Reading — Spectrum & Propagation#
By the end of this lesson you should be able to:
Place a radar in its frequency band and state the tradeoffs of moving up or down the spectrum.
Relate wavelength and frequency through \(\lambda f = c\).
Compute free-space path loss (FSPL) in decibels and apply the 6 dB rules for range and frequency.
Explain how polarization and antenna gain shape the energy that reaches a target.
The spectrum is the battlefield#
Every radar lives at a frequency, and that choice drives everything else: antenna size, resolution, atmospheric loss, and how easy it is to jam. Wavelength and frequency are tied by the speed of light,
A handy form for radar work: at \(f\) in GHz, \(\lambda \approx 30/f\) in centimeters. So X-band at 10 GHz has a 3 cm wavelength; L-band at 1.5 GHz is about 20 cm.
Radar bands#
The standard IEEE letter bands and their rough character:
Band |
Frequency |
Wavelength |
Typical use / character |
|---|---|---|---|
L |
1–2 GHz |
~15–30 cm |
Long-range search; low atmospheric loss, large antennas, coarse resolution. |
S |
2–4 GHz |
~7.5–15 cm |
Search & moderate-range; weather radars. |
C |
4–8 GHz |
~3.7–7.5 cm |
Compromise band; some weather and fire control. |
X |
8–12 GHz |
~2.5–3.7 cm |
Fire control, airborne radar; good resolution, manageable antenna size. |
Ku / K / Ka |
12–40 GHz |
~0.75–2.5 cm |
Seekers, high-resolution imaging; small antennas, higher atmospheric loss. |
The general trend: lower frequency gives longer range and less weather loss but needs a big antenna and gives poor resolution; higher frequency gives fine resolution and small apertures but suffers more atmospheric attenuation. Fire-control radars cluster around X-band because it balances these — which is why the B-21 threat picture is dominated by X-band fire control.
Key Concept
Frequency choice is a chain of consequences. Pick \(f\) and you have implicitly chosen wavelength, antenna size for a given gain, resolution, atmospheric loss, and your exposure to particular jamming techniques.
Free-space path loss#
Energy radiated into free space spreads over an expanding sphere. By the time it reaches range \(R\), the power density has fallen as \(1/R^2\) (one-way). Free-space path loss captures how much a signal weakens between two isotropic antennas. In decibels, with range in kilometers and frequency in GHz, the working formula is
Two things to notice. First, loss grows with both range and frequency. Second, both appear as \(20\log_{10}(\cdot)\) — so each behaves identically in dB.
The 6 dB rules#
Because \(20\log_{10}(2) \approx 6.02\) dB:
Double the range → add 6 dB of loss.
Double the frequency → add 6 dB of loss.
These are worth memorizing because they let you reason about link budgets without a calculator. Going from 50 km to 100 km costs 6 dB. Moving a sensor from X-band (10 GHz) to Ku (20 GHz) costs another 6 dB of one-way path loss before you account for anything else.
Quick Exercise
An X-band radar (\(f = 10\) GHz) sees a target at 50 km. Using the FSPL formula and the 6 dB rules:
What is the one-way FSPL at 50 km?
The target flies out to 200 km. How much additional loss?
A second radar at 20 GHz looks at the same 200 km target. How does its FSPL compare to the 10 GHz radar at 200 km?
Solution
\(L = 20\log_{10}(50) + 20\log_{10}(10) + 92.45 = 33.98 + 20 + 92.45 \approx 146.4\) dB.
50 → 200 km is two doublings (×4), so \(+12\) dB.
Doubling frequency adds one more 6 dB, so the 20 GHz radar sees about 6 dB more one-way path loss than the 10 GHz radar at the same range.
Polarization#
A radar’s electric field oscillates in a particular orientation — its polarization (horizontal, vertical, or circular). The receiving antenna is matched to a polarization too. When transmit and receive polarizations disagree, the captured power drops. A full cross-polarization mismatch (horizontal into a vertical antenna) can cost 20–30 dB.
This cuts both ways in EW. A target’s RCS depends on polarization, so an adversary may choose a polarization that maximizes return. And polarization is a lever for EP and EA: polarization-agile radars resist some jammers, while a jammer that does not match the victim’s polarization wastes most of its power.
Antennas: turning power into gain#
An antenna concentrates radiated power into a beam. Its gain over isotropic is set by its effective aperture \(A_e\) and the wavelength:
For a fixed physical aperture, shorter wavelength → higher gain (and a narrower beam). The half-power beamwidth of an aperture of dimension \(D\) is roughly
So the same dish gives a tighter beam at higher frequency. This is the other half of the frequency tradeoff: high frequency buys gain and angular resolution from a small aperture, which is exactly why missile seekers — where size is at a premium — push into Ku/Ka band.
Key Concept
Antenna gain and beamwidth are two sides of the same coin: \(G \approx 4\pi A_e/\lambda^2\) and \(\theta_\text{BW} \approx \lambda/D\). Concentrating energy into a narrow beam is what makes a radar’s effective transmitted power far exceed what an isotropic radiator would deliver — and it is the \(G_t\) and \(G_r\) terms we meet next.
Wrap-Up#
Frequency sets a radar’s whole personality through \(\lambda f = c\) and the band tradeoffs. Free-space path loss, \(L_\text{fs,dB} = 20\log_{10}(R_\text{km}) + 20\log_{10}(f_\text{GHz}) + 92.45\), thins the signal with both range and frequency, and the 6 dB rules let you reason about it in your head. Polarization mismatch can cost 20–30 dB, and antenna gain \(G \approx 4\pi A_e/\lambda^2\) concentrates power into a beam whose width is \(\theta_\text{BW} \approx \lambda/D\).
These are the pieces of the constant \(K\) from Lesson 1. The next lesson assembles them — transmit power, two antenna gains, target RCS, and the two-way spread — into the radar range equation, and shows where the fourth-power law comes from.