Reading — Spectrum & Propagation

Reading — Spectrum & Propagation#

By the end of this lesson you should be able to:

  1. Place a radar in its frequency band and state the tradeoffs of moving up or down the spectrum.

  2. Relate wavelength and frequency through \(\lambda f = c\).

  3. Compute free-space path loss (FSPL) in decibels and apply the 6 dB rules for range and frequency.

  4. 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,

\[ \lambda f = c, \qquad c \approx 3\times10^{8}\ \text{m/s}. \]

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

\[ \boxed{\,L_\text{fs,dB} = 20\log_{10}(R_\text{km}) + 20\log_{10}(f_\text{GHz}) + 92.45\,} \]

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 K-band (20 GHz) costs another 6 dB of one-way path loss before you account for anything else.

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:

\[ G \approx \frac{4\pi A_e}{\lambda^2}. \]

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

\[ \theta_\text{BW} \approx \frac{\lambda}{D}. \]

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.

Type-along

The FSPL formula says every octave of frequency costs 6 dB. Let’s check what it costs a radar, not an isotropic antenna. Hold the dish fixed at \(A_e = 1\) m\(^2\) and the target at 50 km, and sweep three bands:

c = 3e8;             % m/s
Ae = 1.0;            % m^2 -- same physical antenna at every frequency
R_km = 50;           % fixed range to the target
f_GHz = [5 10 20];   % C-band, X-band, K-band
lam  = c ./ (f_GHz*1e9);
G_dB = 10*log10(4*pi*Ae ./ lam.^2);
L_dB = 20*log10(R_km) + 20*log10(f_GHz) + 92.45;
net  = G_dB - L_dB;
fprintf('  f(GHz)  lam(cm)   G(dBi)  FSPL(dB)    net(dB)\n');
fprintf('%8.1f %8.2f %8.2f %9.2f %10.2f\n', [f_GHz; lam*100; G_dB; L_dB; net]);

Read the last column. What do you notice — and who is lying, the formula or the antenna?

Solution

  f(GHz)  lam(cm)   G(dBi)  FSPL(dB)    net(dB)
     5.0     6.00    35.43    140.41    -104.98
    10.0     3.00    41.45    146.43    -104.98
    20.0     1.50    47.47    152.45    -104.98

FSPL climbs 6.02 dB per octave — and so does the gain, because \(G = 4\pi A_e/\lambda^2\) and \(\lambda = c/f\), so \(G\) carries its own \(+20\log_{10}(f)\). They cancel exactly:

\[ G - L_\text{fs} = 10\log_{10}\!\left(\frac{A_e}{4\pi R^2}\right) = -104.97\ \text{dB}, \]

with no \(f\) in it. Neither is lying. FSPL is defined between isotropic antennas, and a real aperture wins back precisely what that reference charges. The genuine high-frequency penalties are atmospheric absorption, rain, and component loss — none of which are in this formula. In L3 those two gain terms come back as \(G_t\) and \(G_r\).

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.

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