Part 7 · Chapter 29

Antenna Fundamentals and Radiation

Every chapter so far has kept the wave penned in — between conductors, inside a pipe, trapped in a box. An antenna does the opposite: it lets the wave go. A current that merely sits still produces a static field that clings to its source, but a current that changes sheds energy that breaks free and races outward forever as a radiated wave. This chapter follows that escape from first principles — from the retarded potential of an oscillating current element, through the doughnut-shaped pattern it paints on the sky, to the numbers engineers actually quote: radiation resistance, directivity, gain, and the Friis equation that ties two distant antennas into a working radio link.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 56 min
i What you'll learn
  • Why a time-varying current radiates while a steady one does not.
  • The retarded vector potential and how it builds the radiated field.
  • The Hertzian dipole far field, and the near/far field regions.
  • The \(\sin\theta\) radiation pattern and the radiated power.
  • The radiation resistance \(R_{\text{rad}}=80\pi^2(dl/\lambda)^2\).
  • Directivity, gain, effective aperture, the half-wave dipole, and the Friis equation.
Section 29-1

What Makes a Wave Radiate

A constant current produces a static magnetic field that wraps around the wire and stays put — no energy leaves. A current that oscillates is different: it accelerates charge, and accelerating charge radiates. Part of the field near the wire still stores and returns energy each cycle (the reactive near field), but a portion detaches, forms closed loops that pinch off, and propagates away as a self-sustaining electromagnetic wave that never comes back. That escaping part is radiation, and an antenna is simply a structure shaped to maximize it — a transducer between the guided waves of a line and the free-space waves of Part 5.

To compute it we need a tool that respects the finite speed of light: the field at a distant point cannot reflect what the current is doing now, only what it did a travel-time ago.

Section 29-2

Retarded Potentials

The magnetic vector potential of a current distribution, generalized to time-varying sources, is the retarded potential: each contribution is evaluated not at the present instant but at the earlier time \(t-R/u\), where \(R\) is the distance from the source element and \(u\) the wave speed. For a time-harmonic current this retardation becomes a phase factor \(e^{-j\beta R}\):

Retarded vector potential
\[ \vec{A}=\frac{\mu}{4\pi}\int_V \frac{\vec{J}\,e^{-j\beta R}}{R}\,dV' \]

Once \(\vec{A}\) is known, the radiated fields follow from \(\vec{H}=\tfrac{1}{\mu}\nabla\times\vec{A}\) and Maxwell's curl equation for \(\vec{E}\). The single phase factor \(e^{-j\beta R}\) is the whole origin of radiation: without retardation the potential would be the static one and nothing would propagate. Apply this to the simplest possible source and the entire theory of antennas unfolds.

Section 29-3

The Hertzian Dipole

The Hertzian dipole is an infinitesimal current element: a short length \(dl\ll\lambda\) carrying a uniform oscillating current \(I_0\). It is to antennas what the point charge is to electrostatics — the elementary building block from which any real antenna can be assembled by superposition. Evaluating the retarded potential for it and taking the curls produces fields with terms in \(1/r\), \(1/r^2\), and \(1/r^3\). Far from the antenna (\(r\gg\lambda\)) only the \(1/r\) terms survive, and they are the radiation:

Hertzian dipole far field
\[ E_\theta=j\,\frac{\eta_0 I_0\,dl}{2\lambda r}\sin\theta\;e^{-j\beta r}, \qquad H_\phi=\frac{E_\theta}{\eta_0} \]
z I₀ dl r θ P (far) E_θ H_φ
The Hertzian dipole: a far-field point at angle θ sees E_θ and H_φ, perpendicular to each other and to r

In the far field the radiated wave is locally a plane wave: \(E_\theta\) and \(H_\phi\) are perpendicular, in phase, and in the ratio \(\eta_0\), travelling radially outward. The reactive near field (\(r\ll\lambda\), the \(1/r^2\) and \(1/r^3\) terms) stores energy without radiating; the boundary between the two regions sits roughly at \(r=2D^2/\lambda\) for an antenna of size \(D\).

Section 29-4

Radiation Pattern

The far field carries a single angular factor, \(\sin\theta\): the dipole radiates most strongly broadside (\(\theta=90^\circ\), perpendicular to its axis) and not at all along its own axis (\(\theta=0,180^\circ\)). Plotted in three dimensions this is a doughnut wrapped around the wire; in any plane containing the axis it is the familiar figure-of-eight:

dipole axis (null) max (broadside) |E| ∝ sin θ
The short-dipole pattern: a figure-of-eight (doughnut in 3-D) — maximum broadside, null along the axis

The squared factor \(\sin^2\theta\) is the power pattern. Its half-power beamwidth is a broad \(90^\circ\), so the short dipole spreads its energy widely — useful for coverage, poor for reaching far. Sharpening that pattern, by making the antenna larger or arraying many of them, is the route to directivity, taken up shortly and in the next chapter.

Section 29-5

Radiated Power and R_rad

Integrating the time-average Poynting vector \(S_{\text{avg}}=|E_\theta|^2/2\eta_0\) over a large sphere gives the total radiated power. The \(\sin^2\theta\) pattern integrates to a clean constant, leaving:

Radiated power of a Hertzian dipole
\[ P_{\text{rad}}=\frac{\eta_0\pi}{3}\!\left(\frac{I_0\,dl}{\lambda}\right)^2=40\pi^2\!\left(\frac{dl}{\lambda}\right)^2 I_0^2 \]

To the feeding circuit, this lost power looks exactly like power dissipated in a resistor. Defining \(P_{\text{rad}}=\tfrac12 I_0^2 R_{\text{rad}}\) gives the radiation resistance — a fictitious resistance that accounts for radiated (not dissipated) power, and the single most important number for matching an antenna to its feed:

Radiation resistance
\[ R_{\text{rad}}=\frac{2P_{\text{rad}}}{I_0^2}=80\pi^2\!\left(\frac{dl}{\lambda}\right)^2 \]

The quadratic dependence on \(dl/\lambda\) is decisive: a very short element has a tiny radiation resistance, so most of the drive current is wasted in ohmic loss rather than radiated — which is exactly why useful antennas are a sizeable fraction of a wavelength, not arbitrarily small.

Section 29-6

Directivity and Gain

How sharply an antenna concentrates its radiation is captured by a few standard parameters. The radiation intensity \(U=r^2 S_{\text{avg}}\) is power per unit solid angle (independent of \(r\)). The directivity compares the peak intensity to the average — the factor by which the antenna beats an isotropic radiator in its best direction:

Directivity
\[ D=\frac{U_{\max}}{U_{\text{avg}}}=\frac{4\pi U_{\max}}{P_{\text{rad}}} \]

Directivity is a property of the pattern alone. The gain folds in real losses through the radiation efficiency \(e\) (= \(R_{\text{rad}}/(R_{\text{rad}}+R_{\text{loss}})\)), and the effective aperture \(A_e\) — the equivalent collecting area when receiving — is tied to gain by a universal relation:

Gain and effective aperture
\[ G=e\,D, \qquad A_e=\frac{\lambda^2}{4\pi}\,G \]

Directivities are quoted in dBi (decibels over isotropic). The Hertzian dipole has \(D=1.5\) (1.76 dBi) — barely directional. Larger and arrayed antennas reach tens of thousands (dish antennas exceed 40 dBi). Note that an antenna's transmit and receive properties are identical (reciprocity), so the same gain describes it in both roles.

Section 29-7

Half-Wave Dipole and Friis

The practical workhorse is the half-wave dipole, a straight wire of total length \(\lambda/2\) fed at its centre. Its current is a standing-wave cosine, peaking at the feed and vanishing at the tips, and integrating its far field gives a slightly sharper pattern than the short dipole. Two numbers define it:

Half-wave dipole
\[ R_{\text{rad}}\approx 73\ \Omega, \qquad D\approx 1.64\ (2.15\ \text{dBi}), \qquad F(\theta)=\frac{\cos\!\left(\tfrac{\pi}{2}\cos\theta\right)}{\sin\theta} \]
feed I(z)=I₀cos βz λ/2
The half-wave dipole: cosine current, maximum at the centre feed, zero at the tips

Finally, the link between two antennas a distance \(R\) apart in free space is the Friis transmission equation — the foundation of every radio link budget. The received power depends on both antenna gains and an inverse-square spreading term that grows with frequency:

Friis transmission equation
\[ \frac{P_r}{P_t}=G_t\,G_r\!\left(\frac{\lambda}{4\pi R}\right)^2 \]

The bracketed factor is the free-space path loss: signals fade as \(1/R^2\) simply because the radiated power spreads over an ever-larger sphere. Rearranged in decibels it becomes the everyday link equation \(P_r=P_t+G_t+G_r-20\log_{10}(4\pi R/\lambda)\), which sizes everything from a phone link to a deep-space probe.

Section 29-8

Worked Examples

1 Radiation resistance of a short dipole

Problem. A Hertzian dipole has length \(dl=\lambda/50\). Find its radiation resistance.

Solution. Apply \(R_{\text{rad}}=80\pi^2(dl/\lambda)^2\):

Working
\[ R_{\text{rad}}=80\pi^2\!\left(\tfrac{1}{50}\right)^2=\frac{80\pi^2}{2500}\approx 0.316\ \Omega \]
2 Radiated power

Problem. The same dipole carries a peak current \(I_0=2\,\text{A}\). Find the radiated power.

Solution. Use \(P_{\text{rad}}=\tfrac12 I_0^2 R_{\text{rad}}\):

Working
\[ P_{\text{rad}}=\tfrac12(2)^2(0.316)=0.63\ \text{W} \]
3 Far-field strength

Problem. For that dipole at \(\lambda=1\,\text{m}\), \(I_0=2\,\text{A}\), find the peak far-field \(|E_\theta|\) at \(r=1\,\text{km}\) (broadside, \(\theta=90^\circ\)).

Solution. Use \(|E_\theta|=\eta_0 I_0\,dl/(2\lambda r)\) with \(dl=0.02\,\text{m}\):

Working
\[ |E_\theta|=\frac{377(2)(0.02)}{2(1)(1000)}\approx 7.5\ \text{mV/m} \]
4 Directivity to gain

Problem. A half-wave dipole (\(D=1.64\)) has radiation efficiency \(90\%\). Find its gain in dBi.

Solution. \(G=eD\), then convert to decibels:

Working
\[ G=0.9(1.64)=1.48,\qquad G_{\text{dBi}}=10\log_{10}(1.48)\approx 1.7\ \text{dBi} \]
5 Effective aperture

Problem. Find the effective aperture of a half-wave dipole (\(G\approx1.64\)) at \(100\,\text{MHz}\) (\(\lambda=3\,\text{m}\)).

Solution. Use \(A_e=(\lambda^2/4\pi)G\):

Working
\[ A_e=\frac{(3)^2}{4\pi}(1.64)\approx 1.17\ \text{m}^2 \]
6 A Friis link budget

Problem. A \(2.4\,\text{GHz}\) link (\(\lambda=0.125\,\text{m}\)) has \(P_t=10\,\text{W}\), \(G_t=G_r=10\), and \(R=1\,\text{km}\). Find the received power.

Solution. Apply Friis directly:

Working
\[ P_r=10(10)(10)\!\left(\frac{0.125}{4\pi\cdot1000}\right)^2\approx 9.9\times10^{-8}\ \text{W}\approx -40\ \text{dBm} \]
Review

Chapter Summary

Radiation

Only time-varying current radiates; the retarded factor \(e^{-j\beta R}\) is its source.

Hertzian dipole

\(E_\theta=j\dfrac{\eta_0 I_0 dl}{2\lambda r}\sin\theta\,e^{-j\beta r}\), \(H_\phi=E_\theta/\eta_0\).

Pattern & power

\(\sin\theta\) doughnut; \(P_{\text{rad}}=40\pi^2(dl/\lambda)^2 I_0^2\).

Radiation resistance

\(R_{\text{rad}}=80\pi^2(dl/\lambda)^2\); useful antennas are \(\sim\lambda\) in size.

Directivity & gain

\(D=4\pi U_{\max}/P_{\text{rad}}\), \(G=eD\), \(A_e=(\lambda^2/4\pi)G\).

Dipole & Friis

Half-wave: \(R_{\text{rad}}\approx73\,\Omega\), \(D\approx1.64\); \(P_r/P_t=G_tG_r(\lambda/4\pi R)^2\).

Practice

Problems

Track whether a current is the peak or rms value, and keep lengths in wavelengths. Convert to and from decibels confidently. Difficulty rises down the list.

  1. A Hertzian dipole has \(dl=\lambda/100\). Find its radiation resistance.
  2. For \(dl=\lambda/20\) and \(I_0=1\,\text{A}\), find \(P_{\text{rad}}\).
  3. At what fraction \(dl/\lambda\) does a Hertzian dipole reach \(R_{\text{rad}}=1\,\Omega\)?
  4. Find the peak \(|E_\theta|\) of a dipole with \(dl=\lambda/40\), \(I_0=3\,\text{A}\), at \(r=2\,\text{km}\), \(\lambda=2\,\text{m}\), broadside.
  5. Convert directivities of 1.5, 1.64, and 100 to dBi.
  6. An antenna has \(D=20\,\text{dBi}\) and efficiency 0.8. Find its gain in dBi.
  7. Find the effective aperture of a \(G=1.64\) dipole at \(300\,\text{MHz}\).
  8. Show that the directivity of a Hertzian dipole is 1.5 from its \(\sin^2\theta\) pattern.
  9. A \(1\,\text{GHz}\) link has \(P_t=5\,\text{W}\), \(G_t=100\), \(G_r=10\), \(R=10\,\text{km}\). Find \(P_r\).
  10. Express the Friis result of problem 9 in dBm, identifying the path-loss term.
  11. Explain physically why a very short antenna is an inefficient radiator.
  12. Two identical antennas double their separation. By how many dB does the received power drop, and why?
Tip: radiation begins with one idea — a current that changes sheds a wave, and the retarded factor \(e^{-j\beta R}\) is what lets it escape. The Hertzian dipole is the atom of the subject: its far field \(E_\theta\propto\sin\theta\,/\,r\) gives the doughnut pattern, its integrated power gives the radiation resistance \(R_{\text{rad}}=80\pi^2(dl/\lambda)^2\) — small for short elements, the reason real antennas span a fair fraction of a wavelength. From the pattern come the engineering numbers: directivity \(D=4\pi U_{\max}/P_{\text{rad}}\), gain \(G=eD\), aperture \(A_e=(\lambda^2/4\pi)G\); and two antennas a distance apart obey Friis, \(P_r/P_t=G_tG_r(\lambda/4\pi R)^2\). The half-wave dipole (\(73\,\Omega\), \(D\approx1.64\)) is the one to know. Next, Chapter 30 combines many such elements into arrays — steering and sharpening the beam — and surveys the modern antennas built on these same fundamentals.