Question

A cross dipole is shown below, with the two arms centred at the z-axis, aligned along the x- and y-axes, and excited by uniform currents Ix and Iy, respectively. Using superposition and assuming the two dipoles do not interact with each other, derive an expression for the radiation intensity of the cross-dipole antenna using the auxiliary vector potential method. Use the far-field approximations (2.33) and (2.34) for determining the electric and magnetic fields.

Answer 1

Answer:

A very simple radiating element we can study is the ideal dipole, also known as Hertzian dipole and infinitesimal dipole. It is very short (length<< λ), and as such has current uniformly distributed along its length.

Although it is difficult to implement in practice (having a current distribution that is difficult to realize since it is discontinuous), it is highly useful for helping analyze larger wire antennas which can be subdivided into short sections having uniform current (i.e., ideal dipoles). Then, much in the same way as we derived vector potential for a continuous current distribution, we can use superposition to find the fields of a long wire antenna. Let’s orient the ideal dipole along the z-axis and denote the current flowing through the dipole as I. The current has an associated surface current density J.

First, we need to derive the vector potential of the line source. It is a continuous current distribution over its length ∆l = ∆z. Since we only have a z-component of current, A will only have a z-component as well.

Evaluating the integral, we first notice that since ∆z is small, R does not change significantly as we move along the length of the dipole, (i.e. r ≈ R). Now we can find the radiated magnetic field of the dipole:

                                                         H = (1/ µ )∇ × A = (1/ µ) ∇ × Az zˆ

Some important observations:

• E no longer has a radial component; in the far field, it is totally polarized in the θˆ direction;

• E and H are orthogonal to each other and the direction of propagation and hence the resulting wave is TEM (as we expect for a spherical wave);

• The ratio of Eθ/Hφ is: Eθ /Hφ = ωµ /k = root of (µ/ ε) = η

which is also what we found for a plane wave. We shall see that this is a property of radiated fields.

Let’s discuss these observations by examining/deriving the static fields in question. Starting with the first observation, recall from the Biot-Savart Law that for a short (infinitesimal) segment of current, the magnetic field intensity produced is

                                                               dH = (Idl × R )/(4πR^3)

We can also show that these fields are storing energy : both E-field components are in phase quadrature with the H-field component, indicating reactive power. Explicitly evaluating the Poynting vector which is imaginary (power flow/dissipation is always real). Since the power is imaginary, is represents stored energy in the electric/magnetic (near) fields of the antenna. At one point in the cycle, all the energy is stored in charge accumulations at the ends of the antenna (like an electric dipole), and the antenna is acting very much like a capacitor with the dipole ’ends’ acting as plates giving a fringing capacitance. A quarter cycle later, the magnetic field has collapsed producing an EMF that charges the ’capacitor’ back up with the charge polarity reversed, and so on.