potential as: Note the simple final integral form of this. Next we consider Amperes Law for DC current flow: Now these "DC" results give us the shape of the potentials that we need to solve Maxwells Equations in the high frequency (open space) case: The only addition that we have made here is to include the all important exponential phase term! Now there is one very important example where these potential ideas are very easy to apply and that is for the case of an INFINITESMAL dipole element having a CONSTANT current distribution along its length. This is of course just the Hertzian dipole!!! This first line should read: B = H = Curl (A) Now all that we have tried to do here is to give you a `flavour' of how we use "potential" functions to analyse Maxwell's Equations but it has allowed us to develop the extremely important Hertzian Dipole result which we use repeatedly as a `building block' element for analyzing real antennas like the half wave dipole. The general approach to solving field problems numerically in a digital computer is build to up a discrete mesh (which needs to be 3 dimensional in general) which `pictorially' defines the problem and then apply a discrete form of Maxwell's equations at each point in the mesh. At each conducting surface we apply perfect conductor conditions ie Etangential = 0 and Hnormal = 0 and then iteratively force a result at each mesh point. Making the mesh smaller and smaller gets a more accurate result but the number of points can get very large very quickly. It is also possible to surround a problem with an `absorbing' boundary (numerically) which, just like a matched transmission line has no reflected wave! Introduction to the `Method of Moments' This is a computer approach which is used extensively for the analysis of (particularly) wire antennas and it is worthwhile trying to understand the very basic ideas which apply. I could give you a computer program and let you play with it but I just don't think that you will really learn anything useful from that! This is essentially a problem of evaluating V = Q/(4r) which is exactly the expression that we have just met in this lecture developing the potentials. Kraus' develops this (about p 386 in my copy "Example. Charge distribution on wire" ) There he considers the relatively simple problem of evaluating the charge distribution on a thin rod/wire when it has a voltage `placed' on it. He then converts the full `analogue' problem in to a `discrete' one by dividing the rod/wire in to four equal length segments. If we assume a very good (perfect !) conductor we can assume all charge to lie on the surface. We take this one step further and assume the charge to be located around the centre of each segment this is the step of developing a `mesh' of points. He then considers `observation' points along the central axis of the rod/wire. In that case the distance of the observation point to the charge in any segment is the same all around the segment so we can now lump the charge in one segment at single mid-segment points as shown in his Fig 9.23 (in my copy!). It is then very easy to evaluate voltages at `observation' points and completely solve the discretized (ie the mesh) problem. This general approach of surface point charges and onaxis `observation' points is very similar to the approach that we will use when we analyze a dipole antenna. I would like to include a numerical example to emphasize the inherent simplicity of the approach even though it looks complicated it most definitely is very straightforward. It is computationally intensive but that is easily handled with a computer once you understand what is involved AND get the method exactly right! Consider a 1 meter length of wire/rod with 1 Volt placed on it which is divide...

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