Antennas are used for receiving and transmitting the electromagnetic signals. Their size depends upon the operating frequency / wavelength. Higher is the frequency, lower is the size of antenna. They work on Maxwell equations for field theory. They are of various types for different applications like TV transmission, AM transmission, FM transmission and satellite transmission. The waves travel in free space.

Where the wave number comes from. Substituting a time-harmonic field into Maxwell's equations in a source-free lossless medium gives the wave equation

\(\nabla^{2}\mathbf{E}+\omega^{2}\mu\epsilon\,\mathbf{E}=0\)

Comparing with the standard form \(\nabla^{2}\mathbf{E}+k^{2}\mathbf{E}=0\) identifies

\(k=\omega\sqrt{\mu\epsilon}\)

and in free space, with μ₀ and ε₀,

\(k_0=\omega\sqrt{\mu_0\epsilon_0}\)

which is option 1.

Check it against the velocity of light. Since \(c=1/\sqrt{\mu_0\epsilon_0}\),

\(k_0=\dfrac{\omega}{c}=\dfrac{2\pi f}{c}=\dfrac{2\pi}{\lambda}\)

the familiar "phase shift per metre" form. This is the safest confirmation available: the wave number must equal 2π/λ, and only option 1 delivers that.

Check the units. k₀ is a phase constant, so it must come out in radians per metre.

Do not confuse k₀ with η₀. The two free-space constants are built from the same pair of quantities but combined differently:

\(k_0=\omega\sqrt{\mu_0\epsilon_0}=\dfrac{2\pi}{\lambda}, \qquad \eta_0=\sqrt{\dfrac{\mu_0}{\epsilon_0}}=120\pi\approx 377\ \Omega\)

The product under the root gives the wave number; the ratio gives the intrinsic impedance. Options 2 and 4 are built from the ratio and are therefore impedance-like, not phase constants.

Why the passage mentions antenna size. Because \(k_0=2\pi/\lambda\), all antenna dimensions are naturally measured in units of k₀L. A half-wave dipole is the length for which \(k_0L=\pi\), and raising the frequency raises k₀, so the same electrical length is achieved by a physically shorter antenna — exactly the statement made in the passage.

Hence, \(k_0=\omega_0\sqrt{\mu_0\epsilon_0}\).

Start from the definition of directive gain. It compares the power density an antenna produces in a given direction with the power density that would exist if the same total power were radiated equally in all directions:

\(G_d(\theta,\phi)=\dfrac{U(\theta,\phi)}{U_{av}}=\dfrac{4\pi U(\theta,\phi)}{P_{rad}}\)

The denominator, \(U_{av}=P_{rad}/4\pi\), is precisely the radiation intensity of a source that radiates uniformly over the whole sphere — an isotropic antenna. So the isotropic radiator is the reference built into the definition itself.

Why it is chosen even though it cannot exist. A truly isotropic radiator is physically impossible: any real antenna must have a null somewhere, a result of the boundary conditions on the fields (you cannot comb a sphere without a parting). But that does not matter, because the reference is only a normalising constant. Its virtues are that it is unambiguous, frequency-independent, polarisation-independent and identical for every author — which no real antenna is.

The units that follow. Gain referred to an isotropic radiator is written dBi, and it is the standard used in every antenna specification and link budget.

Why the other three fail as references.

Note that "infinitesimal dipole" and "elementary dipole" are two names for the same thing, so neither could be uniquely correct in any case — a useful elimination clue.

The conversion worth remembering.

\(G_{dBi}=G_{dBd}+2.15\)

because the half-wave dipole itself has 2.15 dBi of gain. Confusing the two references is the commonest error in practical link calculations.

Directive gain versus power gain. Directive gain uses radiated power in the denominator; power gain uses input power and therefore includes the radiation efficiency: \(G_p=\eta G_d\). Both are referred to the same isotropic standard.

Hence, the standard reference antenna is the isotropic antenna.

Recognise the form. Rearranged, the given equation reads

\(\nabla^{2}\mathbf{E}_S+k^{2}\mathbf{E}_S=0\)

This is the Helmholtz equation, and because the unknown is a vector field it is the vector Helmholtz equation. Option 1.

Where it comes from. Take the curl of Faraday's law and substitute Ampère's law in a source-free, lossless, linear medium:

\(\nabla\times\nabla\times\mathbf{E}=-j\omega\mu(\nabla\times\mathbf{H})=-j\omega\mu(j\omega\epsilon\mathbf{E})=\omega^{2}\mu\epsilon\mathbf{E}\)

Using the identity \(\nabla\times\nabla\times\mathbf{E}=\nabla(\nabla\cdot\mathbf{E})-\nabla^{2}\mathbf{E}\) with \(\nabla\cdot\mathbf{E}=0\) in a charge-free region gives exactly the equation above, with \(k^{2}=\omega^{2}\mu\epsilon\).

Why the phasor subscript matters. The time-harmonic assumption \(\mathbf{E}(t)=\mathrm{Re}\{\mathbf{E}_Se^{j\omega t}\}\) replaces every \(\partial/\partial t\) by jω, which is what turns the full wave equation — second order in both space and time — into an equation in space alone. Helmholtz is the frequency-domain form of the wave equation.

Distinguish it from the three distractors.

The structural clue. Poisson's equation has a source term on the right and no k, so it cannot be a wave equation. The diffusion equation carries a factor of j on the right-hand side, giving exponential decay rather than propagation. The Coulomb gauge is a condition on the vector potential, \(\nabla\cdot\mathbf{A}=0\), not a field equation at all. Only Helmholtz has the pure \(+k^{2}\) term that produces travelling-wave solutions \(e^{-jkz}\).

Hence, the equation is the vector Helmholtz equation.

Know the two numbers and which is which. For a half-wave dipole the directivity as a ratio is

\(D=1.64\)

and the question asks for it in decibels:

\(D_{dB}=10\log_{10}(1.64)=10\times0.2148\)

\(D_{dB}=2.15\ \text{dB}\)

which is option 3.

The trap is option 4. 1.64 is the correct directivity, but as a plain ratio, not in decibels. The question is testing whether the two forms are kept apart — and 1.64 dB would actually correspond to a ratio of about 1.46, not 1.64. Whenever both a number and its decibel value appear among the choices, check which the question asked for.

Option 1 is the other classic distractor. 1.76 dB is the directivity of the Hertzian (infinitesimal) dipole, whose ratio is 1.5:

\(10\log_{10}(1.5)=1.76\ \text{dB}\)

The half-wave dipole is slightly more directive than the elementary one because its current distribution is sinusoidal rather than uniform, which narrows the beam a little — from a half-power beamwidth of 90° down to 78°.

Where 1.64 comes from. The far field of a half-wave dipole is

\(E_\theta \propto \dfrac{\cos\left(\dfrac{\pi}{2}\cos\theta\right)}{\sin\theta}\)

Integrating \(|E_\theta|^{2}\) over the sphere gives a radiation resistance of 73 Ω and, on forming \(D=4\pi U_{max}/P_{rad}\), the value 1.64.

Why 2.15 dB is worth memorising above all. It is the conversion between the two gain references:

\(G_{dBi}=G_{dBd}+2.15\)

Manufacturers quote antenna gain sometimes against isotropic (dBi) and sometimes against a half-wave dipole (dBd), and 2.15 dB is the constant that reconciles them.

Summary of the standard values. Half-wave dipole: D = 1.64 = 2.15 dBi, R_rad = 73 Ω, length λ/2. Hertzian dipole: D = 1.5 = 1.76 dBi.

Hence, the directivity of a half-wave dipole is 2.15 dB.

The principle that decides every case. An antenna is broadband when its behaviour is governed by angles rather than by a fixed length. If the geometry can be described by angles alone, the structure looks electrically similar at every frequency and the impedance stays roughly constant. If instead the antenna is resonant — defined by a length that must equal a particular fraction of a wavelength — it works only near that one frequency.

Discone — wideband. A disc above a cone is a classic frequency-independent structure: the cone's flare is set by an angle, and the cone presents a progressively larger effective aperture as the wavelength grows. Bandwidths of 10:1 in impedance are routine, which is why discones are the standard antenna on wideband scanning receivers.

Helical — wideband. In axial mode the helix radiates a circularly polarised beam along its axis over roughly a 1.7:1 frequency range, with an almost purely resistive input impedance near 140 Ω. It is the standard antenna for satellite links, where the wide band and circular polarisation both matter.

Folded dipole — relatively wideband. Two parallel conductors joined at the ends behave like a thick radiator; the increased effective diameter lowers the Q of the resonance and flattens the impedance curve, giving several times the bandwidth of a plain dipole. The folding also raises the radiation resistance from 73 Ω to about 292 Ω, which is why it matches 300 Ω twin feeder directly — the familiar television and FM receiving antenna.

Marconi — narrowband, so this is the answer. The Marconi antenna is a quarter-wave vertical monopole worked against a ground plane, the image in the ground completing the equivalent half-wave dipole. Its very definition is a length:

\(L=\dfrac{\lambda}{4}\)

so it is resonant at one frequency, with radiation resistance about 36.5 Ω and a sharply rising reactance either side of resonance. Change the frequency and the antenna is simply the wrong length. It is used for AM broadcast and other long-wave services, where a single assigned channel is all that is needed — the passage's mention of AM transmission is the clue.

The test in one line. Discone, helix and folded dipole all achieve bandwidth through geometry — flare angle, pitch angle, or effective thickness. The Marconi has none of these; it has only a length.

Hence, the antenna that is not wideband is the Marconi.