the radiation pattern of an infinitesimal dipoleIn 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.

point-like current that yields a dipole radiation

where J0 is the amplitude of the current distribution, δ(r) is delta function, and z^ is the unit vector in z direction of the Cartesian coordinate.

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!

  1. Find the  electric and magnetic fields E(r,ω) and H(r,ω) that are generated by the current.
  2. Sketch the far field radiation pattern of the antenna in transverse and vertical planes (e.g. x-y and x-z planes).
  3. 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.

Helmholtz equation for the vector potential in frequency space

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.

electric and magnetic field dependency on vector potential in frequency space

2. Approximate E(r,ω) and H(r,ω) for rλ to find far fields (e.g. Eff(r,ω) and Hff(r,ω)). Then, using far fields calculate the Poynting vector. Finally, sketch the radial component of the Poynting vector versus φ (azimuth angle) in x-y plane, and versus θ (polar angle) in x-z plane.

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.

D(θ,φ)=radiation intensityaverage of radiation intensity over all directions

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.

(1)2A(r,ω)+ω2μεA(r,ω)=μJ0δ(r)z^

We can split this equation into three equations, each in one direction of the Cartesian coordinate (e.g. x^,y^,andz^). If you do that, you will see that for Ax(r,ω) and Ay(r,ω) there is no source term (right hand side of the their governing equations is zero). Therefore, Ax(r,ω) and Ay(r,ω) are zero.

(2)A(r,ω)=Az(r,ω)z^

Using this fact, equation [1] can be simplified as below.

(3)2Az(r,ω)+ω2μεAz(r,ω)=μJ0δ(r)

Since the source term (right hand side of equation [3]) has a spherical symmetry (since it does not depend on θ and φ), Az(r,ω) is also spherically symmetric (there is not any preferred direction for Az(r,ω) ).

Thus, Az(r,ω) is only a function of radial distance from the source point, i.e. Az(r,ω)=Az(r,ω). Due to this spherical symmetry we use spherical coordinates to expand equation [3] as shown below.

(4)1r2r(r2Az(r,ω)r)+ω2μεAz(r,ω)=μJ0δ(r)

The right hand side of equation [4] is only non-zero in the center of coordinate. So, our strategy to tackle equation [4] is to solve it for all space excluding the center, and then satisfying the source condition at center (we will see the second part later). Therefore, we solve the following equation first.

(5)1r2r(r2Az(r,ω)r)+ω2μεAz(r,ω)=0,(r0)

To make equation [5] a little bit simpler, without loss of generality we define Az(r,ω) as below.

(6)Az(r,ω)=ϕ(r,ω)r

By substituting equation [6] in equation [5] we obtain

2ϕ(r,ω)r2+ω2μεϕ(r,ω)=0,(r0)

ϕ(r,ω)=C1eikr+C2eikr

which results in,

Az(r,ω)=C1eikrr+C2eikrr

where, k=ωμε, and C1 and C2 are two arbitrary constants.

Now, we need to determine C1 and C2. Firstly, as we have used eiωt as the time dependency of harmonic fields (see Basics), C1eikr corresponds to an outgoing wave while C2eikr is an incoming wave. Since we assumed that the antenna is in a homogeneous space, electromagnetic waves go away from the antenna. Therefore, there is not any incoming wave. So, C2 is zero.

(7)Az(r,ω)=C1eikrr

Note that equation [7] is valid for r0 and we need to determine C1 in a way that equation [7] satisfies equation [3] at r=0 as well (source condition). This step is a little bit tricky. Since r=0 is a singular point of Az(r,ω), it is not possible to discuss about the value of Az(r,ω) exactly at r=0 and relate it to the source. However, Az(r,ω) is integrable over a volume around r=0, let say a small sphere with radius a. Therefore, we can relate the integral value of A(r,ω) to the integral value of source (J(r,ω)) over this volume to determine C1. To do that, we integrate both sides of the equation [3] over a very small sphere around r=0 with radius a as it approaches zero. For every term in equation [3] we calculate the integral as shown below.

va2Az(r,ω)dv=va(Az(r,ω))dv=saAz(r,ω).ds=saC1eikr(ikr1r2)r2sin(θ)dθdφ=4πC1eika(ika1)

(8)lima0va2Az(r,ω)dv=4πC1

(9)lima0vaω2μεAz(r,ω)dv=ω2μεlima0vaC1eikrrr2sin(θ)dθdφdr=0

(10)lima0vaμJ0δ(r)dv=μJ0lima0vaδ(r)dv=μJ0

By putting the results of equation [8], [9], and [10] together, C1 can be found

C1=μJ04π

and therefore,

Az(r,ω)=μJ04πeikrr

(11)A(r,ω)=μJ04πeikrr

Now, one can calculate the magnetic field easily as follows.

H(r,ω)=1μ×A(r,ω)=1μ×(μJ04πeikrrz^)=J04π(eikrr)×z^=J04πeikr(ikr1r2)r^×z^

(12)H(r,ω)=ikJ04πeikrr(1+ikr)sin(θ)φ^

to calculate the electric field, we do the following.

(A(r,ω))=[μJ04πeikr(ikr1r2)r^z^]=μJ04π[eikr(ikr1r2)cos(θ)]=μJ04πeikr[cos(θ)(k2r2ikr2+2r3)r^+sin(θ)(ikr2+1r3)θ^]

E(r,ω)=iωA(r,ω)(A(r,ω))iωμε(13)=iωμJ04πeikrr[cos(θ)(2ikr+2(kr)2)r^+sin(θ)(1ikr+1(kr)2)θ^]

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 (φ and θ). Since in reality the receiver and transmitter antennas are located very far from each other, the calculation and measurement of radiation patterns are usually done using the fields in far distances form antennas. Hence, the fields are called far fields and the resultant pattern is called far filed radiation pattern or briefly far field pattern (we will see in Problem One how to precisely define the far field zone of an antenna). It should be noted that the radiation pattern is usually normalized to its maximum.
Based on what is mentioned, firstly, we need to calculate far fields. According to equation [12] and [13], for kr1, the electric and magnetic fields can be approximated as below.

(14)Hff(r,ω)ikJ04πeikrrsin(θ)φ^

(15)Eff(r,ω)iωμJ04πeikrrsin(θ)θ^

This approximation is only valid for fields at large distance form the antenna (kr1).This condition can also be rewritten in the following way.

(16)kr1r1k=1ω(με)=cω=λ2πrλ

where λ is the wavelength of the wave in the medium. As can be seen, far field zone for an infinitesimal antenna is all points of space for which the distance to the antenna is much larger than wavelength.

Now, the Poynting vector can be calculated as below.

Sff(r,ω)=12Real{Eff(r,ω)×Hff(r,ω)}(17)=μ32ε12ω2J0232π21r2sin2(θ)r^

As can be seen the intensity drops as 1r2 which is expected from the law of conservation of energy.
The far field radiation pattern can be calculated easily by normalizing equation [17].

(18)Sffn(θ,φ)=sin2(θ)

Here, the superscript 'n' denotes the normalization. The far filed radiation pattern is depicted in figure (1) in three dimensions.

radiation pattern of an infinitesimal dipole
Fig. 1: Radiation pattern of the infinitesimal dipole

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).

radiation pattern infinitesimal dipole plane cut
Fig. 2: Cut through the radiation pattern of the infinitesimal dipole in the x-z plane.

 

infinitesimal dipole radiation pattern in x-y plane, rotational symmetry
Fig. 3: Because of rotational symmetry, the radiation pattern in the x-y plane is not varying.

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 [17], one can calculate the HW of the antenna in x-z plane as below.

μ32ε12ω2J0232π21r2sin2(θ)12×μ32ε12ω2J0232π21r2

sin2(θ)12θ(45,45)

HW=90

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.

(19)D(θ,φ)=radiation intensityaverage of radiation intensity over all directions

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 [19], the directivity of our infinitesimal antenna can be obtained as below.

D(θ,φ)=Sffn(θ,φ)14π02π0πSffn(θ,φ)sin(θ)dθdφ=sin2(θ)14π02π0πsin3(θ)dθdφ(20)=32sin2(θ)

For example the directivity of the antenna in the direction of θ=60,φ=30 is equal to 1.125, which mean that the antenna radiates 1.125 times larger than the average radiation in this direction( note that the maximum directivity for an infinitesimal antenna is 1.5). As can be seen in equation [20], the diectivity is not a function of azimuth angle. Therefore, it is said that the antenna is omni-directional, which means that there is one plane in which the directitity is constant (here x-y plane).

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!



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