Tags: Far Field / Directivity / Poynting Vector / Dipole / Helmholtz equation / Radiation pattern
In this problem we first investigate the radiation properties of a very short and thin filament of current. Although this filament of current as an antenna is not physically realizable, as a mathematical model it will help us to understand the general behavior of antennas. Here we also introduce some of the fundamental antenna parameters.
Problem Statement
A very thin and short antenna which works at frequency ω has the the following current distribution.
where J0 is the amplitude of the current distribution, δ(r) is delta function, and
Assume that the current distribution is placed in a homogeneous environment with permitivity ε and permeability μ, respectively. Remember that this current distribution represents the infinitesimal antenna!
- Find the electric and magnetic fields E(r,ω) and H(r,ω) that are generated by the current.
- Sketch the far field radiation pattern of the antenna in transverse and vertical planes (e.g. x-y and x-z planes).
- Calculate the half power beamwidth (HW) of the antenna in x-z plane and its directivity.
Hints
1. First find vector potential A(r,ω) by solving the equation below.
The latter equation is kind of famous as it resembles the inhomogeneous wave equation in frequency space, i.e. the inhomogeneous Helmholtz equation.
Try to use the symmetries in the problem to simplify this equation. Then, find E(r,ω) and H(r,ω) using the following equations.
2. Approximate E(r,ω) and H(r,ω) for
3. Half power beamwidth (HW), in x-z plane, is defined as the width of the angular interval in which the radiation intensity is more than half of the maximum radiation intensity.
Directivity is defined as below.
Solving the infinitesimal dipole
Let's get dirty and figure out all the mathemics behind this radiation task!
1. The first step to solve this problem is to find the vector potential. Therefore, we start with the equation below.
We can split this equation into three equations, each in one direction of the Cartesian coordinate (e.g.
Using this fact, equation [
Since the source term (right hand side of equation [
Thus,
The right hand side of equation [
To make equation [
By substituting equation [
which results in,
where,
Now, we need to determine
Note that equation [
By putting the results of equation [
and therefore,
Now, one can calculate the magnetic field easily as follows.
to calculate the electric field, we do the following.
2. Lets first clarify what we mean by radiation pattern of an antenna. Generally, the radiation pattern is defined as the radiation intensity (radial component of the Poynting vector) as a function of direction in space (
Based on what is mentioned, firstly, we need to calculate far fields. According to equation [
This approximation is only valid for fields at large distance form the antenna (
where
Now, the Poynting vector can be calculated as below.
As can be seen the intensity drops as
The far field radiation pattern can be calculated easily by normalizing equation [
Here, the superscript 'n' denotes the normalization. The far filed radiation pattern is depicted in figure (1) in three dimensions.
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Conventionally, radiation patterns are drawn in some cut-planes (there are some standard planes which we will define in Problem One), since it is easier to see the details of the pattern which are not visible in a three dimensional figure. Here, due to rotational symmetry, it is sufficient to draw the pattern only in x-z plane. However, to show the rotational symmetry it is better to draw the pattern in x-y plane as well. The far field radiation pattern in x-y and x-z planes are shown in figure (2) and (3).
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3. In this last step we just want to introduce some simple antenna parameters and the way to calculate them, such as half power beamwidth (HW) and "directivity". All of these parameters are calculated using the Poynting vector, therefore, they just provide different pictures of the radiation characteristics of the antenna.
HW, in a determined plane, is defined as the width of the angular interval in which the radiation intensity is more than half of the maximum radiation intensity.
According to equation [
This a measure which means that the antenna radiates effectively over 90 degree in the x-z plane which is a quite large angle. Although HW is an important parameter, it does not indicate in which directions the antenna has higher radiation. Therefore, we define "directivity" to address this question.
Since the radiated intensity of an antenna is a function of the direction of radiation, a criterion is needed to characterize and quantify this dependency. This criterion is the so-called "directivity function" and defined as below.
Physically, this function describes the ability of an antenna to send power in a specific direction and compares it to the average power radiated in all directions.
Using equation [
For example the directivity of the antenna in the direction of
Now this was a lot of insights for antenna theory! We strongly encourage you to do some more calculations on your own to get a feeling for the mathematics behind the infinitesimal dipole antenna!