M0. EM Wave Foundations
Every radar transmission begins as a solution of Maxwell's equations. Before we can reason about pulses, Doppler, or SAR, we need sharp command of the wave equation, polarization, the Larmor radiation formula, the IEEE radar band plan, and how the atmosphere absorbs microwaves.
1. Maxwell Equations and the Wave Equation
In vacuum, the four Maxwell equations in SI units are:
$$\nabla\cdot\mathbf{E} = \frac{\rho}{\varepsilon_0},\qquad \nabla\cdot\mathbf{B} = 0,\qquad \nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t},\qquad \nabla\times\mathbf{B} = \mu_0\mathbf{J} + \mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}.$$
In source-free regions ($\rho=0,\;\mathbf{J}=0$) we take the curl of Faraday's law:
$$\nabla\times(\nabla\times\mathbf{E}) = -\frac{\partial}{\partial t}(\nabla\times\mathbf{B}) = -\mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}.$$
Using the identity $\nabla\times(\nabla\times\mathbf{E})=\nabla(\nabla\cdot\mathbf{E})-\nabla^2\mathbf{E}$and $\nabla\cdot\mathbf{E}=0$ (vacuum) gives the vector wave equation:
$$\boxed{\;\nabla^2\mathbf{E} - \frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = 0\;},\qquad c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 2.998\times10^8\;\text{m/s}.$$
An identical equation holds for $\mathbf{B}$. The finite propagation speed $c$is the origin of radar ranging: the two-way travel time $t = 2R/c$ directly encodes range.
2. Plane-Wave Solutions
Seeking solutions of the form $\mathbf{E}(\mathbf{r},t) = \mathbf{E}_0\,e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$substitutes into the wave equation and yields the dispersion relation:
$$|\mathbf{k}|^2 = \frac{\omega^2}{c^2}\qquad\Longrightarrow\qquad \omega = c|\mathbf{k}|,\;\; \lambda = \frac{2\pi}{k} = \frac{c}{f}.$$
Gauss's law $\nabla\cdot\mathbf{E}=0$ constrains $\mathbf{k}\cdot\mathbf{E}_0=0$: electromagnetic waves are transverse. From Faraday's law we find the magnetic field:
$$\mathbf{B} = \frac{1}{\omega}\,\mathbf{k}\times\mathbf{E},\qquad |\mathbf{B}| = \frac{|\mathbf{E}|}{c}.$$
The time-averaged Poynting vector is the power flux in W/m\u00B2:
$$\langle\mathbf{S}\rangle = \frac{1}{2\mu_0}\,\operatorname{Re}(\mathbf{E}\times\mathbf{B}^{*}) = \frac{|\mathbf{E}_0|^2}{2\,\eta_0}\,\hat{\mathbf{k}},\qquad \eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 376.73\;\Omega.$$
The impedance of free space $\eta_0$ appears throughout antenna theory: aperture efficiency, mismatch, and gain all trace back to this quantity.
Plane-Wave Structure
3. Polarization: Linear, Circular, Elliptical
The most general transverse field can be written (for propagation along $\hat{\mathbf{x}}$):
$$\mathbf{E}(x,t) = \operatorname{Re}\left[\left(E_y\hat{\mathbf{y}} + E_z\,e^{i\delta}\hat{\mathbf{z}}\right)e^{i(kx-\omega t)}\right].$$
- Linear: $\delta = 0$ or $\pi$. The tip of $\mathbf{E}$ traces a line.
- Right circular (RCP): $E_y=E_z,\;\delta=+\pi/2$. The tip rotates clockwise looking along $\hat{\mathbf{k}}$.
- Left circular (LCP): $\delta=-\pi/2$. Tip rotates counterclockwise.
- Elliptical: general case. Described by Stokes parameters $(I,Q,U,V)$.
The Stokes parameters for a fully-polarized wave are:
$$I = |E_y|^2 + |E_z|^2,\; Q = |E_y|^2 - |E_z|^2,\; U = 2\operatorname{Re}(E_y^{*}E_z),\; V = 2\operatorname{Im}(E_y^{*}E_z).$$
Radar systems distinguish HH, VV, HV, VH channels; polarimetric SAR (Module 6) uses the full scattering matrix. Circular polarization is preferred for weather radar because it is insensitive to target orientation.
4. Radiation from Accelerated Charges (Larmor)
A non-relativistic accelerated charge radiates power according to the Larmor formula:
$$\boxed{\;P = \frac{q^2 |\dot{\mathbf{v}}|^2}{6\pi\varepsilon_0 c^3}\;}$$
Derivation sketch: start from the LiΓ©nard-Wiechert potentials, evaluate the Poynting vector in the far field ($1/r$ terms), and integrate over the sphere. The angular distribution is the familiar $\sin^2\theta$ pattern (dipole):
$$\frac{dP}{d\Omega} = \frac{q^2|\dot{\mathbf{v}}|^2}{16\pi^2\varepsilon_0 c^3}\sin^2\theta.$$
For a Hertzian dipole of current $I_0\cos\omega t$ and length $\ell\ll\lambda$, the radiated power is $P = (I_0\ell\omega)^2 /(12\pi\varepsilon_0 c^3)$, giving a radiation resistance of $R_{\text{rad}} = 80\pi^2(\ell/\lambda)^2\,\Omega$. This is the starting point for every transmitting antenna.
5. IEEE Radar Frequency Bands
The IEEE 521-2019 standard divides radar spectrum into lettered bands. Each band has a sweet spot balancing range, resolution, atmospheric transmission, and hardware cost.
| Band | Frequency | $\lambda$ | Typical Use |
|---|---|---|---|
| HF | 3-30 MHz | 10-100 m | OTH (ionospheric) radar |
| VHF | 30-300 MHz | 1-10 m | Very long range, counter-stealth |
| UHF | 300 MHz-1 GHz | 0.3-1 m | Early warning, foliage penetration |
| L | 1-2 GHz | 15-30 cm | ATC (en-route), long-range surveillance |
| S | 2-4 GHz | 7.5-15 cm | ATC terminal, weather (NEXRAD) |
| C | 4-8 GHz | 3.75-7.5 cm | Weather (Europe), Sentinel-1 SAR |
| X | 8-12 GHz | 2.5-3.75 cm | Fire control, marine, TerraSAR |
| Ku | 12-18 GHz | 1.7-2.5 cm | Police speed, terrain mapping |
| K | 18-27 GHz | 1.1-1.7 cm | Rarely used (H\u2082O absorption 22.2 GHz) |
| Ka | 27-40 GHz | 7.5-11 mm | High-resolution imaging, cloud radar |
| V | 40-75 GHz | 4-7.5 mm | Short-range; O\u2082 peak at 60 GHz |
| W | 75-110 GHz | 2.7-4 mm | Automotive 77 GHz, cloud profilers |
Why so many bands? Resolution improves with shorter wavelength but absorption and cost rise. $R^4$ losses dominate at low frequencies, hydrometeor clutter at high. Picking a band is already a system-engineering decision.
6. Atmospheric Attenuation
Microwave absorption is dominated by two molecular resonances:
- Water vapor (H\u2082O): rotational lines at 22.235, 183.31, 325.15 GHz. The 22 GHz line limits K-band.
- Molecular oxygen (O\u2082): a complex of ~44 magnetic-dipole lines near 60 GHz plus 118.75 GHz. Gives >10 dB/km at 60 GHz.
The ITU-R Recommendation P.676 models specific attenuation $\gamma$ [dB/km]. For a two-way path of length $R$, total path loss is $L_{atm} = 2\gamma R$. Typical values at sea level, 20 Β°C, 7.5 g/m\u00B3 humidity:
$$\gamma(10\,\text{GHz}) \approx 0.01\,\text{dB/km},\quad \gamma(35\,\text{GHz}) \approx 0.1,\quad \gamma(60\,\text{GHz}) \approx 15,\quad \gamma(77\,\text{GHz}) \approx 0.5.$$
Rain attenuation adds a term $\gamma_R = a R^b$ (dB/km) where $R$ is the rain rate in mm/hr and the coefficients $(a,b)$ depend on frequency and polarization (ITU-R P.838). Rain absorbs more at higher frequencies β another reason to avoid Ka for long-range weather surveillance.
Electromagnetic Spectrum with Radar Bands
Simulation: Plane Waves, Polarization & Atmospheric Absorption
The Python code below (i) animates an X-band plane wave, (ii) plots polarization ellipses for linear, circular, and elliptical states, (iii) computes specific atmospheric attenuation via the ITU-R P.676 approximation, and (iv) lays out the IEEE radar band allocation chart.
Click Run to execute the Python code
Code will be executed with Python 3 on the server
7. Energy Flow and the Poynting Theorem
The electromagnetic energy density and power flow are defined by:
$$u = \frac{1}{2}\left(\varepsilon_0|\mathbf{E}|^2 + \frac{|\mathbf{B}|^2}{\mu_0}\right),\qquad \mathbf{S} = \frac{1}{\mu_0}\,\mathbf{E}\times\mathbf{B}\;[\text{W/m}^2].$$
From Maxwell's equations one derives Poynting's theorem as a local conservation law:
$$\frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = -\mathbf{J}\cdot\mathbf{E}.$$
For a transmitted radar pulse, the total energy is $E_t = \int P_t(t)\,dt \approx P_{\text{avg}}\cdot\text{PRI}$; it is this energy (and not peak power) that determines SNR through the matched filter.
8. The Helmholtz Equation and Green's Functions
Taking the Fourier transform of the wave equation in time, we get the Helmholtz equation for each frequency component:
$$(\nabla^2 + k^2)\mathbf{E}(\mathbf{r},\omega) = -i\omega\mu_0\mathbf{J}(\mathbf{r},\omega)-\frac{1}{\varepsilon_0}\nabla\rho(\mathbf{r},\omega).$$
The free-space scalar Green's function satisfying $(\nabla^2+k^2)G = -\delta^3(\mathbf{r}-\mathbf{r}')$ is
$$G(\mathbf{r},\mathbf{r}') = \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|},$$
which is the key ingredient for every scattering calculation. Integrating over source currents gives the radiated field used in deriving the radar cross section (Module 1) and in physical optics approximations of complex targets.
9. Reflection at Material Boundaries (Fresnel)
Radar returns arise largely from discontinuities in the complex permittivity. For a plane wave incident at angle $\theta_i$ on a plane interface between vacuum and a dielectric of index $n = \sqrt{\varepsilon_r}$, the Fresnel reflection coefficients are
$$r_\parallel = \frac{n^2\cos\theta_i - \sqrt{n^2-\sin^2\theta_i}}{n^2\cos\theta_i + \sqrt{n^2-\sin^2\theta_i}},\qquad r_\perp = \frac{\cos\theta_i - \sqrt{n^2-\sin^2\theta_i}}{\cos\theta_i + \sqrt{n^2-\sin^2\theta_i}}.$$
A conductor has $|r|\to 1$ (specular reflection) β this is why metal is the ideal radar target. Sea water has $\varepsilon_r \approx 70 + 40i$ at S-band and reflects strongly; dry soil has $\varepsilon_r \approx 3-5$ and reflects weakly. The Brewster angle $\theta_B = \arctan n$ is where $r_\parallel = 0$; this matters for dual-polarization weather radar.
10. Tropospheric Refraction and the 4/3-Earth Approximation
The index of refraction of air differs from unity by a few parts in 10\u2074. Radio engineers express this as refractivity $N = (n-1)\times 10^6$. A useful approximation at microwave is
$$N = \frac{77.6}{T}\left(P + \frac{4810\, e_w}{T}\right),$$
where $T$ is temperature (K), $P$ total pressure (hPa), and $e_w$water-vapor partial pressure (hPa). Typical surface refractivity is $N \approx 315$, decreasing exponentially with height with scale $\sim 7\,\text{km}$. The ray curvature is $\kappa = -dn/dh \approx 4\times 10^{-8}\,\text{m}^{-1}$; an effective Earth radius$a_e = 4a/3$ accounts for this so that rays travel in straight lines in the modified geometry. This is the 4/3 Earth radius approximation, used for low-elevation radar range calculations.
Anomalous propagation (ducting) occurs when $dN/dh < -157$ N-units/km over an elevated layer, trapping rays and creating false long-range returns.
11. Hydrometeor and Rain Attenuation
In addition to gas absorption, liquid water in clouds and rain attenuates radar signals. The Mie-theory specific attenuation in rain is well approximated by the ITU-R P.838 power-law:
$$\gamma_R = k\,R^\alpha\;[\text{dB/km}],$$
where $R$ is rainfall rate in mm/hr. Typical coefficients (horizontal pol): at 10 GHz, $k=0.0101,\alpha=1.276$; at 35 GHz, $k=0.242,\alpha=0.997$; at 94 GHz, $k=1.12,\alpha=0.730$. A 25 mm/hr downpour produces ~3 dB/km at X band and ~15 dB/km at Ka band β catastrophic for long paths.
12. Skin Depth in Conductors
In a good conductor of conductivity $\sigma$, the wave decays exponentially over the skin depth
$$\delta = \sqrt{\frac{2}{\omega\mu\sigma}}.$$
For copper ($\sigma = 5.8\times 10^7\,\text{S/m}$) at 10 GHz, $\delta\approx 0.66\,\mu\text{m}$; this is why the interior of a waveguide must be electroplated smooth. It also explains why thin metallic paint can be used as RAM (radar-absorbent material) only above its cutoff β covered in Module 8 on stealth.
References
- Jackson, J.D. β Classical Electrodynamics, 3rd ed., Wiley (1998), chs. 6-9.
- Balanis, C.A. β Advanced Engineering Electromagnetics, 2nd ed., Wiley (2012).
- IEEE Std 521-2019 β Standard Letter Designations for Radar-Frequency Bands.
- ITU-R Recommendation P.676-12 β Attenuation by atmospheric gases (2019).
- Skolnik, M.I. β Introduction to Radar Systems, 3rd ed., ch. 1.
- Rosenkranz, P.W. β βAbsorption of microwaves by atmospheric gasesβ, in Atmospheric Remote Sensing by Microwave Radiometry, Wiley (1993).
Cross-References
For the underlying math and physics, see: